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Chapter 2

Mathematics

School entrance and scholarship tests past exam questions

Past questions and answers

60 questions set from this chapter. Most repeated first.

  • 2082 · 1 mark

The terminal compound interest on a certain sum xx in tt years at the rate of rr % compound interest is CI4CI_4. What is the relation among x,t,rx, t, r and CI4CI_4? (a) CI4=x[(1+r100)t−1]CI_4 = x\left[\left(1+\frac{r}{100}\right)^t - 1\right] (b) CI4=x[(1+r400)t−1]CI_4 = x\left[\left(1+\frac{r}{400}\right)^t - 1\right] (c) CI4=x[(1+r200)2t−1]CI_4 = x\left[\left(1+\frac{r}{200}\right)^{2t} - 1\right] (d) CI4=x[(1+r400)4t−1]CI_4 = x\left[\left(1+\frac{r}{400}\right)^{4t} - 1\right]

Similar questions: Compound interest (2081)

Answer

(d) CI4=x[(1+r400)4t−1]CI_4 = x\left[\left(1+\frac{r}{400}\right)^{4t} - 1\right]

"Terminal" means quarterly compounding, so the rate per period is r4%\frac{r}{4}\% and there are 4t4t periods: amount =x(1+r400)4t=x\left(1+\frac{r}{400}\right)^{4t}, and interest is amount minus xx.

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  • 2081 · 1 mark

The semi-annual compound interest on a certain sum m in t years at the rate of r % compound interest is CI. What is the relation among m, t, r and CI ? (a) CI=m(1+r100)2tCI = m\left(1 + \frac{r}{100}\right)^{2t} (b) CI=m(1−r100)2tCI = m\left(1 - \frac{r}{100}\right)^{2t} (c) CI=m[(1+r200)2t−1]CI = m\left[\left(1 + \frac{r}{200}\right)^{2t} - 1\right] (d) CI=m[(1−r100)2t−1]CI = m\left[\left(1 - \frac{r}{100}\right)^{2t} - 1\right]

Similar questions: Compound interest (2082)

Answer

(c) CI=m[(1+r200)2t−1]CI = m\left[\left(1 + \frac{r}{200}\right)^{2t} - 1\right]

For half-yearly compounding the rate per period is r2%\frac r2\% and the number of periods is 2t2t, so the amount is m(1+r200)2tm\left(1+\frac{r}{200}\right)^{2t}. Compound interest is amount minus principal.

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  • 2082 · 1 mark

If A and B are the subsets of a universal set U, which of the following notation is used to denote (A−B)∪(B−A)(A - B) \cup (B - A)? (a) AΔBA \Delta B (b) AablaBA abla B (c) A∩B‾\overline{A \cap B} (d) A∪B‾\overline{A \cup B}

Answer

(a) AΔBA \Delta B

(A−B)∪(B−A)(A-B)\cup(B-A) is the symmetric difference of AA and BB, written AΔBA \Delta B.

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  • 2082 · 1 mark

In a group of 150 people, 120 like to play volleyball and 85 like to play football. If 25 like to play none of the games, then how many people like to play both the games? (a) 20 (b) 30 (c) 80 (d) 25

Answer

(c) 80

People playing at least one game =150−25=125=150-25=125.

n(V∪F)=n(V)+n(F)−n(V∩F)n(V\cup F)=n(V)+n(F)-n(V\cap F) 125=120+85−n(V∩F)⇒n(V∩F)=205−125=80125=120+85-n(V\cap F)\Rightarrow n(V\cap F)=205-125=80

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  • 2082 · 1 mark

The policy of bank for same principal, same rate and same time, which of the following policy is more profitable to consumer? (a) Annual Compound Interest (b) Semi-Annual Compound Interest (c) Terminal Compound Interest (d) Monthly Compound Interest

Answer

(d) Monthly Compound Interest

For the same principal, rate and time, the more often interest is compounded, the larger the amount. Monthly compounding (12 times a year) gives the most interest, so it is most profitable to the depositor.

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  • 2082 · 1 mark

The present price of machine is Rs. 800000. If the price of the machine depreciates first year by 10% and then 8% and 5% respectively in the following years, what will be the price of the machine after 3 years? (a) Rs. 6,29,000 (b) Rs. 6,25,280 (c) 7,29,280 (d) 6,29,280

Answer

(d) Rs. 6,29,280

Price=800000×0.90×0.92×0.95\text{Price}=800000\times0.90\times0.92\times0.95 =720000×0.92=662400,662400×0.95=629280=720000\times0.92=662400,\quad 662400\times0.95=629280

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  • 2082 · 1 mark

The price of a good is Rs. 30000. What will be the cost of good after 3 years at the rate of 10% depreciation? (a) Rs. 21087 (b) Rs. 21780 (c) Rs. 21870 (d) Rs. 21708

Answer

(c) Rs. 21870

30000(1−10100)3=30000×0.729=2187030000\left(1-\frac{10}{100}\right)^3=30000\times0.729=21870

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  • 2082 · 1 mark

If I.C. 100 = NRs. 160 and 1 American Dollar = NRs. 120, how many American Dollars are equal to I.C. 7500? (a) Dollar 100 (b) Dollar 10 (c) Dollar 110 (d) I.C. 100

Answer

(a) Dollar 100

I.C. 7500 =7500100×160=Rs 12000=\frac{7500}{100}\times160=\text{Rs }12000.

Dollars =12000120=100=\frac{12000}{120}=100.

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  • 2082 · 1 mark

Find out the area of triangular surfaces of a square based pyramid, if its area of base is 'z' square unit and total surface area is 'y' square unit. (a) yy sq. unit (b) zz sq. unit (c) (y−z)(y-z) sq. unit (d) y−z\sqrt{y-z} sq. unit

Answer

(c) (y−z)(y-z) sq. unit

Total surface area = base area + area of the triangular faces, so the triangular faces have area y−zy-z.

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  • 2082 · 1 mark

In the given figure (an ice-cream cone with a hemisphere on top; slant height of the cone is 25 cm), the slant height of the cone is 25 cm and the sum of the curved surface area is 858 cm2^2. What is the common radius? (a) 7 cm (b) 14 cm (c) 21 cm (d) 15 cm

Answer

(a) 7 cm

The cone and the hemisphere have curved surface areas πrl\pi r l and 2πr22\pi r^2:

πr(25)+2πr2=858⇒πr(25+2r)=858\pi r(25)+2\pi r^2=858\Rightarrow \pi r(25+2r)=858

Try r=7r=7: 227×7×(25+14)=22×39=858\frac{22}{7}\times7\times(25+14)=22\times39=858. So r=7r=7 cm.

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  • 2082 · 1 mark

What is the total surface area of square based pyramid having the slant height 10 cm and vertical height 8 cm? (a) 483 cm2^2 (b) 240 cm2^2 (c) 144 cm2^2 (d) 384 cm2^2

Answer

(d) 384 cm2^2

Slant height l=10l=10 and height h=8h=8, so half the base side is 102−82=6\sqrt{10^2-8^2}=6 cm, giving base side 1212 cm.

TSA=a2+2al=122+2(12)(10)=144+240=384 cm2\text{TSA}=a^2+2al=12^2+2(12)(10)=144+240=384\ \text{cm}^2

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  • 2082 · 1 mark

Which of the following formula is known as the Heron's formula to calculate area of a triangle? (a) A=12bhA = \frac{1}{2}bh (b) A=34a2A = \frac{\sqrt{3}}{4}a^2 (c) A=b44a2−b2A = \frac{b}{4}\sqrt{4a^2 - b^2} (d) A=s(s−a)(s−b)(s−c)A = \sqrt{s(s-a)(s-b)(s-c)}

Answer

(d) A=s(s−a)(s−b)(s−c)A = \sqrt{s(s-a)(s-b)(s-c)}

Heron's formula gives the area of a triangle with sides a,b,ca,b,c and semi-perimeter s=a+b+c2s=\frac{a+b+c}{2}.

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  • 2082 · 1 mark

In the given figure (a triangle with base endpoints marked and a segment drawn from the apex to the base), which segment represents a median of the given triangle? (a) AB (b) BC (c) AC (d) AD

Answer

(d) AD

A median joins a vertex to the midpoint of the opposite side. AD runs from the apex A to the point D on the base BC, so it is the median; AB and AC are sides and BC is the base.

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  • 2082 · 1 mark

If the sum of the first nn terms of arithmetic series is given by Sn=3n2+2nS_n = 3n^2 + 2n, What is the first term of the series? (a) 3 (b) 5 (c) 2 (d) 7

Answer

(b) 5

The first term equals S1=3(1)2+2(1)=5S_1=3(1)^2+2(1)=5.

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  • 2082 · 1 mark

The length of a square room is 10 m and height is 6 m. If the per unit cost of painting the four walls of the room is Rs. 400, find the total cost. (a) Rs. 96000 (b) Rs. 96000 (c) Rs. 69000 (d) Rs. 90600

Answer

(a) Rs. 96000

The room is square with side 10 m, so the four walls have area 2(l+b)h=4×10×6=2402(l+b)h=4\times10\times6=240 m2^2.

Cost=240×400=Rs 96000\text{Cost}=240\times400=\text{Rs }96000

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  • 2082 · 1 mark

What is the volume of cone with radius 3k3k cm and vertical height kπ\frac{k}{\pi} cm? (a) 3k33k^3 (b) 13k3\frac{1}{3}k^3 (c) 13πk3\frac{1}{3}\pi k^3 (d) 3πk33\pi k^3

Answer

(a) 3k33k^3

V=13πr2h=13π(3k)2⋅kπ=13⋅9k2⋅k=3k3V=\frac13\pi r^2h=\frac13\pi(3k)^2\cdot\frac{k}{\pi}=\frac13\cdot9k^2\cdot k=3k^3

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  • 2082 · 1 mark

The present age of husband and wife is 25 years and 22 years. What will be the product of their ages after 3 years? (a) 418 (b) 700 (c) 550 (d) 53

Answer

(b) 700

After 3 years the ages are 25+3=2825+3=28 and 22+3=2522+3=25, and 28×25=70028\times25=700.

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  • 2082 · 1 mark

What is sum of the roots of the quadratic equation: x2−2x−15=0x^2 - 2x - 15 = 0? (a) 2 (b) 3 (c) 4 (d) 5

Answer

(a) 2

For ax2+bx+c=0ax^2+bx+c=0 the sum of roots is −ba=−−21=2-\frac{b}{a}=-\frac{-2}{1}=2. (Check: roots are 55 and −3-3, sum 22.)

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  • 2082 · 1 mark

200 is obtained when subtracted 42 from the double of the square of a natural number, what is the natural number? (a) 7 (b) 9 (c) 11 (d) 13

Answer

(c) 11

2x2−42=200⇒x2=121⇒x=112x^2-42=200\Rightarrow x^2=121\Rightarrow x=11

(natural number, so the negative root is rejected).

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  • 2082 · 1 mark

If 3 is added to the numerator of a fraction, the value of the fraction becomes 1 and when 3 is added to the denominator of the fraction, the value of fraction becomes 14\frac{1}{4}. What is the fraction? (a) 13\frac{1}{3} (b) 23\frac{2}{3} (c) 25\frac{2}{5} (d) 35\frac{3}{5}

Answer

(c) 25\frac{2}{5}

Let the fraction be nd\frac{n}{d}. Then n+3d=1⇒d=n+3\frac{n+3}{d}=1\Rightarrow d=n+3, and nd+3=14⇒4n=d+3=n+6⇒n=2, d=5\frac{n}{d+3}=\frac14\Rightarrow 4n=d+3=n+6\Rightarrow n=2,\ d=5.

Check: 55=1\frac{5}{5}=1 and 28=14\frac{2}{8}=\frac14.

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  • 2082 · 1 mark

What is the value of m in "4m=0.254^m = 0.25"? (a) 0 (b) 1 (c) −1-1 (d) 5

Answer

(c) −1-1

4m=0.25=14=4−1⇒m=−14^m=0.25=\frac14=4^{-1}\Rightarrow m=-1

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  • 2082 · 1 mark

If p(x)p(x) and q(x)q(x) are the two polynomial which have no common factors except 1. What is the L.C.M.? (a) p(x)p(x) (b) q(x)q(x) (c) 1 (d) p(x)×q(x)p(x) \times q(x)

Answer

(d) p(x)×q(x)p(x) \times q(x)

Since p(x)p(x) and q(x)q(x) have no common factor other than 1, their H.C.F. is 1. Using L.C.M.×H.C.F.=p(x)×q(x)\text{L.C.M.} \times \text{H.C.F.} = p(x)\times q(x), the L.C.M. is p(x)×q(x)p(x)\times q(x).

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  • 2082 · 1 mark

In the given figure, ABCD is a parallelogram with AD parallel to BE (B, C, E on one line, with E beyond C), and the area of parallelogram ABCD is 100 sq. cm. What is the area of triangle ABE? (a) 50 sq. cm (b) 100 sq. cm (c) 200 sq. cm (d) 40 sq. cm

Answer

(a) 50 sq. cm

The triangle and the parallelogram stand on the same base and lie between the same parallel lines, so the triangle has half the area of the parallelogram:

12×100=50 sq. cm\frac12 \times 100 = 50 \text{ sq. cm}

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  • 2082 · 1 mark

Which of the following statements is true? (a) Rhombus is a square. (b) Parallelogram is a rectangle. (c) Rectangle is a parallelogram. (d) All squares are not Rectangle.

Answer

(c) Rectangle is a parallelogram.

A rectangle has both pairs of opposite sides parallel and equal, so it is a parallelogram. The other statements are false: a rhombus need not have right angles, a parallelogram need not have right angles, and every square is a rectangle.

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  • 2082 · 1 mark

If A+B=90∘A + B = 90^\circ then A and B are: (a) Supplementary (b) Complementary (c) Adjacent (d) All of the above

Answer

(b) Complementary

Two angles whose sum is 90∘90^\circ are complementary. A sum of 180∘180^\circ would make them supplementary.

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  • 2082 · 1 mark

In the given figure, ABCD is a square and EBCF a parallelogram. If AB = 4 cm, what is the area of parallelogram EBCF? (a) 32 cm2^2 (b) 16 cm2^2 (c) 20 cm2^2 (d) 10 cm2^2

Answer

(b) 16 cm2^2

The square ABCD and the parallelogram EBCF have the same base BC and lie between the same parallels, so their areas are equal:

4×4=16 cm24\times 4 = 16 \text{ cm}^2

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  • 2082 · 1 mark

In the given figure, O is the center of circle. If TAN is tangent then, what is the value of ∠AON+∠ONA\angle AON + \angle ONA? (a) 45∘45^\circ (b) 60∘60^\circ (c) 90∘90^\circ (d) 180∘180^\circ

Answer

(c) 90∘90^\circ

The radius OA is perpendicular to the tangent TAN at A, so ∠OAN=90∘\angle OAN = 90^\circ. In triangle OAN the angle sum is 180∘180^\circ, hence

∠AON+∠ONA=180∘−90∘=90∘\angle AON + \angle ONA = 180^\circ - 90^\circ = 90^\circ

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  • 2082 · 1 mark

What is the value of x in given figure? (A square ABCD is inscribed in a circle with centre O; x is the angle between the diagonal AC and the side from A.) (a) 90∘90^\circ (b) 45∘45^\circ (c) 180∘180^\circ (d) 60∘60^\circ

Answer

(b) 45∘45^\circ

The diagonal AC of a square bisects the right angle at A, so x=90∘2=45∘x = \frac{90^\circ}{2} = 45^\circ.

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  • 2082 · 1 mark

A fair dice is rolled. What is the probability of getting a number greater than 4? (a) 16\frac{1}{6} (b) 13\frac{1}{3} (c) 12\frac{1}{2} (d) 23\frac{2}{3}

Answer

(b) 13\frac{1}{3}

Numbers greater than 4 on a die are 5 and 6, so there are 2 favourable outcomes out of 6:

P=26=13P = \frac{2}{6} = \frac13

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  • 2082 · 1 mark

In a continuous series, if ∑fm=120+8k\sum fm = 120 + 8k and ∑f=15+k\sum f = 15 + k, what is the value of mean? (a) 12 (b) 20 (c) 8 (d) 15

Answer

(c) 8

Mean=∑fm∑f=120+8k15+k=8(15+k)15+k=8\text{Mean} = \frac{\sum fm}{\sum f} = \frac{120+8k}{15+k} = \frac{8(15+k)}{15+k} = 8

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  • 2082 · 1 mark

What is the value of x in given figure? (Right-angled figure: A and B are 2 m apart vertically, with B and C at the bottom, AD = 18 m horizontal and DE perpendicular to AD at D, total height of E above the base 20 m; x is the angle at A between AD and AE.) (a) 45∘45^\circ (b) 60∘60^\circ (c) 90∘90^\circ (d) 30∘30^\circ

Answer

(a) 45∘45^\circ

The height of E above A is 20−2=1820-2 = 18 m and the horizontal distance AD is 18 m, so

tan⁡x=1818=1⇒x=45∘\tan x = \frac{18}{18} = 1 \Rightarrow x = 45^\circ

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  • 2081 · 1 mark

What is the Volume of a Cylinder with radius K and vertical height Y ? (a) πk2y\pi k^2 y (b) 13πk2y\frac{1}{3}\pi k^2 y (c) 23πk2y\frac{2}{3}\pi k^2 y (d) 13kπ2y\frac{1}{3}k\pi^2 y

Answer

(a) πk2y\pi k^2 y

Volume of a cylinder = base area ×\times height = πr2h\pi r^2 h. With r=kr=k and h=yh=y this gives πk2y\pi k^2 y. The factor 13\frac13 belongs to a cone.

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  • 2081 · 1 mark

If 20, x, y, z, 40 are arithmetic sequence then what is the value of y ? (a) 25 (b) 60 (c) 35 (d) 30

Answer

(d) 30

yy is the middle (third) term of the five terms, so it is the mean of the first and last:

y=20+402=30y=\frac{20+40}{2}=30

Check: the common difference is (40−20)/4=5(40-20)/4=5, giving 20, 25, 30, 35, 40.

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  • 2081 · 1 mark

In 5+10+15+⋯5 + 10 + 15 + \cdots which term is 65 ? (a) 10 (b) 11 (c) 12 (d) 13

Answer

(d) 13

Here a=5a=5, d=5d=5, so tn=5+(n−1)5=5nt_n=5+(n-1)5=5n. Setting 5n=655n=65 gives n=13n=13.

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  • 2081 · 1 mark

If 11 is subtracted from the square of a number the result is 38. Find the number. (a) −7-7 (b) 7 (c) ±7\pm 7 (d) −6-6

Answer

(c) ±7\pm 7

x2−11=38⇒x2=49⇒x=±7x^2-11=38 \Rightarrow x^2=49 \Rightarrow x=\pm 7

Both 77 and −7-7 satisfy the condition, since (−7)2−11=38(-7)^2-11=38.

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  • 2081 · 1 mark

The present ages of mother and daughter are 45 years and 15 years respectively. What will be the sum of their ages after 10 years? (a) 50 (b) 60 (c) 70 (d) 80

Answer

(d) 80

After 10 years, mother is 45+10=5545+10=55 and daughter is 15+10=2515+10=25. Sum =55+25=80=55+25=80 years.

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  • 2081 · 1 mark

Which one is a quadratic equation in which the values of y are 2 and −1-1? (a) y2+y−2=0y^2 + y - 2 = 0 (b) y2−y+2=0y^2 - y + 2 = 0 (c) y2−y−2=0y^2 - y - 2 = 0 (d) 2y2−y−1=02y^2 - y - 1 = 0

Answer

(c) y2−y−2=0y^2 - y - 2 = 0

Roots 22 and −1-1: sum =1=1, product =−2=-2. The equation is y2−(sum)y+product=0y^2-(\text{sum})y+\text{product}=0, i.e. y2−y−2=0y^2-y-2=0.

Check: (y−2)(y+1)=y2−y−2(y-2)(y+1)=y^2-y-2.

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  • 2081 · 1 mark

What is the value of (729)−13\sqrt[3]{\sqrt{(729)^{-1}}} ? (a) 13\frac{1}{3} (b) 12\frac{1}{2} (c) 3 (d) 2

Answer

(a) 13\frac{1}{3}

729−1=3−6,3−6=3−3,3−33=3−1=13729^{-1}=3^{-6},\quad \sqrt{3^{-6}}=3^{-3},\quad \sqrt[3]{3^{-3}}=3^{-1}=\frac13

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  • 2081 · 1 mark

If (1am)n=ax\left(\frac{1}{a^m}\right)^n = a^x, what is the value of x ? (a) mm (b) nn (c) mnmn (d) −mn-mn

Answer

(d) −mn-mn

(1am)n=(a−m)n=a−mn\left(\frac{1}{a^m}\right)^n=(a^{-m})^n=a^{-mn}

so x=−mnx=-mn.

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  • 2081 · 1 mark

f(x)f(x) and p(x)p(x) are two polynomials. If there is no common factor to both the polynomials, what will be their lowest common factor (L.C.M.)? (a) f(x)f(x) (b) p(x)p(x) (c) f(x)×p(x)f(x) \times p(x) (d) 1

Answer

(c) f(x)×p(x)f(x)\times p(x)

When two polynomials have no common factor (their H.C.F. is 1), their L.C.M. is the product: L.C.M.=f(x) p(x)H.C.F.=f(x) p(x)\text{L.C.M.}=\dfrac{f(x)\,p(x)}{\text{H.C.F.}}=f(x)\,p(x).

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  • 2081 · 1 mark

Find the value of xx: 3x+1⋅5x−2=273^{x+1} \cdot 5^{x-2} = 27 (a) 2 (b) 3 (c) 4 (d) 5

Answer

(a) 2

Try x=2x=2: 33⋅50=27⋅1=273^{3}\cdot 5^{0}=27\cdot 1=27, which matches.

Check others: x=3x=3 gives 81⋅5=405≠2781\cdot5=405\neq27, and larger values are bigger still, so x=2x=2.

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  • 2081 · 1 mark

In the given figure, the height of the pole including the lamp AF = 19.6 m and the height of the man CE = 1.6 m. If the angle of elevation is 45∘45^\circ when a person looks at the top of the lamp, what is the distance between the person and the lamp post? (a) 1.6 m (b) 18 m (c) 19.6 m (d) 21.2 m

Answer

(b) 18 m

The eye level is at the man's height 1.61.6 m, so the lamp top is 19.6−1.6=1819.6-1.6=18 m above the eye. With elevation 45∘45^\circ:

tan⁡45∘=18d⇒d=18 m\tan 45^\circ=\frac{18}{d}\Rightarrow d=18\text{ m}

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  • 2081 · 1 mark

In the given figure, △RST\triangle RST and square PQRS are standing on the same base and between the same parallel lines SR and PT. If PS = 4 cm, what is the area of △RST\triangle RST ? (a) 4 cm2^2 (b) 8 cm2^2 (c) 16 cm2^2 (d) 32 cm2^2

Answer

(b) 8 cm2^2

Square PQRS has side 44 cm, and △RST\triangle RST has base SR=4SR=4 cm and height 44 cm (the distance between the parallel lines).

Area=12×4×4=8 cm2\text{Area}=\tfrac12\times 4\times 4=8\text{ cm}^2

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  • 2081 · 1 mark

In the given figure, the area of the △ABE\triangle ABE is 108 cm2^2 and AB = 18 cm. What is the height of the parallelogram ABCD ? (a) 9 cm (b) 6 cm (c) 12 cm (d) 18 cm

Answer

(c) 12 cm

△ABE\triangle ABE and parallelogram ABCD stand on the same base AB between the same parallels, so they have the same height hh.

12×18×h=108⇒9h=108⇒h=12 cm\tfrac12\times 18\times h=108\Rightarrow 9h=108\Rightarrow h=12\text{ cm}

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  • 2081 · 1 mark

In the given figure (a circle through P, Q, R, S with a tangent at R), which angle is equal to the ∠QRT\angle QRT ? (a) ∠QRS\angle QRS (b) ∠PQR\angle PQR (c) ∠PSR\angle PSR (d) ∠QPS\angle QPS

Answer

(d) ∠QPS\angle QPS

PQRS is a cyclic quadrilateral, and ∠QRT\angle QRT is an exterior angle at R (T on the extension of SR). An exterior angle of a cyclic quadrilateral equals the interior opposite angle, which is ∠QPS\angle QPS.

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  • 2081 · 1 mark

ABCD is a circle in the given figure where AB = BC. What is the value of ∠ECD\angle ECD ? (a) 100∘100^\circ (b) 70∘70^\circ (c) 50∘50^\circ (d) 80∘80^\circ

Answer

(d) 80∘80^\circ

Since AB = BC, the angles subtended by these equal chords in the circle are equal. The exterior angle ∠ECD\angle ECD of the cyclic quadrilateral ABCD equals the interior opposite angle, which works out to 80∘80^\circ.

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  • 2081 · 1 mark

In a continuous series, mean (xˉ)=18+3m(\bar{x}) = 18 + 3m and ∑fx=126+21m\sum fx = 126 + 21m, find the number of observations. (a) 6 (b) 7 (c) 8 (d) 9

Answer

(b) 7

Since xˉ=∑fxN\bar{x}=\dfrac{\sum fx}{N},

N=126+21m18+3m=21(6+m)3(6+m)=7N=\frac{126+21m}{18+3m}=\frac{21(6+m)}{3(6+m)}=7

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  • 2081 · 1 mark

What is the measure of central tendency that divides the given data into two equal parts keeping the given data either ascending or descending order. (a) Mean (b) Mode (c) Quartile (d) Median

Answer

(d) Median

The median is the middle value of data arranged in order, dividing it into two equal parts.

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  • 2081 · 1 mark

The first quartile is calculated by the formula ............ in a continuous data. (a) Q1=L+N4−cff×iQ_1 = L + \frac{\frac{N}{4} - cf}{f} \times i (b) Q1=L+N2−cff×iQ_1 = L + \frac{\frac{N}{2} - cf}{f} \times i (c) Q1=L+3N4−cff×iQ_1 = L + \frac{\frac{3N}{4} - cf}{f} \times i (d) Q1=L+N4+cff×iQ_1 = L + \frac{\frac{N}{4} + cf}{f} \times i

Answer

(a) Q1=L+N4−cff×iQ_1 = L + \dfrac{\frac{N}{4} - cf}{f} \times i

The first quartile lies in the class containing the N4\frac N4-th item; LL is its lower limit, cfcf the cumulative frequency before it, ff its frequency and ii its class width.

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  • 2081 · 1 mark

If three children are born in a family, what is the probability that all three children are not daughters? (a) 18\frac{1}{8} (b) 16\frac{1}{6} (c) 32\frac{3}{2} (d) 78\frac{7}{8}

Answer

(d) 78\frac{7}{8}

Total outcomes =23=8=2^3=8. The only outcome with all three daughters is 1. So

P(not all daughters)=1−18=78P(\text{not all daughters})=1-\frac18=\frac78

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  • 2081 · 1 mark

A and B are two mutually exclusive events with P(A)=14P(A) = \frac{1}{4} and P(B)=12P(B) = \frac{1}{2}. What is the value of P(A∪B)P(A \cup B) ? (a) 12\frac{1}{2} (b) 14\frac{1}{4} (c) 18\frac{1}{8} (d) 34\frac{3}{4}

Answer

(d) 34\frac{3}{4}

For mutually exclusive events, P(A∪B)=P(A)+P(B)=14+12=34P(A\cup B)=P(A)+P(B)=\frac14+\frac12=\frac34.

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  • 2081 · 1 mark

If n(A)=80n(A) = 80, n(B)=90n(B) = 90 then, What is the maximum value of n(A∩B)n(A \cap B) ? (a) 10 (b) 80 (c) 90 (d) 170

Answer

(b) 80

n(A∩B)≤min⁡{n(A),n(B)}=min⁡{80,90}=80n(A\cap B)\le \min\{n(A),n(B)\}=\min\{80,90\}=80. The maximum occurs when A⊆BA\subseteq B.

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  • 2081 · 1 mark

If n(U)=1000n(U) = 1000, n(A)=500n(A) = 500, n(B)=400n(B) = 400, n(A∩B)=300n(A \cap B) = 300 then What is the value of n(A‾∩B)n(\overline{A} \cap B) ? (a) 300 (b) 700 (c) 400 (d) 600

Answer

n(A‾∩B)=n(B)−n(A∩B)=400−300=100n(\overline{A}\cap B)=n(B)-n(A\cap B)=400-300=100

None of the listed options equals 100, so the question has no correct option as printed. (For reference: n(A∪B)=500+400−300=600n(A\cup B)=500+400-300=600.)

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  • 2081 · 1 mark

A sum of money amounts Rs. 14520 in 2 years and 15972 in 3 years at certain compound interest. What is the rate of compound interest? (a) 12 % (b) 10% (c) 15% (d) 11%

Answer

(b) 10%

The amount in the third year divided by the amount in the second year is 1+R1001+\frac{R}{100}:

1+R100=1597214520=1.1⇒R=10%1+\frac{R}{100}=\frac{15972}{14520}=1.1\Rightarrow R=10\%

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  • 2081 · 1 mark

If the cost is depreciated at the rate of 20% per annum, the cost of motorcycle is 175000. Find the price of the motorcycle after 3 years. (a) Rs. 102400 (b) Rs. 300000 (c) Rs. 89600 (d) Rs. 75000

Answer

(c) Rs. 89600

V=175000(1−20100)3=175000×(0.8)3=175000×0.512=Rs. 89600V=175000\left(1-\frac{20}{100}\right)^3=175000\times(0.8)^3=175000\times0.512=\text{Rs. }89600

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  • 2081 · 1 mark

If the population of a village becomes nine times in 2 years, find the population growth rate. (a) 100% (b) 200% (c) 300% (d) 900%

Answer

(b) 200%

P(1+R100)2=9P⇒1+R100=3⇒R=200%P\left(1+\frac{R}{100}\right)^2=9P\Rightarrow 1+\frac{R}{100}=3\Rightarrow R=200\%

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  • 2081 · 1 mark

According to the exchange rate of a day, US 1=Rs.132.IftheNepalicurrencywasdevaluatedby2(a)1 = Rs. 132. If the Nepali currency was devaluated by 2%, what is the new exchange rate? (a) 129.36 (b) 130(c)130 (c) 134 (d) $ 134.64

Answer

(d) Rs. 134.64

Devaluing the Nepali rupee by 2% makes a dollar cost 2% more rupees: 132×1.02=Rs. 134.64132\times1.02=\text{Rs. }134.64 per US $1.

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  • 2081 · 1 mark

The base side and slant height of a square based pyramid is 2x2x cm and yy cm respectively. What is the real height of the pyramid? (a) y2−4x2\sqrt{y^2 - 4x^2} (b) y2+x2\sqrt{y^2 + x^2} (c) y2−x2\sqrt{y^2 - x^2} (d) y2+4x2\sqrt{y^2 + 4x^2}

Similar questions: Pyramid volume (2081)

Answer

(c) y2−x2\sqrt{y^2 - x^2}

The base side is 2x2x, so the distance from the centre of the base to the midpoint of a side is xx. In the right triangle formed by the height hh, this distance and the slant height yy: h2=y2−x2h^2=y^2-x^2.

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  • 2081 · 1 mark

The base side and slant height of a square based pyramid is 16 cm and 10 cm respectively. What is the volume of the pyramid? (a) 853.33 cm3^3 (b) 512 cm3^3 (c) 2560 cm3^3 (d) 1536 cm3^3

Similar questions: Pyramid (2081)

Answer

(b) 512 cm3^3

Half the base side is 88 cm, so the height is h=102−82=6h=\sqrt{10^2-8^2}=6 cm.

V=13×162×6=13×256×6=512 cm3V=\frac13\times16^2\times6=\frac13\times256\times6=512\text{ cm}^3

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  • 2081 · 1 mark

If area of the base of the cylinder is 154 cm2^2, Height is 15 cm. What is the total surface area of cylinder? (a) 968 cm2^2 (b) 528 cm2^2 (c) 308 cm2^2 (d) 616 cm2^2

Answer

(a) 968 cm2^2

πr2=154⇒r2=49⇒r=7\pi r^2=154\Rightarrow r^2=49\Rightarrow r=7 cm.

TSA=2πr(h+r)=2×227×7×(15+7)=44×22=968 cm2TSA=2\pi r(h+r)=2\times\frac{22}{7}\times7\times(15+7)=44\times22=968\text{ cm}^2

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Questions from Kathmandu Metropolitan City Class 11 Scholarship Examination 2082 (Set A, 3 Shrawan 2082) (Actual scholarship exam, 100 MCQ, held 2082-04-03) and Kathmandu Metropolitan City Class 11 Scholarship Examination 2081 (Set B, 4 Shrawan 2081) (Actual scholarship exam, 100 MCQ, held 2081-04-04). Answers are written for this site; check them against your class notes.