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Chapter 2

Mathematics

+2 college entrance (after SEE) past exam questions

Past questions and answers

161 questions set from this chapter, 41 of them more than once; 15 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 3 of 8 exams
  • Asked 3 times
  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set B · 1 mark

If n(A−B)=327n(A - B) = 327, n(B−A)=217n(B - A) = 217 and n(A∪B)=600n(A \cup B) = 600, find n(A∩B)n(A \cap B). (a) 56 (b) 46 (c) 66 (d) 36

Answer

(a) 56

The union splits into three disjoint parts:

n(A∪B)=n(A−B)+n(B−A)+n(A∩B)n(A \cup B) = n(A - B) + n(B - A) + n(A \cap B)

600=327+217+n(A∩B)⇒n(A∩B)=600−544=56600 = 327 + 217 + n(A \cap B) \Rightarrow n(A \cap B) = 600 - 544 = 56

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • Asked 3 times
  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set D · 1 mark

If yy is the measure of an acute angle such that sin⁡y=a5\sin y = \frac{a}{5}, then tan⁡y=\tan y = (a) 25−a25\frac{\sqrt{25-a^2}}{5} (b) a25−a2\frac{a}{\sqrt{25-a^2}} (c) 1a2+25\frac{1}{\sqrt{a^2+25}} (d) 525−a2\frac{5}{\sqrt{25-a^2}}

Answer

(b) a25−a2\frac{a}{\sqrt{25-a^2}}

With hypotenuse 55 and opposite side aa, the adjacent side is 25−a2\sqrt{25-a^2}, so

tan⁡y=a25−a2\tan y = \frac{a}{\sqrt{25-a^2}}

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • Asked 3 times
  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set B · 1 mark
  • 2083 · Science Practice Set A · 1 mark

If the graphs of the equations 2x+5y=72x + 5y = 7 and 6x+cy=116x + cy = 11 are parallel lines, then what is cc? (a) 3 (b) 5 (c) 10 (d) 15

Answer

(d) 15

Parallel lines have equal slopes, so the coefficients are proportional: 62=c5\frac{6}{2}=\frac{c}{5}, giving c=15c=15. The constants are not in the same ratio (117≠3\frac{11}{7}\neq 3), so the lines are distinct and truly parallel.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • Asked 3 times
  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set D · 1 mark

If m%m\% of 300 is 150, what is m%m\% of 180? (a) 45 (b) 90 (c) 120 (d) 360

Answer

(b) 90

m100×300=150\frac{m}{100}\times 300=150 gives m=50m=50. Then 50%50\% of 180=90180=90.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • Asked 3 times
  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set B · 1 mark
  • 2083 · Management Practice Set C · 1 mark

If 1x+y=2\frac{1}{x} + y = 2 and x+1y=3x + \frac{1}{y} = 3, the ratio of xx to yy is equal to (a) 1:2 (b) 2:3 (c) 3:1 (d) 3:2

Answer

(d) 3:2

From the first equation y=2x−1xy=\frac{2x-1}{x}. Substituting in the second: x+x2x−1=3x+\frac{x}{2x-1}=3, which simplifies to 2x2−6x+3=02x^2-6x+3=0, so x2=3x−32x^2=3x-\frac{3}{2}.

Then xy=x22x−1=3x−322x−1=32.\frac{x}{y}=\frac{x^2}{2x-1}=\frac{3x-\frac{3}{2}}{2x-1}=\frac{3}{2}. So x:y=3:2x:y=3:2.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • Asked 3 times
  • 2083 · Management Practice Set B · 1 mark
  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set B · 1 mark

The population of a town increases at the rate of 5% per annum and the present population of the town is 185220. What was the population of the town 3 years ago? (a) 160000 (b) 170000 (c) 180000 (d) 150000

Answer

(a) 160000

P(1+5100)3=185220⇒P=1852201.157625=160000P\left(1+\tfrac{5}{100}\right)^3=185220\Rightarrow P=\frac{185220}{1.157625}=160000

(Check: 2120\frac{21}{20} cubed is 92618000\frac{9261}{8000}, and 185220×80009261=160000185220\times\frac{8000}{9261}=160000.)

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • Asked 3 times
  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set C · 1 mark
  • 2083 · Science Practice Set D · 1 mark

If the volume of a cone is 24π24\pi cubic feet and its height is 2 feet, what is the diameter of the base in feet? (a) 3 (b) 6 (c) 12 (d) 24

Answer

(c) 12

24π=13πr2(2)  ⟹  r2=36  ⟹  r=6,24\pi=\tfrac13\pi r^2(2) \implies r^2=36 \implies r=6,

so the diameter is 2r=122r=12 feet.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • Asked 2 times
  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set C · 1 mark

A man can do a job in hh hours alone and his son can do it in 2h2h hours alone. Together, how many hours will it take them to do the job? (a) 3h3h (b) h3\frac{h}{3} (c) 3h4\frac{3h}{4} (d) 2h3\frac{2h}{3}

Similar questions: A man can do a job in h hours alone and his son (2083 · Science Practice Set D)

Answer

(d) 2h3\frac{2h}{3}

Combined rate:

1h+12h=32h\frac{1}{h} + \frac{1}{2h} = \frac{3}{2h}

Time together =2h3= \dfrac{2h}{3} hours.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • Asked 2 times
  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set C · 1 mark

If xx is a positive acute angle and tan⁡x=p\tan x = p, then what is cos⁡x\cos x? (a) 1p2+1\frac{1}{\sqrt{p^2+1}} (b) pp2+1\frac{p}{\sqrt{p^2+1}} (c) pp2−1\frac{p}{\sqrt{p^2-1}} (d) p2+1\sqrt{p^2+1}

Similar questions: If x is a positive acute angle and tan x = p, (2083 · Management Practice Set B)

Answer

(a) 1p2+1\frac{1}{\sqrt{p^2+1}}

Since tan⁡x=p/1\tan x = p/1, take opposite =p= p and adjacent =1= 1, so hypotenuse =p2+1= \sqrt{p^2+1}.

cos⁡x=1p2+1\cos x = \frac{1}{\sqrt{p^2+1}}

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • Asked 2 times
  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set C · 1 mark

If the radius of a circle is decreased by 30%, by what percent will the area of the circular region be decreased? (a) 70% (b) 60% (c) 51% (d) 49%

Similar questions: If the radius of a circle is decreased by 30%, by (2083 · Science Practice Set D)

Answer

(c) 51%

New radius =0.7r=0.7r, so new area =π(0.7r)2=0.49πr2=\pi(0.7r)^2=0.49\pi r^2. The area decreases by 100−49=51%100-49=51\%.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • Asked 2 times
  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set C · 1 mark

If nn is an odd integer, which one of the following is an even integer? (a) 2n+n2n + n (b) 2n+12n + 1 (c) n3n^3 (d) n(n+3)n(n+3)

Similar questions: If n is an even integer, which one of the (2083 · Management Practice Set B)

Answer

(d) n(n+3)n(n+3)

For odd nn: 2n+n=3n2n+n=3n is odd, 2n+12n+1 is odd, and n3n^3 is odd. But n+3n+3 is even (odd plus odd), so n(n+3)n(n+3) is even.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • 2083 · Management Practice Set B · 1 mark

If xx is a positive acute angle and tan⁡x=p\tan x = p, then what is cos⁡x\cos x? (a) pp2+1\frac{p}{\sqrt{p^2+1}} (b) 1p2+1\frac{1}{\sqrt{p^2+1}} (c) pp2−1\frac{p}{\sqrt{p^2-1}} (d) p2+1\sqrt{p^2+1}

Similar questions: If x is a positive acute angle and tan x = p, (2083 · Management Practice Set A)

Answer

(b) 1p2+1\frac{1}{\sqrt{p^2+1}}

sec⁡2x=1+tan⁡2x=1+p2\sec^2x=1+\tan^2x=1+p^2, and cos⁡x>0\cos x>0 for an acute angle, so cos⁡x=1p2+1\cos x=\frac{1}{\sqrt{p^2+1}}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • 2083 · Management Practice Set B · 1 mark

If nn is an even integer, which one of the following must be an odd integer? (a) 3n−23n - 2 (b) 3(n+1)3(n+1) (c) n−2n - 2 (d) n3\frac{n}{3}

Similar questions: If n is an odd integer, which one of the (2083 · Management Practice Set C)

Answer

(b) 3(n+1)3(n+1)

If nn is even, n+1n+1 is odd, and 3 times an odd number is odd. 3n−23n-2 and n−2n-2 are even, and n3\frac n3 need not even be an integer.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • 2083 · Science Practice Set D · 1 mark

A man can do a job in hh hours alone and his son can do it in 2h2h hours alone. Together, how many hours will it take them to do the job? (a) 13h\frac{1}{3h} (b) h3\frac{h}{3} (c) 3h4\frac{3h}{4} (d) 2h3\frac{2h}{3}

Similar questions: A man can do a job in h hours alone and his son (2083 · Management Practice Set A)

Answer

(d) 2h3\frac{2h}{3}

Their combined rate per hour is 1h+12h=32h\frac1h+\frac1{2h}=\frac{3}{2h}, so together they take 2h3\frac{2h}{3} hours.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 8 exams
  • 2083 · Science Practice Set D · 1 mark

If the radius of a circle is decreased by 30%, by what percent will the area of the circular region be decreased? (a) 70% (b) 60% (c) 49% (d) 51%

Similar questions: If the radius of a circle is decreased by 30%, by (2083 · Management Practice Set A)

Answer

(d) 51%

The new radius is 0.7r0.7r, so the new area is π(0.7r)2=0.49πr2\pi(0.7r)^2=0.49\pi r^2. The area decreases by 100−49=51%100-49=51\%.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set C · 1 mark

A survey of students in a class shows that 45 students like mathematics and 41 students like English. If 12 students like both subjects, how many students like either Mathematics or English? (a) 66 (b) 75 (c) 98 (d) 74

Answer

(d) 74

n(M∪E)=n(M)+n(E)−n(M∩E)=45+41−12=74n(M \cup E) = n(M) + n(E) - n(M \cap E) = 45 + 41 - 12 = 74

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  • Asked 2 times
  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set C · 1 mark

How many numbers in the set {−8,−5,0,10,20}\{-8, -5, 0, 10, 20\} satisfy the condition ∣x−5∣<11|x - 5| < 11? (a) One (b) Two (c) Three (d) Four

Answer

(c) Three

∣x−5∣<11|x-5|<11 means −11<x−5<11-11 < x-5 < 11, i.e. −6<x<16-6 < x < 16.

Checking the set: −8-8 fails, −5-5 passes, 00 passes, 1010 passes, 2020 fails. So three numbers satisfy it.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set C · 1 mark

What is the simplified expression for 2x−44x+20⋅x2−25x2−7x+10\frac{2x-4}{4x+20}\cdot\frac{x^2-25}{x^2-7x+10}? (a) 14\frac{1}{4} (b) 12\frac{1}{2} (c) 2 (d) 1

Answer

(b) 12\frac{1}{2}

Factorise each part:

2(x−2)4(x+5)⋅(x−5)(x+5)(x−5)(x−2)=24=12\frac{2(x-2)}{4(x+5)} \cdot \frac{(x-5)(x+5)}{(x-5)(x-2)} = \frac{2}{4} = \frac{1}{2}

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  • 2083 · Management Practice Set A · 1 mark
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The legs of a right-angled triangle are in the ratio of 1 : 2. If the area of the triangle is 25, what is the hypotenuse? (a) 535\sqrt{3} (b) 555\sqrt{5} (c) 10310\sqrt{3} (d) 10510\sqrt{5}

Answer

(b) 555\sqrt{5}

Let the legs be kk and 2k2k. Area: 12⋅k⋅2k=k2=25\tfrac12 \cdot k \cdot 2k = k^2 = 25, so k=5k = 5 and the legs are 55 and 1010.

Hypotenuse=52+102=125=55\text{Hypotenuse} = \sqrt{5^2 + 10^2} = \sqrt{125} = 5\sqrt{5}

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  • 2083 · Management Practice Set A · 1 mark
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What is the solution of the equation p=ryr+yp = \frac{ry}{r+y} for yy? (a) r+prp\frac{r+p}{rp} (b) rpr−p\frac{rp}{r-p} (c) r+pr−p\frac{r+p}{r-p} (d) r−prp\frac{r-p}{rp}

Answer

(b) rpr−p\frac{rp}{r-p}

p(r+y)=ry⇒pr+py=ry⇒pr=y(r−p)⇒y=rpr−pp(r+y) = ry \Rightarrow pr + py = ry \Rightarrow pr = y(r-p) \Rightarrow y = \frac{rp}{r-p}

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  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Science Practice Set A · 1 mark

In the xy-plane, line ll is perpendicular to the line 4x−y=14x - y = 1 and contains the point (0,2)(0, 2). Which one of the following is the equation of line ll? (a) x+4y=8x + 4y = 8 (b) 4x−y=24x - y = 2 (c) 4x+y=−24x + y = -2 (d) 4x+y=84x + y = 8

Answer

(a) x+4y=8x + 4y = 8

The line 4x−y=14x-y=1 has slope 44, so a perpendicular line has slope −14-\frac{1}{4}. Through (0,2)(0,2): y=−14x+2⇒x+4y=8.y=-\frac{1}{4}x+2 \Rightarrow x+4y=8.

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  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set C · 1 mark

In the xy-plane, the lines with the equations x+y=15x + y = 15 and y=3x−1y = 3x - 1 intersect at (a,b)(a, b). What is the value of a−ba - b? (a) 11 (b) 4 (c) −7-7 (d) −6-6

Answer

(c) −7-7

Substitute y=3x−1y=3x-1 into x+y=15x+y=15: 4x−1=154x-1=15, so x=4x=4 and y=11y=11. Hence a−b=4−11=−7a-b=4-11=-7.

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  • 2083 · Management Practice Set A · 1 mark
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20% of what number is equal to 23\frac{2}{3} of 90? (a) 30 (b) 120 (c) 300 (d) 600

Answer

(c) 300

23×90=60\frac{2}{3}\times 90=60. Let the number be NN: 0.2N=600.2N=60, so N=300N=300.

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  • 2083 · Management Practice Set A · 1 mark
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If 4x+16=244x + 16 = 24, what is the value of x+8x + 8? (a) 14 (b) 12 (c) 10 (d) 8

Answer

(c) 10

4x=24−16=84x=24-16=8, so x=2x=2 and x+8=10x+8=10.

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  • 2083 · Management Practice Set A · 1 mark
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A couple plans to have three children. What is the probability that exactly two will be boys? (a) 23\frac{2}{3} (b) 38\frac{3}{8} (c) 28\frac{2}{8} (d) 18

Answer

(b) 38\frac{3}{8}

There are 23=82^3=8 equally likely outcomes. Exactly two boys can occur in (32)=3\binom{3}{2}=3 ways (BBG, BGB, GBB), so the probability is 38\frac{3}{8}.

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  • 2083 · Management Practice Set A · 1 mark
  • 2083 · Management Practice Set C · 1 mark

A bag contains 60 marbles. A marble is to be chosen from the bag. If the probability of a blue marble being chosen is 0.35, how many marbles in the bag are blue? (a) 21 (b) 25 (c) 35 (d) 39

Answer

(a) 21

Number of blue marbles =0.35×60=21=0.35\times 60=21.

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  • 2083 · Management Practice Set A · 1 mark
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If 13,y1,y2,9\frac{1}{3}, y_1, y_2, 9 are in G.P., then what is the value of y2y_2? (a) 1 (b) 3 (c) 6 (d) 152\frac{15}{2}

Answer

(b) 3

Let the common ratio be rr. The fourth term is 13r3=9\frac{1}{3}r^3=9, so r3=27r^3=27 and r=3r=3. Then y1=1y_1=1 and y2=1×3=3y_2=1\times 3=3.

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The sum of the roots of 3x2−4x+4=03x^2 - 4x + 4 = 0 is (a) 43\frac{4}{3} (b) 34\frac{3}{4} (c) −43-\frac{4}{3} (d) 4

Answer

(a) 43\frac{4}{3}

For ax2+bx+c=0ax^2+bx+c=0 the sum of the roots is −ba=−−43=43-\frac{b}{a}=-\frac{-4}{3}=\frac{4}{3}.

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If x=1+2x = 1 + \sqrt{2}, then x2−2x+1x^2 - 2x + 1 is equal to (a) 1+21 + \sqrt{2} (b) −1-1 (c) 2\sqrt{2} (d) 2

Answer

(d) 2

x2−2x+1=(x−1)2=(2)2=2x^2-2x+1=(x-1)^2=(\sqrt{2})^2=2.

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  • 2083 · Management Practice Set A · 1 mark
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The average of 34 students in a class is 13 years. If the age of a teacher is included, the average is increased by 1 year. What is the age of the teacher in years? (a) 40 (b) 48 (c) 55 (d) 65

Answer

(b) 48

Total age of the students =34×13=442=34\times 13=442. With the teacher there are 35 people with average 14, so total =35×14=490=35\times 14=490. Teacher's age =490−442=48=490-442=48 years.

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If 6 Japanese yen = 5 Swedish kroner and 4 Swedish kroner = 3 Danish kroner, how many Japanese yen can be exchanged for 48 Danish kroner? (a) 70.8 Yen (b) 76.8 Yen (c) 78.8 Yen (d) 77.8 Yen

Answer

(b) 76.8 Yen

Convert Danish kroner to Swedish kroner: 48×43=6448 \times \frac{4}{3} = 64 Swedish kroner. Then convert to yen: 64×65=76.864 \times \frac{6}{5} = 76.8 yen.

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  • 2083 · Management Practice Set B · 1 mark
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A fraction becomes 45\frac{4}{5} if 2 is added to the numerator. If 1 is subtracted from the denominator, the fraction becomes 34\frac{3}{4}. What was the original fraction? (a) 1825\frac{18}{25} (b) 2025\frac{20}{25} (c) 2425\frac{24}{25} (d) 2518\frac{25}{18}

Answer

(a) 1825\frac{18}{25}

Let the fraction be xy\frac{x}{y}. Then 5(x+2)=4y5(x+2)=4y and 4x=3(y−1)4x=3(y-1), i.e. 5x+10=4y5x+10=4y and 4x=3y−34x=3y-3.

From the first, y=5x+104y=\frac{5x+10}{4}. Substituting: 16x=15x+30−1216x=15x+30-12, so x=18x=18 and y=25y=25.

Check: 2025=45\frac{20}{25}=\frac45 and 1824=34\frac{18}{24}=\frac34.

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  • 2083 · Management Practice Set B · 1 mark
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If pp lb of salt are dissolved in qq lb of water, the percent of salt in the resulting solution is (a) 100pp+q\frac{100p}{p+q} (b) pp+q\frac{p}{p+q} (c) 100 pq\frac{100\,p}{q} (d) 100 qpq\frac{100\,q}{pq}

Answer

(a) 100pp+q\frac{100p}{p+q}

The solution weighs p+qp+q lb and contains pp lb of salt, so the percentage is pp+q×100\frac{p}{p+q}\times100.

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  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set B · 1 mark

2x+3y=12002x + 3y = 1200, 3x+2y=13003x + 2y = 1300. Based on the system of equations above, what is the value of 5x+5y5x + 5y? (a) 500 (b) 1000 (c) 2300 (d) 2500

Answer

(d) 2500

Add the two equations:

(2x+3y)+(3x+2y)=1200+1300  ⟹  5x+5y=2500.(2x+3y)+(3x+2y)=1200+1300 \implies 5x+5y=2500.

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  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set B · 1 mark

If u+t=5u + t = 5 and u−t=2u - t = 2, what is the value of (u−t)(u2−t2)(u - t)(u^2 - t^2)? (a) 10 (b) 20 (c) 25 (d) 50

Answer

(b) 20

u2−t2=(u+t)(u−t)=5×2=10u^2-t^2=(u+t)(u-t)=5\times 2=10, so (u−t)(u2−t2)=2×10=20(u-t)(u^2-t^2)=2\times 10=20.

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  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set B · 1 mark

A right circular cone has a volume of 24π24\pi cubic inches. If the height of the cone is 2 inches, what is the radius, in inches, of the base of the cone? (a) 232\sqrt{3} (b) 6 (c) 12 (d) 36

Answer

(b) 6

V=13πr2h  ⟹  24π=13πr2(2)  ⟹  r2=36  ⟹  r=6.V=\tfrac13\pi r^2h \implies 24\pi=\tfrac13\pi r^2(2) \implies r^2=36 \implies r=6.

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  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set B · 1 mark

Which of the following gives the radius of the sphere in terms of the volume of the sphere? (a) V4π\sqrt{\frac{V}{4\pi}} (b) 3V4π\sqrt{\frac{3V}{4\pi}} (c) 3V4π3\sqrt[3]{\frac{3V}{4\pi}} (d) V4π3\frac{V}{\sqrt[3]{4\pi}}

Answer

(c) 3V4π3\sqrt[3]{\frac{3V}{4\pi}}

From V=43πr3V=\dfrac{4}{3}\pi r^3 we get r3=3V4πr^3=\dfrac{3V}{4\pi}, so r=3V4π3r=\sqrt[3]{\dfrac{3V}{4\pi}}.

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  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set B · 1 mark

Circles O and O' are internally tangent to each other. Circle O' passes through the center of O. If the area of circle O is 16, the area of circle O' is (a) 222\sqrt{2} (b) 2 (c) 2π2\sqrt{\pi} (d) 4

Answer

(d) 4

Since O′O' is internally tangent to OO and passes through the centre of OO, the diameter of O′O' equals the radius RR of OO, so its radius is R/2R/2. Area scales with the square of radius, so area of O′O' is 14\tfrac14 of the area of OO: 14×16=4\tfrac14\times 16=4.

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  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set B · 1 mark

Given △PQR\triangle PQR, ∠P=70∘\angle P = 70^\circ, QS bisects ∠Q\angle Q, and RS bisects ∠R\angle R (S is the point where the two bisectors meet). ∠S=\angle S = (a) 125∘125^\circ (b) 120∘120^\circ (c) 115∘115^\circ (d) 110∘110^\circ

Answer

(a) 125∘125^\circ

In △QSR\triangle QSR: ∠S=180∘−12(∠Q+∠R)\angle S=180^\circ-\tfrac12(\angle Q+\angle R). Since ∠Q+∠R=180∘−70∘=110∘\angle Q+\angle R=180^\circ-70^\circ=110^\circ,

∠S=180∘−55∘=125∘.\angle S=180^\circ-55^\circ=125^\circ.

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  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set B · 1 mark

If f(x)=5x−3f(x) = 5x - 3 and f(t)=7f(t) = 7, what is the value of tt? (a) 0 (b) 1 (c) 2 (d) 32

Answer

(c) 2

f(t)=5t−3=7  ⟹  5t=10  ⟹  t=2f(t)=5t-3=7 \implies 5t=10 \implies t=2.

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  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set B · 1 mark

Let xx be the angle formed by the diagonal of the face of a cube and the diagonal of the cube, drawn from the same vertex. cos⁡2x=\cos^2 x = (a) 23\frac{2}{3} (b) 13\frac{1}{3} (c) 12\frac{1}{2} (d) 1

Answer

(a) 23\frac{2}{3}

Take a cube of side 1. The face diagonal has length 2\sqrt2 and the cube diagonal has length 3\sqrt3. The face diagonal, the edge perpendicular to that face and the cube diagonal form a right triangle (right angle at the end of the face diagonal), so

cos⁡x=23  ⟹  cos⁡2x=23.\cos x=\frac{\sqrt2}{\sqrt3}\implies \cos^2x=\frac23.

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  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set B · 1 mark

A flask contains 40 grams of solution that is 25% sugar by weight. How much sugar must be added to the flask so that the resulting solution is 40% sugar by weight? (a) 6 (b) 8 (c) 10 (d) 15

Answer

(c) 10

The flask has 0.25×40=100.25\times40=10 g sugar. Adding xx g of sugar:

10+x40+x=0.4  ⟹  10+x=16+0.4x  ⟹  0.6x=6  ⟹  x=10.\frac{10+x}{40+x}=0.4 \implies 10+x=16+0.4x \implies 0.6x=6 \implies x=10.

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  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set C · 1 mark

ABCD is a parallelogram with A(4,5)A(4, 5), D(2,−3)D(2, -3) and C(6,−3)C(6, -3). What is the x-coordinate of point B? (a) 2 (b) 5 (c) 6 (d) 8

Answer

(d) 8

The diagonals of a parallelogram bisect each other, so the midpoint of ACAC equals the midpoint of BDBD: B=A+C−DB=A+C-D.

B=(4+6−2,  5−3+3)=(8,5).B=(4+6-2,\;5-3+3)=(8,5).

The x-coordinate is 88.

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  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set C · 1 mark

ABCD is a trapezoid (AB parallel to DC, BC perpendicular to AB) with area 50 square units. If the length of AB is 6 units and the length of BC is 5 units, what is the length of CD? (a) 4 (b) 7 (c) 10 (d) 14

Answer

(d) 14

BCBC is perpendicular to the parallel sides, so it is the height. Then

12(AB+CD)⋅BC=50  ⟹  12(6+CD)(5)=50  ⟹  6+CD=20  ⟹  CD=14.\tfrac12(AB+CD)\cdot BC=50 \implies \tfrac12(6+CD)(5)=50 \implies 6+CD=20 \implies CD=14.

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  • 2083 · Management Practice Set C · 1 mark
  • 2083 · Science Practice Set B · 1 mark

In the xy-plane, the graph of the function f(x)=x2+5x+4f(x) = x^2 + 5x + 4 has two x-intercepts. What is the distance between the x-intercepts? (a) 1 (b) 2 (c) 3 (d) 4

Answer

(c) 3

Setting f(x)=0f(x)=0: x2+5x+4=(x+1)(x+4)=0x^2+5x+4=(x+1)(x+4)=0, so the x-intercepts are x=−1x=-1 and x=−4x=-4. The distance between them is ∣−1−(−4)∣=3|-1-(-4)|=3.

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  • 2083 · Management Practice Set B · 1 mark

The selling price of a TV is Rs. 18000. If there is a discount of 10%, find the marked price. (a) Rs. 19000 (b) Rs. 20000 (c) Rs. 17000 (d) Rs. 21000

Similar questions: The selling price of a TV is Rs. 18000. If there (2083 · Science Practice Set C)

Answer

(b) Rs. 20000

SP is 90% of MP, so

MP=180000.9=Rs. 20000MP=\frac{18000}{0.9}=\text{Rs. }20000

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  • 2083 · Management Practice Set B · 1 mark

A car drives a distance of dd miles at 30 mph and returns at 60 mph. What is the average rate for the round trip? (a) 40 mph (b) 43 mph (c) 2d35\frac{2d}{35} (d) 352d\frac{35}{2d}

Similar questions: A car drives a distance of d miles at 30 mph and (2083 · Science Practice Set A)

Answer

(a) 40 mph

Average speed =total distancetotal time=2dd30+d60=2dd20=40=\dfrac{\text{total distance}}{\text{total time}}=\dfrac{2d}{\frac{d}{30}+\frac{d}{60}}=\dfrac{2d}{\frac{d}{20}}=40 mph.

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  • 2083 · Management Practice Set C · 1 mark

A book was on sale at 40% off its original price. If the sale price of the book was Rs. 18, what was the original price of the book? (a) Rs. 7.20 (b) Rs. 10.80 (c) Rs. 30.00 (d) Rs. 45.00

Similar questions: A book was on sale for 40% off its original (2083 · Science Practice Set B)

Answer

(c) Rs. 30.00

The sale price is 60%60\% of the original price PP: 0.6P=180.6P=18, so P=180.6=30P=\dfrac{18}{0.6}=30.

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  • 2083 · Management Practice Set C · 1 mark

A boy grew one year from a height of xx inches to a height of yy inches. The percent of increase was (a) 100(y−x)y\frac{100(y-x)}{y} (b) 100(y−x)x\frac{100(y-x)}{x} (c) y−xx\frac{y-x}{x} (d) y−x100\frac{y-x}{100}

Similar questions: A boy grew in one year from a height of x inches (2083 · Science Practice Set B)

Answer

(b) 100(y−x)x\frac{100(y-x)}{x}

Percent increase is the increase divided by the original value, times 100: y−xx×100\dfrac{y-x}{x}\times100.

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  • 2083 · Science Practice Set A · 1 mark

Let a⃗\vec{a} and b⃗\vec{b} be two given vectors such that ∣a⃗∣=3|\vec{a}| = \sqrt{3}, ∣b⃗∣=2|\vec{b}| = 2 and a⃗⋅b⃗=6\vec{a}\cdot\vec{b} = \sqrt{6}, then what is the angle between the vectors? (a) π3\frac{\pi}{3} (b) π4\frac{\pi}{4} (c) π2\frac{\pi}{2} (d) 0

Similar questions: Let a and b be two given vectors such that |a| = (2011 · Science Entrance Sample Paper)

Answer

(b) π4\frac{\pi}{4}

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=623=22\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\frac{\sqrt6}{2\sqrt3}=\frac{\sqrt2}{2}

So θ=π4\theta=\frac{\pi}{4}.

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  • 2083 · Science Practice Set A · 1 mark

A car drives a distance of dd miles at 30 mph and returns at 60 mph. What is its average rate for the round trip? (a) 40 mph (b) 43 mph (c) 2d35\frac{2d}{35} mph (d) 352d\frac{35}{2d} mph

Similar questions: A car drives a distance of d miles at 30 mph and (2083 · Management Practice Set B)

Answer

(a) 40 mph

Average speed=2dd30+d60=2d3d60=40 mph\text{Average speed}=\frac{2d}{\frac d{30}+\frac d{60}}=\frac{2d}{\frac{3d}{60}}=40\text{ mph}

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  • 2083 · Science Practice Set B · 1 mark

A book was on sale for 40% off its original price. If the sale price of the book was Rs 18, what was the original price of the book? (a) Rs. 7.20 (b) Rs. 10.80 (c) Rs. 30.00 (d) Rs. 45.00

Similar questions: A book was on sale at 40% off its original price. (2083 · Management Practice Set C)

Answer

(c) Rs 30.00

The sale price is 60%60\% of the original price PP.

0.6P=18⇒P=180.6=300.6P = 18 \Rightarrow P = \frac{18}{0.6} = 30

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  • 2083 · Science Practice Set B · 1 mark

A boy grew in one year from a height of xx inches to a height of yy inches. The percent of increase was: (a) 100(y−x)y\frac{100(y-x)}{y} (b) 100(y−x)x\frac{100(y-x)}{x} (c) y−xx\frac{y-x}{x} (d) y−x100\frac{y-x}{100}

Similar questions: A boy grew one year from a height of x inches to (2083 · Management Practice Set C)

Answer

(b) 100(y−x)x\frac{100(y-x)}{x}

Percent increase is the increase divided by the original value, times 100100. The increase is y−xy-x and the original height is xx, so the percent increase is 100(y−x)x\frac{100(y-x)}{x}.

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  • 2083 · Science Practice Set C · 1 mark

The selling price of a TV is Rs. 18000. If there is a discount of 10%, what is the marked price? (a) Rs. 19000 (b) Rs. 20000 (c) Rs. 17000 (d) Rs. 21000

Similar questions: The selling price of a TV is Rs. 18000. If there (2083 · Management Practice Set B)

Answer

(b) Rs. 20000

Selling price =90%=90\% of marked price, so MP=180000.9=Rs. 20000.MP=\frac{18000}{0.9}=\text{Rs. }20000.

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  • 2011 · Science Entrance Sample Paper · 1 mark

Let a⃗\vec{a} and b⃗\vec{b} be two given vectors such that ∣a⃗∣=3|\vec{a}| = \sqrt{3}, ∣b⃗∣=2|\vec{b}| = 2 and a⃗⋅b⃗=6\vec{a}\cdot\vec{b} = \sqrt{6}, then the angle between a⃗\vec{a} and b⃗\vec{b} is (a) π3\frac{\pi}{3} (b) π4\frac{\pi}{4} (c) π2\frac{\pi}{2} (d) 0

Similar questions: Let a and b be two given vectors such that |a| = (2083 · Science Practice Set A)

Answer

(b) π4\frac{\pi}{4}

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=63⋅2=22,\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{\sqrt{6}}{\sqrt{3}\cdot 2} = \frac{\sqrt{2}}{2},

so θ=π4\theta = \frac{\pi}{4}.

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  • 2083 · Management Practice Set A · 1 mark

If nn is a positive integer, which of the following is always odd? (a) 19n+619n + 6 (b) 19n+519n + 5 (c) 19n2+519n^2 + 5 (d) 18n+518n + 5

Answer

(d) 18n+518n + 5

18n18n is always even, and even + 5 is always odd. The others can be even: for odd nn, 19n+519n+5 and 19n2+519n^2+5 are even, and for even nn, 19n+619n+6 is even.

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  • 2083 · Management Practice Set A · 1 mark

2x+7y=92x + 7y = 9, 8x+28y=a8x + 28y = a. In the given system of equations, aa is a constant. If the system has infinitely many solutions, what is the value of aa? (a) 4 (b) 9 (c) 36 (d) 54

Answer

(c) 36

Infinitely many solutions need the second equation to be a multiple of the first. Multiplying 2x+7y=92x+7y=9 by 4 gives 8x+28y=368x+28y=36, so a=36a=36.

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  • 2083 · Management Practice Set A · 1 mark

The equation y=7x−1y = 7^x - 1 is graphed in the xy-plane. What is the y-intercept of the graph? (a) (0,−1)(0, -1) (b) (0,7)(0, 7) (c) (0,6)(0, 6) (d) (0,0)(0, 0)

Answer

(d) (0,0)(0, 0)

Put x=0x=0: y=70−1=1−1=0y=7^0-1=1-1=0, so the y-intercept is (0,0)(0,0).

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  • 2083 · Management Practice Set A · 1 mark

At what time does a sum of money double itself at the rate of 8% p.a.? (a) 8 years (b) 10 years (c) 12.5 years (d) 16 years

Answer

(c) 12.5 years

For simple interest, doubling means the interest equals the principal: P=P×8×T100P=\frac{P\times 8\times T}{100}, so T=1008=12.5T=\frac{100}{8}=12.5 years.

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  • 2083 · Management Practice Set A · 1 mark

Which one of the following is the compound interest of amount PP after nn years at the rate of r%r\%? (a) P[(1+r100)n−1]P\left[\left(1+\frac{r}{100}\right)^n - 1\right] (b) P[(1+r100)n+1]P\left[\left(1+\frac{r}{100}\right)^n + 1\right] (c) [(1+r100)n−1]\left[\left(1+\frac{r}{100}\right)^n - 1\right] (d) P[(1−r100)n+1]P\left[\left(1-\frac{r}{100}\right)^n + 1\right]

Answer

(a) P[(1+r100)n−1]P\left[\left(1+\frac{r}{100}\right)^n - 1\right]

The amount is A=P(1+r100)nA=P\left(1+\frac{r}{100}\right)^n, and compound interest is the amount minus the principal, A−PA-P.

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  • 2083 · Management Practice Set A · 1 mark

If x=−8x = -8, the value of x2/3+2x0x^{2/3} + 2x^0 is (a) −2-2 (b) 4 (c) 6 (d) −6-6

Answer

(c) 6

x2/3=((−8)1/3)2=(−2)2=4x^{2/3}=\left((-8)^{1/3}\right)^2=(-2)^2=4, and 2x0=22x^0=2. The sum is 4+2=64+2=6.

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  • 2083 · Management Practice Set A · 1 mark

If 23x+10=(14)x2^{3x+10} = \left(\frac{1}{4}\right)^x, then x=x = (a) −2-2 (b) 1 (c) −1-1 (d) 2

Answer

(a) −2-2

Write (14)x=2−2x\left(\frac{1}{4}\right)^x=2^{-2x}. Then 3x+10=−2x3x+10=-2x, so 5x=−105x=-10 and x=−2x=-2.

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  • 2083 · Management Practice Set A · 1 mark

If 3x=4y3x = 4y, the value of (x+y)2:(x−y)2(x+y)^2 : (x-y)^2 is (a) 4 (b) 9 (c) 49 (d) 7

Answer

(c) 49

Let x=4kx=4k, y=3ky=3k. Then (x+y)2:(x−y)2=(7k)2:(k)2=49:1(x+y)^2:(x-y)^2=(7k)^2:(k)^2=49:1, so the value is 4949.

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  • 2083 · Management Practice Set A · 1 mark

How many cubes with side 4 cm can be made from a cube with 16 cm in length? (a) 4 (b) 16 (c) 64 (d) 404

Answer

(c) 64

Number of cubes =16343=(164)3=43=64=\frac{16^3}{4^3}=\left(\frac{16}{4}\right)^3=4^3=64.

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  • 2083 · Management Practice Set A · 1 mark

If dd is the diameter of a sphere, what is the surface area of the sphere? (a) 4πd24\pi d^2 (b) 2πd22\pi d^2 (c) πd24\frac{\pi d^2}{4} (d) πd2\pi d^2

Answer

(d) πd2\pi d^2

Surface area =4πr2=4\pi r^2 with r=d2r=\frac{d}{2}, so 4πd24=πd24\pi\frac{d^2}{4}=\pi d^2.

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  • 2083 · Management Practice Set A · 1 mark

Two identical solid cubes of side aa are joined end to end. Then what is the volume of the resulting cuboid? (a) a3a^3 (b) 2a32a^3 (c) 8a38a^3 (d) 10a210a^2

Answer

(b) 2a32a^3

Each cube has volume a3a^3, and joining them preserves total volume: a3+a3=2a3a^3+a^3=2a^3 (a cuboid 2a×a×a2a\times a\times a).

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  • 2083 · Management Practice Set A · 1 mark

If 3n−2=1813^{n-2} = \frac{1}{81}, then the value of nn is equal to (a) 0 (b) 1 (c) 2 (d) −2-2

Answer

(d) −2-2

181=3−4\frac{1}{81}=3^{-4}, so n−2=−4n-2=-4 and n=−2n=-2.

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  • 2083 · Management Practice Set B · 1 mark

If n(A)=40n(A) = 40, n(B)=50n(B) = 50, n(U)=200n(U) = 200, n(A∩B)=20n(A \cap B) = 20, then n(A∪B‾)=n(\overline{A \cup B}) = (a) 20 (b) 130 (c) 40 (d) 100

Answer

(b) 130

n(A∪B)=40+50−20=70n(A\cup B)=40+50-20=70

n(A∪B‾)=200−70=130n(\overline{A\cup B})=200-70=130

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  • 2083 · Management Practice Set B · 1 mark

Which one of the following is the defined statement for A∪BA \cup B? (a) {x:x∈A and x∈B}\{x : x \in A \text{ and } x \in B\} (b) {x:x∉A and x∉B}\{x : x \notin A \text{ and } x \notin B\} (c) {x:x∈A and x∉B}\{x : x \in A \text{ and } x \notin B\} (d) {x:x∈A or x∈B}\{x : x \in A \text{ or } x \in B\}

Answer

(d) {x:x∈A or x∈B}\{x : x \in A \text{ or } x \in B\}

The union contains the elements that belong to at least one of the two sets.

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  • 2083 · Management Practice Set B · 1 mark

When kk is subtracted from 10, and the difference is divided by 2, the result is 3. What is the value of kk? (a) 3 (b) 4 (c) 6 (d) 10

Answer

(b) 4

10−k2=3⇒10−k=6⇒k=4\frac{10-k}{2}=3 \Rightarrow 10-k=6 \Rightarrow k=4

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  • 2083 · Management Practice Set B · 1 mark

The sum of the digits of a two-digit number is 8. If 18 is subtracted from the number the digits are reversed. What is the original number? (a) 35 (b) 44 (c) 53 (d) 62

Answer

(c) 53

Let the digits be aa (tens) and bb (units), so a+b=8a+b=8. Reversing the digits gives (10a+b)−(10b+a)=9(a−b)=18(10a+b)-(10b+a)=9(a-b)=18, so a−b=2a-b=2. Hence a=5a=5, b=3b=3 and the number is 53 (check: 53−18=3553-18=35).

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  • 2083 · Management Practice Set B · 1 mark

The area of an isosceles triangle is 12 cm212\ \text{cm}^2 and its base is 8 cm. Find the length of the equal sides of the triangle. (a) 6 cm (b) 8 cm (c) 5 cm (d) 7 cm

Answer

(c) 5 cm

12×8×h=12⇒h=3 cm\tfrac12\times 8\times h=12 \Rightarrow h=3\text{ cm}

The height bisects the base, so each equal side is 42+32=5\sqrt{4^2+3^2}=5 cm.

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  • 2083 · Management Practice Set B · 1 mark

The perimeter of rectangle ABCD (with length yy and width xx) is equal to pp, and x=23yx = \frac{2}{3}y. What is the value of yy in terms of pp? (a) p10\frac{p}{10} (b) 3p10\frac{3p}{10} (c) p3\frac{p}{3} (d) 2p5\frac{2p}{5}

Answer

(b) 3p10\frac{3p}{10}

p=2(x+y)=2(23y+y)=10y3⇒y=3p10p=2(x+y)=2\left(\tfrac{2}{3}y+y\right)=\tfrac{10y}{3}\Rightarrow y=\tfrac{3p}{10}

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  • 2083 · Management Practice Set B · 1 mark

If the average measure of two angles in a parallelogram is y∘y^\circ, what is the average degree measure of the other two angles? (a) 180−y180 - y (b) 180−y2180 - \frac{y}{2} (c) 360−2y360 - 2y (d) 360+y360 + y

Answer

(a) 180−y180 - y

The four angles of a parallelogram add up to 360∘360^\circ. The two given angles total 2y2y, so the other two total 360−2y360-2y, with average 180−y180-y.

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  • 2083 · Management Practice Set B · 1 mark

If 0.1% of mm is equal to 10% of nn, then mm is what percent of 10n10n? (a) 11000%\frac{1}{1000}\% (b) 1% (c) 10% (d) 1000%

Answer

(d) 1000%

0.1100m=10100n⇒m=100n\frac{0.1}{100}m=\frac{10}{100}n\Rightarrow m=100n

m10n×100=100n10n×100=1000%\frac{m}{10n}\times100=\frac{100n}{10n}\times100=1000\%

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  • 2083 · Management Practice Set B · 1 mark

A conical tent is 5 m tall and its diameter is 14 m. What is the cost of cloth required to make the tent at Rs. 200 per square meter? (a) Rs. 37850 (b) Rs. 37800 (c) Rs. 37875 (d) Rs. 37880

Answer

(a) Rs. 37850

Radius r=7r=7 m, height h=5h=5 m, so slant height l=49+25=74≈8.602l=\sqrt{49+25}=\sqrt{74}\approx 8.602 m.

CSA=πrl=227×7×8.602≈189.25 m2\text{CSA}=\pi r l=\tfrac{22}{7}\times7\times8.602\approx189.25\ \text{m}^2

Cost=189.25×200≈Rs. 37850\text{Cost}=189.25\times200\approx\text{Rs. }37850

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  • 2083 · Management Practice Set B · 1 mark

If a+2b=10a + 2b = 10 and b−a=2b - a = 2, what is the value of bb? (a) 10 (b) 8 (c) 6 (d) 4

Answer

(d) 4

From b−a=2b-a=2, a=b−2a=b-2. Then (b−2)+2b=10⇒3b=12⇒b=4(b-2)+2b=10\Rightarrow 3b=12\Rightarrow b=4.

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  • 2083 · Management Practice Set B · 1 mark

If a>ba > b and a2−2ab+b2=169a^2 - 2ab + b^2 = 169, what is the value of a−ba - b? (a) 11 (b) 12 (c) 13 (d) 15

Answer

(c) 13

a2−2ab+b2=(a−b)2=169a^2-2ab+b^2=(a-b)^2=169, so a−b=±13a-b=\pm13. Since a>ba>b, a−b=13a-b=13.

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  • 2083 · Management Practice Set B · 1 mark

Which of the following is a factor of 3x3−11x2−42x3x^3 - 11x^2 - 42x? (a) x−7x - 7 (b) xx (c) 3x−63x - 6 (d) x+6x + 6

Answer

(b) xx

3x3−11x2−42x=x(3x2−11x−42)=x(3x+7)(x−6)3x^3-11x^2-42x = x(3x^2-11x-42) = x(3x+7)(x-6), so xx is a factor. None of x−7x-7, 3x−63x-6, x+6x+6 divides it.

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  • 2083 · Management Practice Set B · 1 mark

If the perimeter of a square is 28, what is the length of the diagonal? (a) 2142\sqrt{14} (b) 727\sqrt{2} (c) 737\sqrt{3} (d) 14

Answer

(b) 727\sqrt{2}

Side =28/4=7=28/4=7, and the diagonal of a square is side×2=72\text{side}\times\sqrt2 = 7\sqrt2.

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  • 2083 · Management Practice Set B · 1 mark

If am=ana^m = a^n, then what is the relation between mm and nn? (a) m+n=1m + n = 1 (b) m−n=0m - n = 0 (c) mn=0mn = 0 (d) mn=0\frac{m}{n} = 0

Answer

(b) m−n=0m - n = 0

Equal powers of the same base give equal exponents, so m=nm=n, i.e. m−n=0m-n=0 (taking a≠0,±1a\ne 0,\pm1).

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  • 2083 · Management Practice Set B · 1 mark

If 103y−3=10.00110^{3y-3} = \frac{1}{0.001}, then what will be the value of yy? (a) 3 (b) 4 (c) 1 (d) 2

Answer

(d) 2

10.001=1000=103\frac{1}{0.001}=1000=10^3, so 3y−3=33y-3=3, giving y=2y=2.

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  • 2083 · Management Practice Set B · 1 mark

Which one of the following is the simplest form for 2n+4−2(2n)2(2n+3)\frac{2^{n+4} - 2(2^n)}{2(2^{n+3})}? (a) 12\frac{1}{2} (b) 14\frac{1}{4} (c) 78\frac{7}{8} (d) 58\frac{5}{8}

Answer

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

2n+4−2⋅2n2⋅2n+3=2n(16−2)2n⋅16=1416=78\frac{2^{n+4}-2\cdot2^n}{2\cdot2^{n+3}}=\frac{2^n(16-2)}{2^{n}\cdot 16}=\frac{14}{16}=\frac78

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  • 2083 · Management Practice Set B · 1 mark

What is the total surface area of a cone of radius 7 cm and height 24 cm? (a) 710 cm2710\ \text{cm}^2 (b) 704 cm2704\ \text{cm}^2 (c) 700 cm2700\ \text{cm}^2 (d) 725 cm2725\ \text{cm}^2

Answer

(b) 704 cm2704\ \text{cm}^2

Slant height l=72+242=25l=\sqrt{7^2+24^2}=25 cm.

TSA=πr(l+r)=227×7×(25+7)=22×32=704 cm2\text{TSA}=\pi r(l+r)=\frac{22}{7}\times7\times(25+7)=22\times32=704\ \text{cm}^2

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  • 2083 · Management Practice Set B · 1 mark

1 litre = --------- cubic centimeter. (a) 1000 (b) 100 (c) 10 (d) 1

Answer

(a) 1000

1 L=1 dm3=(10 cm)3=1000 cm31\ \text{L}=1\ \text{dm}^3=(10\ \text{cm})^3=1000\ \text{cm}^3.

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  • 2083 · Management Practice Set B · 1 mark

If the radius of a sphere is doubled, what is the percent increase in the surface area? (a) 100 (b) 200 (c) 300 (d) 400

Answer

(c) 300

Surface area =4πr2=4\pi r^2. With radius 2r2r it becomes 16πr216\pi r^2, which is 4 times the original, so the increase is 16−44×100=300%\frac{16-4}{4}\times100=300\%.

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  • 2083 · Management Practice Set B · 1 mark

In a cube, the ratio of the longest diagonal to a diagonal of a base is (a) 2:3\sqrt{2} : \sqrt{3} (b) 3:2\sqrt{3} : \sqrt{2} (c) 2:12 : 1 (d) 6:2\sqrt{6} : \sqrt{2}

Answer

(b) 3:2\sqrt{3} : \sqrt{2}

For edge aa, the longest (space) diagonal is a3a\sqrt3 and a base diagonal is a2a\sqrt2, so the ratio is 3:2\sqrt3:\sqrt2.

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  • 2083 · Management Practice Set B · 1 mark

In a bag there are 7 red and 5 blue colour balls. What is the probability that two different coloured balls are selected when the balls are drawn without replacement? (a) 35132\frac{35}{132} (b) 3566\frac{35}{66} (c) 1235\frac{12}{35} (d) 511\frac{5}{11}

Answer

(b) 3566\frac{35}{66}

P=712⋅511+512⋅711=70132=3566P=\frac{7}{12}\cdot\frac{5}{11}+\frac{5}{12}\cdot\frac{7}{11}=\frac{70}{132}=\frac{35}{66}

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  • 2083 · Management Practice Set B · 1 mark

A laptop which is priced at Rs. 45000 is sold at a discount of 12%. If the selling price of the laptop including VAT is Rs. 44748, what is the rate of VAT? (a) 13% (b) 14% (c) 15% (d) 12%

Answer

(a) 13%

Price after discount =45000×0.88=Rs 39600=45000\times0.88=\text{Rs }39600. VAT amount =44748−39600=Rs 5148=44748-39600=\text{Rs }5148.

VAT rate=514839600×100=13%\text{VAT rate}=\frac{5148}{39600}\times100=13\%

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  • 2083 · Management Practice Set B · 1 mark

If the graphs of x−ay=10x - ay = 10 and 2x−y=32x - y = 3 are perpendicular lines, then the value of a=a = (a) 2 (b) −1-1 (c) −2-2 (d) 1

Answer

(c) −2-2

Slope of x−ay=10x-ay=10 is 1a\frac1a; slope of 2x−y=32x-y=3 is 22. For perpendicular lines the product is −1-1: 2a=−1\frac{2}{a}=-1, so a=−2a=-2.

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  • 2083 · Management Practice Set B · 1 mark

Opposite angles of a cyclic quadrilateral are (a) equal (b) complementary (c) supplementary (d) double of each other

Answer

(c) supplementary

In a cyclic quadrilateral the opposite angles add up to 180∘180^\circ.

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  • 2083 · Management Practice Set B · 1 mark

The radius of the circle passing through the point (6,2)(6, 2) whose two diameters are x+y=6x + y = 6 and x+2y=4x + 2y = 4 is (a) 252\sqrt{5} (b) 6 (c) 4 (d) 10

Answer

(a) 252\sqrt{5}

The centre is the intersection of the diameters. Subtracting x+y=6x+y=6 from x+2y=4x+2y=4 gives y=−2y=-2, so x=8x=8; centre (8,−2)(8,-2).

r=(8−6)2+(−2−2)2=4+16=25r=\sqrt{(8-6)^2+(-2-2)^2}=\sqrt{4+16}=2\sqrt5

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  • 2083 · Management Practice Set B · 1 mark

A circle of radius xx has an area twice that of a square of side aa. The equation used to find the radius of the circle is (a) 2πx2=a2π2\pi x^2 = a^2\pi (b) πx2=2a2\pi x^2 = 2a^2 (c) a2=2x2a^2 = 2x^2 (d) πa2=4x2\pi a^2 = 4x^2

Answer

(b) πx2=2a2\pi x^2 = 2a^2

Circle area πx2\pi x^2 is twice the square area a2a^2, so πx2=2a2\pi x^2=2a^2.

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  • 2083 · Management Practice Set B · 1 mark

If the graph of the equation x+y−8+4k=0x + y - 8 + 4k = 0 passes through the origin, then the value of kk is equal to (a) −2-2 (b) 2 (c) 0 (d) 1

Answer

(b) 2

Putting (0,0)(0,0) in x+y−8+4k=0x+y-8+4k=0 gives −8+4k=0-8+4k=0, so k=2k=2.

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  • 2083 · Management Practice Set B · 1 mark

If the nthn^{th} term of a sequence is b3n−3b^{3n} - 3 and the second term of the sequence is 61, which one of the following could be the value of bb? (a) 1 (b) 2 (c) 3 (d) 4

Answer

(b) 2

Second term: b6−3=61b^{6}-3=61, so b6=64b^6=64 and b=2b=2 (for positive bb; 26=642^6=64).

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  • 2083 · Management Practice Set C · 1 mark

If t1,t2,t3t_1, t_2, t_3 are in AP, then t1k,t2k,t3k,…\frac{t_1}{k}, \frac{t_2}{k}, \frac{t_3}{k}, \dots are in (a) AP (b) GP (c) HP (d) none

Answer

(a) AP

If the common difference of the original AP is dd, then dividing every term by the constant kk gives consecutive differences t2k−t1k=dk\dfrac{t_2}{k}-\dfrac{t_1}{k}=\dfrac dk, which are equal. So the new sequence is an AP.

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  • 2083 · Science Practice Set A · 1 mark

If n(A)=40n(A) = 40, n(B)=50n(B) = 50, n(U)=200n(U) = 200 and n(A∩B)=20n(A \cap B) = 20, then n(A∪B‾)=n(\overline{A \cup B}) = (a) 20 (b) 130 (c) 40 (d) 100

Answer

(b) 130

n(A∪B)=40+50−20=70n(A\cup B)=40+50-20=70

n(A∪B‾)=n(U)−n(A∪B)=200−70=130n(\overline{A\cup B})=n(U)-n(A\cup B)=200-70=130

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  • 2083 · Science Practice Set A · 1 mark

In a week a company produced 10,000 computer disks, of which 30 are found to be defective. At this rate, if the factory produced 1,000,000 computer disks, how many would be defective? (a) 30 (b) 300 (c) 3,000 (d) 30,000

Answer

(c) 3,000

The rate is 3010000\frac{30}{10000}. The new number is 100 times larger, so the defective disks are 30×100=300030\times100=3000.

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  • 2083 · Science Practice Set A · 1 mark

A bag contains 4 red balls and 10 blue balls. If two balls are removed from the bag at random and no ball is returned to the bag after removal, what is the probability that both balls will be blue? (a) 2549\frac{25}{49} (b) 2449\frac{24}{49} (c) 4598\frac{45}{98} (d) 4591\frac{45}{91}

Answer

(d) 4591\frac{45}{91}

There are 14 balls in total. Without replacement:

P(both blue)=1014×913=90182=4591P(\text{both blue})=\frac{10}{14}\times\frac{9}{13}=\frac{90}{182}=\frac{45}{91}

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  • 2083 · Science Practice Set A · 1 mark

If f(x)=x2f(x) = x^2 and g(x)=x−1g(x) = x - 1, what is the value of f∘g(2x+1)f\circ g(2x+1)? (a) 2x22x^2 (b) 4x2+4x4x^2 + 4x (c) 4x24x^2 (d) (2x−1)2(2x-1)^2

Answer

(c) 4x24x^2

g(2x+1)=(2x+1)−1=2xg(2x+1)=(2x+1)-1=2x

f∘g(2x+1)=f(2x)=(2x)2=4x2f\circ g(2x+1)=f(2x)=(2x)^2=4x^2

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  • 2083 · Science Practice Set A · 1 mark

What is the radius of the circle whose equation is x2+y2−6x+8y=0x^2 + y^2 - 6x + 8y = 0? (a) 1 (b) 2 (c) 3 (d) 5

Answer

(d) 5

Complete the squares:

(x−3)2+(y+4)2=9+16=25(x-3)^2+(y+4)^2=9+16=25

So the centre is (3,−4)(3,-4) and the radius is 25=5\sqrt{25}=5.

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  • 2083 · Science Practice Set A · 1 mark

AC and AB are radii of the circle with centre A, and AB⊥ACAB \perp AC. If △ABC\triangle ABC has area 18, what is the circumference of the circle? (a) 6π6\pi (b) 9π9\pi (c) 12π12\pi (d) 18π18\pi

Answer

(c) 12π12\pi

△ABC\triangle ABC is right-angled at AA with legs rr, so 12r2=18\frac12 r^2=18, giving r=6r=6.

C=2πr=12πC=2\pi r=12\pi

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  • 2083 · Science Practice Set A · 1 mark

A cylindrical log of wood is cut into two equal halves vertically from its diameter. If its length and diameter are 200 cm and 140 cm respectively, calculate the total surface area of one half. (a) 87000 cm287000\ \text{cm}^2 (b) 87200 cm287200\ \text{cm}^2 (c) 87400 cm287400\ \text{cm}^2 (d) 87600 cm287600\ \text{cm}^2

Answer

(c) 87400 cm287400\ \text{cm}^2

Radius r=70r=70 cm, length l=200l=200 cm, using π=227\pi=\frac{22}{7}. One half has a curved half-surface, two half-circular ends and one flat rectangular face.

Curved=πrl=227×70×200=44000\text{Curved}=\pi r l=\frac{22}{7}\times70\times200=44000

Two half-ends=πr2=227×4900=15400\text{Two half-ends}=\pi r^2=\frac{22}{7}\times4900=15400

Flat face=2r×l=140×200=28000\text{Flat face}=2r\times l=140\times200=28000

Total=44000+15400+28000=87400 cm2\text{Total}=44000+15400+28000=87400\ \text{cm}^2

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  • 2083 · Science Practice Set A · 1 mark

A researcher found that the number of bacteria in a certain sample doubles every hour. If there were 6 bacteria in the sample at the beginning, how many bacteria were there after 9 hours? (a) 54 (b) 512 (c) 1536 (d) 3072

Answer

(d) 3072

Doubling each hour gives 6×29=6×512=30726\times2^9=6\times512=3072.

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  • 2083 · Science Practice Set A · 1 mark

A value that satisfies the equation cos⁡2x−2cos⁡x=0\cos^2 x - 2\cos x = 0 in degrees is (a) 0 (b) 30 (c) 60 (d) 90

Answer

(d) 90

cos⁡x(cos⁡x−2)=0\cos x(\cos x-2)=0

Since cos⁡x=2\cos x=2 is impossible, cos⁡x=0\cos x=0, so x=90∘x=90^\circ.

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  • 2083 · Science Practice Set A · 1 mark

cot⁡θ−cos⁡2θsin⁡θcos⁡θ\cot\theta - \frac{\cos 2\theta}{\sin\theta\cos\theta} is equal to (a) sin⁡θ\sin\theta (b) cos⁡θ\cos\theta (c) tan⁡θ\tan\theta (d) sec⁡θ\sec\theta

Answer

(c) tan⁡θ\tan\theta

cot⁡θ−cos⁡2θsin⁡θcos⁡θ=cos⁡2θ−(cos⁡2θ−sin⁡2θ)sin⁡θcos⁡θ=sin⁡2θsin⁡θcos⁡θ=tan⁡θ\cot\theta-\frac{\cos2\theta}{\sin\theta\cos\theta}=\frac{\cos^2\theta-(\cos^2\theta-\sin^2\theta)}{\sin\theta\cos\theta}=\frac{\sin^2\theta}{\sin\theta\cos\theta}=\tan\theta

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  • 2083 · Science Practice Set A · 1 mark

If m>1m > 1, the maximum value of 2msin⁡2x2m\sin 2x is equal to (a) 2 (b) mm (c) 2m2m (d) 4m4m

Answer

(c) 2m2m

The maximum of sin⁡2x\sin2x is 1, so the maximum of 2msin⁡2x2m\sin2x is 2m2m.

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  • 2083 · Science Practice Set A · 1 mark

If AA is a matrix of 3×43 \times 4, BB is another matrix such that ABAB and BABA are both possible, the order of matrix BB is (a) 3×33 \times 3 (b) 3×43 \times 4 (c) 4×34 \times 3 (d) 4×44 \times 4

Answer

(c) 4×34\times3

ABAB exists only if BB has 4 rows, and BABA exists only if the columns of BB equal the 3 rows of AA. So BB is 4×34\times3.

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  • 2083 · Science Practice Set A · 1 mark

sin⁡(135∘+x)⋅sin⁡(135∘−x)\sin(135^\circ + x)\cdot\sin(135^\circ - x) equals (a) 2sin⁡x\sqrt{2}\sin x (b) 2cos⁡x\sqrt{2}\cos x (c) 2cos⁡2x2\cos 2x (d) cos⁡2x2\frac{\cos 2x}{2}

Answer

(d) cos⁡2x2\frac{\cos 2x}{2}

Using sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\sin A\sin B=\frac12[\cos(A-B)-\cos(A+B)]:

sin⁡(135∘+x)sin⁡(135∘−x)=12[cos⁡2x−cos⁡270∘]=cos⁡2x2\sin(135^\circ+x)\sin(135^\circ-x)=\frac12[\cos2x-\cos270^\circ]=\frac{\cos2x}{2}

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  • 2083 · Science Practice Set A · 1 mark

The first, second and last terms of an A.P. are aa, bb and 2a2a, then the number of terms in the series is (a) bb−a\frac{b}{b-a} (b) bb+a\frac{b}{b+a} (c) ab−a\frac{a}{b-a} (d) bb+a\frac{b}{b+a}

Answer

(a) bb−a\frac{b}{b-a}

The common difference is d=b−ad=b-a. The last term gives 2a=a+(n−1)(b−a)2a=a+(n-1)(b-a), so n−1=ab−an-1=\frac{a}{b-a}.

n=ab−a+1=bb−an=\frac{a}{b-a}+1=\frac{b}{b-a}

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  • 2083 · Science Practice Set B · 1 mark

−3x+y=6-3x + y = 6, ax+2y=4ax + 2y = 4. In the system of equations above, aa is a constant. For which of the following values of aa does the system have no solution? (a) −6-6 (b) −3-3 (c) 3 (d) 6

Answer

(a) −6-6

Write each equation as y=mx+cy = mx + c. The first gives y=3x+6y = 3x + 6 with slope 33. The second gives y=−a2x+2y = -\frac{a}{2}x + 2 with slope −a2-\frac{a}{2}.

No solution requires parallel distinct lines, so −a2=3-\frac{a}{2} = 3, which gives a=−6a = -6. The intercepts 66 and 22 differ, so the lines are distinct and the system has no solution.

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  • 2083 · Science Practice Set B · 1 mark

tan⁡A2+cot⁡A2\tan\frac{A}{2} + \cot\frac{A}{2} is equivalent to (a) 2sin⁡A2\sin A (b) 2cos⁡A2\cos A (c) 2sec⁡A2\sec A (d) 2csc⁡A2\csc A

Answer

(d) 2csc⁡A2\csc A

tan⁡A2+cot⁡A2=sin⁡2A2+cos⁡2A2sin⁡A2cos⁡A2=1sin⁡A2cos⁡A2=2sin⁡A=2csc⁡A\tan\frac{A}{2}+\cot\frac{A}{2}=\frac{\sin^2\frac{A}{2}+\cos^2\frac{A}{2}}{\sin\frac{A}{2}\cos\frac{A}{2}}=\frac{1}{\sin\frac{A}{2}\cos\frac{A}{2}}=\frac{2}{\sin A}=2\csc A

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  • 2083 · Science Practice Set B · 1 mark

For 0≤x≤π0 \le x \le \pi and cos⁡x=−1\cos x = -1, what is the value of cos⁡x2\cos\frac{x}{2}? (a) −32-\frac{\sqrt{3}}{2} (b) −12-\frac{1}{2} (c) 0 (d) 32\frac{\sqrt{3}}{2}

Answer

(c) 00

In 0≤x≤π0 \le x \le \pi, cos⁡x=−1\cos x = -1 gives x=πx = \pi. Then cos⁡x2=cos⁡π2=0\cos\frac{x}{2} = \cos\frac{\pi}{2} = 0.

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  • 2083 · Science Practice Set B · 1 mark

If two unbiased dice are rolled once, what is the probability that the number of dots on the two faces will be 4? (a) 14\frac{1}{4} (b) 112\frac{1}{12} (c) 118\frac{1}{18} (d) 136\frac{1}{36}

Answer

(b) 112\frac{1}{12}

Taking the total on the two faces to be 44, the favourable outcomes are (1,3),(2,2),(3,1)(1,3),(2,2),(3,1), which is 33 of the 3636 equally likely outcomes.

P=336=112P=\frac{3}{36}=\frac{1}{12}

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  • 2083 · Science Practice Set B · 1 mark

If t1,t2,t3t_1, t_2, t_3 are in AP, then t1k,t2k,t3k,…\frac{t_1}{k}, \frac{t_2}{k}, \frac{t_3}{k}, \dots are in (a) AP (b) GP (c) HP (d) None of them

Answer

(a) AP

If t2−t1=t3−t2=dt_2 - t_1 = t_3 - t_2 = d, then dividing each term by kk gives a common difference dk\frac{d}{k} between consecutive terms. The new sequence is therefore also an AP.

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  • 2083 · Science Practice Set B · 1 mark

If A=(214−1)A = \begin{pmatrix} 2 & 1 \\ 4 & -1 \end{pmatrix}, then A2−AA^2 - A is equal to (a) II (b) 2I2I (c) 6I6I (d) OO

Answer

(c) 6I6I

A2=(214−1)(214−1)=(8145)A^2=\begin{pmatrix}2&1\\4&-1\end{pmatrix}\begin{pmatrix}2&1\\4&-1\end{pmatrix}=\begin{pmatrix}8&1\\4&5\end{pmatrix}

A2−A=(8−21−14−45+1)=(6006)=6IA^2-A=\begin{pmatrix}8-2&1-1\\4-4&5+1\end{pmatrix}=\begin{pmatrix}6&0\\0&6\end{pmatrix}=6I

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  • 2083 · Science Practice Set C · 1 mark

If AA and BB are any two sets such that n(A)=17n(A) = 17, n(A∪B)=38n(A \cup B) = 38, n(A∩B)=2n(A \cap B) = 2, then n(A−B)=n(A - B) = (a) 21 (b) 15 (c) 23 (d) 51

Answer

(b) 15

n(A−B)=n(A)−n(A∩B)=17−2=15n(A-B)=n(A)-n(A\cap B)=17-2=15

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  • 2083 · Science Practice Set C · 1 mark

What is the area of the circle whose equation is (x−3)2+(y+5)2=18(x-3)^2 + (y+5)^2 = 18? (a) 9π9\pi (b) 18π18\pi (c) 72π72\pi (d) 81π81\pi

Answer

(b) 18π18\pi

The equation is of the form (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2, so r2=18r^2=18.

Area=πr2=18π\text{Area}=\pi r^2=18\pi

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  • 2083 · Science Practice Set C · 1 mark

A sphere is inscribed inside a cube. What is the probability that a point that is inside the cube is also inside the sphere? (a) 1π\frac{1}{\pi} (b) π8\frac{\pi}{8} (c) π6\frac{\pi}{6} (d) π4\frac{\pi}{4}

Answer

(c) π6\frac{\pi}{6}

Let the cube have side 2r2r, so the sphere has radius rr.

P=43πr3(2r)3=4πr3/38r3=π6P=\frac{\frac43\pi r^3}{(2r)^3}=\frac{4\pi r^3/3}{8r^3}=\frac{\pi}{6}

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  • 2083 · Science Practice Set C · 1 mark

John's six test scores are given. What is the positive difference between the mean of the scores and the median of the scores: 86, 72, 92, 62, 99, 93? (a) 5 (b) 6 (c) 7 (d) 37

Answer

(a) 5

Mean:

86+72+92+62+99+936=5046=84\frac{86+72+92+62+99+93}{6}=\frac{504}{6}=84

Ordered scores: 62, 72, 86, 92, 93, 99, so the median is 86+922=89\frac{86+92}{2}=89.

Difference: 89−84=589-84=5.

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  • 2083 · Science Practice Set C · 1 mark

If tan⁡α=12\tan\alpha = \frac{1}{2}, tan⁡β=13\tan\beta = \frac{1}{3}, then α+β\alpha + \beta is equal to (a) π/2\pi/2 (b) π/3\pi/3 (c) π/4\pi/4 (d) π/6\pi/6

Answer

(c) π/4\pi/4

tan⁡(α+β)=12+131−12⋅13=5/65/6=1\tan(\alpha+\beta)=\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}=\frac{5/6}{5/6}=1

Hence α+β=π/4\alpha+\beta=\pi/4.

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  • 2083 · Science Practice Set C · 1 mark

Two poles stand on opposite sides of a level street that is 12 meters wide. If one pole is 5 meters taller than the other, what is the angle of elevation between the tops of the poles? (a) sin⁡−1512\sin^{-1}\frac{5}{12} (b) sin⁡−115\sin^{-1}\frac{1}{5} (c) tan⁡−1512\tan^{-1}\frac{5}{12} (d) tan⁡−1513\tan^{-1}\frac{5}{13}

Answer

(c) tan⁡−1512\tan^{-1}\frac{5}{12}

Joining the two tops gives a right triangle with horizontal side 12 m (the street) and vertical side 5 m (the difference in heights). So

tan⁡θ=512⇒θ=tan⁡−1512\tan\theta=\frac{5}{12}\Rightarrow \theta=\tan^{-1}\frac{5}{12}

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  • 2083 · Science Practice Set C · 1 mark

If f(x)=3x+5f(x) = 3x + 5 and f(g(x))=6x−4f(g(x)) = 6x - 4, what is g(x)g(x)? (a) 2x−92x - 9 (b) 2x−32x - 3 (c) 3x−93x - 9 (d) 3x−33x - 3

Answer

(b) 2x−32x-3

f(g(x))=3g(x)+5=6x−4⇒g(x)=6x−93=2x−3f(g(x))=3g(x)+5=6x-4\Rightarrow g(x)=\frac{6x-9}{3}=2x-3

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  • 2083 · Science Practice Set C · 1 mark

What is the tenth term of the arithmetic sequence whose first term is xx and its third term is x+6ax + 6a? (a) 33a33a (b) x+24ax + 24a (c) x+27ax + 27a (d) x+30ax + 30a

Answer

(c) x+27ax+27a

a3=x+2d=x+6aa_3=x+2d=x+6a, so d=3ad=3a.

a10=x+9d=x+27aa_{10}=x+9d=x+27a

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  • 2083 · Science Practice Set C · 1 mark

If A=[K14K]A = \begin{bmatrix} K & 1 \\ 4 & K \end{bmatrix} is a singular matrix, then the value of KK is (a) 1 (b) 2 (c) 3 (d) 4

Answer

(b) 2

A matrix is singular when its determinant is zero: K⋅K−1⋅4=K2−4=0⇒K=±2.K\cdot K-1\cdot 4=K^2-4=0\Rightarrow K=\pm 2. Of the options given, K=2K=2 is the one listed.

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  • 2083 · Science Practice Set C · 1 mark

5(x+1)(x−1)(x+4)=Ax−1+3x+4\frac{5(x+1)}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{3}{x+4}, the value of AA is (a) 1 (b) 2 (c) 3 (d) −3-3

Answer

(b) 2

Multiply through by (x−1)(x+4)(x-1)(x+4): 5(x+1)=A(x+4)+3(x−1)5(x+1)=A(x+4)+3(x-1).

Put x=1x=1: 10=5A10=5A, so A=2A=2. Check with x=−4x=-4: −15=−5⋅3-15=-5\cdot 3, which holds.

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  • 2083 · Science Practice Set C · 1 mark

If 2x+4=322^{x+4} = 32, what is the value of 3−x3^{-x}? (a) 13\frac{1}{3} (b) 3 (c) 19\frac{1}{9} (d) 1

Answer

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

2x+4=32=252^{x+4}=32=2^5, so x+4=5x+4=5 and x=1x=1. Then 3−x=3−1=133^{-x}=3^{-1}=\frac{1}{3}.

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  • 2083 · Science Practice Set C · 1 mark

If OACB is a parallelogram with OC→=a⃗\overrightarrow{OC} = \vec{a} and AB→=b⃗\overrightarrow{AB} = \vec{b}, then OA→=\overrightarrow{OA} = … (a) a⃗+b⃗\vec{a} + \vec{b} (b) a⃗−b⃗\vec{a} - \vec{b} (c) 12(a⃗+b⃗)\frac{1}{2}(\vec{a} + \vec{b}) (d) 12(a⃗−b⃗)\frac{1}{2}(\vec{a} - \vec{b})

Answer

(d) 12(a⃗−b⃗)\frac{1}{2}(\vec{a}-\vec{b})

In parallelogram OACB, OC→\overrightarrow{OC} and AB→\overrightarrow{AB} are the diagonals. Then OC→=OA→+OB→=a⃗\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{OB}=\vec a and AB→=OB→−OA→=b⃗\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\vec b.

Subtracting: 2OA→=a⃗−b⃗2\overrightarrow{OA}=\vec a-\vec b, so OA→=12(a⃗−b⃗)\overrightarrow{OA}=\frac12(\vec a-\vec b).

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  • 2083 · Science Practice Set C · 1 mark

If Rs 300 is invested at 20%, compounded continuously, how long (to the nearest year) will it take for the money to double? (a) 3 (b) 4 (c) 5 (d) 6

Answer

(a) 3

Doubling: 300e0.2t=600300e^{0.2t}=600, so e0.2t=2e^{0.2t}=2 and t=ln⁡20.2=0.69310.2≈3.47≈3 years.t=\frac{\ln 2}{0.2}=\frac{0.6931}{0.2}\approx 3.47\approx 3\text{ years}.

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  • 2083 · Science Practice Set D · 1 mark

If AA and BB are any two non-empty sets, then A∩(A∪B)′A \cap (A \cup B)' is equal to (a) BB (b) AA (c) ϕ\phi (d) A∩BA \cap B

Answer

(c) ϕ\phi

By De Morgan's law, (A∪B)′=A′∩B′(A\cup B)'=A'\cap B'. So A∩(A∪B)′=A∩A′∩B′=ϕ∩B′=ϕ.A\cap(A\cup B)'=A\cap A'\cap B'=\phi\cap B'=\phi.

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  • 2083 · Science Practice Set D · 1 mark

If a=b+12a = b + \frac{1}{2}, b=2c+12b = 2c + \frac{1}{2} and c=3d+12c = 3d + \frac{1}{2}, which of the following is an expression for dd in terms of aa? (a) a−26\frac{a-2}{6} (b) 2a−36\frac{2a-3}{6} (c) 2a−312\frac{2a-3}{12} (d) 3a−218\frac{3a-2}{18}

Answer

(a) a−26\frac{a-2}{6}

From a=b+12a=b+\frac12 we get b=a−12b=a-\frac12. Then c=b−122=a−12c=\frac{b-\frac12}{2}=\frac{a-1}{2}. Then d=c−123=a−12−123=a−26.d=\frac{c-\frac12}{3}=\frac{\frac{a-1}{2}-\frac12}{3}=\frac{a-2}{6}.

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  • 2083 · Science Practice Set D · 1 mark

The relationship between the area AA of a circle and its circumference is given by the formula A=kC2A = kC^2, where kk is a constant. What is the value of kk? (a) 14π\frac{1}{4\pi} (b) 12π\frac{1}{2\pi} (c) 14\frac{1}{4} (d) 4π24\pi^2

Answer

(a) 14π\frac{1}{4\pi}

With A=πr2A=\pi r^2 and C=2πrC=2\pi r, we have C2=4π2r2C^2=4\pi^2r^2, so k=AC2=πr24π2r2=14π.k=\frac{A}{C^2}=\frac{\pi r^2}{4\pi^2r^2}=\frac{1}{4\pi}.

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  • 2083 · Science Practice Set D · 1 mark

What is the ratio of the area of a square to the area of a circle if the square of 10 unit length is inscribed in the circle? (a) π:2\pi : 2 (b) 1:π1 : \pi (c) 1:21 : 2 (d) 2:π2 : \pi

Answer

(d) 2:π2 : \pi

The square has side 10, so its area is 100100. Its diagonal 10210\sqrt2 is the circle's diameter, so r=52r=5\sqrt2 and the circle's area is π⋅50=50π\pi\cdot 50=50\pi. The ratio is 100:50π=2:π100:50\pi=2:\pi.

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  • 2083 · Science Practice Set D · 1 mark

If the area of the triangle with vertices (0,0)(0, 0), (5,0)(5, 0) and (0,y)(0, y) is 30 square units, what is the value of yy? (a) 5 (b) 6 (c) 10 (d) 12

Answer

(d) 12

The base along the x-axis is 5 and the height is yy, so 12⋅5⋅y=30⇒y=12.\tfrac12\cdot5\cdot y=30\Rightarrow y=12.

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  • 2083 · Science Practice Set D · 1 mark

If f(2x)=x+5f(2x) = x + 5 and f(g(6))=13f(g(6)) = 13, then 2g(6)=2g(6) = (a) 6 (b) 16 (c) 32 (d) 36

Answer

(c) 32

Let t=2xt=2x. Then f(t)=t2+5f(t)=\frac t2+5. Since f(g(6))=13f(g(6))=13, we get g(6)2+5=13\frac{g(6)}{2}+5=13, so g(6)=16g(6)=16 and 2g(6)=322g(6)=32.

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  • 2083 · Science Practice Set D · 1 mark

If the sum of the first 6 terms of a G.P. is 9 times the sum of its first three terms, then the common ratio of the series is (a) 1 (b) 32\frac{3}{2} (c) 2 (d) −2-2

Answer

(c) 2

S6=9S3S_6=9S_3 gives a(r6−1)r−1=9⋅a(r3−1)r−1\frac{a(r^6-1)}{r-1}=9\cdot\frac{a(r^3-1)}{r-1}. Dividing by a(r3−1)r−1\frac{a(r^3-1)}{r-1} gives r3+1=9r^3+1=9, so r3=8r^3=8 and r=2r=2.

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  • 2083 · Science Practice Set D · 1 mark

If 6,9,14,2a−8,2a+10,38,47,506, 9, 14, 2a - 8, 2a + 10, 38, 47, 50 are in ascending order and their median is 25, then the value of aa will be (a) 13 (b) 12 (c) 11 (d) 14

Answer

(b) 12

There are 8 terms, so the median is the mean of the 4th and 5th terms: (2a−8)+(2a+10)2=2a+1=25⇒a=12.\frac{(2a-8)+(2a+10)}{2}=2a+1=25\Rightarrow a=12. Check: the terms are 6,9,14,16,34,38,47,506,9,14,16,34,38,47,50, which are ascending.

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  • 2083 · Science Practice Set D · 1 mark

A bag contains 4 white and 6 black balls. Two balls are drawn in succession. Find the probability that both the balls are differently coloured if the first ball is not replaced. (a) 815\frac{8}{15} (b) 315\frac{3}{15} (c) 415\frac{4}{15} (d) 124\frac{1}{24}

Answer

(a) 815\frac{8}{15}

P=410⋅69+610⋅49=24+2490=815.P=\frac{4}{10}\cdot\frac{6}{9}+\frac{6}{10}\cdot\frac{4}{9}=\frac{24+24}{90}=\frac{8}{15}.

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  • 2083 · Science Practice Set D · 1 mark

The value of 1−sin⁡x1+sin⁡x\sqrt{\frac{1-\sin x}{1+\sin x}} is equal to (a) tan⁡(x2+π4)\tan\left(\frac{x}{2}+\frac{\pi}{4}\right) (b) tan⁡(x2−π4)\tan\left(\frac{x}{2}-\frac{\pi}{4}\right) (c) cot⁡(x2+π)\cot\left(\frac{x}{2}+\pi\right) (d) tan⁡(π4−x2)\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)

Answer

(d) tan⁡(π4−x2)\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)

Write 1−sin⁡x=(cos⁡x2−sin⁡x2)21-\sin x=\left(\cos\frac x2-\sin\frac x2\right)^2 and 1+sin⁡x=(cos⁡x2+sin⁡x2)21+\sin x=\left(\cos\frac x2+\sin\frac x2\right)^2. Then 1−sin⁡x1+sin⁡x=cos⁡x2−sin⁡x2cos⁡x2+sin⁡x2=1−tan⁡x21+tan⁡x2=tan⁡(π4−x2).\sqrt{\frac{1-\sin x}{1+\sin x}}=\frac{\cos\frac x2-\sin\frac x2}{\cos\frac x2+\sin\frac x2}=\frac{1-\tan\frac x2}{1+\tan\frac x2}=\tan\left(\frac{\pi}{4}-\frac{x}{2}\right).

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  • 2083 · Science Practice Set D · 1 mark

The value of 4sin⁡Acos⁡3A−4cos⁡Asin⁡3A=4\sin A\cos^3 A - 4\cos A\sin^3 A = (a) cos⁡8A\cos 8A (b) sin⁡2A\sin 2A (c) cos⁡4A\cos 4A (d) sin⁡4A\sin 4A

Answer

(d) sin⁡4A\sin 4A

4sin⁡Acos⁡A(cos⁡2A−sin⁡2A)=2sin⁡2Acos⁡2A=sin⁡4A.4\sin A\cos A(\cos^2A-\sin^2A)=2\sin2A\cos2A=\sin4A.

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  • 2083 · Science Practice Set D · 1 mark

If 32x+1=813^{2x+1} = 81, then x=x = (a) 1 (b) 32\frac{3}{2} (c) 2 (d) 52\frac{5}{2}

Answer

(b) 32\frac{3}{2}

Since 81=3481=3^4, we have 2x+1=42x+1=4, so x=32x=\frac32.

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  • 2083 · Science Practice Set D · 1 mark

Which of the following matrices does not have an inverse matrix? (a) (1001)\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} (b) (5317)\begin{pmatrix} 5 & 3 \\ 1 & 7 \end{pmatrix} (c) (2−43−5)\begin{pmatrix} 2 & -4 \\ 3 & -5 \end{pmatrix} (d) (2346)\begin{pmatrix} 2 & 3 \\ 4 & 6 \end{pmatrix}

Answer

(d) (2346)\begin{pmatrix} 2 & 3 \\ 4 & 6 \end{pmatrix}

A matrix has no inverse when its determinant is zero. Here 2⋅6−3⋅4=02\cdot6-3\cdot4=0. The other determinants are 11, 35−3=3235-3=32 and −10+12=2-10+12=2, all non-zero.

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  • 2083 · Science Practice Set D · 1 mark

PQ and PT are tangent to circle O. If angle P measures 70∘70^\circ, how many degrees are in the minor arc QT? (a) 110 (b) 120 (c) 125 (d) 140

Answer

(a) 110

Let the minor arc QTQT be mm. Then the major arc is 360∘−m360^\circ-m, and the angle between two tangents is half the difference of the arcs: 70=12((360−m)−m)⇒140=360−2m⇒m=110∘.70=\tfrac12\big((360-m)-m\big)\Rightarrow 140=360-2m\Rightarrow m=110^\circ.

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  • 2011 · Science Entrance Sample Paper · 1 mark

Grace has 16 jellybeans in her pocket. She has 8 red ones, 4 green ones, and 4 blue ones. What is the minimum number of jellybeans she must take out of her pocket to ensure that she has one of each color? (a) 4 (b) 8 (c) 12 (d) 13

Answer

(d) 13

In the worst case she first takes all 8 red and all 4 green jellybeans, which is 12 and still has no blue one. The 13th jellybean must be blue, so 13 guarantees one of each colour.

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  • 2011 · Science Entrance Sample Paper · 1 mark

The function f(x)=−3x+12f(x) = -3x + 12 on R\mathbb{R} is (a) Increasing (b) Decreasing (c) Constant (d) None of these

Answer

(b) Decreasing

f′(x)=−3<0f'(x) = -3 < 0 for all x∈Rx \in \mathbb{R}, so the function always decreases as xx increases.

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  • 2011 · Science Entrance Sample Paper · 1 mark

The system of equations x+2y=−1x + 2y = -1, −2x−4y=2-2x - 4y = 2 has (a) Only one solution (b) Finitely many solutions (c) No solution (d) Infinitely many solutions

Answer

(d) Infinitely many solutions

Multiplying the first equation by −2-2 gives −2x−4y=2-2x - 4y = 2, which is exactly the second equation. The two equations represent the same line, so every point on it is a solution.

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  • 2011 · Science Entrance Sample Paper · 1 mark

The area of the triangle formed by the lines y−x=0y - x = 0, y+x=0y + x = 0 and x=cx = c is (a) cc sq. units (b) c2c^2 sq. units (c) 12c\frac{1}{2}c sq. units (d) 12c2\frac{1}{2}c^2 sq. units

Answer

(b) c2c^2 sq. units

The lines y=xy = x and y=−xy = -x meet at the origin, and x=cx = c cuts them at (c,c)(c, c) and (c,−c)(c, -c). The base along x=cx = c is 2c2c and the height is cc:

Area=12(2c)(c)=c2.\text{Area} = \tfrac{1}{2}(2c)(c) = c^2.

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  • 2011 · Science Entrance Sample Paper · 1 mark

If AA is an m×nm \times n matrix and BB is a matrix given in such a way that ABAB and BABA are both defined, then the size of the BB matrix is (a) m×nm \times n (b) n×nn \times n (c) m×nm \times n (d) n×mn \times m

Answer

(d) n×mn \times m

ABAB is defined only if the number of columns of AA (nn) equals the number of rows of BB. BABA is defined only if the number of columns of BB equals the number of rows of AA (mm). So BB is n×mn \times m.

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  • 2011 · Science Entrance Sample Paper · 1 mark

The value of pp for which the vectors a⃗=pi⃗+j⃗−2k⃗\vec{a} = p\vec{i} + \vec{j} - 2\vec{k} and b⃗=i⃗+j⃗−3k⃗\vec{b} = \vec{i} + \vec{j} - 3\vec{k} are perpendicular to each other is (a) 9 (b) −9-9 (c) 7 (d) −7-7

Answer

(d) −7-7

Perpendicular vectors have zero dot product:

a⃗⋅b⃗=p+1+6=0⇒p=−7.\vec{a}\cdot\vec{b} = p + 1 + 6 = 0 \Rightarrow p = -7.

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  • 2011 · Science Entrance Sample Paper · 1 mark

A can solve 90% of the problems given in a book and B can solve 70%. What is the probability that at least one of them will solve a problem selected at random from the book? (a) 0.63 (b) 0.16 (c) 0.97 (d) 1.60

Answer

(c) 0.97

P(A∪B)=P(A)+P(B)−P(A)P(B)=0.9+0.7−0.63=0.97.P(A \cup B) = P(A) + P(B) - P(A)P(B) = 0.9 + 0.7 - 0.63 = 0.97.

Check: 1−(0.1)(0.3)=0.971 - (0.1)(0.3) = 0.97.

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  • 2011 · Science Entrance Sample Paper · 1 mark

The number of non-empty proper subsets of order 6 is (a) 64 (b) 63 (c) 62 (d) 60

Answer

(c) 62

A set of 6 elements has 26=642^6 = 64 subsets. Removing the empty set and the set itself leaves 64−2=6264 - 2 = 62 non-empty proper subsets.

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  • 2011 · Science Entrance Sample Paper · 1 mark

If the common difference is 5, then the total number of arithmetic means between 10 and 50 is (a) 8 (b) 9 (c) 5 (d) 7

Answer

(d) 7

Let there be nn arithmetic means between 10 and 50. Then 50 is the (n+2)(n+2)th term:

50=10+(n+1)⋅5⇒n+1=8⇒n=7.50 = 10 + (n+1)\cdot 5 \Rightarrow n + 1 = 8 \Rightarrow n = 7.

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  • 2011 · Science Entrance Sample Paper · 1 mark

The area of the triangle formed by the line 2x+3y=122x + 3y = 12 with the co-ordinate axes is (a) 24 (b) 12 (c) 6 (d) 3

Answer

(b) 12

The intercepts of 2x+3y=122x + 3y = 12 are x=6x = 6 and y=4y = 4.

Area=12⋅6⋅4=12.\text{Area} = \tfrac{1}{2}\cdot 6 \cdot 4 = 12.

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  • 2011 · Science Entrance Sample Paper · 1 mark

In a race of 1 km, A beats B by 100 m, B beats C by 50 m, C beats D by 40 m, then A beats D by (a) 190 m (b) 820.8 m (c) 179.2 m (d) none

Answer

(c) 179.2 m

When A runs 1000 m, B runs 900 m. When B runs 1000 m, C runs 950 m. When C runs 1000 m, D runs 960 m.

So when A runs 1000 m, B runs 900 m, C runs 900×9501000=855900 \times \frac{950}{1000} = 855 m, and D runs 855×9601000=820.8855 \times \frac{960}{1000} = 820.8 m. A beats D by 1000−820.8=179.21000 - 820.8 = 179.2 m.

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  • 2011 · Science Entrance Sample Paper · 1 mark

If the mthm^{th} term of an AP is 1n\frac{1}{n} and its nthn^{th} term is 1m\frac{1}{m}, then its (mn)th(mn)^{th} term is (a) mnmn (b) 1mn\frac{1}{mn} (c) 1 (d) mn\frac{m}{n}

Answer

(c) 1

Let the first term be aa and common difference dd. Then a+(m−1)d=1na + (m-1)d = \frac{1}{n} and a+(n−1)d=1ma + (n-1)d = \frac{1}{m}. Subtracting:

(m−n)d=1n−1m=m−nmn⇒d=1mn.(m-n)d = \frac{1}{n} - \frac{1}{m} = \frac{m-n}{mn} \Rightarrow d = \frac{1}{mn}.

Then a=1n−m−1mn=1mna = \frac{1}{n} - \frac{m-1}{mn} = \frac{1}{mn}. Hence

Tmn=a+(mn−1)d=1mn+mn−1mn=1.T_{mn} = a + (mn-1)d = \frac{1}{mn} + \frac{mn-1}{mn} = 1.

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  • 2011 · Science Entrance Sample Paper · 1 mark

If the AM and GM of the roots of a quadratic equation are 8 and 5 respectively, then the equation is (a) x2+6x+25=0x^2 + 6x + 25 = 0 (b) x2−16x+25=0x^2 - 16x + 25 = 0 (c) x2−16x−25=0x^2 - 16x - 25 = 0 (d) x2−13x+40=0x^2 - 13x + 40 = 0

Answer

(b) x2−16x+25=0x^2 - 16x + 25 = 0

AM =8= 8 gives sum of roots =16= 16. GM =5= 5 gives product of roots =25= 25. The equation x2−(sum)x+product=0x^2 - (\text{sum})x + \text{product} = 0 is

x2−16x+25=0.x^2 - 16x + 25 = 0.

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  • 2011 · Science Entrance Sample Paper · 1 mark

The value of cos⁡11∘+sin⁡11∘cos⁡11∘−sin⁡11∘\frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ} is equal to the value of (a) tan⁡56∘\tan 56^\circ (b) sin⁡56∘\sin 56^\circ (c) cot⁡56∘\cot 56^\circ (d) tan⁡79∘\tan 79^\circ

Answer

(a) tan⁡56∘\tan 56^\circ

Divide numerator and denominator by cos⁡11∘\cos 11^\circ:

1+tan⁡11∘1−tan⁡11∘=tan⁡45∘+tan⁡11∘1−tan⁡45∘tan⁡11∘=tan⁡(45∘+11∘)=tan⁡56∘.\frac{1 + \tan 11^\circ}{1 - \tan 11^\circ} = \frac{\tan 45^\circ + \tan 11^\circ}{1 - \tan 45^\circ \tan 11^\circ} = \tan(45^\circ + 11^\circ) = \tan 56^\circ.

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  • 2011 · Science Entrance Sample Paper · 1 mark

Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking? (a) 2 hours (b) 4124\frac{1}{2} hours (c) 5345\frac{3}{4} hours (d) 6 hours

Answer

(b) 4124\frac{1}{2} hours

When the second cyclist starts, the first is already 6×3=186 \times 3 = 18 miles ahead. The second gains 10−6=410 - 6 = 4 miles per hour, so

t=184=412 hours.t = \frac{18}{4} = 4\tfrac{1}{2} \text{ hours}.

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  • 2011 · Science Entrance Sample Paper · 1 mark

The sum of the first nn terms of the series n−1n,n−2n,n−3n,…\frac{n-1}{n}, \frac{n-2}{n}, \frac{n-3}{n}, \dots is (a) n−12n\frac{n-1}{2n} (b) n−12\frac{n-1}{2} (c) n(n+1)2\frac{n(n+1)}{2} (d) n2\frac{n}{2}

Answer

(b) n−12\frac{n-1}{2}

The series is an AP with first term n−1n\frac{n-1}{n}, common difference −1n-\frac{1}{n} and last (nnth) term n−nn=0\frac{n-n}{n} = 0.

Sn=n2(n−1n+0)=n−12.S_n = \frac{n}{2}\left(\frac{n-1}{n} + 0\right) = \frac{n-1}{2}.

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  • 2011 · Science Entrance Sample Paper · 1 mark

If (a+b+c)(a−b+c)=3ac(a+b+c)(a-b+c) = 3ac, then angle BB is equal to (a) π9\frac{\pi}{9} (b) π2\frac{\pi}{2} (c) π3\frac{\pi}{3} (d) π4\frac{\pi}{4}

Answer

(c) π3\frac{\pi}{3}

(a+c+b)(a+c−b)=(a+c)2−b2=3ac(a+c+b)(a+c-b) = (a+c)^2 - b^2 = 3ac, so a2+c2−b2=aca^2 + c^2 - b^2 = ac. By the cosine rule,

cos⁡B=a2+c2−b22ac=ac2ac=12⇒B=π3.\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{ac}{2ac} = \frac{1}{2} \Rightarrow B = \frac{\pi}{3}.

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  • 2011 · Science Entrance Sample Paper · 1 mark

The circles x2+y2+2ax+c=0x^2 + y^2 + 2ax + c = 0 and x2+y2+2by+c=0x^2 + y^2 + 2by + c = 0 will touch each other if (a) 1a2−1b2=1c2\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} (b) 1a2+1b2+1c2=0\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = 0 (c) 1a2+1b2=1c2\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} (d) 1a+1b=1c\frac{1}{a} + \frac{1}{b} = \frac{1}{c}

Answer

(c) 1a2+1b2=1c2\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}

The centres are (−a,0)(-a, 0) and (0,−b)(0, -b), so the distance between them satisfies d2=a2+b2d^2 = a^2 + b^2. Writing the constant term as c2c^2 (the standard form of this problem), the radii satisfy r12=a2−c2r_1^2 = a^2 - c^2 and r22=b2−c2r_2^2 = b^2 - c^2.

For the circles to touch, d2=(r1±r2)2d^2 = (r_1 \pm r_2)^2, which gives r1r2=±c2r_1 r_2 = \pm c^2. Squaring:

(a2−c2)(b2−c2)=c4⇒a2b2=c2(a2+b2).(a^2 - c^2)(b^2 - c^2) = c^4 \Rightarrow a^2 b^2 = c^2(a^2 + b^2).

Dividing by a2b2c2a^2 b^2 c^2:

1a2+1b2=1c2.\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}.

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Questions from Trinity International College Practice Questions for +2 Entrance Exam, Management Set A (College-published practice booklet (April 2026) for the +2 Management entrance: 100 MCQs in English, Mathematics and General Knowledge; practice paper, not a recalled board paper), Trinity International College Practice Questions for +2 Entrance Exam, Management Set B (College-published practice booklet (April 2026) for the +2 Management entrance: 100 MCQs in English, Mathematics and General Knowledge; practice paper, not a recalled board paper), Trinity International College Practice Questions for +2 Entrance Exam, Management Set C (College-published practice booklet (April 2026) for the +2 Management entrance: 100 MCQs in English, Mathematics and General Knowledge; practice paper, not a recalled board paper), Trinity International College Practice Questions for +2 Entrance Exam, Science Set A (College-published practice booklet (April 2026) for the +2 Science entrance: 100 MCQs in Physics, Chemistry, Biology, Mathematics, English and General Knowledge; practice paper, not a recalled board paper), Trinity International College Practice Questions for +2 Entrance Exam, Science Set B (College-published practice booklet (April 2026) for the +2 Science entrance: 100 MCQs in Physics, Chemistry, Biology, Mathematics, English and General Knowledge; practice paper, not a recalled board paper), Trinity International College Practice Questions for +2 Entrance Exam, Science Set C (College-published practice booklet (April 2026) for the +2 Science entrance: 100 MCQs in Physics, Chemistry, Biology, Mathematics, English and General Knowledge; practice paper, not a recalled board paper), Trinity International College Practice Questions for +2 Entrance Exam, Science Set D (College-published practice booklet (April 2026) for the +2 Science entrance: 100 MCQs in Physics, Chemistry, Biology, Mathematics, English and General Knowledge; practice paper, not a recalled board paper) and Trinity International College Entrance Test 2011, +2 Science (sample test paper) (College sample test paper for the +2 Science entrance, 100 MCQs (Science 50, Mathematics 20, English 20, General Knowledge 10), 2011). Answers are written for this site; check them against your class notes.