Mathematics — Class 10, Chapter 4: Algebra

Most repeated · 8 of 45 exams
Asked 8 times
2075 Chaitra · Set AP · 4 marks
2075 Supplementary · 4 marks
2075 Chaitra · Set BP · 4 marks
2075 Chaitra · Set CP · 4 marks
2075 Chaitra · Set EP · 4 marks
2075 Chaitra · Set FP · 4 marks
2075 Chaitra · Set GP · 4 marks
2075 Chaitra · Set DP · 4 marks

Simplify: $\dfrac{1}{1-p+p^2}-\dfrac{1}{1+p+p^2}-\dfrac{2p}{1-p^2+p^4}$

Answer

First combine the first two fractions:

$\frac{1}{1 - p + p^2} - \frac{1}{1 + p + p^2} = \frac{(1 + p + p^2) - (1 - p + p^2)}{(1 + p^2)^2 - p^2} = \frac{2p}{1 + p^2 + p^4}$

since $(1 + p^2)^2 - p^2 = 1 + p^2 + p^4$ .

Now subtract the third fraction:

$\frac{2p}{1 + p^2 + p^4} - \frac{2p}{1 - p^2 + p^4} = 2p \cdot \frac{(1 - p^2 + p^4) - (1 + p^2 + p^4)}{(1 + p^2 + p^4)(1 - p^2 + p^4)} = \frac{2p(-2p^2)}{(1 + p^4)^2 - p^4}$

The denominator is $(1 + p^4)^2 - p^4 = 1 + 2p^4 + p^8 - p^4 = 1 + p^4 + p^8$ .

$\therefore \text{ the result is } \frac{-4p^3}{1 + p^4 + p^8}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Most repeated · 6 of 45 exams
Asked 6 times
2075 Chaitra · Set BP · 1 mark
2075 Chaitra · Set CP · 1 mark
2075 Chaitra · Set EP · 1 mark
2075 Chaitra · Set FP · 1 mark
2075 Chaitra · Set GP · 1 mark
2075 Chaitra · Set DP · 1 mark

What is the value of $(4a)^0$ ? Write it.

Answer

$(4a)^0 = 1$ (for $a \neq 0$ ), since any non-zero quantity raised to the power zero is $1$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 1 MCQ

Loading…

Most repeated · 6 of 45 exams
Asked 2 times
2075 Chaitra · Set BP · 2 marks
2075 Chaitra · Set DP · 2 marks

If the sum of two consecutive even numbers is 26, then find the numbers.

Similar questions: Two consecutive even numbers with sum 42 (2075 Chaitra · Set AP) · Two consecutive even numbers with sum 34 (2075 Chaitra · Set CP) · Two consecutive even numbers with sum 14 (2075 Chaitra · Set EP)

Answer

Let the numbers be $x$ and $x + 2$ .

$x + (x+2) = 26 \Rightarrow 2x = 24 \Rightarrow x = 12$

The numbers are 12 and 14.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Most repeated · 6 of 45 exams
2075 Chaitra · Set AP · 2 marks

If the sum of two consecutive even numbers is 42, find the numbers.

Similar questions: Two consecutive even numbers with sum 34 (2075 Chaitra · Set CP) · Two consecutive even numbers with sum 14 (2075 Chaitra · Set EP) · Two consecutive even numbers with sum 26 (2075 Chaitra · Set BP)

Answer

Let the two consecutive even numbers be $x$ and $x+2$ .

$x+(x+2)=42 \Rightarrow 2x=40 \Rightarrow x=20$

The numbers are 20 and 22.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Most repeated · 6 of 45 exams
2075 Chaitra · Set AP · 5 marks

The area of a rectangular field is 1440 sq. metre and its perimeter is 152 metre. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Find it.

Similar questions: Rectangle of area 720 and perimeter 108 made a square (2075 Chaitra · Set BP) · Rectangle of area 3000 and perimeter 220 made a square (2075 Chaitra · Set CP) · Rectangle of area 2000 and perimeter 180 made a square (2075 Chaitra · Set EP)

Answer

Let the length be $l$ m and the breadth be $b$ m.

Perimeter: $2(l + b) = 152 \Rightarrow l + b = 76$ .

Area: $lb = 1440$ .

So $l$ and $b$ are the roots of $x^2 - 76x + 1440 = 0$ :

$x^2 - 40x - 36x + 1440 = 0 \Rightarrow (x - 40)(x - 36) = 0$

Hence the length is $40$ m and the breadth is $36$ m.

To make the field a square, the longer side must be reduced to $36$ m (reducing the breadth cannot work, as it would make the sides even more unequal). So the length is to be decreased.

$\text{Decrease} = 40 - 36 = 4 \text{ m}, \qquad \text{Percentage} = \frac{4}{40} \times 100\% = 10\%$

The length should be decreased by $10\%$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Most repeated · 6 of 45 exams
2075 Chaitra · Set BP · 5 marks

The area of a rectangular land is 720 sq. metre and its perimeter is 108 metre. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Find it.

Similar questions: Rectangle of area 1440 and perimeter 152 made a square (2075 Chaitra · Set AP) · Rectangle of area 3000 and perimeter 220 made a square (2075 Chaitra · Set CP) · Rectangle of area 2000 and perimeter 180 made a square (2075 Chaitra · Set EP)

Answer

Let the length be $l$ and the breadth be $b$ .

Perimeter: $2(l+b)=108 \Rightarrow l+b=54$ .

Area: $lb=720$ .

Now $(l-b)^2=(l+b)^2-4lb=54^2-4\times720=2916-2880=36$ , so $l-b=6$ .

Solving $l+b=54$ and $l-b=6$ :

$l=30\text{ m},\qquad b=24\text{ m}.$

A square has equal sides, so the longer side (the length) must be decreased from $30$ m to $24$ m, a decrease of $6$ m.

$\text{Percentage decrease}=\frac{6}{30}\times100\%=20\%.$

Hence the length should be decreased by 20% to make the land a square of side $24$ m.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Most repeated · 6 of 45 exams
2075 Chaitra · Set CP · 2 marks

If the sum of two consecutive even numbers is 34, find the numbers.

Similar questions: Two consecutive even numbers with sum 42 (2075 Chaitra · Set AP) · Two consecutive even numbers with sum 14 (2075 Chaitra · Set EP) · Two consecutive even numbers with sum 26 (2075 Chaitra · Set BP)

Answer

Let the two consecutive even numbers be $x$ and $x+2$ .

$x+(x+2)=34 \Rightarrow 2x=32 \Rightarrow x=16.$

The numbers are 16 and 18 (check: $16+18=34$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Most repeated · 6 of 45 exams
2075 Chaitra · Set CP · 5 marks

The area of a rectangular land is 3000 sq. metres and its perimeter is 220 metres. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Find it.

Similar questions: Rectangle of area 2000 and perimeter 180 made a square (2075 Chaitra · Set EP) · Rectangle of area 660 and perimeter 104 made a square (2075 Chaitra · Set FP) · Rectangle of area 500 and perimeter 90 made a square (2075 Chaitra · Set GP)

Answer

Let the length be $l$ and the breadth be $b$ .

Perimeter: $2(l+b)=220 \Rightarrow l+b=110$ .

Area: $lb=3000$ .

$(l-b)^2=(l+b)^2-4lb=110^2-4\times3000=12100-12000=100 \Rightarrow l-b=10.$

Solving $l+b=110$ and $l-b=10$ :

$l=60\text{ m},\qquad b=50\text{ m}.$

To make it a square, the length must be decreased from $60$ m to $50$ m, a decrease of $10$ m.

$\text{Percentage decrease}=\frac{10}{60}\times100\%=16\frac23\%.$

Hence the length should be decreased by $16\frac23\%$ (about $16.67\%$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Most repeated · 6 of 45 exams
2075 Chaitra · Set EP · 2 marks

If the sum of two consecutive even numbers is 14, find the numbers.

Similar questions: Two consecutive even numbers with sum 42 (2075 Chaitra · Set AP) · Two consecutive even numbers with sum 34 (2075 Chaitra · Set CP) · Two consecutive even numbers with sum 26 (2075 Chaitra · Set BP)

Answer

Let the numbers be $x$ and $x+2$ .

$x + (x+2) = 14 \Rightarrow 2x = 12 \Rightarrow x = 6$ .

The numbers are 6 and 8 (check: $6+8=14$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Most repeated · 6 of 45 exams
2075 Chaitra · Set EP · 5 marks

The area of a rectangular land is 2000 sq. metres and its perimeter is 180 metres. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Why? Find it.

Similar questions: Rectangle of area 660 and perimeter 104 made a square (2075 Chaitra · Set FP) · Rectangle of area 500 and perimeter 90 made a square (2075 Chaitra · Set GP) · Rectangle of area 3000 and perimeter 220 made a square (2075 Chaitra · Set CP)

Answer

Let the length be $l$ and breadth $b$ .

$2(l+b) = 180 \Rightarrow l + b = 90$ and $lb = 2000$ .

$(l-b)^2 = (l+b)^2 - 4lb = 8100 - 8000 = 100 \Rightarrow l - b = 10$ .

Solving $l+b=90$ and $l-b=10$ : $l = 50$ m and $b = 40$ m.

To make the land a square, all four sides must be equal. The length is the larger side, so it is the length that must be decreased, from $50$ m to $40$ m, to match the breadth. (Decreasing the breadth would only make the sides more unequal.)

Decrease $= 50 - 40 = 10$ m

$\text{Percentage decrease} = \frac{10}{50}\times 100\% = \textbf{20\%}$

So the length is to be decreased by $20\%$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Most repeated · 6 of 45 exams
2075 Chaitra · Set FP · 5 marks

The area of a rectangular land is 660 sq. metres and its perimeter is 104 metres. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Why? Find it.

Similar questions: Rectangle of area 2000 and perimeter 180 made a square (2075 Chaitra · Set EP) · Rectangle of area 500 and perimeter 90 made a square (2075 Chaitra · Set GP) · Rectangle of area 3000 and perimeter 220 made a square (2075 Chaitra · Set CP)

Answer

Let length $= l$ m and breadth $= b$ m.

2(l + b) = 104 \Rightarrow l + b = 52, \qquad lb = 660

Then $l - b = \sqrt{(l + b)^2 - 4lb} = \sqrt{2704 - 2640} = \sqrt{64} = 8$ .

So $l = \dfrac{52 + 8}{2} = 30$ m and $b = \dfrac{52 - 8}{2} = 22$ m.

A square has equal sides. The length is the larger side, so it is the length that must be decreased, from 30 m to 22 m (decreasing the breadth cannot make the sides equal, since the breadth is already the smaller one).

\text{Decrease} = 30 - 22 = 8 \text{ m}

\text{Percentage decrease} = \frac{8}{30} \times 100\% = 26\tfrac23\% \approx 26.67\%

Answer: the length should be decreased by $26\frac23\%$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Most repeated · 6 of 45 exams
2075 Chaitra · Set GP · 2 marks

Find the two consecutive even numbers whose sum is 30.

Similar questions: Two consecutive even numbers with sum 42 (2075 Chaitra · Set AP) · Two consecutive even numbers with sum 34 (2075 Chaitra · Set CP) · Two consecutive even numbers with sum 14 (2075 Chaitra · Set EP)

Answer

Let the numbers be $x$ and $x + 2$ .

x + (x + 2) = 30 \Rightarrow 2x = 28 \Rightarrow x = 14

Answer: the numbers are 14 and 16.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 1 MCQ

Loading…

Most repeated · 6 of 45 exams
2075 Chaitra · Set GP · 5 marks

The area of a rectangular land is 500 sq. metres and its perimeter is 90 metres. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Why? Find it.

Similar questions: Rectangle of area 2000 and perimeter 180 made a square (2075 Chaitra · Set EP) · Rectangle of area 660 and perimeter 104 made a square (2075 Chaitra · Set FP) · Rectangle of area 3000 and perimeter 220 made a square (2075 Chaitra · Set CP)

Answer

Let the length be $l$ m and the breadth $b$ m.

$lb = 500, \qquad 2(l + b) = 90 \Rightarrow l + b = 45$

Then $b = 45 - l$ , so $l(45 - l) = 500 \Rightarrow l^2 - 45l + 500 = 0 \Rightarrow (l - 25)(l - 20) = 0$

So the length is $25$ m and the breadth is $20$ m (the larger is the length).

To make the land a square, the two sides must be equal. The breadth cannot be decreased, because it is already the smaller side (decreasing it would make the difference larger). So the length must be decreased from $25$ m to $20$ m.

$\text{Decrease} = \frac{25 - 20}{25} \times 100\% = 20\%$

The length should be decreased by $20\%$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Most repeated · 3 of 45 exams
Asked 3 times
2081 Chaitra · Gandaki · 3 marks
2082 Chaitra · Koshi · 3 marks
2082 Pre-qualifying · PABSON Kathmandu · 2 marks

Solve: $2^x+\dfrac{1}{2^x}=2\dfrac{1}{2}$

Answer

Given $2^x+\dfrac{1}{2^x}=\dfrac{5}{2}$ .

Let $2^x=y$ . Then $y+\dfrac1y=\dfrac52$

2y^2-5y+2=0

2y^2-4y-y+2=0 \Rightarrow (y-2)(2y-1)=0

So $y=2$ or $y=\dfrac12$ .

If $2^x=2=2^1$ , then $x=1$ .
If $2^x=\dfrac12=2^{-1}$ , then $x=-1$ .

Hence $x=1$ or $x=-1$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Most repeated · 3 of 45 exams
Asked 3 times
2080 Chaitra · Bagmati · 2 marks
2082 Preparation · Bhaktapur · Bhaktapur Municipality · 2 marks
2082 Preparation · Bhaktapur · 2 marks

Solve: $3^{2x} - 4 \times 3^{x} + 3 = 0$

Answer

$3^{2x}-4\times3^x+3=0$ . Let $3^x=y$ .

y^2-4y+3=0 \Rightarrow (y-1)(y-3)=0

$3^x=1=3^0 \Rightarrow x=0$
$3^x=3 \Rightarrow x=1$

Answer: $x=0$ or $x=1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Asked 2 times
2080 Chaitra · Karnali · 2 marks
2081 Chaitra · Lumbini · 2 marks

Simplify: $\frac{m^3+1}{m^2-m+1}+\frac{m^3-1}{m^2+m+1}$

Answer

Use $m^3+1=(m+1)(m^2-m+1)$ and $m^3-1=(m-1)(m^2+m+1)$ .

\frac{(m+1)(m^2-m+1)}{m^2-m+1}+\frac{(m-1)(m^2+m+1)}{m^2+m+1}=(m+1)+(m-1)=2m

Answer: $2m$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Asked 2 times
2082 Chaitra · Karnali · 3 marks
2080 Model · Set 2 · 2 marks

Solve: $2^{x-2} + 2^{3-x} = 3$

Answer

$2^{x-2}+2^{3-x}=3 \Rightarrow \dfrac{2^x}{4}+\dfrac{8}{2^x}=3$

Let $2^x=y$ : $\dfrac y4+\dfrac8y=3 \Rightarrow y^2+32=12y$

y^2-12y+32=0 \Rightarrow (y-4)(y-8)=0

$2^x=4 \Rightarrow x=2$
$2^x=8 \Rightarrow x=3$

Answer: $x=2$ or $x=3$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Asked 2 times
2081 Chaitra · Sudurpashchim · 3 marks
2080 Model · Set 3 · 3 marks

If $x^2 = 3^{\frac{2}{3}} + 3^{\frac{-2}{3}} - 2$ , prove that: $3x^3 + 9x = 8$

Answer

Given $x^2=3^{\frac23}+3^{-\frac23}-2$ .

Let $a=3^{\frac13}$ , $b=3^{-\frac13}$ . Then $ab=1$ , $a^3=3$ , $b^3=\dfrac13$ .

$x^2=a^2+b^2-2ab=(a-b)^2 \Rightarrow x=a-b=3^{\frac13}-3^{-\frac13}$ (taking the positive root, since $a>b$ , as the result needs $x>0$ ).

Cubing: $x^3=a^3-b^3-3ab(a-b)$

x^3=3-\frac13-3x \Rightarrow x^3+3x=\frac83

Multiplying by 3:

3x^3+9x=8

Hence proved.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Asked 2 times
2075 Chaitra · Set CP · 2 marks
2075 Chaitra · Set FP · 2 marks

Find the L.C.M. of: $p^2-q^2,\ (p+q)^2$ .

Answer

Factorise each expression:

$p^2-q^2=(p+q)(p-q),\qquad (p+q)^2=(p+q)(p+q).$

Taking each factor with its highest power:

$\text{L.C.M.}=(p+q)^2(p-q).$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Supplementary · 2 marks

If the sum of two consecutive odd numbers is 20, find the numbers.

Similar questions: Two consecutive odd numbers with sum 12 (2075 Chaitra · Set FP)

Answer

Let the two consecutive odd numbers be $x$ and $x+2$ .

$x+(x+2)=20 \Rightarrow 2x=18 \Rightarrow x=9$

The numbers are 9 and 11.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set FP · 2 marks

If the sum of two consecutive odd numbers is 12, find the numbers.

Similar questions: Two consecutive odd numbers with sum 20 (2075 Supplementary)

Answer

Let the numbers be $x$ and $x+2$ (consecutive odd numbers).

$x + (x+2) = 12 \Rightarrow 2x = 10 \Rightarrow x = 5$ .

The numbers are 5 and 7 (check: $5+7=12$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Koshi · 1+2+2 marks

Ramila opened account on behalf of her daughter in a finance and the amounts deposited monthly is increased in common difference are presented below in the table.

Months 1st 2nd 3rd 4th 5th
Amount (Rs.) 2000 4000 6000 8000 10000
(a) What type of sequence is formed on saving for daughter of Ramila? Write it. (1)
(b) How much total amount did Ramila deposit in 1 year? Find it. (2)
(c) How many months will Ramila require to deposit Rs.4,20,000 in the account of her daughter? Find it. (2)

Months	1st	2nd	3rd	4th	5th
Amount (Rs.)	2000	4000	6000	8000	10000

Answer

(a) The amounts 2000, 4000, 6000, ... have a constant difference of Rs 2000, so they form an arithmetic sequence (AP).

(b) Here $a=2000$ , $d=2000$ , $n=12$ months.

S_n=\frac n2[2a+(n-1)d]

S_{12}=\frac{12}{2}[2\times2000+11\times2000]=6\times26000=156000

Total deposit in 1 year $=$ Rs 1,56,000.

(c) Let $S_n=420000$ .

\frac n2[4000+(n-1)2000]=420000

1000\,n(n+1)=420000 \Rightarrow n^2+n-420=0

(n+21)(n-20)=0 \Rightarrow n=20 \ (n=-21\text{ rejected})

She needs 20 months.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Koshi · 1+3 marks

In a two digit number, the product of digits is 20 and their difference is 1.
(a) Write the standard form of a number of two digits. (1)
(b) Find the number. (3)

Answer

(a) If $x$ is the tens digit and $y$ the units digit, the number is $10x+y$ .

(b) Let the digits be $x$ and $y$ with $xy=20$ and $x-y=1$ , so $y=x-1$ .

x(x-1)=20 \Rightarrow x^2-x-20=0

(x-5)(x+4)=0 \Rightarrow x=5 \ (x=-4\text{ rejected})

Then $y=4$ , and the digits are 5 and 4 (product 20, difference 1).

So the number is 54 (tens digit larger). If the units digit is the larger one ( $y-x=1$ ), the number is 45.

Check: $5\times4=20$ , $5-4=1$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Koshi · 1 mark

Solve: $5^{\sqrt{x}} = 25$

Answer

$5^{\sqrt x}=25=5^2$

Equal bases, so $\sqrt x=2$ . Squaring, $x=4$ .

Answer: $x=4$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 1 MCQ

Loading…

2082 Chaitra · Koshi · 2 marks

Simplify: $\dfrac{1}{p^2 - pq + q^2} - \dfrac{p^2+q^2}{p^4 + p^2q^2 + q^4}$

Answer

Factorise $p^4+p^2q^2+q^4=(p^2+pq+q^2)(p^2-pq+q^2)$ .

\frac{1}{p^2-pq+q^2}-\frac{p^2+q^2}{(p^2+pq+q^2)(p^2-pq+q^2)} =\frac{(p^2+pq+q^2)-(p^2+q^2)}{(p^2+pq+q^2)(p^2-pq+q^2)}

=\frac{pq}{p^4+p^2q^2+q^4}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Madhesh · 1+2+2 marks

A teacher made a video of a lesson which he/she taught and uploaded in a social media. Daily viewers of this video are increased by the same ratio as per the table given below.

Days 1 2 3 4 5 ......
New Students 100 200 400 800 1600 ......
(a) In which sequence new students are being increased every day? Write it. (1)
(b) How many total new students will view the video till 10th day? Find it. (2)
(c) The ratio of students who viewed the video up to the 3rd day and up to other any day is 7:255, which day is that day? Find it. (2)
Figure: Table of days 1 to 5 with new students 100, 200, 400, 800, 1600.

Days	1	2	3	4	5	......
New Students	100	200	400	800	1600	......

Answer

(a) 100, 200, 400, 800, ... have a constant ratio 2, so the viewers increase in a geometric sequence.

(b) $a=100$ , $r=2$ , $n=10$ .

S_n=\frac{a(r^n-1)}{r-1} \Rightarrow S_{10}=\frac{100(2^{10}-1)}{2-1}=100\times1023=102300

Total new students $=$ 1,02,300.

(c) $S_3=100(2^3-1)=700$ .

Let the required day be $n$ . $\dfrac{S_3}{S_n}=\dfrac{7}{255}$

S_n=\frac{700\times255}{7}=25500

100(2^n-1)=25500 \Rightarrow 2^n=256=2^8 \Rightarrow n=8

The day is the 8th day.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Madhesh · 1+2+2 marks

In a natural number of two digits, the product of digits is 24. When 45 is added to the number, the places of the digits of the number are interchanged.
(a) Write the standard form of a quadratic equation. (1)
(b) Make a quadratic equation from the given verbal problem. (2)
(c) Find the natural number. (2)

Answer

(a) The standard form of a quadratic equation is $ax^2+bx+c=0$ , where $a\neq0$ .

(b) Let the tens digit be $x$ and the units digit $y$ . Number $=10x+y$ .

Product of digits: $xy=24$ .

Adding 45 interchanges the digits: $10x+y+45=10y+x \Rightarrow 9y-9x=45 \Rightarrow y=x+5$ .

Substituting, $x(x+5)=24$

x^2+5x-24=0

(c) $x^2+8x-3x-24=0 \Rightarrow (x+8)(x-3)=0$

$x=3$ (since $x=-8$ is rejected), $y=8$ .

The number is 38. Check: $3\times8=24$ and $38+45=83$ (digits of 38 reversed). Correct.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Madhesh · 2 marks

Simplify: $\dfrac{1}{1+a^{-1}} + \dfrac{1}{1-a^{-1}}$

Answer

\frac{1}{1+a^{-1}}+\frac{1}{1-a^{-1}}=\frac{1}{1+\frac1a}+\frac{1}{1-\frac1a} =\frac{a}{a+1}+\frac{a}{a-1}

=\frac{a(a-1)+a(a+1)}{(a+1)(a-1)}=\frac{2a^2}{a^2-1}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Madhesh · 3 marks

Solve: $2^{2x} - 3.2^{x+1} + 8 = 0$

Answer

$2^{2x}-3\cdot2^{x+1}+8=0$

$(2^x)^2-6\cdot2^x+8=0$ . Let $2^x=y$ .

y^2-6y+8=0 \Rightarrow (y-2)(y-4)=0

$2^x=2 \Rightarrow x=1$
$2^x=4=2^2 \Rightarrow x=2$

Answer: $x=1$ or $x=2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Bagmati · 1+2+2 marks

The first term of an arithmetic series is 12 and the sum of its first five terms is 50.
(a) What is the sum of first $n$ terms of an arithmetic series having first term $a$ and common difference $d$ ? Write it. (1)
(b) Find the common difference of the series. (2)
(c) Out of first three terms of the series, if 5 and 7 are subtracted from the second and third terms respectively, the terms form a geometric series. Verify with reason. (2)

Answer

(a) $S_n=\dfrac n2[2a+(n-1)d]$

(b) $a=12$ , $S_5=50$ .

\frac52[2\times12+4d]=50 \Rightarrow 24+4d=20 \Rightarrow d=-1

Common difference $=-1$ .

(c) The first three terms are $12,\ 11,\ 10$ .

After subtracting 5 from the second and 7 from the third: $12,\ 11-5=6,\ 10-7=3$ .

Ratios: $\dfrac{6}{12}=\dfrac12$ and $\dfrac36=\dfrac12$ .

The ratio is the same, so 12, 6, 3 form a geometric series with common ratio $\dfrac12$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Bagmati · 1+2+2 marks

In 2075 B.S. the age of a father was 6 times the age of his son and in 2080 B.S. the numerical product of their ages was 350.
(a) Define quadratic equation. (1)
(b) Make a quadratic equation according to the given condition. (2)
(c) In which year the son was born? Find it. (2)

Answer

(a) An equation of the form $ax^2+bx+c=0$ where $a,b,c$ are real numbers and $a\neq0$ (the highest power of the variable is 2) is called a quadratic equation.

(b) Let the son's age in 2075 B.S. be $x$ years; the father's age is $6x$ .

In 2080 B.S. (5 years later): ages $6x+5$ and $x+5$ .

(6x+5)(x+5)=350 \Rightarrow 6x^2+35x+25=350

6x^2+35x-325=0

(c) Discriminant $=35^2+4\times6\times325=1225+7800=9025=95^2$

x=\frac{-35\pm95}{12}=5\ \text{or}\ -\frac{130}{12}

Age cannot be negative, so $x=5$ . The son was 5 years old in 2075 B.S.

Born in $2075-5=$ 2070 B.S.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Bagmati · 2 marks

Simplify: $\frac{1}{x-y} - \frac{x+y}{x^2-y^2}$

Answer

\frac{1}{x-y}-\frac{x+y}{x^2-y^2}=\frac{1}{x-y}-\frac{x+y}{(x+y)(x-y)}=\frac{1}{x-y}-\frac{1}{x-y}=0

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Bagmati · 3 marks

Solve: $25^x - 6 \times 5^{x+1} + 125 = 0$

Answer

$25^x-6\times5^{x+1}+125=0$

$(5^x)^2-30\cdot5^x+125=0$ . Let $5^x=y$ .

y^2-30y+125=0 \Rightarrow (y-5)(y-25)=0

$5^x=5 \Rightarrow x=1$
$5^x=25 \Rightarrow x=2$

Answer: $x=1$ or $x=2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Gandaki · 6 marks

The monthly salary of an employee is Rs.45,000 at present. He receives a grade of Rs.1,500 on his monthly salary every year.
(a) Write the name of sequence formed from above context. (1)
(b) What will be his total income in 4 years when the number of grades increases yearly at the same rate? Calculate it. (3)
(c) In how many years, his total income will be Rs.35,10,000 when the number of grades increases yearly by the same rate? Find it. (2)

Answer

(a) The monthly salary increases by a constant Rs 1,500 each year, so it forms an arithmetic sequence.

(b) Yearly income: year 1 $=12\times45000=$ Rs 5,40,000; year 2 $=12\times46500=$ Rs 5,58,000; and so on, an AP with $a=540000$ , $d=12\times1500=18000$ , $n=4$ .

S_4=\frac42[2\times540000+3\times18000]=2[1080000+54000]=2268000

Total income in 4 years $=$ Rs 22,68,000.

(c) $S_n=3510000$

\frac n2[1080000+(n-1)18000]=3510000

n[540000+9000(n-1)]=3510000 \Rightarrow n(531000+9000n)=3510000

Dividing by 9000: $n^2+59n-390=0$

(n+65)(n-6)=0 \Rightarrow n=6

It takes 6 years.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Gandaki · 5 marks

The sum of the present ages of two brothers is 28 years and the product of their ages is 187 numerically.
(a) In the quadratic equation of $ax^2 + bx + c = 0$ , write the formula to find the value of $x$ . (1)
(b) Find their present ages. (2)
(c) After how many years will the product of their ages be 247? Find it. (2)

Answer

(a) $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

(b) Let one brother's age be $x$ ; the other's is $28-x$ .

x(28-x)=187 \Rightarrow x^2-28x+187=0

(x-17)(x-11)=0 \Rightarrow x=17\ \text{or}\ 11

Their present ages are 17 years and 11 years.

(c) After $t$ years: $(17+t)(11+t)=247$

t^2+28t+187-247=0 \Rightarrow t^2+28t-60=0

(t+30)(t-2)=0 \Rightarrow t=2

After 2 years (check: $19\times13=247$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Gandaki · 2 marks

Simplify: $\dfrac{1}{m+n} - \dfrac{1}{m-n}$

Answer

\frac{1}{m+n}-\frac{1}{m-n}=\frac{(m-n)-(m+n)}{(m+n)(m-n)}=\frac{-2n}{m^2-n^2}=\frac{2n}{n^2-m^2}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Gandaki · 2 marks

Solve: $3^{y+1} + 3^{y} = 36$

Answer

$3^{y+1}+3^y=36$

$3^y\cdot3+3^y=36 \Rightarrow 3^y(3+1)=36 \Rightarrow 3^y=9=3^2$

Answer: $y=2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Lumbini · 1+2+2 marks

The fifth term and tenth term of an arithmetic sequence are 23 and 43 respectively.
(a) What is the arithmetic mean between two numbers $a$ and $b$ ? Write it. (1)
(b) Find the common difference and first term of the given arithmetic sequence. (2)
(c) Find the sum of the first 10 terms of the sequence. (2)

Answer

(a) The arithmetic mean between $a$ and $b$ is $\dfrac{a+b}{2}$ .

(b) $t_5=a+4d=23$ and $t_{10}=a+9d=43$ .

Subtracting: $5d=20 \Rightarrow d=4$ . Then $a=23-16=7$ .

Common difference $=4$ , first term $=7$ .

(c)

S_{10}=\frac{10}{2}[2\times7+9\times4]=5[14+36]=250

Sum of the first 10 terms $=250$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Lumbini · 1+2+2 marks

The area of a rectangular field is 525 square meter. The breadth of the field is 4 meter shorter than its length.
(a) Write the roots of quadratic equation $ax^2 + bx + c = 0$ ; $a \neq 0$ . (1)
(b) Find the length and breadth of the field. (2)
(c) How many maximum numbers of pieces of land can be made with size 4 meter $\times$ 4 meter in that rectangular field? Also present it in a diagram. (2)

Answer

(a) For $ax^2+bx+c=0$ , $a\neq0$ , the roots are $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ .

(b) Let the length be $l$ m; the breadth is $(l-4)$ m.

l(l-4)=525 \Rightarrow l^2-4l-525=0

l=\frac{4\pm\sqrt{16+2100}}{2}=\frac{4\pm46}{2}=25\ \text{(rejecting }-21)

Length $=25$ m, breadth $=21$ m.

(c) Along the length: $25\div4=6$ pieces (remainder 1 m). Along the breadth: $21\div4=5$ pieces (remainder 1 m).

Maximum pieces $=6\times5=$ 30.

 <-------------------- 25 m -------------------->
 +-----+-----+-----+-----+-----+-----+
 | 4x4 | 4x4 | 4x4 | 4x4 | 4x4 | 4x4 |
 +-----+-----+-----+-----+-----+-----+
 | 4x4 | 4x4 | 4x4 | 4x4 | 4x4 | 4x4 |
 +-----+-----+-----+-----+-----+-----+
 | 4x4 | 4x4 | 4x4 | 4x4 | 4x4 | 4x4 |
 +-----+-----+-----+-----+-----+-----+
 | 4x4 | 4x4 | 4x4 | 4x4 | 4x4 | 4x4 |
 +-----+-----+-----+-----+-----+-----+
 | 4x4 | 4x4 | 4x4 | 4x4 | 4x4 | 4x4 |
 +-----+-----+-----+-----+-----+-----+
 (21 m side: 5 pieces of 4 m, 1 m left over; 25 m side: 6 pieces, 1 m left over)

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Lumbini · 3 marks

Simplify: $\dfrac{1}{a+2b} - \dfrac{1}{a-2b} + \dfrac{2a}{4b^2-a^2}$

Answer

\frac{1}{a+2b}-\frac{1}{a-2b}+\frac{2a}{4b^2-a^2}

Since $4b^2-a^2=-(a^2-4b^2)$ :

=\frac{(a-2b)-(a+2b)}{a^2-4b^2}-\frac{2a}{a^2-4b^2} =\frac{-4b-2a}{a^2-4b^2}

=\frac{-2(a+2b)}{(a+2b)(a-2b)}=\frac{-2}{a-2b}=\frac{2}{2b-a}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Lumbini · 2 marks

Solve: $4^{x-2} = 0.5$

Answer

$4^{x-2}=0.5=\dfrac12$

$2^{2(x-2)}=2^{-1} \Rightarrow 2x-4=-1 \Rightarrow 2x=3$

Answer: $x=\dfrac32$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Karnali · 1+2+2 marks

The number of shoes sold by a shop in the first three days are given below in the table. The shop is selling double number of shoes each day compared to the previous day.

Days 1st 2nd 3rd
Pair of shoes 4 8 16
(a) The number of geometric means between the first term (a) and last term (b) is (n), write the formula to find common ratio (r). (1)
(b) How many pair of shoes will be sold up to 8 days as given in the above table? Calculate it by using formula. (2)
(c) In how many days the shop sells 4092 pair of shoes? Find it. (2)
Figure: Table of pairs of shoes sold: day 1 = 4, day 2 = 8, day 3 = 16.

Days	1st	2nd	3rd
Pair of shoes	4	8	16

Answer

(a) If $n$ geometric means are inserted between $a$ and $b$ , there are $n+2$ terms and $b=ar^{n+1}$ , so

r=\left(\frac ba\right)^{\frac1{n+1}}

(b) Sales 4, 8, 16, ... form a GP with $a=4$ , $r=2$ , $n=8$ .

S_8=\frac{a(r^n-1)}{r-1}=\frac{4(2^8-1)}{2-1}=4\times255=1020

1020 pairs are sold up to 8 days.

(c) $S_n=4092$

4(2^n-1)=4092 \Rightarrow 2^n-1=1023 \Rightarrow 2^n=1024=2^{10}

$n=10$ . It takes 10 days.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Karnali · 1+2+2 marks

The sum of the present ages of two sisters is 40 and the product is 391 numerically.
(a) If the present age of elder sister is $x$ years, then express the age of younger sister in terms of $x$ . (1)
(b) Find the present ages of both sisters. (2)
(c) After how many years, if the square of that number of years is added to the present age, will the new ratio become 13:11? (2)

Answer

(a) The younger sister's age $=(40-x)$ years.

(b) $x(40-x)=391 \Rightarrow x^2-40x+391=0$

x=\frac{40\pm\sqrt{1600-1564}}{2}=\frac{40\pm6}{2}=23\ \text{or}\ 17

Elder sister is 23 years, younger sister is 17 years.

(c) After $t$ years, add $t^2$ to the present ages (as stated):

\frac{23+t^2}{17+t^2}=\frac{13}{11}

253+11t^2=221+13t^2 \Rightarrow 2t^2=32 \Rightarrow t^2=16

$t=4$ . After 4 years. (Check: $\frac{39}{33}=\frac{13}{11}$ .)

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Karnali · 2 marks

Simplify: $\frac{x}{x-y} - \frac{x}{x+y}$

Answer

\frac{x}{x-y}-\frac{x}{x+y}=\frac{x(x+y)-x(x-y)}{(x-y)(x+y)}=\frac{2xy}{x^2-y^2}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Sudurpashchim · 1+2+2 marks

The monthly salary of an officer working in an institution is Rs.60,000. His/her monthly salary increases by Rs.2,000 in every year as a grade up to 10 years.
(a) If 'a', 'm' and 'b' are in arithmetic sequence, what is the value of arithmetic mean 'm'? Write it. (1)
(b) How much amount does he/she earn up to 7 years? Find it. (2)
(c) In how many years he/she can earn Rs.1,02,24,000? Is it possible? Give logic with calculation. (2)

Answer

(a) $m=\dfrac{a+b}{2}$

(b) Yearly income forms an AP: first year $=12\times60000=720000$ , common difference $=12\times2000=24000$ , $n=7$ .

S_7=\frac72[2\times720000+6\times24000]=\frac72[1440000+144000]=\frac72\times1584000=5544000

He earns Rs 55,44,000 in 7 years.

(c) $S_n=10224000$

\frac n2[1440000+(n-1)24000]=10224000

n[720000+12000(n-1)]=10224000 \Rightarrow n(708000+12000n)=10224000

Dividing by 12000: $n^2+59n-852=0$

n=\frac{-59\pm\sqrt{3481+3408}}{2}=\frac{-59\pm83}{2}=12

The calculation gives 12 years. But the grade (increment) is given only up to 10 years, so the AP holds for 10 years only. Hence it is not possible to earn Rs 1,02,24,000 under this grade scheme in 12 years (the AP total for 10 years is only Rs 82,80,000).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Sudurpashchim · 1+2+2 marks

The length of a rectangular field is 3 meter less than twice of its breath. The area of that field is 135 square meter.
(a) Write the roots of quadratic equation $ax^2 + bx + c = 0;\ a \neq 0$ . (1)
(b) Find the length of that field by making quadratic equation. (2)
(c) How many maximum number of rectangular pieces having dimension 4 meter $\times$ 4 meter can be prepared in that field? Also present it in diagram. (2)

Answer

(a) $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

(b) Let the breadth be $b$ m; the length is $(2b-3)$ m.

b(2b-3)=135 \Rightarrow 2b^2-3b-135=0

b=\frac{3\pm\sqrt{9+1080}}{4}=\frac{3\pm33}{4}=9

Breadth $=9$ m, so the length $=2\times9-3=$ 15 m.

(c) Field $15\text{ m}\times9\text{ m}$ . Along 15 m: $15\div4=3$ pieces; along 9 m: $9\div4=2$ pieces.

Maximum pieces $=3\times2=$ 6.

 <------------- 15 m ------------->
 +-----+-----+-----+
 | 4x4 | 4x4 | 4x4 |
 +-----+-----+-----+
 | 4x4 | 4x4 | 4x4 |
 +-----+-----+-----+
 (9 m side: 2 pieces, 1 m left over; 15 m side: 3 pieces, 3 m left over; 3 x 2 = 6)

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Sudurpashchim · 2 marks

Simplify: $\dfrac{p}{x-y} - \dfrac{p}{x+y}$

Answer

\frac{p}{x-y}-\frac{p}{x+y}=\frac{p(x+y)-p(x-y)}{x^2-y^2}=\frac{2py}{x^2-y^2}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Chaitra · Sudurpashchim · 3 marks

Solve: $2^x + 2^{-x} - 1 = \dfrac{3}{2}$

Answer

$2^x+2^{-x}-1=\dfrac32 \Rightarrow 2^x+\dfrac1{2^x}=\dfrac52$

Let $2^x=y$ : $2y^2-5y+2=0 \Rightarrow (y-2)(2y-1)=0$

$y=2 \Rightarrow x=1$
$y=\frac12 \Rightarrow 2^x=2^{-1} \Rightarrow x=-1$

Answer: $x=\pm1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Koshi · 1+2+2 marks

Ramesh deposits the amount in a co-operative for 7 days by increasing the amount every day double than the previous day. He deposited Rs. 10 on the first day, Rs. 20 on the second day, Rs. 40 on the third day and so on till the 7th day.
(a) What type of series is formed from the deposit amount according to above context? Write it. (1)
(b) How much amount will Ramesh deposit by the end of 7 days? Find it using formula. (2)
(c) If Ramesh withdraws the amount deposited by 4 days, how much will he receive at the end of the 7th days? Find it. (2)

Answer

(a) 10, 20, 40, ... have a common ratio 2, so it is a geometric series.

(b) $a=10$ , $r=2$ , $n=7$ .

S_7=\frac{a(r^n-1)}{r-1}=\frac{10(2^7-1)}{2-1}=10\times127=1270

He deposits Rs 1,270.

(c) Deposit in the first 4 days: $S_4=10(2^4-1)=150$ .

Amount he receives at the end of the 7th day (the remaining deposits of days 5 to 7) $=1270-150=$ Rs 1,120.

(Check: $160+320+640=1120$ .)

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Koshi · 1+2+2 marks

The longer side of a rectangular field is 40 m more than the shorter side and its diagonal is 40 m more than its longer side.
(a) Write the relation among the length ( $l$ ), breadth ( $b$ ) and diagonal ( $d$ ) of the field according to the above context. (1)
(b) Find the length of the shorter side and longer side of the rectangular filed. (2)
(c) How many maximum numbers of plots of size 30 m $\times$ 20 m can be made from the rectangular filed? Find it. (2)
Figure: A rectangle with length $l$ (horizontal), breadth $b$ (vertical) and a diagonal $d$ drawn from the top-left to the bottom-right corner.

Answer

(a) By the Pythagoras theorem, $d^2=l^2+b^2$ .

(b) Let the shorter side be $x$ m. Longer side $=(x+40)$ m, diagonal $=(x+80)$ m.

x^2+(x+40)^2=(x+80)^2

2x^2+80x+1600=x^2+160x+6400 \Rightarrow x^2-80x-4800=0

x=\frac{80\pm\sqrt{6400+19200}}{2}=\frac{80\pm160}{2}=120

Shorter side $=120$ m, longer side $=160$ m. (Diagonal $=200$ m.)

(c) Area of the field $=160\times120=19200\ \text{m}^2$ . Area of a plot $=30\times20=600\ \text{m}^2$ .

Placing the 30 m side along the 120 m side ( $120\div30=4$ ) and the 20 m side along the 160 m side ( $160\div20=8$ ) uses the field completely.

Maximum plots $=4\times8=$ 32.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Koshi · 2 marks

Simplify: $\frac{p+q}{pq}-\frac{q+r}{qr}-\frac{r+p}{rp}$

Answer

\frac{p+q}{pq}=\frac1q+\frac1p,\quad \frac{q+r}{qr}=\frac1r+\frac1q,\quad \frac{r+p}{rp}=\frac1p+\frac1r

\left(\frac1p+\frac1q\right)-\left(\frac1q+\frac1r\right)-\left(\frac1p+\frac1r\right)=-\frac2r

Answer: $-\dfrac{2}{r}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Koshi · 3 marks

Solve: $3^{y}+3^{-y}=9\frac{1}{9}$

Answer

$3^y+3^{-y}=9\frac19=\dfrac{82}{9}$

Let $3^y=z$ : $z+\dfrac1z=\dfrac{82}9 \Rightarrow 9z^2-82z+9=0$

9z^2-81z-z+9=0 \Rightarrow (z-9)(9z-1)=0

$z=9 \Rightarrow 3^y=3^2 \Rightarrow y=2$
$z=\frac19 \Rightarrow 3^y=3^{-2} \Rightarrow y=-2$

Answer: $y=\pm2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Madhesh · 1+2+1+2 marks

Hari deposited Rs.1000, Rs.2,000, Rs.3,000 in bank on his son Aashish's first, second and third birthday respectively. In this way, he increases the deposit by Rs. 1,000 on every birthday.
(a) Define mean in arithmetic series. (1)
(b) How much total money is deposited upto 10th birthday? Find it. (2)
(c) Is Rs.66,000 deposited on Aashish account by 11th birthday? Give reasons. (1)
(d) In how many years, will Rs.1,05,000 be deposited in the account of Aashish? Find it. (2)

Answer

(a) The terms lying between two given terms $a$ and $b$ of an arithmetic series are called arithmetic means. In particular, the single mean between $a$ and $b$ is $\dfrac{a+b}{2}$ .

(b) $a=1000$ , $d=1000$ , $n=10$ .

S_{10}=\frac{10}2[2\times1000+9\times1000]=5\times11000=55000

Total deposit up to the 10th birthday $=$ Rs 55,000.

(c)

S_{11}=\frac{11}2[2000+10\times1000]=\frac{11}2\times12000=66000

Yes, Rs 66,000 is deposited by the 11th birthday, because $S_{11}=66000$ .

(d) $S_n=105000$

\frac n2[2000+(n-1)1000]=105000

Simplifying: $\dfrac n2\times1000(n+1)=105000 \Rightarrow n^2+n=210$

n^2+n-210=0 \Rightarrow (n+15)(n-14)=0 \Rightarrow n=14

It takes 14 years.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Madhesh · 1+3 marks

Kriti wants to fence her field having length twice of breadth. The area of the field is 800 square feet.
(a) Write down the standard form of quadratic equation. (1)
(b) How much the length of wire is required to fence the field once with wire? Find it. (3)

Answer

(a) $ax^2+bx+c=0$ , where $a\neq0$ .

(b) Let the breadth be $x$ ft; the length is $2x$ ft.

x\times2x=800 \Rightarrow 2x^2=800 \Rightarrow x^2-400=0

$x=20$ (rejecting $-20$ ). So breadth $=20$ ft and length $=40$ ft.

Wire needed to fence once = perimeter:

P=2(l+b)=2(40+20)=120\text{ ft}

120 feet of wire is required.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Madhesh · 2 marks

Simplify: $\dfrac{x}{xy-y^2}+\dfrac{y}{xy-x^2}$

Answer

$xy-y^2=y(x-y)$ and $xy-x^2=-x(x-y)$ .

\frac{x}{y(x-y)}-\frac{y}{x(x-y)}=\frac{x^2-y^2}{xy(x-y)}=\frac{(x+y)(x-y)}{xy(x-y)}=\frac{x+y}{xy}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Madhesh · 3 marks

Solve: $2^x+\dfrac{16}{2^x}=10$

Answer

$2^x+\dfrac{16}{2^x}=10$ . Let $2^x=y$ .

y^2+16=10y \Rightarrow y^2-10y+16=0 \Rightarrow (y-2)(y-8)=0

$2^x=2 \Rightarrow x=1$
$2^x=8 \Rightarrow x=3$

Answer: $x=1$ or $x=3$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Bagmati · 1+2+2 marks

A shop sells shoes and slippers (in pairs) during winter season according to the following table.

Day 1st day 2nd day 3rd day 4th day 5th day
No. of shoes 2 4 8 16 ..........
No. of slippers 3 6 9 12 ..........
(a) Write the relationship between arithmetic mean and geometric mean. (1)
(b) How many slippers are sold up to 8th days? Find it. (2)
(c) Compare the total number of shoes and slippers sold up to 8th days. (2)
Figure: Table of shoes (2, 4, 8, 16, blank) and slippers (3, 6, 9, 12, blank) sold on days 1 to 5.

Day	1st day	2nd day	3rd day	4th day	5th day
No. of shoes	2	4	8	16	..........
No. of slippers	3	6	9	12	..........

Answer

(a) For two positive numbers $a$ and $b$ : $AM=\dfrac{a+b}2$ , $GM=\sqrt{ab}$ , and $AM\ge GM$ (equal only when $a=b$ ). Also $GM^2=a\times b$ .

(b) Slippers: 3, 6, 9, 12, ... is an AP with $a=3$ , $d=3$ , $n=8$ .

S_8=\frac82[2\times3+7\times3]=4\times27=108

108 pairs of slippers are sold.

(c) Shoes: 2, 4, 8, 16, ... is a GP with $a=2$ , $r=2$ , $n=8$ .

S_8=\frac{2(2^8-1)}{2-1}=2\times255=510

Shoes sold $=510$ pairs; slippers sold $=108$ pairs.

More shoes were sold than slippers, by $510-108=$ 402 pairs.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Bagmati · 3+2 marks

The area and perimeter of a rectangular field are 150 sq.m and 50 m respectively.
(a) What is the length and breadth of the field? Find it by making quadratic equation. (3)
(b) How much equal part should be subtracted from the length and breadth of the land to get the area 84 square meter? Solve it. (2)

Answer

(a) $l+b=\dfrac{50}2=25$ and $lb=150$ .

Let the length be $x$ ; breadth $=25-x$ .

x(25-x)=150 \Rightarrow x^2-25x+150=0

(x-15)(x-10)=0 \Rightarrow x=15\ \text{or}\ 10

Length $=15$ m, breadth $=10$ m.

(b) Let $y$ m be subtracted from both.

(15-y)(10-y)=84 \Rightarrow y^2-25y+150-84=0 \Rightarrow y^2-25y+66=0

(y-3)(y-22)=0 \Rightarrow y=3\ \text{or}\ 22

$y=22$ is impossible (larger than the breadth), so 3 m should be subtracted. (Check: $12\times7=84$ .)

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Bagmati · 1 mark

Write $\frac{1}{x^{-6}}$ in the positive index of $x$ .

Answer

$\dfrac1{x^{-6}}=x^{6}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 1 MCQ

Loading…

2081 Chaitra · Bagmati · 2 marks

Simplify: $\frac{a^2}{a-2b}+\frac{4b^2}{2b-a}$

Answer

\frac{a^2}{a-2b}+\frac{4b^2}{2b-a}=\frac{a^2}{a-2b}-\frac{4b^2}{a-2b}=\frac{a^2-4b^2}{a-2b}=\frac{(a+2b)(a-2b)}{a-2b}=a+2b

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Bagmati · 2 marks

Solve: $4^{x-2}=0.25$

Answer

$4^{x-2}=0.25=\dfrac14=4^{-1}$

$x-2=-1 \Rightarrow x=1$

Answer: $x=1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 1 MCQ

Loading…

2081 Chaitra · Gandaki · 1+2+2 marks

There are 7 arithmetic means between 3 and 27.
(a) Write the formula to calculate arithmetic mean between $a$ and $b$ .
(b) What is the $5^{th}$ mean of the given sequence? Find it.
(c) Which one is greater by how much in arithmetic mean and geometric mean between 3 and 27? Compare it.

Answer

(a) The arithmetic mean between $a$ and $b$ is $\dfrac{a+b}2$ .

(b) With 7 means, there are 9 terms: $a=3$ , $t_9=27$ .

27=3+8d \Rightarrow d=3

The 5th mean is the 6th term: $3+5\times3=$ 18.

(c) $AM=\dfrac{3+27}2=15$ , $GM=\sqrt{3\times27}=\sqrt{81}=9$ .

The arithmetic mean is greater than the geometric mean by $15-9=$ 6.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Gandaki · 1+2+2 marks

The perimeter and area of a rectangular ground are 44 meter and 120 square meters respectively.
(a) Write the formula to solve the quadratic equation $ax^2+bx+c=0,\ a\neq 0$ .
(b) Find the length and breadth of the ground.
(c) If the ground is made a square by reducing the length side, by what percent the area will be increased or decreased? Find it.

Answer

(a) $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

(b) $l+b=22$ and $lb=120$ . Let the length be $x$ , breadth $22-x$ .

x(22-x)=120 \Rightarrow x^2-22x+120=0 \Rightarrow (x-12)(x-10)=0

Length $=12$ m, breadth $=10$ m.

(c) Reducing the length to 10 m gives a square of side 10 m.

New area $=100\ \text{m}^2$ ; old area $=120\ \text{m}^2$ .

\text{Decrease}=\frac{120-100}{120}\times100\%=16\frac23\%

The area will decrease by $16\frac23\%$ (about 16.67%).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Gandaki · 2 marks

Simplify: $\dfrac{a}{a-b}+\dfrac{b}{b-a}$

Answer

\frac{a}{a-b}+\frac{b}{b-a}=\frac{a}{a-b}-\frac{b}{a-b}=\frac{a-b}{a-b}=1

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Lumbini · 1+2+2 marks

The first and last terms of an arithmetic series are 5 and 329 respectively. The sum of all terms is 4676.
(a) What is called mean in arithmetic series? Write it. (1)
(b) Find the common difference of the series. (2)
(c) How many terms should be removed from last in the series to make the sum 51? Find it by calculation. (2)

Answer

(a) In an arithmetic series, the arithmetic mean is the average of the first and last terms: $\text{A.M.}=\frac{a+l}{2}=\frac{S_n}{n}.$ Here, $\text{A.M.}=\frac{5+329}{2}=167.$

(b) $a=5$ , $l=329$ , $S_n=4676$ .

S_n=\frac n2(a+l) \Rightarrow 4676=\frac n2(5+329)=\frac n2\times334=167n \Rightarrow n=28

l=a+(n-1)d \Rightarrow 329=5+(28-1)d \Rightarrow 329=5+27d \Rightarrow 27d=324 \Rightarrow d=12

Common difference $=12$ .

(c) Suppose the first $m$ terms remain and have sum 51.

\frac m2[2a+(m-1)d]=51

\frac m2[10+(m-1)12]=51

\frac m2(12m-2)=51 \Rightarrow m(6m-1)=51

6m^2-m-51=0

m=\frac{1\pm\sqrt{1+1224}}{12}=\frac{1\pm35}{12}

Taking the positive value, $m=3.$

Check: $5+17+29=51$ .

Total terms $=28$ , so terms to be removed from the last: $28-3=25.$

25 terms should be removed from the last.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Lumbini · 1+2+2 marks

In a positive number of two digits, the product of two digits is 27. When 54 is subtracted from the number, the places of the digits are interchanged.
(a) If the two digits number be $10x + y$ , then write the number obtained by interchanging its digits. (1)
(b) Make a quadratic equation from the given verbal problem. (2)
(c) Find the number. (2)

Answer

(a) The number obtained by interchanging the digits is $10y+x$ .

(b) Tens digit $x$ , units digit $y$ . Then $xy=27$ .

Subtracting 54 interchanges the digits: $10x+y-54=10y+x \Rightarrow 9x-9y=54 \Rightarrow y=x-6$ .

x(x-6)=27 \Rightarrow x^2-6x-27=0

(c) $(x-9)(x+3)=0 \Rightarrow x=9$ (rejecting $-3$ ). Then $y=3$ .

The number is 93. Check: $9\times3=27$ and $93-54=39$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Lumbini · 3 marks

Solve: $5^x + \dfrac{1}{5^x} = 5\dfrac{1}{5}$

Answer

$5^x+\dfrac1{5^x}=5\frac15=\dfrac{26}5$ . Let $5^x=z$ .

5z^2-26z+5=0 \Rightarrow 5z^2-25z-z+5=0 \Rightarrow (z-5)(5z-1)=0

$z=5 \Rightarrow x=1$
$z=\frac15 \Rightarrow 5^x=5^{-1} \Rightarrow x=-1$

Answer: $x=\pm1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Karnali · 1+2+2 marks

The number of words learned by a child daily in the double than previous day is shown in the following table.

Day 1st 2nd 3rd 4th .... ....
Number of words 3 6 12 24 .... ....
(a) In which sequence is the child learning words? Write it. (1)
(b) How many words does the child learn upto 8 days? Find it using formula. (2)
(c) In how many days will the child learn 6141 words? Find it. (2)

Day	1st	2nd	3rd	4th	....	....
Number of words	3	6	12	24	....	....

Answer

(a) 3, 6, 12, 24, ... have a common ratio 2, so the child learns in a geometric sequence.

(b) $a=3$ , $r=2$ , $n=8$ .

S_8=\frac{3(2^8-1)}{2-1}=3\times255=765

The child learns 765 words up to 8 days.

(c) $S_n=6141$

3(2^n-1)=6141 \Rightarrow 2^n-1=2047 \Rightarrow 2^n=2048=2^{11}

$n=11$ . It takes 11 days.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Karnali · 1+2+2 marks

A two digit number is four times the sum of digits and three times the product of digits.
(a) If one's place digit be $y$ and ten's place digit is $x$ , write the two digit number in algebraic form. (1)
(b) Make a quadratic equation in terms of $x$ according to given conditions. (2)
(c) Find the number. (2)

Answer

(a) The number is $10x+y$ .

(b) The number is four times the sum of digits: $10x+y=4(x+y) \Rightarrow 6x=3y \Rightarrow y=2x$ .

The number is three times the product of digits: $10x+y=3xy$ .

Substituting $y=2x$ : $10x+2x=3x(2x) \Rightarrow 12x=6x^2$

6x^2-12x=0 \quad\text{or}\quad x^2-2x=0

(c) $x(x-2)=0 \Rightarrow x=2$ (since $x=0$ gives no two-digit number). Then $y=4$ .

The number is 24. Check: $4\times(2+4)=24$ and $3\times(2\times4)=24$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Karnali · 2 marks

Simplify: $\frac{1}{b-1} - \frac{1}{b+1}$

Answer

\frac1{b-1}-\frac1{b+1}=\frac{(b+1)-(b-1)}{(b-1)(b+1)}=\frac{2}{b^2-1}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Karnali · 3 marks

Solve: $7^x + 7^{-x} = 7\frac{1}{7}$

Answer

$7^x+\dfrac1{7^x}=7\frac17=\dfrac{50}7$ . Let $7^x=z$ .

7z^2-50z+7=0 \Rightarrow 7z^2-49z-z+7=0 \Rightarrow (z-7)(7z-1)=0

$z=7 \Rightarrow x=1$
$z=\frac17 \Rightarrow x=-1$

Answer: $x=\pm1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Sudurpashchim · 1+2+2 marks

Hira collected following sum of money in the first 5 days of month Baisakh.

Baisakh -1 Baisakh -2 Baisakh -3 Baisakh -4 Baisakh -5
Rs. 10 Rs. 20 Rs. 40 Rs. 80 Rs. 160
(a) What is the mean value of the amounts collected on 2nd Baisakh and 4th Baisakh? Write it. (1)
(b) How much money will be collected by 10th days? Find using formula. (2)
(c) Up to how many days of Baisakh can Rs. 1,63,830 be collected? Find it. (2)

Baisakh -1	Baisakh -2	Baisakh -3	Baisakh -4	Baisakh -5
Rs. 10	Rs. 20	Rs. 40	Rs. 80	Rs. 160

Answer

(a) The collected amounts 10, 20, 40, 80, 160 form a GP. The mean between the 2nd (Rs 20) and 4th (Rs 80) amounts is the geometric mean:

GM=\sqrt{20\times80}=\sqrt{1600}=\text{Rs }40

(this is the 3rd day's amount). The arithmetic mean would be $\frac{20+80}2=$ Rs 50.

(b) $a=10$ , $r=2$ , $n=10$ .

S_{10}=\frac{10(2^{10}-1)}{2-1}=10\times1023=10230

Rs 10,230 is collected by the 10th day.

(c) $S_n=163830$

10(2^n-1)=163830 \Rightarrow 2^n=16384=2^{14}

$n=14$ . Up to the 14th day of Baisakh.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Sudurpashchim · 1+2+2 marks

The length of rectangular field is twice of its breadth and its area is 200 square meter.
(a) Write the standard form of quadratic equation. (1)
(b) Find the length and breadth of the rectangular field. (2)
(c) How many maximum numbers of pieces having the size $5\text{ m}\times 4\text{ m}$ can be made in the field? Also present it in diagram. (2)

Answer

(a) $ax^2+bx+c=0$ , where $a\neq0$ .

(b) Let the breadth be $x$ m; the length is $2x$ m.

2x^2=200 \Rightarrow x^2=100 \Rightarrow x=10

Breadth $=10$ m, length $=20$ m.

(c) Place the 4 m side along the 20 m length ( $20\div4=5$ ) and the 5 m side along the 10 m breadth ( $10\div5=2$ ).

Maximum pieces $=5\times2=$ 10 (area check: $200\div20=10$ ).

 <----------- 20 m ----------->
 +----+----+----+----+----+   ^
 | 4x5| 4x5| 4x5| 4x5| 4x5|   |
 +----+----+----+----+----+   | 10 m
 | 4x5| 4x5| 4x5| 4x5| 4x5|   |
 +----+----+----+----+----+   v

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Chaitra · Sudurpashchim · 2 marks

Simplify:
$\frac{1}{x-y}-\frac{1}{x+y}$

Answer

\frac1{x-y}-\frac1{x+y}=\frac{(x+y)-(x-y)}{(x-y)(x+y)}=\frac{2y}{x^2-y^2}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Koshi · 1+2+2 marks

More than 650 species of birds are found in Arun Valley. Let among them, the total number of birds of 10 species is 2,04,600 where one species is double of the former species.
(a) To which sequence of the given context is related with? Write it. (1)
(b) Find the number of birds of minimum first species. (2)
(c) How many times is the number of fourth species of the birds than that of second species? Calculate it. (2)

Answer

(a) Each species has double the birds of the former, so the numbers form a geometric sequence (series) with $r=2$ .

(b) $S_{10}=204600$ , $r=2$ , $n=10$ ; let the first species have $a$ birds.

\frac{a(2^{10}-1)}{2-1}=204600 \Rightarrow 1023\,a=204600 \Rightarrow a=200

The minimum (first) species has 200 birds.

(c) 2nd species $=ar=400$ ; 4th species $=ar^3=1600$ .

\frac{ar^3}{ar}=r^2=4

The fourth species has 4 times as many birds as the second.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Koshi · 2+2 marks

The product of two consecutive positive numbers is 420.
(a) What are the numbers? Find it. (2)
(b) What should be subtracted from the numbers to get their product as 182? Find it. (2)

Answer

(a) Let the numbers be $x$ and $x+1$ .

x(x+1)=420 \Rightarrow x^2+x-420=0 \Rightarrow (x+21)(x-20)=0

$x=20$ (positive). The numbers are 20 and 21.

(b) Let $k$ be subtracted from each.

(20-k)(21-k)=182 \Rightarrow k^2-41k+420-182=0 \Rightarrow k^2-41k+238=0

k=\frac{41\pm\sqrt{1681-952}}{2}=\frac{41\pm27}{2}=7\ \text{or}\ 34

$k=34$ gives negative numbers, so reject it. 7 should be subtracted. (Check: $13\times14=182$ .)

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Koshi · 1 mark

What is the value of $\left(\frac{x}{y}\right)^0$ ? Write it.

Answer

$\left(\dfrac xy\right)^0=1$ (any non-zero quantity raised to the power 0 equals 1).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 1 MCQ

Loading…

2080 Chaitra · Koshi · 3 marks

Simplify: $\frac{3x+1}{9x^2+3x+1}+\frac{3x-1}{9x^2-3x+1}+\frac{2}{81x^4+9x^2+1}$

Answer

Let $t=3x$ . Then $9x^2=t^2$ and $81x^4=t^4$ .

\frac{t+1}{t^2+t+1}+\frac{t-1}{t^2-t+1}+\frac2{t^4+t^2+1}

Since $t^4+t^2+1=(t^2+t+1)(t^2-t+1)$ , the common denominator is $t^4+t^2+1$ .

\frac{(t+1)(t^2-t+1)+(t-1)(t^2+t+1)+2}{t^4+t^2+1} =\frac{(t^3+1)+(t^3-1)+2}{t^4+t^2+1}=\frac{2t^3+2}{(t^2+t+1)(t^2-t+1)}

=\frac{2(t+1)(t^2-t+1)}{(t^2+t+1)(t^2-t+1)}=\frac{2(t+1)}{t^2+t+1}

Putting back $t=3x$ :

\frac{2(3x+1)}{9x^2+3x+1}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Koshi · 2 marks

Solve by completing square: $x^2-5x+6=0$

Answer

$x^2-5x+6=0 \Rightarrow x^2-5x=-6$

Add $\left(\dfrac52\right)^2=\dfrac{25}4$ to both sides:

x^2-5x+\frac{25}4=-6+\frac{25}4 \Rightarrow \left(x-\frac52\right)^2=\frac14

x-\frac52=\pm\frac12 \Rightarrow x=\frac52\pm\frac12

Answer: $x=3$ or $x=2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Madhesh · 5 marks

Urvasi made a music video and uploaded in a social media. In an interval of every hour, new viewers are observed the video increasing equally as below.

Hours 1 2 3 ...
New viewers 500 550 600 .....
(a) In which sequence new viewers are increasing? (1)
(b) How many total new viewers will be there till $10^{th}$ hour? Find it. (2)
(c) Does the music video make 7250 viewers in 10 hours? Find it. (2)
Figure: Table of hours 1, 2, 3, ... against new viewers 500, 550, 600, .

Hours	1	2	3	...
New viewers	500	550	600	.....

Answer

(a) 500, 550, 600, ... increase by a constant 50, so the new viewers increase in an arithmetic sequence.

(b) $a=500$ , $d=50$ , $n=10$ .

S_{10}=\frac{10}2[2\times500+9\times50]=5[1000+450]=7250

There will be 7,250 new viewers till the 10th hour.

(c) Yes. From (b), the total number of viewers in 10 hours is exactly 7250, so the video makes 7250 viewers in 10 hours.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Madhesh · 6 marks

(a) What is the value of $\frac{x}{x-7} - \frac{x}{x-7}$ ? Write it. (1)
(b) Simplify: $\frac{4x^2+y^2}{4x^2-y^2} - \frac{2x-y}{2x+y}$ (2)
(c) The present ages of Kamal and Reena are 20 years and 16 years respectively. In how many years, the numerical product of their ages will be 480? Find it. (3)

Answer

(a) $\dfrac{x}{x-7}-\dfrac{x}{x-7}=0$

(b)

\frac{4x^2+y^2}{4x^2-y^2}-\frac{2x-y}{2x+y} =\frac{4x^2+y^2}{4x^2-y^2}-\frac{(2x-y)(2x-y)}{(2x+y)(2x-y)}

Write the second fraction with denominator $4x^2-y^2$ : $\dfrac{(2x-y)^2}{4x^2-y^2}$ .

=\frac{4x^2+y^2-(4x^2-4xy+y^2)}{4x^2-y^2}=\frac{4xy}{4x^2-y^2}

(c) After $t$ years the ages are $(20+t)$ and $(16+t)$ .

(20+t)(16+t)=480 \Rightarrow t^2+36t+320=480 \Rightarrow t^2+36t-160=0

t=\frac{-36\pm\sqrt{1296+640}}{2}=\frac{-36\pm44}{2}=4

After 4 years. (Check: $24\times20=480$ .)

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Madhesh · 4 marks

An exponential equation is given below.
$3^{2x} - 4 \times 3^{x+1} + 27 = 0$ (a) Find the value of $x$ solving the above equation. (2)
(b) Do the both values of $x$ satisfy the equation $4^x - 6 \times 2^{x+1} + 32 = 0$ ? Justify it. (2)

Answer

(a) $3^{2x}-4\times3^{x+1}+27=0 \Rightarrow (3^x)^2-12\cdot3^x+27=0$ . Let $3^x=y$ .

y^2-12y+27=0 \Rightarrow (y-3)(y-9)=0

$3^x=3 \Rightarrow x=1$
$3^x=9 \Rightarrow x=2$

So $x=1$ or $x=2$ .

(b) Test $4^x-6\times2^{x+1}+32=0$ , i.e. $(2^x)^2-12\cdot2^x+32=0$ .

For $x=1$ : $4-6\times4+32=4-24+32=12\neq0$ .

For $x=2$ : $16-6\times8+32=16-48+32=0$ . Satisfied.

No, both values do not satisfy the equation. Only $x=2$ does; $x=1$ does not. (The equation's roots are $x=2$ and $x=3$ .)

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Bagmati · 1+2+2 marks

A table tennis ball is dropped on a wooden board and returned till the height of 1024 mm in the first bounce. The ball bounces just half of the preceding height continuously.
a) In which series the above context can be expressed? Write it. (1)
b) What is the total distance of heights, the ball may reach till the fifth bounce? Calculate it. (2)
c) Find all mean heights between the first and fifth bounces. (2)

Answer

(a) The heights 1024, 512, 256, ... have a constant ratio $\frac12$ , so they form a geometric series.

(b) $a=1024$ mm, $r=\dfrac12$ , $n=5$ .

S_5=\frac{a(1-r^5)}{1-r}=\frac{1024\left(1-\frac1{32}\right)}{\frac12}=2048\times\frac{31}{32}=1984

Total of the heights $=$ 1984 mm.

(c) Heights: 1st $=1024$ , 5th $=1024\times\left(\frac12\right)^4=64$ . The means between them are the 2nd, 3rd and 4th terms:

1024\times\tfrac12=512,\quad 256,\quad 128

The mean heights are 512 mm, 256 mm and 128 mm.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Bagmati · 1+2+2 marks

The product of two positive numbers is 128 and the first number is double of the second number.
a) If the second number be $x$ , what is the first number? Write it. (1)
b) Find the two numbers. (2)
c) If the first number was trebled of the second number, what would be the difference between their products? Calculate it. (2)

Answer

(a) The first number $=2x$ .

(b) $2x\cdot x=128 \Rightarrow x^2=64 \Rightarrow x=8$ (positive).

The numbers are 16 and 8.

(c) If the first number is three times the second (second remains 8), the first is $3\times8=24$ .

New product $=24\times8=192$ .

Difference $=192-128=$ 64.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Bagmati · 2 marks

Simplify:
$\frac{x+y}{x-y} - \frac{x-y}{x+y} - \frac{2xy}{x^2-y^2}$

Answer

\frac{x+y}{x-y}-\frac{x-y}{x+y}-\frac{2xy}{x^2-y^2} =\frac{(x+y)^2-(x-y)^2-2xy}{x^2-y^2}

=\frac{4xy-2xy}{x^2-y^2}=\frac{2xy}{x^2-y^2}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Gandaki · 1+2+2 marks

A student bought a Nepali dictionary from the Sajha Pustak Bhandar. He decided to learn new words in arithmetic sequence every day. On the first day he learned 20 new words and on the fifth day 40 new words.
a) How many terms are there in an arithmetic sequence having five means? (1)
b) How many new words did he learn in the succeeding day? Find it. (2)
c) How many new words will he learn till tenth day? Find it. (2)

Answer

(a) With five means between two terms there are $5+2=$ 7 terms.

(b) $a=20$ , $t_5=40$ .

20+4d=40 \Rightarrow d=5

The words learned on successive days are 20, 25, 30, 35, 40, 45, ... So on the succeeding day (the 6th day, after the fifth) he learns $40+5=$ 45 words.

(c)

S_{10}=\frac{10}2[2\times20+9\times5]=5[40+45]=425

He learns 425 new words till the tenth day.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Gandaki · 1+2+2 marks

The present age of father is 50 years and the age of his daughter is 13 years.
a) What were the ages of father and his daughter before x years? Write it. (1)
b) How many years before, the numerical product of their ages was 360? Calculate it by forming quadratic equation. (2)
c) Does the product of their ages after 12 years, a perfect square? Give reason. (2)

Answer

(a) Before $x$ years: father $=(50-x)$ years, daughter $=(13-x)$ years.

(b)

(50-x)(13-x)=360 \Rightarrow x^2-63x+650-360=0 \Rightarrow x^2-63x+290=0

x=\frac{63\pm\sqrt{3969-1160}}{2}=\frac{63\pm53}{2}=5\ \text{or}\ 58

$x=58$ is impossible (daughter's age would be negative). So 5 years ago. (Check: $45\times8=360$ .)

(c) After 12 years: father $=62$ , daughter $=25$ . Product $=62\times25=1550$ .

$39^2=1521$ and $40^2=1600$ , so 1550 lies between two consecutive squares. No, it is not a perfect square.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Gandaki · 2 marks

Simplify: $\dfrac{1}{x-y} - \dfrac{y}{xy+y^2}$

Answer

$xy+y^2=y(x+y)$

\frac1{x-y}-\frac{y}{y(x+y)}=\frac1{x-y}-\frac1{x+y}=\frac{(x+y)-(x-y)}{x^2-y^2}=\frac{2y}{x^2-y^2}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Gandaki · 3 marks

If $x = 2^{\frac{1}{3}} - 2^{-\frac{1}{3}}$ , prove that: $2x^3 + 6x - 3 = 0$

Answer

Let $a=2^{\frac13}$ , $b=2^{-\frac13}$ . Then $ab=1$ , $a^3=2$ , $b^3=\dfrac12$ , and $x=a-b$ .

Cubing: $x^3=a^3-b^3-3ab(a-b)$

x^3=2-\frac12-3x \Rightarrow x^3+3x=\frac32

Multiplying by 2: $2x^3+6x=3$

2x^3+6x-3=0

Hence proved.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Lumbini · 1+2+3 marks

A civil servant decided to deposit certain amount of his salary in a cooperative in the arithmetic sequence. He deposited Rs 6000 in the first month, Rs 6100 in the second month, Rs 6200 in the third month and so on.
a) Write the formula to find the sum of the terms of an arithmetic sequence. (1)
b) How much amount does he save within a year in the cooperative? Find it. (2)
c) How many months will he take to deposit Rs 64,500? Calculate it. (3)

Answer

(a) $S_n=\dfrac n2[2a+(n-1)d]$ (or $S_n=\dfrac n2(a+l)$ ).

(b) $a=6000$ , $d=100$ , $n=12$ .

S_{12}=\frac{12}2[2\times6000+11\times100]=6[12000+1100]=78600

He saves Rs 78,600 in a year.

(c) $S_n=64500$

\frac n2[12000+(n-1)100]=64500 \Rightarrow n(11900+100n)=129000

Dividing by 100: $n^2+119n-1290=0$

n=\frac{-119\pm\sqrt{14161+5160}}{2}=\frac{-119\pm139}{2}=10

(Check: $S_{10}=5[12000+900]=64500$ .)

He takes 10 months.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Lumbini · 1+1+2 marks

The ages of father and his daughter were 48 years and 17 years respectively in 2079 B.S.
a) What will be the age of the father and his daughter after $x$ years? Write it. (1)
b) Prove that the product of the ages of father in 3 years ago and the age of daughter 3 years hence is a square number. (1)
c) In which year (B.S.), the numerical product of their ages will be 1020? Calculate it by forming quadratic equation. (2)

Answer

(a) After $x$ years: father $=(48+x)$ years, daughter $=(17+x)$ years.

(b) Father 3 years ago $=48-3=45$ ; daughter 3 years hence $=17+3=20$ .

45\times20=900=30^2

So the product is a perfect square. Hence proved.

(c) After $x$ years from 2079 B.S.:

(48+x)(17+x)=1020 \Rightarrow x^2+65x+816-1020=0 \Rightarrow x^2+65x-204=0

x=\frac{-65\pm\sqrt{4225+816}}{2}=\frac{-65\pm71}{2}=3

(Check: $51\times20=1020$ .)

Year $=2079+3=$ 2082 B.S.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Lumbini · 2 marks

Simplify: $\frac{m+n}{m-n}-\frac{m-n}{m+n}$

Answer

\frac{m+n}{m-n}-\frac{m-n}{m+n}=\frac{(m+n)^2-(m-n)^2}{m^2-n^2}=\frac{4mn}{m^2-n^2}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Lumbini · 3 marks

Solve: $6^{x}+6^{-x}=6\frac{1}{6}$

Answer

$6^x+\dfrac1{6^x}=6\frac16=\dfrac{37}6$ . Let $6^x=z$ .

6z^2-37z+6=0 \Rightarrow 6z^2-36z-z+6=0 \Rightarrow (z-6)(6z-1)=0

$z=6 \Rightarrow x=1$
$z=\frac16 \Rightarrow x=-1$

Answer: $x=\pm1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Karnali · 1+2+2 marks

The workers in a brick factory laid down bricks in rows every day in the sun. They laid 300 bricks in the first row in the first day, 320 bricks in the second row in the second day, 340 bricks in the third row in the third day and so on increasing in equal number.
(a) In which sequence did they lay down the bricks? Write it. (1)
(b) How many total bricks will they lay down within 15 days? Find it. (2)
(c) Compare the number of bricks laid down till the fifth rows and till the second rows. (2)

Answer

(a) 300, 320, 340, ... increase by a constant 20, so the bricks are laid in an arithmetic sequence.

(b) $a=300$ , $d=20$ , $n=15$ .

S_{15}=\frac{15}2[2\times300+14\times20]=\frac{15}2\times880=6600

6,600 bricks within 15 days.

(c)

S_5=\frac52[600+4\times20]=\frac52\times680=1700,\qquad S_2=\frac22[600+20]=620

Bricks till the 5th row $=1700$ , till the 2nd row $=620$ . Difference $=1700-620=1080$ ; the number till the fifth row is about 2.7 times that till the second row ( $1700:620=85:31$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Karnali · 1+2+2 marks

The sum of present ages of two brothers is 20 years. Four years ago, the product of their ages in numbers was 48.
(a) Define quadratic equation. (1)
(b) Find the quadratic equation formed from the given statement. (2)
(c) If possible, find their present ages. If not, give reason. (2)

Answer

(a) An equation of the form $ax^2+bx+c=0$ , where $a,b,c$ are real and $a\neq0$ , is called a quadratic equation.

(b) Let one brother's present age be $x$ ; the other's is $20-x$ .

Four years ago: $(x-4)(20-x-4)=48$

(x-4)(16-x)=48 \Rightarrow -x^2+20x-64=48 \Rightarrow x^2-20x+112=0

(c) Discriminant $=b^2-4ac=400-448=-48<0$ .

The discriminant is negative, so the equation has no real roots. It is not possible to find their present ages (such ages do not exist).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Karnali · 3 marks

Solve: $9^x - 10\times 3^x + 9 = 0$

Answer

$9^x-10\times3^x+9=0 \Rightarrow (3^x)^2-10\cdot3^x+9=0$ . Let $3^x=y$ .

y^2-10y+9=0 \Rightarrow (y-1)(y-9)=0

$3^x=1 \Rightarrow x=0$
$3^x=9 \Rightarrow x=2$

Answer: $x=0$ or $x=2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Sudurpashchim · 1+2+2 marks

There are 7 terms in an arithmetic sequence. The first term is 7 and the last term is $-17$ .
(a) If the first term of an arithmetic sequence is $a$ and the third term is $b$ , write the formula to find the mean. (1)
(b) Find all the arithmetic means of the sequence. (2)
(c) If 5 terms are added to that sequence, then calculate the sum of the terms of the new sequence. (2)

Answer

(a) The terms $a$ and $b$ are the first and third terms, so the term between them is the arithmetic mean: $m=\dfrac{a+b}2$ .

(b) $a=7$ , $t_7=-17$ .

-17=7+6d \Rightarrow d=-4

The terms are $7,\ 3,\ -1,\ -5,\ -9,\ -13,\ -17$ . The five arithmetic means are 3, -1, -5, -9, -13.

(c) After adding 5 terms (continuing the sequence), $n=12$ , $a=7$ , $d=-4$ .

S_{12}=\frac{12}2[2\times7+11\times(-4)]=6[14-44]=-180

The sum of the new sequence is -180.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Sudurpashchim · 1+2+2 marks

Ramesh purchased a rectangular field to make a house in Surkhet. The area of the field is 180 square meter and its perimeter is 56 meter.
(a) Write the standard form of the quadratic equation. (1)
(b) What is the breadth of the field? Find it. (2)
(c) Can he make a 21 meter long straight wall in the field? Write with reason. (2)

Answer

(a) $ax^2+bx+c=0$ , where $a\neq0$ .

(b) $l+b=28$ and $lb=180$ . Let the breadth be $x$ m; length $=(28-x)$ m.

x(28-x)=180 \Rightarrow x^2-28x+180=0

(x-10)(x-18)=0 \Rightarrow x=10\ \text{or}\ 18

Breadth $=10$ m (the smaller side), length $=18$ m.

(c) The longest straight line in a rectangle is its diagonal.

d=\sqrt{18^2+10^2}=\sqrt{324+100}=\sqrt{424}\approx20.6\text{ m}

Since $20.6\text{ m}<21\text{ m}$ , he cannot make a 21 m long straight wall in the field.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Sudurpashchim · 2 marks

Simplify: $\frac{1}{a+2b} + \frac{2a}{a^2-4b^2}$

Answer

\frac1{a+2b}+\frac{2a}{a^2-4b^2}=\frac{(a-2b)+2a}{(a+2b)(a-2b)}=\frac{3a-2b}{a^2-4b^2}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Chaitra · Sudurpashchim · 3 marks

Solve: $4 \times 3^{x+1} - 9^x = 27$

Answer

$4\times3^{x+1}-9^x=27 \Rightarrow 12\cdot3^x-(3^x)^2=27$ . Let $3^x=y$ .

y^2-12y+27=0 \Rightarrow (y-3)(y-9)=0

$3^x=3 \Rightarrow x=1$
$3^x=9 \Rightarrow x=2$

Answer: $x=1$ or $x=2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Shrawan · Grade increment · 1+2+2 marks

The third term and sixth term of a geometric series are 27 and 729 respectively.
(a) What is the geometric mean of $a$ and $b$ ? Write it. (1)
(b) Find the common ratio of the series. (2)
(c) Find the sum of first five terms of the series. (2)

Answer

(a) The geometric mean of $a$ and $b$ is $\sqrt{ab}$ .

(b) $t_3=ar^2=27$ and $t_6=ar^5=729$ .

\frac{ar^5}{ar^2}=r^3=\frac{729}{27}=27 \Rightarrow r=3

(c) $a=\dfrac{27}{r^2}=\dfrac{27}9=3$ .

S_5=\frac{a(r^5-1)}{r-1}=\frac{3(243-1)}{2}=\frac{3\times242}2=363

Sum of the first five terms $=363$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Shrawan · Grade increment · 1+2+2 marks

The length of the rectangular garden is 2 m. more than the breadth and the area of the garden is 48 square meter.
(a) Write the formula to solve the quadratic equation $ax^2 + bx + c = 0$ . (1)
(b) Find the length and breadth of the rectangular garden. (2)
(c) How many maximum numbers of plots having size 3 m × 2 m can be made on that garden? Find it. (2)

Answer

(a) $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

(b) Let the breadth be $b$ m; length $=(b+2)$ m.

b(b+2)=48 \Rightarrow b^2+2b-48=0 \Rightarrow (b+8)(b-6)=0

$b=6$ (rejecting $-8$ ). Breadth $=6$ m, length $=8$ m.

(c) Plot $3\text{ m}\times2\text{ m}$ . Put the 2 m side along the 8 m length ( $8\div2=4$ ) and the 3 m side along the 6 m breadth ( $6\div3=2$ ).

Maximum plots $=4\times2=$ 8 (area check: $48\div6=8$ ).

 <-------- 8 m -------->
 +----+----+----+----+   ^
 |2x3 |2x3 |2x3 |2x3 |   | 6 m
 +----+----+----+----+   |
 |2x3 |2x3 |2x3 |2x3 |   v
 +----+----+----+----+

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Shrawan · Grade increment · 2 marks

Solve: $3^{x+1} + 3^{x+2} = 36$

Answer

$3^{x+1}+3^{x+2}=36 \Rightarrow 3^x(3+9)=36$

3^x=3 \Rightarrow x=1

Answer: $x=1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2081 Shrawan · Grade increment · 3 marks

Simplify: $\frac{1}{x+1} + \frac{1}{x-1} + \frac{2x}{x^2+1}$

Answer

\frac1{x+1}+\frac1{x-1}=\frac{(x-1)+(x+1)}{x^2-1}=\frac{2x}{x^2-1}

\frac{2x}{x^2-1}+\frac{2x}{x^2+1}=2x\cdot\frac{(x^2+1)+(x^2-1)}{(x^2-1)(x^2+1)}=\frac{2x\cdot2x^2}{x^4-1}

=\frac{4x^3}{x^4-1}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Grade increment · 1+2+2 marks

In a Business expo, daily sells of number of mobiles in two stalls are given below.

Stalls/Days 1st Day 2nd Day 3rd Day .......
Stall - A 20 25 30 .......
Stall - B 3 6 12 .......
(a) Write the number of sales of mobiles of Stall-A in series form. (1)
(b) How many total mobiles did Stall-A sell in a week? Find it. (2)
(c) Compare the sales of mobile phones sold by the stalls in a week. (2)
Figure: Table of daily mobile sales: Stall-A 20, 25, 30, ...; Stall-B 3, 6, 12, ... for days 1, 2, 3, .

Stalls/Days	1st Day	2nd Day	3rd Day	.......
Stall - A	20	25	30	.......
Stall - B	3	6	12	.......

Answer

(a) Stall-A: $20+25+30+\cdots$ (an arithmetic series with $a=20$ , $d=5$ ).

(b) $n=7$ days.

S_7=\frac72[2\times20+6\times5]=\frac72\times70=245

Stall-A sold 245 mobiles in a week.

(c) Stall-B: 3, 6, 12, ... is a GP with $a=3$ , $r=2$ .

S_7=\frac{3(2^7-1)}{2-1}=3\times127=381

Stall-B sold 381 and Stall-A sold 245. Stall-B sold more, by $381-245=136$ mobiles.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Grade increment · 1+3+1 marks

The numerical product of the present ages of two sisters Rita and Gita is 160. Four years ago Rita was twice as old as Gita.
(a) If the age of Gita 4 years ago was $x$ years then what was the age of Rita? Write it. (1)
(b) Find the present ages of Rita and Gita. (3)
(c) If both live on, then what will be their ages after 10 years? Find it. (1)

Answer

(a) Rita's age 4 years ago $=2x$ years.

(b) Present ages: Gita $=(x+4)$ , Rita $=(2x+4)$ .

(x+4)(2x+4)=160 \Rightarrow 2x^2+12x+16=160 \Rightarrow x^2+6x-72=0

(x+12)(x-6)=0 \Rightarrow x=6

Gita $=6+4=$ 10 years, Rita $=2\times6+4=$ 16 years. (Check: $10\times16=160$ .)

(c) After 10 years: Gita $=20$ years, Rita $=26$ years.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Grade increment · 2 marks

Simplify: $\dfrac{1}{x-1}-\dfrac{2}{x^2-1}$

Answer

\frac1{x-1}-\frac2{x^2-1}=\frac{(x+1)-2}{(x-1)(x+1)}=\frac{x-1}{(x-1)(x+1)}=\frac1{x+1}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Grade increment · 3 marks

Solve: $2^x+\dfrac{1}{2^x}=4\dfrac{1}{4}$

Answer

$2^x+\dfrac1{2^x}=4\frac14=\dfrac{17}4$ . Let $2^x=y$ .

4y^2-17y+4=0 \Rightarrow 4y^2-16y-y+4=0 \Rightarrow (y-4)(4y-1)=0

$y=4 \Rightarrow x=2$
$y=\frac14 \Rightarrow 2^x=2^{-2} \Rightarrow x=-2$

Answer: $x=\pm2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Supplementary · 1 mark

What is the value of $11a^0$ ? Write it.

Answer

Any non-zero quantity to the power $0$ is $1$ , so $11a^0 = 11\times 1 = 11$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 1 MCQ

Loading…

2075 Chaitra · Set AP · 2 marks

Prove that: $\dfrac{7^{x+2}+7^x}{50\times 7^x}=1$ .

Answer

$\text{LHS}=\frac{7^{x+2}+7^x}{50\times 7^x}=\frac{7^x\cdot 7^2+7^x}{50\times 7^x}=\frac{7^x(49+1)}{50\times 7^x}=\frac{50}{50}=1=\text{RHS}$

Hence proved.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Supplementary · 2 marks

Prove that: $\dfrac{7^{x+1}-7^x}{12\times 7^x}=\dfrac{1}{2}$ .

Answer

$\text{LHS}=\frac{7^{x+1}-7^x}{12\times 7^x}=\frac{7^x\cdot 7-7^x}{12\times 7^x}=\frac{7^x(7-1)}{12\times 7^x}=\frac{6}{12}=\frac12=\text{RHS}$

Hence proved.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set AP · 2 marks

Rationalize the denominator of: $\dfrac{\sqrt{2}+1}{\sqrt{2}-1}$ .

Answer

Multiply the numerator and denominator by the conjugate $\sqrt2+1$ of the denominator.

$\frac{\sqrt2+1}{\sqrt2-1}=\frac{(\sqrt2+1)(\sqrt2+1)}{(\sqrt2-1)(\sqrt2+1)}=\frac{(\sqrt2+1)^2}{2-1}=\frac{2+2\sqrt2+1}{1}=3+2\sqrt2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Supplementary · 2 marks

Rationalize the denominator of: $\dfrac{1}{2+\sqrt{3}}$ .

Answer

Multiply the numerator and denominator by the conjugate $2-\sqrt3$ .

$\frac{1}{2+\sqrt3}=\frac{2-\sqrt3}{(2+\sqrt3)(2-\sqrt3)}=\frac{2-\sqrt3}{4-3}=2-\sqrt3$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set AP · 2 marks

Find the L.C.M. of: $x^2-y^2,\ (x+y)^2$ .

Answer

$x^2-y^2=(x+y)(x-y)$ and $(x+y)^2=(x+y)(x+y)$ .

Taking each factor with its highest power:

$\text{L.C.M.}=(x+y)^2(x-y)$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Supplementary · 2 marks

Find the L.C.M. of: $z^2-y^2,\ (z-y)^2$ .

Answer

$z^2-y^2=(z+y)(z-y)$ and $(z-y)^2=(z-y)(z-y)$ .

Taking each factor with its highest power:

$\text{L.C.M.}=(z+y)(z-y)^2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set AP · 2 marks

Solve: $\sqrt{x-2}=\sqrt[3]{125}$ .

Answer

$\sqrt[3]{125}=5$ , so

$\sqrt{x-2}=5$

Squaring both sides: $x-2=25 \Rightarrow x=27$ .

Check: $\sqrt{27-2}=\sqrt{25}=5=\sqrt[3]{125}$ . So $x=27$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Supplementary · 2 marks

Solve: $\sqrt{x+11}=6$ .

Answer

$\sqrt{x+11}=6$

Squaring both sides: $x+11=36 \Rightarrow x=25$ .

Check: $\sqrt{25+11}=\sqrt{36}=6$ . So $x=25$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set AP · 4 marks

Find the H.C.F. of: $b^2(b^2+4bc+4c^2)$ , $b^5+8b^2c^3$ and $3b^4+b^3c-10b^2c^2$ .

Answer

Factorise each expression.

First: $b^2(b^2 + 4bc + 4c^2) = b^2(b + 2c)^2$ .

Second: $b^5 + 8b^2c^3 = b^2(b^3 + 8c^3) = b^2(b + 2c)(b^2 - 2bc + 4c^2)$ .

Third: $3b^4 + b^3c - 10b^2c^2 = b^2(3b^2 + bc - 10c^2) = b^2(3b - 5c)(b + 2c)$ .

Common factors in all three are $b^2$ and $(b + 2c)$ .

$\text{H.C.F.} = b^2(b + 2c)$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Supplementary · 4 marks

Find the H.C.F. of: $a^3+4a^2b+4ab^2$ , $a^4+8ab^3$ and $3a^4+a^3b-10a^2b^2$ .

Answer

Factorise each expression.

First: $a^3 + 4a^2b + 4ab^2 = a(a^2 + 4ab + 4b^2) = a(a + 2b)^2$ .

Second: $a^4 + 8ab^3 = a(a^3 + 8b^3) = a(a + 2b)(a^2 - 2ab + 4b^2)$ .

Third: $3a^4 + a^3b - 10a^2b^2 = a^2(3a^2 + ab - 10b^2) = a^2(3a - 5b)(a + 2b)$ .

Common factors in all three are $a$ and $(a + 2b)$ .

$\text{H.C.F.} = a(a + 2b)$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Supplementary · 5 marks

In 2060 BS, the age of a father was three times the age of his son. If the age of the father will be two times the age of his son in 2080 BS, in which year was the father born? Find it.

Answer

Let the son's age in 2060 BS be $x$ years. Then the father's age is $3x$ years.

In 2080 BS (after $20$ years), the son is $x + 20$ and the father is $3x + 20$ .

According to the question:

$3x + 20 = 2(x + 20) \Rightarrow 3x + 20 = 2x + 40 \Rightarrow x = 20$

So in 2060 BS the son was $20$ years old and the father was $3 \times 20 = 60$ years old.

Check: in 2080 BS the father is $80$ and the son is $40$ , and $80 = 2 \times 40$ .

$\text{Father's birth year} = 2060 - 60 = 2000 \text{ BS}$

The father was born in 2000 BS.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Grade increment · 1+2+2 marks

The first term of an arithmetic series is 24 and the sum of its first five terms is 100.
(a) Write the formula to find the sum of the first $n$ terms of an arithmetic series whose first term is $a$ and common difference is $d$ . (1)
(b) Find the common difference of the series. (2)
(c) How many terms should be taken from the first to make the sum of the series 156? Find using formula. (2)

Answer

(a) $S_n = \dfrac{n}{2}\left[2a + (n-1)d\right]$ .

(b) Given $a = 24$ , $S_5 = 100$ .

$\frac{5}{2}\left[2 \times 24 + 4d\right] = 100 \Rightarrow 48 + 4d = 40 \Rightarrow d = -2$

The common difference is $-2$ .

(c) Put $S_n = 156$ with $a = 24$ , $d = -2$ :

$\frac{n}{2}\left[48 + (n-1)(-2)\right] = 156 \Rightarrow n(25 - n) = 156 \Rightarrow n^2 - 25n + 156 = 0$

$(n - 12)(n - 13) = 0 \Rightarrow n = 12 \text{ or } 13$

So either 12 or 13 terms give the sum 156. (The 13th term is $24 + 12(-2) = 0$ , so adding it does not change the sum.)

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Grade increment · 1+2+2 marks

The numerical product of the present ages of a mother and a daughter is 1,500. Ten years ago the mother was twice as old as the daughter.
(a) Write the roots of the quadratic equation $ax^2+bx+c=0;\ a\neq 0$ . (1)
(b) Make a quadratic equation according to the given condition. (2)
(c) Find the present ages of the mother and the daughter. (2)

Answer

(a) $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .

(b) Let the daughter's present age be $x$ years and the mother's be $y$ years. Then $xy = 1500$ .

Ten years ago: $y - 10 = 2(x - 10) \Rightarrow y = 2x - 10$ .

Substituting: $x(2x - 10) = 1500 \Rightarrow 2x^2 - 10x - 1500 = 0$ , that is

$x^2 - 5x - 750 = 0$

(c) Factorise: $x^2 - 30x + 25x - 750 = 0 \Rightarrow (x - 30)(x + 25) = 0$ .

So $x = 30$ or $x = -25$ . Age cannot be negative, so $x = 30$ and $y = 2 \times 30 - 10 = 50$ .

The daughter is 30 years old and the mother is 50 years old. (Check: $30 \times 50 = 1500$ ; ten years ago they were 20 and 40.)

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Grade increment · 2 marks

Simplify: $\dfrac{1}{m-n}-\dfrac{m-n}{m^2-n^2}$

Answer

$\frac{1}{m-n} - \frac{m-n}{m^2-n^2} = \frac{1}{m-n} - \frac{m-n}{(m-n)(m+n)} = \frac{1}{m-n} - \frac{1}{m+n}$

$= \frac{(m+n) - (m-n)}{(m-n)(m+n)} = \frac{2n}{m^2 - n^2}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Grade increment · 3 marks

Solve: $16^x-5\times 4^{x+1}+64=0$

Answer

Write $16^x = (4^x)^2$ and $4^{x+1} = 4 \cdot 4^x$ . Let $y = 4^x$ .

$y^2 - 20y + 64 = 0 \Rightarrow (y - 4)(y - 16) = 0 \Rightarrow y = 4 \text{ or } 16$

If $4^x = 4$ , then $x = 1$ . If $4^x = 16 = 4^2$ , then $x = 2$ .

So $x = 1$ or $x = 2$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2079 Chaitra · Bagmati · 1 mark

What is the order of the surd $\sqrt[5]{y^4}$ ?

Answer

The order of the surd $\sqrt[5]{y^4}$ is $5$ (the index of the root).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 1 MCQ

Loading…

2079 Chaitra · Bagmati · 2 marks

Find the L.C.M. of: $p^4-p,\ p^2+p+1$ .

Answer

$p^4 - p = p(p^3 - 1) = p(p - 1)(p^2 + p + 1)$

The second expression is $p^2 + p + 1$ , which is already a factor of the first.

$\text{L.C.M.} = p(p - 1)(p^2 + p + 1) = p^4 - p$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2079 Chaitra · Bagmati · 2 marks

Solve: $7^{x^2}=2401$ .

Answer

$2401 = 7^4$ , so $7^{x^2} = 7^4$ .

Equating the powers: $x^2 = 4 \Rightarrow x = \pm 2$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2079 Chaitra · Bagmati · 2 marks

Evaluate: $\left(\dfrac{125}{27}\right)^{-\frac{1}{3}}\div\left(\dfrac{25}{9}\right)^{-\frac{1}{2}}$

Answer

$\left(\frac{125}{27}\right)^{-\frac{1}{3}} = \left(\frac{27}{125}\right)^{\frac{1}{3}} = \frac{3}{5}, \qquad \left(\frac{25}{9}\right)^{-\frac{1}{2}} = \left(\frac{9}{25}\right)^{\frac{1}{2}} = \frac{3}{5}$

$\frac{3}{5} \div \frac{3}{5} = 1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2079 Chaitra · Bagmati · 2 marks

Simplify: $\dfrac{a-1}{a^2-3a+2}+\dfrac{a-2}{a^2-5a+6}$

Answer

Factorise each denominator: $a^2-3a+2=(a-1)(a-2)$ and $a^2-5a+6=(a-2)(a-3)$ .

$\frac{a-1}{(a-1)(a-2)}+\frac{a-2}{(a-2)(a-3)}=\frac{1}{a-2}+\frac{1}{a-3}=\frac{(a-3)+(a-2)}{(a-2)(a-3)}=\frac{2a-5}{(a-2)(a-3)}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2079 Chaitra · Bagmati · 4 marks

Find the H.C.F. of: $x^2+2xy+y^2,\ x^2-y^2,\ x^3+y^3$ .

Answer

Factorise each expression:

$x^2+2xy+y^2=(x+y)^2=(x+y)(x+y)$

$x^2-y^2=(x+y)(x-y)$

$x^3+y^3=(x+y)(x^2-xy+y^2)$

The only factor common to all three is $(x+y)$ .

$\text{H.C.F.}=x+y$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2079 Chaitra · Bagmati · 4 marks

Solve: $\dfrac{x-1}{\sqrt{x}+1}=1+\dfrac{\sqrt{x}-1}{2}$

Answer

Since $x-1=(\sqrt x)^2-1^2=(\sqrt x-1)(\sqrt x+1)$ :

$\frac{x-1}{\sqrt x+1}=\sqrt x-1$

The equation becomes

$\sqrt x-1=1+\frac{\sqrt x-1}{2}$

Multiply by 2: $2\sqrt x-2=2+\sqrt x-1$

$\sqrt x=3 \Rightarrow x=9$

Check: LHS $=\frac{9-1}{3+1}=2$ ; RHS $=1+\frac{3-1}{2}=2$ . So $x=9$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2079 Chaitra · Bagmati · 5 marks

In 2050 B.S., the sum of the ages of Madan Bahadur and Hari Bahadur was 40 years. If in 2065 B.S. the ratio of their ages was $3:4$ , find their ages in 2080 B.S.

Answer

Let the ages in 2050 B.S. be $x$ years (Madan Bahadur) and $(40-x)$ years (Hari Bahadur).

From 2050 B.S. to 2065 B.S. is 15 years, so in 2065 B.S. their ages are $(x+15)$ and $(55-x)$ .

$\frac{x+15}{55-x}=\frac34$

$4x+60=165-3x \Rightarrow 7x=105 \Rightarrow x=15$

In 2050 B.S.: Madan Bahadur $=15$ years, Hari Bahadur $=25$ years.

From 2050 B.S. to 2080 B.S. is 30 years, so in 2080 B.S.:

Madan Bahadur $=15+30=$ 45 years
Hari Bahadur $=25+30=$ 55 years

Check: in 2065 B.S. their ages are $30$ and $40$ , and $30:40=3:4$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Model · Set 1 · 1+2+1 marks

There are 4 geometric means between 3 and 243.
(a) Given the first term $a$ , the last term $b$ and the number of geometric means $n$ , write the formula for the calculation of the common ratio $(r)$ in the given condition. (1)
(b) What is the third mean of the given series? Find it. (2)
(c) In the arithmetic mean and the geometric mean between 3 and 243, which one is greater and by how much? Compare it. (1)

Answer

Here $a=3$ , $b=243$ and $n=4$ . With 4 means there are 6 terms in all.

(a) $r=\left(\frac ba\right)^{\frac1{n+1}}$

(b) $r=\left(\frac{243}{3}\right)^{\frac15}=81^{\frac15}=3^{\frac45}\approx2.408$

The third mean is the 4th term:

$ar^3=3\times\left(3^{\frac45}\right)^3=3^{1+\frac{12}5}=3^{\frac{17}5}\approx \textbf{41.90}$

(Note: the numbers are tidy, with $r=3$ and means $9, 27, 81$ , only when there are 3 means; with 4 means the ratio is $3^{4/5}$ .)

(c) Arithmetic mean $=\frac{3+243}2=123$ .

Geometric mean $=\sqrt{3\times243}=\sqrt{729}=27$ .

The AM is greater, by $123-27=\mathbf{96}$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Model · Set 1 · 1+2+1 marks

The perimeter and area of a rectangular ground are 66 m and 260 sq. m respectively.
(a) Illustrate the roots of $x$ in the quadratic equation $ax^2+bx+c=0,\ a\neq 0$ . (1)
(b) Find the length and breadth of the given ground. (2)
(c) How many pieces of land with dimension $(13\times 4)$ square meter can be made from that rectangular field? Calculate it. (1)

Answer

(a) For $ax^2+bx+c=0$ , $a\neq0$ , the roots are given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

(b) Let the length be $l$ and breadth $b$ .

Perimeter: $2(l+b)=66 \Rightarrow l+b=33$ . Area: $lb=260$ .

So $l$ and $b$ are the roots of $x^2-33x+260=0$ .

$x^2-33x+260=0 \Rightarrow (x-20)(x-13)=0 \Rightarrow x=20 \text{ or } 13$

(Check: $20\times13=260$ and $20+13=33$ .)

Length $=$ 20 m, breadth $=$ 13 m.

(c) Area of one piece $=13\times4=52\ \text{m}^2$ .

$\text{Number of pieces}=\frac{260}{52}=\textbf{5}$

This fits exactly: five strips of width 4 m along the 20 m side, each 13 m long.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Model · Set 1 · 2 marks

Simplify: $\dfrac{x+y}{x-y}-\dfrac{x-y}{x+y}$

Answer

$\frac{x+y}{x-y}-\frac{x-y}{x+y}=\frac{(x+y)^2-(x-y)^2}{(x-y)(x+y)}=\frac{(x^2+2xy+y^2)-(x^2-2xy+y^2)}{x^2-y^2}=\frac{4xy}{x^2-y^2}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Model · Set 1 · 3 marks

Solve: $4^x+\dfrac{1}{4^x}=16\dfrac{1}{16}$

Answer

Here $16\dfrac{1}{16}=\dfrac{257}{16}=16+\dfrac{1}{16}$ . Put $a=4^x$ , so the equation becomes

$a+\frac{1}{a}=16+\frac{1}{16}.$

The function $a+\frac1a$ takes the same value at $a$ and at $\frac1a$ , so $a=16$ or $a=\frac{1}{16}$ . Equivalently, $16a^2-257a+16=0$ , which factorises as $(a-16)(16a-1)=0$ .

If $4^x=16=4^2$ , then $x=2$ .
If $4^x=\dfrac{1}{16}=4^{-2}$ , then $x=-2$ .

Hence $x=2$ or $x=-2$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Model · Set 2 · 1+2+2 marks

The first and last terms of an arithmetic series having some terms are 4 and 40 respectively. The sum of all terms is 220.
(a) Write the formula to calculate the sum of the first $n$ terms of the series. (1)
(b) Find the total number of terms in the series. (2)
(c) What should be added to the third term of the series so that the first three terms form a geometric series? Find it. (2)

Answer

(a) $S_n=\dfrac n2(a+l)$ , or equivalently $S_n=\dfrac n2\left[2a+(n-1)d\right]$ .

(b) Here $a=4$ , $l=40$ , $S_n=220$ :

$220=\frac n2(4+40)=22n\Rightarrow n=10.$

There are 10 terms.

(c) Find the common difference from $l=a+(n-1)d$ : $40=4+9d\Rightarrow d=4$ . The series is $4+8+12+\dots$ , so the third term is $12$ .

Let $x$ be added to the third term so that $4,\ 8,\ 12+x$ is a geometric series. Then

$\frac{8}{4}=\frac{12+x}{8}\Rightarrow 12+x=16\Rightarrow x=4.$

So 4 should be added (the geometric series is $4,8,16$ with common ratio 2).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Model · Set 2 · 1+3+1 marks

The length of a rectangular plot is 8 m more than its breadth. The area of the plot is 384 sq. m.
(a) How many roots does the quadratic equation $ax^2+bx+c=0,\ a\neq 0$ have? Write it. (1)
(b) What are the length and breadth of the plot? Find it. (3)
(c) How long should the plot be decreased from its length to form it as a square plot? Calculate it. (1)

Answer

(a) A quadratic equation has two roots (real and distinct, real and equal, or a pair of complex roots).

(b) Let the breadth be $x$ m, so the length is $(x+8)$ m. Then

$x(x+8)=384\Rightarrow x^2+8x-384=0\Rightarrow (x+24)(x-16)=0.$

So $x=16$ (the value $x=-24$ is rejected because a length cannot be negative).

Breadth $=16$ m and length $=16+8=24$ m. (Check: $24\times16=384$ .)

(c) To make the plot a square of side 16 m, the length must be decreased by $24-16=8$ m.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Model · Set 2 · 3 marks

Simplify: $\dfrac{1}{x^2-5x+6}+\dfrac{2}{4x-x^2-3}$

Answer

Factorise the denominators:

$x^2-5x+6=(x-2)(x-3),\qquad 4x-x^2-3=-(x^2-4x+3)=-(x-1)(x-3).$

Therefore

\frac{1}{(x-2)(x-3)}-\frac{2}{(x-1)(x-3)} =\frac{(x-1)-2(x-2)}{(x-1)(x-2)(x-3)} =\frac{3-x}{(x-1)(x-2)(x-3)}.

Since $3-x=-(x-3)$ , cancelling $(x-3)$ gives

$\boxed{-\frac{1}{(x-1)(x-2)}}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Model · Set 3 · 1+2+2 marks

Mahendra collected the following sum of money in the first five days of the month Baisakh.

Baisakh 1 Baisakh 2 Baisakh 3 Baisakh 4 Baisakh 5
Rs. 20 Rs. 40 Rs. 80 Rs. 160 Rs. 320
(a) Whether the above sequence is arithmetic or geometric on the basis of the deposited money in each day? Write with reason. (1)
(b) How much money will be deposited by the tenth day? Find using formula. (2)
(c) Based on the above sequence, are extra 2 days enough after 10 days to collect the total amount Rs. 81000? Write with reason. (2)

Baisakh 1	Baisakh 2	Baisakh 3	Baisakh 4	Baisakh 5
Rs. 20	Rs. 40	Rs. 80	Rs. 160	Rs. 320

Answer

(a) The sequence $20,40,80,160,320$ is geometric, because each term is obtained by multiplying the previous one by the same number: $\dfrac{40}{20}=\dfrac{80}{40}=\dots=2$ . (The differences $20,40,80,\dots$ are not constant, so it is not arithmetic.)

(b) $a=20$ , $r=2$ , and $t_n=ar^{\,n-1}$ :

$t_{10}=20\times2^{9}=20\times512=\text{Rs }10{,}240.$

(c) Total after $n$ days: $S_n=\dfrac{a(r^n-1)}{r-1}$ .

After 10 days: $S_{10}=20(2^{10}-1)=20\times1023=20{,}460$ .

After 12 days: $S_{12}=20(2^{12}-1)=20\times4095=81{,}900$ .

Since $81{,}900>81{,}000$ , yes, the extra 2 days are enough. (After 11 days the total is only $20\times2047=40{,}940$ , so the second extra day is needed.)

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Model · Set 3 · 1+3+1 marks

A positive number of two digits is 18 more than its reversed number and the product of the digits is 8.
(a) If $x$ is the digit of tens place and $y$ is the digit of one's place, express the number algebraically. (1)
(b) Find the number. (3)
(c) What is the ratio of the number and its reversed number? Find it. (1)

Answer

(a) The number is $10x+y$ .

(b) The reversed number is $10y+x$ . By the first condition,

$(10x+y)-(10y+x)=18\Rightarrow9(x-y)=18\Rightarrow x-y=2\Rightarrow y=x-2.$

By the second condition $xy=8$ :

$x(x-2)=8\Rightarrow x^2-2x-8=0\Rightarrow(x-4)(x+2)=0.$

A digit cannot be negative, so $x=4$ and $y=2$ . The number is $42$ . (Check: $42-24=18$ and $4\times2=8$ .)

(c) The reversed number is $24$ .

$42:24=7:4.$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2080 Model · Set 3 · 2 marks

Simplify: $\dfrac{p-q}{p+q}+\dfrac{p+q}{p-q}$

Answer

Take the common denominator $(p+q)(p-q)=p^2-q^2$ :

$\frac{p-q}{p+q}+\frac{p+q}{p-q}=\frac{(p-q)^2+(p+q)^2}{(p+q)(p-q)}=\frac{2p^2+2q^2}{p^2-q^2}=\frac{2(p^2+q^2)}{p^2-q^2}.$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set BP · 2 marks

Prove that: $\dfrac{3^{x+1}+3^x}{4\times 3^x}=1$ .

Answer

$\text{LHS} = \frac{3^{x+1} + 3^x}{4\times 3^x} = \frac{3\cdot 3^x + 3^x}{4\times 3^x} = \frac{3^x(3+1)}{4\times 3^x} = \frac{4\times 3^x}{4\times 3^x} = 1 = \text{RHS}$

Hence proved.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set BP · 2 marks

Rationalize the denominator of: $\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ .

Answer

Multiply the numerator and denominator by $\sqrt{5} - \sqrt{3}$ :

$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} = \frac{(\sqrt{5}-\sqrt{3})^2}{(\sqrt{5})^2 - (\sqrt{3})^2} = \frac{5 + 3 - 2\sqrt{15}}{5 - 3} = \frac{8 - 2\sqrt{15}}{2} = 4 - \sqrt{15}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set BP · 2 marks

Find the L.C.M. of: $a^2-b^2,\ (a-b)^2$ .

Answer

Factorise: $a^2 - b^2 = (a-b)(a+b)$ and $(a-b)^2 = (a-b)(a-b)$ .

$\text{L.C.M.} = (a-b)^2(a+b)$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set BP · 2 marks

Solve: $\sqrt{x+1}=\sqrt[3]{8}$ .

Answer

$\sqrt[3]{8} = 2$ , so $\sqrt{x+1} = 2$ .

Squaring both sides: $x + 1 = 4 \Rightarrow x = 3$ .

Check: $\sqrt{3+1} = 2$ . True.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set BP · 4 marks

Find the H.C.F. of: $p^3+4p^2q+4pq^2$ , $p^4+8pq^3$ and $3p^4+p^3q-10p^2q^2$ .

Answer

Factorise each expression:

$p^3 + 4p^2q + 4pq^2 = p(p^2 + 4pq + 4q^2) = p(p+2q)^2$
$p^4 + 8pq^3 = p(p^3 + 8q^3) = p(p+2q)(p^2 - 2pq + 4q^2)$
$3p^4 + p^3q - 10p^2q^2 = p^2(3p^2 + pq - 10q^2) = p^2(3p - 5q)(p + 2q)$

Common factors: $p$ and $(p+2q)$ .

$\text{H.C.F.} = p(p+2q)$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set CP · 2 marks

Prove that: $\dfrac{4^{a+2}+4^a}{17\times 4^a}=1$ .

Answer

$\text{LHS}=\frac{4^{a+2}+4^a}{17\times4^a}=\frac{4^a\times4^2+4^a}{17\times4^a}=\frac{4^a(16+1)}{17\times4^a}=\frac{17\times4^a}{17\times4^a}=1=\text{RHS}.$

Hence proved.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set CP · 2 marks

Rationalize the denominator of: $\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ .

Answer

Multiply the numerator and denominator by the conjugate $\sqrt3+\sqrt2$ :

$\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}=\frac{(\sqrt3+\sqrt2)^2}{(\sqrt3)^2-(\sqrt2)^2}=\frac{3+2+2\sqrt6}{3-2}=\mathbf{5+2\sqrt6}.$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set CP · 2 marks

Solve: $\sqrt{w+4}=\sqrt[3]{64}$ .

Answer

$\sqrt[3]{64}=4$ , so

$\sqrt{w+4}=4.$

Squaring both sides: $w+4=16 \Rightarrow w=12$ .

Check: $\sqrt{12+4}=\sqrt{16}=4=\sqrt[3]{64}$ . So $w=$ 12.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set CP · 4 marks

Find the H.C.F. of: $p^2+4pq+4q^2$ , $p^4+8pq^3$ and $3p^4-10p^2q^2+p^3q$ .

Answer

Factorise each expression.

First: $p^2+4pq+4q^2=(p+2q)^2$ .

Second: $p^4+8pq^3=p(p^3+8q^3)=p(p+2q)(p^2-2pq+4q^2)$ .

Third: $3p^4-10p^2q^2+p^3q=p^2(3p^2+pq-10q^2)=p^2(3p-5q)(p+2q)$ ,

since $(3p-5q)(p+2q)=3p^2+6pq-5pq-10q^2=3p^2+pq-10q^2$ .

The only factor common to all three is $(p+2q)$ .

$\text{H.C.F.}=p+2q.$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set EP · 2 marks

Prove that: $\dfrac{4^{x+1}+4^x}{5\cdot 4^x}=1$ .

Answer

$\text{LHS} = \frac{4^{x+1}+4^x}{5\cdot 4^x} = \frac{4^x\cdot 4 + 4^x}{5\cdot 4^x} = \frac{4^x(4+1)}{5\cdot 4^x} = \frac{5\cdot 4^x}{5\cdot 4^x} = 1 = \text{RHS}$

Hence proved.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set EP · 2 marks

Rationalize the denominator of: $\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}$ .

Answer

Multiply the numerator and denominator by the conjugate $\sqrt7-\sqrt3$ :

$\frac{\sqrt7-\sqrt3}{\sqrt7+\sqrt3} = \frac{(\sqrt7-\sqrt3)^2}{(\sqrt7)^2-(\sqrt3)^2} = \frac{7+3-2\sqrt{21}}{7-3} = \frac{10-2\sqrt{21}}{4} = \frac{5-\sqrt{21}}{2}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set EP · 2 marks

Find the L.C.M. of: $m^2-n^2,\ (m+n)^2$ .

Answer

Factorise: $m^2-n^2 = (m+n)(m-n)$ and $(m+n)^2 = (m+n)(m+n)$ .

Take each factor with its highest power:

$\text{L.C.M.} = (m+n)^2(m-n)$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set EP · 2 marks

Solve: $\sqrt{x-1}=\sqrt[3]{27}$ .

Answer

$\sqrt[3]{27} = 3$ , so $\sqrt{x-1} = 3$ .

Squaring both sides: $x-1 = 9$ , so $x = 10$ .

Check: $\sqrt{10-1} = 3 = \sqrt[3]{27}$ . So $x =$ 10.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set EP · 4 marks

Find the H.C.F. of: $a^2+4ab+4b^2$ , $a^4+8ab^3$ and $3a^4-10a^2b^2+a^3b$ .

Answer

Factorise each expression:

$a^2+4ab+4b^2 = (a+2b)^2$
$a^4+8ab^3 = a(a^3+8b^3) = a(a+2b)(a^2-2ab+4b^2)$
$3a^4-10a^2b^2+a^3b = a^2(3a^2+ab-10b^2) = a^2(3a-5b)(a+2b)$

The only factor common to all three is $(a+2b)$ .

$\text{H.C.F.} = a+2b$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set FP · 2 marks

Prove that: $\dfrac{5^{x+1}+5^x}{6\times 5^x}=1$ .

Answer

$\text{LHS} = \frac{5^{x+1}+5^x}{6\times 5^x} = \frac{5^x\cdot 5 + 5^x}{6\times 5^x} = \frac{5^x(5+1)}{6\times 5^x} = \frac{6\times 5^x}{6\times 5^x} = 1 = \text{RHS}$

Hence proved.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set FP · 2 marks

Rationalize the denominator of: $\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ .

Answer

Multiply the numerator and denominator by the conjugate $\sqrt5+\sqrt2$ :

$\frac{\sqrt5+\sqrt2}{\sqrt5-\sqrt2} = \frac{(\sqrt5+\sqrt2)^2}{(\sqrt5)^2-(\sqrt2)^2} = \frac{5+2+2\sqrt{10}}{5-2} = \frac{7+2\sqrt{10}}{3}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set FP · 2 marks

Solve: $\sqrt{x-1}=\sqrt[3]{8}$ .

Answer

$\sqrt[3]{8} = 2$ , so $\sqrt{x-1} = 2$ .

Squaring both sides: $x-1 = 4$ , so $x = 5$ .

Check: $\sqrt{5-1} = 2 = \sqrt[3]{8}$ . So $x =$ 5.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set FP · 4 marks

Find the H.C.F. of: $x^3+4x^2y+4xy^2$ , $x^4+8xy^3$ and $3x^4+x^3y-10x^2y^2$ .

Answer

Factorise each expression.

x^3 + 4x^2y + 4xy^2 = x(x^2 + 4xy + 4y^2) = x(x + 2y)^2

x^4 + 8xy^3 = x(x^3 + 8y^3) = x(x + 2y)(x^2 - 2xy + 4y^2)

3x^4 + x^3y - 10x^2y^2 = x^2(3x^2 + xy - 10y^2) = x^2(3x - 5y)(x + 2y)

The common factors are $x$ and $(x + 2y)$ .

Answer: H.C.F. $= x(x + 2y)$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set GP · 2 marks

Prove that: $\dfrac{2^{a+1}+2^a}{3\times 2^a}=1$ .

Answer

\text{LHS} = \frac{2^{a+1} + 2^a}{3 \times 2^a} = \frac{2^a \times 2 + 2^a}{3 \times 2^a} = \frac{2^a(2 + 1)}{3 \times 2^a} = \frac{3 \times 2^a}{3 \times 2^a} = 1 = \text{RHS}

Proved.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set GP · 2 marks

Rationalize the denominator of: $\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$ .

Answer

Multiply the numerator and denominator by the conjugate $\sqrt3 + 1$ :

\frac{\sqrt3 + 1}{\sqrt3 - 1} \times \frac{\sqrt3 + 1}{\sqrt3 + 1} = \frac{(\sqrt3 + 1)^2}{(\sqrt3)^2 - 1^2} = \frac{3 + 2\sqrt3 + 1}{2} = \frac{4 + 2\sqrt3}{2} = 2 + \sqrt3

Answer: $2 + \sqrt3$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set GP · 2 marks

Find the L.C.M. of: $y^2-z^2,\ (y-z)^2$ .

Answer

y^2 - z^2 = (y - z)(y + z), \qquad (y - z)^2 = (y - z)(y - z)

Taking each factor to its highest power:

L.C.M. $= (y - z)^2(y + z)$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set GP · 2 marks

Solve: $\sqrt{y-1}=\sqrt[3]{1}$ .

Answer

$\sqrt[3]{1} = 1$ , so

\sqrt{y - 1} = 1

Squaring both sides: $y - 1 = 1 \Rightarrow y = 2$ .

Check: $\sqrt{2 - 1} = 1 = \sqrt[3]{1}$ .

Answer: $y = 2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set GP · 4 marks

Find the H.C.F. of: $m^5+8m^2n^3$ , $3m^4+m^3n-10m^2n^2$ and $m^3+4m^2n+4mn^2$ .

Answer

Factorise each expression.

$m^5 + 8m^2n^3 = m^2(m^3 + 8n^3) = m^2(m + 2n)(m^2 - 2mn + 4n^2)$

$3m^4 + m^3n - 10m^2n^2 = m^2(3m^2 + mn - 10n^2) = m^2(3m - 5n)(m + 2n)$

$m^3 + 4m^2n + 4mn^2 = m(m^2 + 4mn + 4n^2) = m(m + 2n)^2$

Common factors: $m$ (lowest power) and $(m + 2n)$ .

$\text{H.C.F.} = m(m + 2n)$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set DP · 2 marks

Prove that: $\dfrac{11^{x+2}+11^x}{122\times 11^x}=1$ .

Answer

$\text{LHS} = \frac{11^{x+2} + 11^x}{122 \times 11^x} = \frac{11^x \cdot 11^2 + 11^x}{122 \times 11^x} = \frac{11^x(121 + 1)}{122 \times 11^x} = \frac{122 \times 11^x}{122 \times 11^x} = 1 = \text{RHS}$

Hence proved.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set DP · 2 marks

Rationalize the denominator of: $\dfrac{\sqrt{5}+\sqrt{4}}{\sqrt{5}-\sqrt{4}}$ .

Answer

Here $\sqrt{4} = 2$ . Multiply numerator and denominator by the conjugate $\sqrt{5} + 2$ :

$\frac{\sqrt{5} + 2}{\sqrt{5} - 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{(\sqrt{5} + 2)^2}{5 - 4} = \frac{5 + 4\sqrt{5} + 4}{1} = 9 + 4\sqrt{5}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set DP · 2 marks

Find the L.C.M. of: $w^2-y^2,\ (w+y)^2$ .

Answer

$w^2 - y^2 = (w - y)(w + y), \qquad (w + y)^2 = (w + y)(w + y)$

Taking each factor with its highest power:

$\text{L.C.M.} = (w - y)(w + y)^2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set DP · 2 marks

Solve: $\sqrt{y-5}=\sqrt[3]{216}$ .

Answer

$\sqrt[3]{216} = 6$ , so $\sqrt{y - 5} = 6$ Squaring both sides: $y - 5 = 36 \Rightarrow y = 41$ .

Check: $\sqrt{41 - 5} = \sqrt{36} = 6$ ✓. So $y = 41$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2075 Chaitra · Set DP · 4 marks

Find the H.C.F. of: $v^2(v^2+4vw+4w^2)$ , $v^5+8v^2w^3$ and $3v^4+v^3w-10v^2w^2$ .

Answer

Factorise each expression.

$v^2(v^2 + 4vw + 4w^2) = v^2(v + 2w)^2$

$v^5 + 8v^2w^3 = v^2(v^3 + 8w^3) = v^2(v + 2w)(v^2 - 2vw + 4w^2)$

$3v^4 + v^3w - 10v^2w^2 = v^2(3v^2 + vw - 10w^2) = v^2(3v - 5w)(v + 2w)$

Common factors: $v^2$ and $(v + 2w)$ .

$\text{H.C.F.} = v^2(v + 2w)$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2078 Baisakh · Set A · 1 mark

What does $n$ represent in the surd $\sqrt[n]{x}$ ? Write it.

Answer

$n$ is the order (index) of the surd, that is, the root to be taken. $\sqrt[n]{x}$ is the $n$ th root of $x$ , where $x$ is the radicand.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2078 Baisakh · Set A · 2 marks

Find the H.C.F. of: $y^3-1$ and $y^2+y+1$ .

Answer

$y^3-1=(y-1)(y^2+y+1)$

$y^2+y+1$ is already one factor of $y^3-1$ and cannot be factorised further, so it divides both expressions.

$\text{H.C.F.}=\mathbf{y^2+y+1}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2078 Baisakh · Set A · 2 marks

Simplify: $\sqrt[3]{32}+\sqrt[3]{108}-8\sqrt[3]{4}$

Answer

$\sqrt[3]{32}=\sqrt[3]{8\times4}=2\sqrt[3]{4}$ and $\sqrt[3]{108}=\sqrt[3]{27\times4}=3\sqrt[3]{4}$

$2\sqrt[3]{4}+3\sqrt[3]{4}-8\sqrt[3]{4}=(2+3-8)\sqrt[3]{4}=\mathbf{-3\sqrt[3]{4}}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2078 Baisakh · Set A · 2 marks

Solve: $5^{y^2-1}=1$

Answer

$5^{y^2-1}=1=5^0$

Equating the powers (same base): $y^2-1=0\Rightarrow y^2=1$

$y=\pm1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2078 Baisakh · Set A · 2 marks

If one fourth of the square of a positive number is 9, find the number.

Answer

Let the number be $x$ .

$\frac14x^2=9\Rightarrow x^2=36\Rightarrow x=\pm6$

Since the number is positive, the number is 6.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2078 Baisakh · Set A · 2 marks

Find the L.C.M. of: $p^2+2pq+q^2$ and $p+q$ .

Answer

$p^2+2pq+q^2=(p+q)^2,\qquad p+q=(p+q)$

The highest power of the factor $(p+q)$ is $(p+q)^2$ .

$\text{L.C.M.}=\mathbf{(p+q)^2}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2078 Baisakh · Set A · 5 marks

The total cost of 5 kg rice and 6 kg sugar is Rs. 940. If the rate of rice increases by $20\%$ and the rate of sugar decreases by $10\%$ , the total cost of 4 kg rice and 3 kg sugar will be Rs. 627. By what percent is the cost of 1 kg rice more or less than the cost of 1 kg sugar? Find it.

Answer

Let the price of 1 kg rice be Rs $x$ and 1 kg sugar be Rs $y$ .

First condition: $5x+6y=940$ ... (i)

New rates: rice $1.2x$ , sugar $0.9y$ .

$4(1.2x)+3(0.9y)=627\Rightarrow4.8x+2.7y=627$

Multiplying by 10: $48x+27y=6270$ , i.e. $16x+9y=2090$ ... (ii)

Multiply (i) by 3 and (ii) by 2 to match $y$ :

$15x+18y=2820,\qquad 32x+18y=4180$

Subtract: $17x=1360\Rightarrow x=80$

From (i): $400+6y=940\Rightarrow y=90$

Rice is Rs 80 per kg and sugar is Rs 90 per kg.

$\text{Rice is cheaper by }\frac{90-80}{90}\times100=\textbf{11.11\%}\ \left(11\tfrac19\%\right)\text{ than sugar.}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Chaitra · Set CC · 2 marks

Find the H.C.F. of: $p^5+p^2$ and $p^2-p+1$ .

Answer

$p^5 + p^2 = p^2(p^3+1) = p^2(p+1)(p^2-p+1)$

$p^2 - p + 1$ is already irreducible (cannot be factorised further).

The only common factor is $p^2-p+1$ , so

H.C.F. $= p^2 - p + 1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Chaitra · Set CC · 2 marks

Simplify: $4\sqrt[3]{250}-8\sqrt[3]{128}+4\sqrt[3]{54}$

Answer

$\sqrt[3]{250} = \sqrt[3]{125\times 2} = 5\sqrt[3]{2}$ , $\ \sqrt[3]{128} = \sqrt[3]{64\times 2} = 4\sqrt[3]{2}$ , $\ \sqrt[3]{54} = \sqrt[3]{27\times 2} = 3\sqrt[3]{2}$

$4\sqrt[3]{250}-8\sqrt[3]{128}+4\sqrt[3]{54} = 20\sqrt[3]{2} - 32\sqrt[3]{2} + 12\sqrt[3]{2} = (20-32+12)\sqrt[3]{2} = 0$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Chaitra · Set CC · 2 marks

If $a=5^x$ , $b=5^y$ and $a^y\cdot b^x=25$ , then prove that: $xy=1$ .

Answer

Given $a = 5^x$ and $b = 5^y$ , so

$a^y\cdot b^x = (5^x)^y\cdot(5^y)^x = 5^{xy}\cdot 5^{xy} = 5^{2xy}$

Since $a^y\cdot b^x = 25 = 5^2$ , we have $5^{2xy} = 5^2$ .

Equating the powers of the same base: $2xy = 2$ , so $xy = 1$ . $\quad$ Proved.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Chaitra · Set CC · 2 marks

Simplify: $\dfrac{1}{m-n}-\dfrac{m+n}{m^2-n^2}$

Answer

$\frac{1}{m-n}-\frac{m+n}{m^2-n^2} = \frac{1}{m-n}-\frac{m+n}{(m+n)(m-n)} = \frac{1}{m-n}-\frac{1}{m-n} = 0$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Chaitra · Set CC · 2 marks

Simplify: $\dfrac{c^2+cd+d^2}{c+d}+\dfrac{c^2-cd+d^2}{c-d}$

Answer

$\frac{c^2+cd+d^2}{c+d}+\frac{c^2-cd+d^2}{c-d} = \frac{(c^2+cd+d^2)(c-d)+(c^2-cd+d^2)(c+d)}{(c+d)(c-d)}$

Using $(c^2+cd+d^2)(c-d) = c^3-d^3$ and $(c^2-cd+d^2)(c+d) = c^3+d^3$ :

$= \frac{(c^3-d^3)+(c^3+d^3)}{c^2-d^2} = \frac{2c^3}{c^2-d^2}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Chaitra · Set CC · 4 marks

Find the L.C.M. of: $m^4+m^3-m^2-m,\ m^4-m$ and $4m^2-4$ .

Answer

Factorise each expression:

$m^4+m^3-m^2-m = m\,(m^3+m^2-m-1) = m\,[m^2(m+1)-(m+1)] = m(m+1)(m^2-1) = m(m+1)^2(m-1)$

$m^4-m = m(m^3-1) = m(m-1)(m^2+m+1)$

$4m^2-4 = 4(m^2-1) = 4(m-1)(m+1)$

Taking each factor at its highest power (and the L.C.M. of the numerical coefficients, 4):

$\text{L.C.M.} = 4\,m\,(m-1)(m+1)^2(m^2+m+1)$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Chaitra · Set CC · 4 marks

The sum of the digits in a two digit number is 6. If 36 is subtracted from the number, the places of the digits are interchanged. Find the initial number.

Answer

Let the tens digit be $x$ and the units digit be $y$ . The number is $10x+y$ .

Sum of digits: $x+y=6 \quad\cdots(i)$

On subtracting 36 the digits are interchanged: $10x+y-36 = 10y+x$

$9x-9y = 36 \implies x-y = 4 \quad\cdots(ii)$

Adding (i) and (ii): $2x=10$ , so $x=5$ . From (i), $y=1$ .

The initial number is 51.

Check: $51-36=15$ , which is 51 with its digits interchanged. $\checkmark$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Chaitra · Set CC · 5 marks

Simplify: $\dfrac{\left(d^2-\dfrac{1}{e^2}\right)^{d}\left(d-\dfrac{1}{e}\right)^{e-d}}{\left(e^2-\dfrac{1}{d^2}\right)^{e}\left(e+\dfrac{1}{d}\right)^{d-e}}$

Answer

Factorise each difference of squares:

$d^2-\frac{1}{e^2}=\left(d-\frac1e\right)\left(d+\frac1e\right),\qquad e^2-\frac{1}{d^2}=\left(e-\frac1d\right)\left(e+\frac1d\right)$

Numerator:

$\left(d-\tfrac1e\right)^d\left(d+\tfrac1e\right)^d\left(d-\tfrac1e\right)^{e-d}=\left(d-\tfrac1e\right)^{e}\left(d+\tfrac1e\right)^{d}$

Denominator:

$\left(e-\tfrac1d\right)^e\left(e+\tfrac1d\right)^e\left(e+\tfrac1d\right)^{d-e}=\left(e-\tfrac1d\right)^{e}\left(e+\tfrac1d\right)^{d}$

Now $d-\dfrac1e=\dfrac{de-1}{e}$ , $d+\dfrac1e=\dfrac{de+1}{e}$ , $e-\dfrac1d=\dfrac{de-1}{d}$ , $e+\dfrac1d=\dfrac{de+1}{d}$ . So

$\text{Expression}=\frac{\left(\frac{de-1}{e}\right)^{e}\left(\frac{de+1}{e}\right)^{d}}{\left(\frac{de-1}{d}\right)^{e}\left(\frac{de+1}{d}\right)^{d}}=\left(\frac{d}{e}\right)^{e}\left(\frac{d}{e}\right)^{d}=\left(\frac{d}{e}\right)^{d+e}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Model question · 2 marks

Simplify: $x^{a-b}\times x^{b-a}$

Answer

$x^{a-b}\times x^{b-a}=x^{(a-b)+(b-a)}=x^{0}=1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Model question · 2 marks

Simplify: $2\sqrt{98}-8\sqrt{32}+3\sqrt{72}$

Answer

$\sqrt{98}=7\sqrt2,\quad\sqrt{32}=4\sqrt2,\quad\sqrt{72}=6\sqrt2$

$2\sqrt{98}-8\sqrt{32}+3\sqrt{72}=14\sqrt2-32\sqrt2+18\sqrt2=(14-32+18)\sqrt2=0$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Model question · 2 marks

Solve: $9-\sqrt{x-4}=5$

Answer

$9-\sqrt{x-4}=5\Rightarrow\sqrt{x-4}=4$

Squaring both sides: $x-4=16$ , so $x=20$ .

Check: $9-\sqrt{20-4}=9-4=5$ . So $x=20$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Model question · 2 marks

Evaluate: $\dfrac{3\times 4^x+4^{x-1}}{13\times 4^x}$

Answer

Write $4^{x-1}=\dfrac{4^x}{4}$ and take $4^x$ common in the numerator:

$\frac{3\times4^x+\frac{4^x}{4}}{13\times4^x}=\frac{4^x\left(3+\frac14\right)}{13\times4^x}=\frac{\frac{13}{4}}{13}=\frac14$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Model question · 2 marks

If 5 is added to the square of a positive number, the sum is 69. Find the number.

Answer

Let the positive number be $x$ . Then

$x^2+5=69\Rightarrow x^2=64\Rightarrow x=\pm8.$

Since the number is positive, the number is 8.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Model question · 2 marks

Simplify: $\dfrac{1}{x-y}-\dfrac{1}{x+y}+\dfrac{2x}{x^2-y^2}$

Answer

Using $x^2-y^2=(x-y)(x+y)$ as the common denominator:

$\frac{1}{x-y}-\frac{1}{x+y}+\frac{2x}{x^2-y^2}=\frac{(x+y)-(x-y)+2x}{(x-y)(x+y)}=\frac{2y+2x}{(x-y)(x+y)}$

$=\frac{2(x+y)}{(x-y)(x+y)}=\frac{2}{x-y}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Model question · 4 marks

Find the H.C.F. of: $a^3+1+2a^2+2a,\ a^3-1$ and $a^4+a^2+1$ .

Answer

Factorise each expression.

$a^3+2a^2+2a+1=(a^3+1)+2a(a+1)=(a+1)(a^2-a+1)+2a(a+1)=(a+1)(a^2+a+1)$

$a^3-1=(a-1)(a^2+a+1)$

$a^4+a^2+1=(a^2+1)^2-a^2=(a^2+a+1)(a^2-a+1)$

The only factor common to all three is $a^2+a+1$ .

$\text{H.C.F.}=a^2+a+1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2074 Model question · 5 marks

Three years ago, the sum of the ages of the father and son was 48 years. After three years the ratio of the ages of the father and son will be $3:1$ . How much older is the father than the son? Find it.

Answer

Three years ago the sum of their ages was 48, so the present sum is $48 + 2\times 3 = 54$ years. After three years the sum will be $54 + 6 = 60$ years.

Let the ages after three years be father $=3x$ and son $=x$ (ratio $3:1$ ).

3x + x = 60 \Rightarrow x = 15

Father after 3 years $=45$ years and son $=15$ years.

Present ages: father $=42$ years, son $=12$ years (check: $42+12=54$ ).

Difference of ages $=45-15=30$ .

The father is 30 years older than the son.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Pre-board · PABSON Set A · 1+2+2 marks

Harka and Heera, working in the same company, decided to invest certain amounts each month starting from the same date. Harka invested Rs. 4,000 in the first month and increased his investment in succeeding months by Rs. 100 per month. However, Heera invested Rs. 5,000 in the first month and decreased her investment in succeeding months by Rs. 100 per month.
(a) Write the formula to find the sum of the first $n$ terms of an arithmetic sequence. (1)
(b) Who invested more amount up to 10 months? (2)
(c) In which month did they invest equal amounts? (2)

Answer

(a) $S_n=\dfrac n2\left[2a+(n-1)d\right]$

(b) Harka: $a=4000$ , $d=100$ , $n=10$ .

S_{10}=\frac{10}{2}\left[2\times4000+9\times100\right]=5\times8900=\text{Rs }44{,}500

Heera: $a=5000$ , $d=-100$ , $n=10$ .

S_{10}=\frac{10}{2}\left[2\times5000+9\times(-100)\right]=5\times9100=\text{Rs }45{,}500

Heera invested more (by Rs 1,000).

(c) Equal monthly amounts in the $n$ th month:

4000+(n-1)100=5000-(n-1)100

200(n-1)=1000\Rightarrow n-1=5\Rightarrow n=6

They invested equal amounts (Rs 4,500 each) in the 6th month.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Pre-board · PABSON Set A · 2+2+1 marks

The present age of a father is three years more than three times the age of his son. The product of their ages 7 years ago was 470.
(a) Express the above information in mathematical form. (2)
(b) Find the present ages of the father and the son. (2)
(c) Find the difference of the ages of the father and the son. (1)

Answer

(a) Let the present age of the son be $x$ years. Then the father's age is $(3x+3)$ years.

Ages 7 years ago: son $=x-7$ , father $=3x-4$ . So

(3x-4)(x-7)=470

(b) $3x^2-25x+28=470\Rightarrow3x^2-25x-442=0$

x=\frac{25\pm\sqrt{625+5304}}{6}=\frac{25\pm77}{6}

Taking the positive value, $x=17$ .

Son $=17$ years, father $=3\times17+3=54$ years.

Check: $(54-7)(17-7)=47\times10=470$ .

(c) Difference $=54-17=\mathbf{37}$ years.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Pre-board · PABSON Set A

Simplify: $\dfrac{m}{m-n}-\dfrac{m}{m+n}+\dfrac{2mn}{m^2+n^2}$

Answer

\frac{m}{m-n}-\frac{m}{m+n}=\frac{m(m+n)-m(m-n)}{m^2-n^2}=\frac{2mn}{m^2-n^2}

\frac{2mn}{m^2-n^2}+\frac{2mn}{m^2+n^2}=\frac{2mn\left[(m^2+n^2)+(m^2-n^2)\right]}{(m^2-n^2)(m^2+n^2)}=\frac{2mn\times2m^2}{m^4-n^4}

Answer: $\dfrac{4m^3n}{m^4-n^4}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Pre-board · PABSON Set A

Solve: $13^x+\dfrac{1}{13^x}=169\dfrac{1}{169}$

Answer

Since $169\dfrac1{169}=169+\dfrac1{169}=13^2+\dfrac1{13^2}$ , the equation is

13^x+\frac1{13^x}=13^2+\frac1{13^2}

Let $y=13^x$ . Then $y+\dfrac1y=169+\dfrac1{169}$ , which gives $y=169$ or $y=\dfrac1{169}$ .

If $13^x=13^2$ , then $x=2$ . If $13^x=13^{-2}$ , then $x=-2$ .

Answer: $x=2$ or $x=-2$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Pre-board · PABSON Set B · 1+2+2 marks

There are three arithmetic means between 6 and 54.
(a) If the first term is $a$ , the last term is $b$ and the number of arithmetic means is $n$ , write the formula for finding the common difference. (1)
(b) Find all the arithmetic means of the sequence formed. (2)
(c) How many terms should be added after 54 so that the sum of the new sequence becomes 600? Find it by calculation. (2)

Answer

(a) $d=\dfrac{b-a}{n+1}$

(b) $a=6$ , $b=54$ , $n=3$ :

d=\frac{54-6}{3+1}=12

The three arithmetic means are $6+12=\mathbf{18}$ , $\mathbf{30}$ and $\mathbf{42}$ . The sequence is $6,18,30,42,54$ .

(c) Let the new sequence have $n$ terms, with $a=6$ , $d=12$ and sum $600$ :

S_n=\frac n2\left[2\times6+(n-1)12\right]=\frac n2\times12n=6n^2

6n^2=600\Rightarrow n^2=100\Rightarrow n=10

The new sequence has 10 terms, so the number of terms to be added after 54 is $10-5=\mathbf{5}$ .

Check: the 10th term is $6+9\times12=114$ and $S_{10}=5(6+114)=600$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Pre-board · PABSON Set B · 1+3+1 marks

In a right-angled triangle, the hypotenuse is 6 m more than twice the length of its shortest side. The remaining side is 2 m less than the hypotenuse.
(a) If the shortest side of the triangle is $x$ , then write the length of the remaining sides of the triangle in terms of $x$ . (1)
(b) Find all the sides of the right-angled triangle. (3)
(c) A new square is constructed such that its diagonal is equal to the hypotenuse of the triangle. Find how much more the area of the square is than that of the triangle. (1)

Answer

(a) Hypotenuse $=(2x+6)$ m and the remaining side $=(2x+6)-2=(2x+4)$ m.

(b) By the Pythagoras theorem:

x^2+(2x+4)^2=(2x+6)^2

x^2+4x^2+16x+16=4x^2+24x+36\Rightarrow x^2-8x-20=0

(x-10)(x+2)=0\Rightarrow x=10\ (x=-2\text{ rejected})

The sides are $10$ m, $24$ m and $26$ m.

(c) Diagonal of the square $=26$ m, so its area $=\dfrac{d^2}{2}=\dfrac{26^2}{2}=338$ m $^2$ .

Area of the triangle $=\tfrac12\times10\times24=120$ m $^2$ .

Difference $=338-120=\mathbf{218\text{ m}^2}$ . The square is 218 m $^2$ larger.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Pre-board · PABSON Set B · 3 marks

Solve: $3^{x+3}+\dfrac{1}{3^x}=28$

Answer

3^{x+3}+\frac1{3^x}=28\Rightarrow27\cdot3^x+\frac1{3^x}=28

Let $y=3^x$ :

27y+\frac1y=28\Rightarrow27y^2-28y+1=0\Rightarrow(27y-1)(y-1)=0

So $y=\dfrac1{27}$ or $y=1$ .

If $3^x=3^{-3}$ , then $x=-3$ . If $3^x=3^0$ , then $x=0$ .

Check: $x=0$ gives $27+1=28$ ; $x=-3$ gives $1+27=28$ .

Answer: $x=0$ or $x=-3$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Pre-board · PABSON Set B · 2 marks

Simplify: $\dfrac{2}{1+x^2}-\dfrac{4}{1-x^4}$

Answer

\frac{2}{1+x^2}-\frac{4}{1-x^4}=\frac{2(1-x^2)-4}{(1+x^2)(1-x^2)}=\frac{-2-2x^2}{(1+x^2)(1-x^2)}=\frac{-2(1+x^2)}{(1+x^2)(1-x^2)}

Answer: $-\dfrac{2}{1-x^2}=\dfrac{2}{x^2-1}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Preparatory · PABSON Kaski · 1+3+2 marks

Two stationary shops $A$ and $B$ at Palikhe Chok, Pokhara sell the books in the following pattern.

Day / Shop Baisakh-1 Baisakh-2 Baisakh-3 Baisakh-4 ...
$A$ 1000 1200 1400 1600 ...
$B$ 1 2 4 8 ...
(a) How do you identify any sequence to be arithmetic or geometric? Write. (1)
(b) Which shop sold more books in total in the first 15 days of Baisakh? Calculate it. (3)
(c) Compare the total number of books sold by shop $A$ in the first 15 days to the number of books sold by shop $B$ in the 17th day only. (2)

Day / Shop	Baisakh-1	Baisakh-2	Baisakh-3	Baisakh-4	...
$A$	1000	1200	1400	1600	...
$B$	1	2	4	8	...

Answer

(a) Find the differences and the ratios of consecutive terms. If the difference $t_{n+1} - t_n$ is constant, the sequence is arithmetic. If the ratio $\dfrac{t_{n+1}}{t_n}$ is constant, it is geometric.

Here $A$ : 1000, 1200, 1400 has common difference 200 (arithmetic); $B$ : 1, 2, 4, 8 has common ratio 2 (geometric).

(b) Shop $A$ : $a = 1000$ , $d = 200$ , $n = 15$ .

$S_{15} = \dfrac{15}{2}\left[2\times 1000 + 14\times 200\right] = \dfrac{15}{2}\times 4800 = 36,000$

Shop $B$ : $a = 1$ , $r = 2$ , $n = 15$ .

$S_{15} = \dfrac{1\,(2^{15} - 1)}{2 - 1} = 32,768 - 1 = 32,767$

Since $36,000 > 32,767$ , shop $A$ sold more books in the first 15 days.

(c) Shop $B$ on the 17th day: $t_{17} = 1\times 2^{16} = 65,536$ books.

Shop $A$ 's 15-day total is 36,000. So shop $B$ sold $65,536 - 36,000 = 29,536$ more books on the 17th day alone than shop $A$ sold in 15 days (about 1.82 times as many).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Preparatory · PABSON Kaski · 1+2+2 marks

The length of a rectangular futsal ground is 5 m more than its breadth. Its area is 500 m $^2$ .
(a) What is the condition for the equation $ax^2+bx+c=0$ to be a quadratic equation? (1)
(b) Find the length and breadth of the futsal ground. (2)
(c) The owner of the ground wishes to fence all the sides of the ground by thin plastic net to make the height of the fence as 8 m on each side around it. Find the area of the net required and also the cost of it at the rate of Rs. 10 per sq. meter. (2)

Answer

(a) The coefficient $a$ of $x^2$ must not be zero ( $a \neq 0$ ), and $a$ , $b$ , $c$ are real numbers.

(b) Let the breadth be $x$ m, so the length is $(x + 5)$ m.

$x(x + 5) = 500 \Rightarrow x^2 + 5x - 500 = 0$

$x^2 + 25x - 20x - 500 = 0 \Rightarrow (x + 25)(x - 20) = 0$

$x = 20$ (since $x = -25$ is not possible).

Breadth $= 20$ m, length $= 25$ m.

(c) Perimeter $= 2(25 + 20) = 90$ m.

Area of net $= 90\times 8 = 720$ m $^2$ .

Cost $= 720\times 10 =$ Rs 7,200.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Preparatory · PABSON Kaski · 2+2 marks

(a) Simplify the expression $\dfrac{1}{4(k-8)}-\dfrac{1}{4(k-4)}$ and prove it to be $\dfrac{1}{k^2-12k+32}$ . (2)
(b) Find the value of $x$ if $k=2^x$ and $k^2-12k+32=0$ . (2)

Answer

(a)

$\dfrac{1}{4(k-8)} - \dfrac{1}{4(k-4)} = \dfrac{(k-4) - (k-8)}{4(k-8)(k-4)} = \dfrac{4}{4(k-8)(k-4)} = \dfrac{1}{(k-8)(k-4)}$

$(k-8)(k-4) = k^2 - 12k + 32$

So the expression equals $\dfrac{1}{k^2 - 12k + 32}$ . Proved.

(b) $k^2 - 12k + 32 = 0 \Rightarrow (k-8)(k-4) = 0 \Rightarrow k = 8$ or $k = 4$ .

Since $k = 2^x$ :

$2^x = 8 = 2^3 \Rightarrow x = 3$
$2^x = 4 = 2^2 \Rightarrow x = 2$

So $x = 2$ or $x = 3$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Preparation · Bhaktapur · Bhaktapur Municipality · 1+2+2 marks

A student saves money every month in two different ways.

In Plan A, he saves Rs. 500 in the first month and increases his saving by Rs. 100 every month.

In Plan B, he saves Rs. 500 in the first month and each month his saving becomes twice the previous month's saving.

On the basis of the above information, answer the following questions:
(a) State which plan represents an Arithmetic Sequence and which represents a Geometric Sequence. (1)
(b) Find the amount saved in the 6th month in Plan A. (2)
(c) In how many months will the total saving amount be Rs. 1,27,500 in Plan B? (2)

Answer

(a) Plan A (a fixed amount of Rs 100 added each month) is an arithmetic sequence: 500, 600, 700, ...

Plan B (each month is twice the previous) is a geometric sequence: 500, 1000, 2000, ...

(b) Plan A: $a = 500$ , $d = 100$ , $n = 6$ .

$t_6 = a + 5d = 500 + 5\times 100 = \text{Rs } 1,000$

(c) Plan B: $a = 500$ , $r = 2$ , $S_n = 127500$ .

$S_n = \dfrac{a(r^n - 1)}{r - 1} \Rightarrow 127500 = 500(2^n - 1) \Rightarrow 2^n - 1 = 255 \Rightarrow 2^n = 256 = 2^8$

$n = 8$ . It takes 8 months.

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Preparation · Bhaktapur · Bhaktapur Municipality · 1+2+2 marks

The present age of Ram is 50 years and the present age of Sita is 45 years.
(a) What was the age of Ram and Sita before $x$ years? (1)
(b) If the product of their ages before $x$ years was 750, what is the value of $x$ ? (2)
(c) How old will they be when the ratio of their ages is $13:12$ ? Calculate it. (2)

Answer

(a) Before $x$ years, Ram's age was $(50 - x)$ years and Sita's age was $(45 - x)$ years.

(b) $(50 - x)(45 - x) = 750$

$2250 - 95x + x^2 = 750 \Rightarrow x^2 - 95x + 1500 = 0$

$x^2 - 75x - 20x + 1500 = 0 \Rightarrow (x - 75)(x - 20) = 0$

$x = 20$ (since $x = 75$ would give negative ages).

Check: $30\times 25 = 750$ .

(c) After $y$ years, ages are $(50 + y)$ and $(45 + y)$ .

$\dfrac{50 + y}{45 + y} = \dfrac{13}{12} \Rightarrow 600 + 12y = 585 + 13y \Rightarrow y = 15$

After 15 years, Ram will be 65 years old and Sita 60 years old (ratio $65 : 60 = 13 : 12$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Preparation · Bhaktapur · Bhaktapur Municipality · 3 marks

Simplify: $\dfrac{x+y}{x-y}-\dfrac{x-y}{x+y}-\dfrac{2xy}{x^2-y^2}$

Answer

Since $x^2-y^2=(x-y)(x+y)$ , take it as the common denominator.

$\frac{x+y}{x-y}-\frac{x-y}{x+y}-\frac{2xy}{x^2-y^2}=\frac{(x+y)^2-(x-y)^2-2xy}{x^2-y^2}$

Now $(x+y)^2-(x-y)^2=4xy$ , so the numerator is $4xy-2xy=2xy$ .

$=\frac{2xy}{x^2-y^2}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Pre-qualifying · PABSON Kathmandu · 1+2+2 marks

The 3rd term and the 9th term of a geometric sequence are 16 and 1024 respectively.
(a) What is the geometric mean between two numbers $a$ and $b$ ? (1)
(b) Find the common ratio and the first term. (2)
(c) Find the sum of the first five terms of the sequence. (2)

Answer

(a) The geometric mean between $a$ and $b$ is $\sqrt{ab}$ .

(b) Let the first term be $a$ and the common ratio $r$ . Then $ar^2=16,\qquad ar^8=1024$ Dividing: $r^6=\dfrac{1024}{16}=64\Rightarrow r=2$ .

From $ar^2=16$ : $4a=16\Rightarrow a=4$ .

The common ratio is $2$ and the first term is $4$ . (Since $r^6=64$ also gives $r=-2$ , the sequence $4, -8, 16,\dots$ also fits the two terms; the usual answer is $r=2$ .)

(c) $S_5=\frac{a(r^5-1)}{r-1}=\frac{4(2^5-1)}{2-1}=4\times31=124$ Check: $4+8+16+32+64=124$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Pre-qualifying · PABSON Kathmandu · 1+3+1 marks

The sum of the present ages of Sita and Gita is 21 years. After 2 years, the product of their ages will be 144.
(a) What are the roots of $x$ in the quadratic equation $ax^2+bx+c=0,\ a\neq 0$ ? (1)
(b) Find the present ages of Sita and Gita. (3)
(c) After how many years will the sum of their ages be 31? (1)

Answer

(a) The roots are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

(b) Let Sita's present age be $x$ years. Then Gita's age is $(21-x)$ years. After 2 years their ages are $(x+2)$ and $(23-x)$ . $(x+2)(23-x)=144$ $23x-x^2+46-2x=144\Rightarrow x^2-21x+98=0$ $(x-7)(x-14)=0\Rightarrow x=7\text{ or }14$ So the present ages are $7$ years and $14$ years (check: $9\times16=144$ ).

(c) The sum of their ages increases by $2$ every year. Let it be $31$ after $n$ years: $21+2n=31\Rightarrow n=5$ After $5$ years the sum of their ages will be $31$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

2082 Pre-qualifying · PABSON Kathmandu · 3 marks

Simplify: $\dfrac{1}{(x-1)^2}+\dfrac{1}{(x-2)^2}-\dfrac{1}{(x-1)^2(x-2)^2}$

Answer

Common denominator is $(x-1)^2(x-2)^2$ .

$\frac{1}{(x-1)^2}+\frac{1}{(x-2)^2}-\frac{1}{(x-1)^2(x-2)^2}=\frac{(x-2)^2+(x-1)^2-1}{(x-1)^2(x-2)^2}$

Numerator: $(x^2-4x+4)+(x^2-2x+1)-1=2x^2-6x+4=2(x-1)(x-2)$ .

$=\frac{2(x-1)(x-2)}{(x-1)^2(x-2)^2}=\frac{2}{(x-1)(x-2)}$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

CDC Model Question 2080, Grade 10 Compulsory Mathematics · 1+2+2 marks

Anju deposits money to her daughter's piggy bank (Khutruke) as double of previous day. The table shows the amount of money she has deposited in first five days.

First day Second day Third day Fourth day Fifth day
Rs 5 Rs 10 Rs 20 Rs 40 Rs 80
(a) Is the sequence made by the amount Anju deposited in her daughter's piggy bank (Khutruke) arithmetic or geometric?
(b) If the money is deposited in the same fashion, how much money will be deposited by the 10th day?
(c) Based on the above sequence, is 2 days enough after 10 days of deposit to collect a total of Rs 20000? Write it with reason.

First day	Second day	Third day	Fourth day	Fifth day
Rs 5	Rs 10	Rs 20	Rs 40	Rs 80

Answer

(a) 5, 10, 20, 40, 80 have a common ratio $r=2$ , so it is a geometric sequence.

(b) $a=5$ , $r=2$ , $n=10$ .

S_{10}=\frac{5(2^{10}-1)}{2-1}=5\times1023=5115

Rs 5,115 will be deposited by the 10th day.

(c) Two more days after day 10 means 12 days in total.

S_{12}=5(2^{12}-1)=5\times4095=20475

Since $20475>20000$ (and $S_{11}=5\times2047=10235<20000$ ), yes, 2 more days are enough: on days 11 and 12 she deposits $5120+10240=15360$ , and $5115+15360=20475\ge20000$ .

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

CDC Model Question 2080, Grade 10 Compulsory Mathematics · 1+3+1 marks

The length of a rectangular field is twice the width and its area is 72 square meter.
(a) Write the standard form of the quadratic equation.
(b) Find the length and breadth of that rectangular field.
(c) How many pieces of land can be made with dimension (6×3) square metre in that rectangular field? Calculate it.

Answer

(a) $ax^2+bx+c=0$ , where $a\neq0$ .

(b) Let the breadth be $x$ m; the length is $2x$ m.

2x^2=72 \Rightarrow x^2=36 \Rightarrow x=6

Breadth $=6$ m, length $=12$ m.

(c) Piece $6\times3$ . Along 12 m: $12\div6=2$ ; along 6 m: $6\div3=2$ .

Number of pieces $=2\times2=$ 4 (area check: $72\div18=4$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

CDC Model Question 2080, Grade 10 Compulsory Mathematics · 3 marks

Solve: $25^x - 6 \times 5^x + 5 = 0$

Answer

$25^x-6\times5^x+5=0 \Rightarrow (5^x)^2-6\cdot5^x+5=0$ . Let $5^x=y$ .

y^2-6y+5=0 \Rightarrow (y-1)(y-5)=0

$5^x=1 \Rightarrow x=0$
$5^x=5 \Rightarrow x=1$

Answer: $x=0$ or $x=1$

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

CDC Model Question 2080, Grade 10 Compulsory Mathematics · 2 marks

Simplify: $\frac{m+b}{m-b} + \frac{m-b}{m+b} - \frac{m^2+b^2}{m^2-b^2}$

Answer

\frac{m+b}{m-b}+\frac{m-b}{m+b}-\frac{m^2+b^2}{m^2-b^2} =\frac{(m+b)^2+(m-b)^2-(m^2+b^2)}{m^2-b^2}

=\frac{2m^2+2b^2-m^2-b^2}{m^2-b^2}=\frac{m^2+b^2}{m^2-b^2}

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Past questions and answers

1.Simplify: 11−p+p2−11+p+p2−2p1−p2+p4\dfrac{1}{1-p+p^2}-\dfrac{1}{1+p+p^2}-\dfrac{2p}{1-p^2+p^4}1−p+p21​−1+p+p21​−1−p2+p42p​

2.What is the value of (4a)0(4a)^0(4a)0? Write it.

3.If the sum of two consecutive even numbers is 26, then find the numbers.

4.If the sum of two consecutive even numbers is 42, find the numbers.

5.The area of a rectangular field is 1440 sq. metre and its perimeter is 152 metre. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Find it.

6.The area of a rectangular land is 720 sq. metre and its perimeter is 108 metre. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Find it.

7.If the sum of two consecutive even numbers is 34, find the numbers.

8.The area of a rectangular land is 3000 sq. metres and its perimeter is 220 metres. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Find it.

9.If the sum of two consecutive even numbers is 14, find the numbers.

10.The area of a rectangular land is 2000 sq. metres and its perimeter is 180 metres. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Why? Find it.

11.The area of a rectangular land is 660 sq. metres and its perimeter is 104 metres. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Why? Find it.

12.Find the two consecutive even numbers whose sum is 30.

13.The area of a rectangular land is 500 sq. metres and its perimeter is 90 metres. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Why? Find it.

14.Solve: 2x+12x=2122^x+\dfrac{1}{2^x}=2\dfrac{1}{2}2x+2x1​=221​

15.Solve: 32x−4×3x+3=03^{2x} - 4 \times 3^{x} + 3 = 032x−4×3x+3=0

16.Simplify: m3+1m2−m+1+m3−1m2+m+1\frac{m^3+1}{m^2-m+1}+\frac{m^3-1}{m^2+m+1}m2−m+1m3+1​+m2+m+1m3−1​

17.Solve: 2x−2+23−x=32^{x-2} + 2^{3-x} = 32x−2+23−x=3

18.If x2=323+3−23−2x^2 = 3^{\frac{2}{3}} + 3^{\frac{-2}{3}} - 2x2=332​+33−2​−2, prove that: 3x3+9x=83x^3 + 9x = 83x3+9x=8

19.Find the L.C.M. of: p2−q2, (p+q)2p^2-q^2,\ (p+q)^2p2−q2, (p+q)2.

20.If the sum of two consecutive odd numbers is 20, find the numbers.

21.If the sum of two consecutive odd numbers is 12, find the numbers.

23.In a two digit number, the product of digits is 20 and their difference is 1. (a) Write the standard form of a number of two digits. (1) (b) Find the number. (3)

24.Solve: 5x=255^{\sqrt{x}} = 255x​=25

25.Simplify: 1p2−pq+q2−p2+q2p4+p2q2+q4\dfrac{1}{p^2 - pq + q^2} - \dfrac{p^2+q^2}{p^4 + p^2q^2 + q^4}p2−pq+q21​−p4+p2q2+q4p2+q2​

28.Simplify: 11+a−1+11−a−1\dfrac{1}{1+a^{-1}} + \dfrac{1}{1-a^{-1}}1+a−11​+1−a−11​

29.Solve: 22x−3.2x+1+8=02^{2x} - 3.2^{x+1} + 8 = 022x−3.2x+1+8=0

31.In 2075 B.S. the age of a father was 6 times the age of his son and in 2080 B.S. the numerical product of their ages was 350. (a) Define quadratic equation. (1) (b) Make a quadratic equation according to the given condition. (2) (c) In which year the son was born? Find it. (2)

32.Simplify: 1x−y−x+yx2−y2\frac{1}{x-y} - \frac{x+y}{x^2-y^2}x−y1​−x2−y2x+y​

33.Solve: 25x−6×5x+1+125=025^x - 6 \times 5^{x+1} + 125 = 025x−6×5x+1+125=0

36.Simplify: 1m+n−1m−n\dfrac{1}{m+n} - \dfrac{1}{m-n}m+n1​−m−n1​

37.Solve: 3y+1+3y=363^{y+1} + 3^{y} = 363y+1+3y=36

40.Simplify: 1a+2b−1a−2b+2a4b2−a2\dfrac{1}{a+2b} - \dfrac{1}{a-2b} + \dfrac{2a}{4b^2-a^2}a+2b1​−a−2b1​+4b2−a22a​

41.Solve: 4x−2=0.54^{x-2} = 0.54x−2=0.5

44.Simplify: xx−y−xx+y\frac{x}{x-y} - \frac{x}{x+y}x−yx​−x+yx​

47.Simplify: px−y−px+y\dfrac{p}{x-y} - \dfrac{p}{x+y}x−yp​−x+yp​

48.Solve: 2x+2−x−1=322^x + 2^{-x} - 1 = \dfrac{3}{2}2x+2−x−1=23​

51.Simplify: p+qpq−q+rqr−r+prp\frac{p+q}{pq}-\frac{q+r}{qr}-\frac{r+p}{rp}pqp+q​−qrq+r​−rpr+p​

52.Solve: 3y+3−y=9193^{y}+3^{-y}=9\frac{1}{9}3y+3−y=991​

54.Kriti wants to fence her field having length twice of breadth. The area of the field is 800 square feet. (a) Write down the standard form of quadratic equation. (1) (b) How much the length of wire is required to fence the field once with wire? Find it. (3)

55.Simplify: xxy−y2+yxy−x2\dfrac{x}{xy-y^2}+\dfrac{y}{xy-x^2}xy−y2x​+xy−x2y​

56.Solve: 2x+162x=102^x+\dfrac{16}{2^x}=102x+2x16​=10

59.Write 1x−6\frac{1}{x^{-6}}x−61​ in the positive index of xxx.

60.Simplify: a2a−2b+4b22b−a\frac{a^2}{a-2b}+\frac{4b^2}{2b-a}a−2ba2​+2b−a4b2​

61.Solve: 4x−2=0.254^{x-2}=0.254x−2=0.25

62.There are 7 arithmetic means between 3 and 27. (a) Write the formula to calculate arithmetic mean between aaa and bbb. (b) What is the 5th5^{th}5th mean of the given sequence? Find it. (c) Which one is greater by how much in arithmetic mean and geometric mean between 3 and 27? Compare it.

64.Simplify: aa−b+bb−a\dfrac{a}{a-b}+\dfrac{b}{b-a}a−ba​+b−ab​

67.Solve: 5x+15x=5155^x + \dfrac{1}{5^x} = 5\dfrac{1}{5}5x+5x1​=551​

70.Simplify: 1b−1−1b+1\frac{1}{b-1} - \frac{1}{b+1}b−11​−b+11​

71.Solve: 7x+7−x=7177^x + 7^{-x} = 7\frac{1}{7}7x+7−x=771​

74.Simplify: 1x−y−1x+y\frac{1}{x-y}-\frac{1}{x+y}x−y1​−x+y1​

76.The product of two consecutive positive numbers is 420. (a) What are the numbers? Find it. (2) (b) What should be subtracted from the numbers to get their product as 182? Find it. (2)

77.What is the value of (xy)0\left(\frac{x}{y}\right)^0(yx​)0? Write it.

78.Simplify: 3x+19x2+3x+1+3x−19x2−3x+1+281x4+9x2+1\frac{3x+1}{9x^2+3x+1}+\frac{3x-1}{9x^2-3x+1}+\frac{2}{81x^4+9x^2+1}9x2+3x+13x+1​+9x2−3x+13x−1​+81x4+9x2+12​

79.Solve by completing square: x2−5x+6=0x^2-5x+6=0x2−5x+6=0

85.Simplify: x+yx−y−x−yx+y−2xyx2−y2\frac{x+y}{x-y} - \frac{x-y}{x+y} - \frac{2xy}{x^2-y^2}x−yx+y​−x+yx−y​−x2−y22xy​

88.Simplify: 1x−y−yxy+y2\dfrac{1}{x-y} - \dfrac{y}{xy+y^2}x−y1​−xy+y2y​

89.If x=213−2−13x = 2^{\frac{1}{3}} - 2^{-\frac{1}{3}}x=231​−2−31​, prove that: 2x3+6x−3=02x^3 + 6x - 3 = 02x3+6x−3=0

92.Simplify: m+nm−n−m−nm+n\frac{m+n}{m-n}-\frac{m-n}{m+n}m−nm+n​−m+nm−n​

93.Solve: 6x+6−x=6166^{x}+6^{-x}=6\frac{1}{6}6x+6−x=661​

95.The sum of present ages of two brothers is 20 years. Four years ago, the product of their ages in numbers was 48. (a) Define quadratic equation. (1) (b) Find the quadratic equation formed from the given statement. (2) (c) If possible, find their present ages. If not, give reason. (2)

96.Solve: 9x−10×3x+9=09^x - 10\times 3^x + 9 = 09x−10×3x+9=0

99.Simplify: 1a+2b+2aa2−4b2\frac{1}{a+2b} + \frac{2a}{a^2-4b^2}a+2b1​+a2−4b22a​

100.Solve: 4×3x+1−9x=274 \times 3^{x+1} - 9^x = 274×3x+1−9x=27

101.The third term and sixth term of a geometric series are 27 and 729 respectively. (a) What is the geometric mean of aaa and bbb? Write it. (1) (b) Find the common ratio of the series. (2) (c) Find the sum of first five terms of the series. (2)

103.Solve: 3x+1+3x+2=363^{x+1} + 3^{x+2} = 363x+1+3x+2=36

104.Simplify: 1x+1+1x−1+2xx2+1\frac{1}{x+1} + \frac{1}{x-1} + \frac{2x}{x^2+1}x+11​+x−11​+x2+12x​

107.Simplify: 1x−1−2x2−1\dfrac{1}{x-1}-\dfrac{2}{x^2-1}x−11​−x2−12​

108.Solve: 2x+12x=4142^x+\dfrac{1}{2^x}=4\dfrac{1}{4}2x+2x1​=441​

109.What is the value of 11a011a^011a0? Write it.

110.Prove that: 7x+2+7x50×7x=1\dfrac{7^{x+2}+7^x}{50\times 7^x}=150×7x7x+2+7x​=1.

111.Prove that: 7x+1−7x12×7x=12\dfrac{7^{x+1}-7^x}{12\times 7^x}=\dfrac{1}{2}12×7x7x+1−7x​=21​.

112.Rationalize the denominator of: 2+12−1\dfrac{\sqrt{2}+1}{\sqrt{2}-1}2​−12​+1​.

113.Rationalize the denominator of: 12+3\dfrac{1}{2+\sqrt{3}}2+3​1​.

114.Find the L.C.M. of: x2−y2, (x+y)2x^2-y^2,\ (x+y)^2x2−y2, (x+y)2.

115.Find the L.C.M. of: z2−y2, (z−y)2z^2-y^2,\ (z-y)^2z2−y2, (z−y)2.

116.Solve: x−2=1253\sqrt{x-2}=\sqrt[3]{125}x−2​=3125​.

117.Solve: x+11=6\sqrt{x+11}=6x+11​=6.

118.Find the H.C.F. of: b2(b2+4bc+4c2)b^2(b^2+4bc+4c^2)b2(b2+4bc+4c2), b5+8b2c3b^5+8b^2c^3b5+8b2c3 and 3b4+b3c−10b2c23b^4+b^3c-10b^2c^23b4+b3c−10b2c2.

119.Find the H.C.F. of: a3+4a2b+4ab2a^3+4a^2b+4ab^2a3+4a2b+4ab2, a4+8ab3a^4+8ab^3a4+8ab3 and 3a4+a3b−10a2b23a^4+a^3b-10a^2b^23a4+a3b−10a2b2.

Simplify: $\dfrac{1}{1-p+p^2}-\dfrac{1}{1+p+p^2}-\dfrac{2p}{1-p^2+p^4}$

What is the value of $(4a)^0$ ? Write it.

If the sum of two consecutive even numbers is 26, then find the numbers.

If the sum of two consecutive even numbers is 42, find the numbers.

The area of a rectangular field is 1440 sq. metre and its perimeter is 152 metre. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Find it.

The area of a rectangular land is 720 sq. metre and its perimeter is 108 metre. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Find it.

If the sum of two consecutive even numbers is 34, find the numbers.

The area of a rectangular land is 3000 sq. metres and its perimeter is 220 metres. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Find it.

If the sum of two consecutive even numbers is 14, find the numbers.

The area of a rectangular land is 2000 sq. metres and its perimeter is 180 metres. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Why? Find it.

The area of a rectangular land is 660 sq. metres and its perimeter is 104 metres. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Why? Find it.

Find the two consecutive even numbers whose sum is 30.

The area of a rectangular land is 500 sq. metres and its perimeter is 90 metres. Out of the length or the breadth, which one is to be decreased by what percentage to make it a square? Why? Find it.

Solve: $2^x+\dfrac{1}{2^x}=2\dfrac{1}{2}$

Solve: $3^{2x} - 4 \times 3^{x} + 3 = 0$

Simplify: $\frac{m^3+1}{m^2-m+1}+\frac{m^3-1}{m^2+m+1}$

Solve: $2^{x-2} + 2^{3-x} = 3$

If $x^2 = 3^{\frac{2}{3}} + 3^{\frac{-2}{3}} - 2$ , prove that: $3x^3 + 9x = 8$

Find the L.C.M. of: $p^2-q^2,\ (p+q)^2$ .

If the sum of two consecutive odd numbers is 20, find the numbers.

If the sum of two consecutive odd numbers is 12, find the numbers.

In a two digit number, the product of digits is 20 and their difference is 1.
(a) Write the standard form of a number of two digits. (1)
(b) Find the number. (3)

Solve: $5^{\sqrt{x}} = 25$

Simplify: $\dfrac{1}{p^2 - pq + q^2} - \dfrac{p^2+q^2}{p^4 + p^2q^2 + q^4}$

Simplify: $\dfrac{1}{1+a^{-1}} + \dfrac{1}{1-a^{-1}}$

Solve: $2^{2x} - 3.2^{x+1} + 8 = 0$

In 2075 B.S. the age of a father was 6 times the age of his son and in 2080 B.S. the numerical product of their ages was 350.
(a) Define quadratic equation. (1)
(b) Make a quadratic equation according to the given condition. (2)
(c) In which year the son was born? Find it. (2)

Simplify: $\frac{1}{x-y} - \frac{x+y}{x^2-y^2}$

Solve: $25^x - 6 \times 5^{x+1} + 125 = 0$

Simplify: $\dfrac{1}{m+n} - \dfrac{1}{m-n}$

Solve: $3^{y+1} + 3^{y} = 36$

Simplify: $\dfrac{1}{a+2b} - \dfrac{1}{a-2b} + \dfrac{2a}{4b^2-a^2}$

Solve: $4^{x-2} = 0.5$

Simplify: $\frac{x}{x-y} - \frac{x}{x+y}$

Simplify: $\dfrac{p}{x-y} - \dfrac{p}{x+y}$

Solve: $2^x + 2^{-x} - 1 = \dfrac{3}{2}$

Simplify: $\frac{p+q}{pq}-\frac{q+r}{qr}-\frac{r+p}{rp}$

Solve: $3^{y}+3^{-y}=9\frac{1}{9}$

Kriti wants to fence her field having length twice of breadth. The area of the field is 800 square feet.
(a) Write down the standard form of quadratic equation. (1)
(b) How much the length of wire is required to fence the field once with wire? Find it. (3)

Simplify: $\dfrac{x}{xy-y^2}+\dfrac{y}{xy-x^2}$

Solve: $2^x+\dfrac{16}{2^x}=10$

Write $\frac{1}{x^{-6}}$ in the positive index of $x$ .

Simplify: $\frac{a^2}{a-2b}+\frac{4b^2}{2b-a}$

Solve: $4^{x-2}=0.25$

There are 7 arithmetic means between 3 and 27.
(a) Write the formula to calculate arithmetic mean between $a$ and $b$ .
(b) What is the $5^{th}$ mean of the given sequence? Find it.
(c) Which one is greater by how much in arithmetic mean and geometric mean between 3 and 27? Compare it.

Simplify: $\dfrac{a}{a-b}+\dfrac{b}{b-a}$

Solve: $5^x + \dfrac{1}{5^x} = 5\dfrac{1}{5}$

Simplify: $\frac{1}{b-1} - \frac{1}{b+1}$

Solve: $7^x + 7^{-x} = 7\frac{1}{7}$

Simplify:
$\frac{1}{x-y}-\frac{1}{x+y}$

The product of two consecutive positive numbers is 420.
(a) What are the numbers? Find it. (2)
(b) What should be subtracted from the numbers to get their product as 182? Find it. (2)

What is the value of $\left(\frac{x}{y}\right)^0$ ? Write it.

Simplify: $\frac{3x+1}{9x^2+3x+1}+\frac{3x-1}{9x^2-3x+1}+\frac{2}{81x^4+9x^2+1}$

Solve by completing square: $x^2-5x+6=0$

Simplify:
$\frac{x+y}{x-y} - \frac{x-y}{x+y} - \frac{2xy}{x^2-y^2}$

Simplify: $\dfrac{1}{x-y} - \dfrac{y}{xy+y^2}$

If $x = 2^{\frac{1}{3}} - 2^{-\frac{1}{3}}$ , prove that: $2x^3 + 6x - 3 = 0$

Simplify: $\frac{m+n}{m-n}-\frac{m-n}{m+n}$

Solve: $6^{x}+6^{-x}=6\frac{1}{6}$

The sum of present ages of two brothers is 20 years. Four years ago, the product of their ages in numbers was 48.
(a) Define quadratic equation. (1)
(b) Find the quadratic equation formed from the given statement. (2)
(c) If possible, find their present ages. If not, give reason. (2)

Solve: $9^x - 10\times 3^x + 9 = 0$

Simplify: $\frac{1}{a+2b} + \frac{2a}{a^2-4b^2}$

Solve: $4 \times 3^{x+1} - 9^x = 27$

The third term and sixth term of a geometric series are 27 and 729 respectively.
(a) What is the geometric mean of $a$ and $b$ ? Write it. (1)
(b) Find the common ratio of the series. (2)
(c) Find the sum of first five terms of the series. (2)

Solve: $3^{x+1} + 3^{x+2} = 36$

Simplify: $\frac{1}{x+1} + \frac{1}{x-1} + \frac{2x}{x^2+1}$

Simplify: $\dfrac{1}{x-1}-\dfrac{2}{x^2-1}$

Solve: $2^x+\dfrac{1}{2^x}=4\dfrac{1}{4}$

What is the value of $11a^0$ ? Write it.

Prove that: $\dfrac{7^{x+2}+7^x}{50\times 7^x}=1$ .

Prove that: $\dfrac{7^{x+1}-7^x}{12\times 7^x}=\dfrac{1}{2}$ .

Rationalize the denominator of: $\dfrac{\sqrt{2}+1}{\sqrt{2}-1}$ .

Rationalize the denominator of: $\dfrac{1}{2+\sqrt{3}}$ .

Find the L.C.M. of: $x^2-y^2,\ (x+y)^2$ .

Find the L.C.M. of: $z^2-y^2,\ (z-y)^2$ .

Solve: $\sqrt{x-2}=\sqrt[3]{125}$ .

Solve: $\sqrt{x+11}=6$ .

Find the H.C.F. of: $b^2(b^2+4bc+4c^2)$ , $b^5+8b^2c^3$ and $3b^4+b^3c-10b^2c^2$ .

Find the H.C.F. of: $a^3+4a^2b+4ab^2$ , $a^4+8ab^3$ and $3a^4+a^3b-10a^2b^2$ .

In 2060 BS, the age of a father was three times the age of his son. If the age of the father will be two times the age of his son in 2080 BS, in which year was the father born? Find it.