For Parents For Educators & Experts For Institutions For Employers

SSC CHSL 13 November 2025 · Shift 2 · Quantitative Aptitude — questions with answers · ShikshaSphere

Chapter 15 of 200

13 November 2025 · Shift 2

Quantitative Aptitude

25 questions ~25 min readFree

25 questions

P, Q, and R enter into a partnership. P invests Rs.

1, 00, 000, Q

invests Rs. 80,000, and R invests Rs. 60,000. After 6 months, Q increases his capital by

20 %

, while R decreases his capital by

20 %

. At the end of 1 year, the total profit is Rs .

1, 40, 000

. What is R's share in the profit? (approximate value)

A.₹50,000
B.₹31,240
C.₹40,000
D.₹65,780

Solution

Partnership with changing capital Profit in a partnership is divided in the ratio of (capital

\times

time), so count each partner's money month by month. P keeps ₹1,00,000 for the whole year:

100000 \times 12 = 1200000

. Q keeps ₹80,000 for 6 months and then

80000 \times \frac{120}{100} = 96000

for 6 months:

80000 \times 6 + 96000 \times 6 = 1056000

. R keeps ₹60,000 for 6 months and then

60000 \times \frac{80}{100} = 48000

for 6 months:

60000 \times 6 + 48000 \times 6 = 648000

Ratio =1200000:1056000:648000=50:44:27, so the profit has 50+44+27=121 parts. R's share

= 140000 \times \frac{27}{121} = \frac{3780000}{121} \approx 31240

. Hence R gets about ₹31,240 and option (b) is correct.

\cos A = \frac{8}{17}

and

\sin B = \frac{15}{17}

, find

\sin A \cos B - \cos A \sin B

A. $\frac{240}{289}$
B. $\frac{15}{17}$
C.0
D. $\frac{120}{289}$

Solution

Trigonometric identity From

\cos A = \frac{8}{17}

\sin A = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}

. From

\sin B = \frac{15}{17}

\cos B = \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{64}{289}} = \frac{8}{17}

. So

\sin A = \sin B = \frac{15}{17}

and

\cos A = \cos B = \frac{8}{17}

(both angles come from the same 8, 15, 17 triangle). Now

\sin A \cos B - \cos A \sin B = \frac{15}{17} \times \frac{8}{17} - \frac{8}{17} \times \frac{15}{17} = \frac{120}{289} - \frac{120}{289} = 0

. The expression is exactly

\sin (A - B)

, and here A=B, so its value is 0; option (c) is correct.

A trader buys 60 kg of rice at ₹60 per kg and 40 kg of a different variety at ₹70 per kg. He mixes the two varieties and sells the entire mixture at ₹65 per kg. What is his total profit or loss?

A.₹200 profit
B.₹200 loss
C.₹100 profit
D.₹100 loss

Solution

Mixture, profit and loss Cost of the first variety

= 60 \times 60 = 3600

, i.e. ₹3,600. Cost of the second variety

= 40 \times 70 = 2800

, i.e. ₹2,800. Total cost price =3600+2800=6400, i.e. ₹6,400 for 60+40=100 kg of the mixture.

Selling price of the mixture

= 100 \times 65 = 6500

, i.e. ₹6,500. Since SP is greater than CP, profit =6500-6400=100. Hence he makes a profit of ₹100 and option (c) is correct.

Two numbers are respectively 30% less and 40% more than a third number. What is the ratio of the two numbers?

A.1 : 2
B.2 : 1
C.7 : 10
D. $10 : 7$

Solution

Percentage and ratio Let the third number be 100 (any convenient value works, because only a ratio is asked). First number is 30% less:

100 \times \frac{70}{100} = 70

. Second number is 40% more:

100 \times \frac{140}{100} = 140

. Required ratio =70:140=1:2. Hence option (a) is correct.

The average of 10 numbers is 55. If each of the first 5 numbers is increased by 6 and each of the last 5 numbers is decreased by 4, what will be the new average ?

A.55
B.57
C.56
D.58

Solution

Average after change Sum of the 10 numbers

= 10 \times 55 = 550

. The first five numbers add

5 \times 6 = 30

to this sum. The last five numbers take away

5 \times 4 = 20

from it. New sum =550+30-20=560. New average

= \frac{560}{10} = 56

. Hence option (c) is correct.

The HCF of

2^{5} \times 3^{3} \times 5^{2} \times 11^{2}

and

31 \times 5^{2} \times 13 \times 3

is:

A.31
B.11
C.5
D.75

Solution

HCF from prime factors The HCF keeps only those primes that occur in both numbers, each raised to the smaller of its two powers. First number

= 2^{5} \times 3^{3} \times 5^{2} \times 11^{2}

; second number

= 31 \times 5^{2} \times 13 \times 3 = 3 \times 5^{2} \times 13 \times 31

. Common prime 3: powers are 3 and 1, so take

3^{1}

. Common prime 5: powers are 2 and 2, so take

5^{2}

. The primes 2, 11, 13 and 31 appear in only one of the two numbers, so they are dropped. HCF

= 3 \times 5^{2} = 3 \times 25 = 75

, so option (d) is correct.

The population of a town increases by 25% in the first year, decreases by 20% in the second year, and again increases by 30% in the third year. If the population at the beginning was 60,000 , what will it be after 3 years?

A.64,000
B.53,200
C.78,000
D.84,000

Solution

Successive percentage change A rise of 25% multiplies the population by

\frac{125}{100}

, a fall of 20% by

\frac{80}{100}

and a rise of 30% by

\frac{130}{100}

. After the first year:

60000 \times \frac{125}{100} = 75000

. After the second year:

75000 \times \frac{80}{100} = 60000

. After the third year:

60000 \times \frac{130}{100} = 78000

So the population after 3 years is 78,000 and option (c) is correct.

A thief escapes from a bank at 12:00 noon at a speed of

30 km / h

. A police car starts chasing him at 12:30 PM at a speed of

45 km / h

. At what time will the police car catch the thief?

A.01:00 PM
B.01:30 PM
C.02:00 PM
D.02:30 PM

Solution

Relative speed chase From 12:00 noon to 12:30 PM the thief runs alone, i.e. for

\frac{1}{2}

hour. Head start

= 30 \times \frac{1}{2} = 15

km. Both then move in the same direction, so the gap closes at the relative speed =45-30=15 km/h. Time needed to close the gap

= \frac{15}{15} = 1

hour.

So the police car catches the thief at 12:30 PM + 1 hour = 01:30 PM, and option (b) is correct.

A can do a work in 30 days, B in 45 days. They start together but A leaves after 10 days. In how many more days will B complete the remaining work?

A.20 days
B.16 days
C.10 days
D.18 days

Solution

Time and work Take the total work as the LCM of 30 and 45, i.e. 90 units. A's one day work

= \frac{90}{30} = 3

units and B's one day work

= \frac{90}{45} = 2

units.

Working together for 10 days they finish

(3 + 2) \times 10 = 50

units. Work left =90-50=40 units, and this is done by B alone. Extra days needed

= \frac{40}{2} = 20

days. Hence option (a) is correct.

A jeweler marks a necklace at 40% above its cost price. She gives a cash discount of ₹200 on the marked price. If the necklace is sold for ₹2000 and she earns a profit of

25 %

, what is the cost price of the necklace?

A.₹1800
B.₹1500
C.₹1650
D.₹1600

Solution

Cost price from profit Let the cost price be x. A profit of 25% means the selling price is

\frac{125}{100}

of the cost price, i.e. SP

= \frac{125}{100} x

. The necklace is actually sold for ₹2,000, so

\frac{125}{100} x = 2000

x = 2000 \times \frac{100}{125} = 1600

. Hence the cost price of the necklace is ₹1,600 and option (d) is correct.

A sum of money triple itself in 10 years at a certain rate of simple interest. In how many years will it become six times itself at the same rate ?

A.20 years
B.21 years
C.25 years
D.20 years

Solution

Simple interest — time Let the sum be P. 'Triples itself' means the amount after 10 years is 3P, so the simple interest earned is 3P-P=2P. Formula:

S I = \frac{P \times R \times T}{100}

2 P = \frac{P \times R \times 10}{100} \Rightarrow R = 20 %

per annum. To become six times, the interest required is 6P-P=5P.

5 P = \frac{P \times 20 \times T}{100} \Rightarrow T = \frac{5 \times 100}{20} = 25

Hence the sum becomes six times itself in 25 years, so option (c) is correct.

Shyam invests a portion of a total sum of ₹A in Scheme P for 6 years and the remaining amount in Scheme Q for 7 years. Scheme P offers a simple interest rate of r% per annum, while Scheme Q offers a simple interest rate of

(r + 3) %

per annum. The amount invested in Scheme P is

40 %

more than the amount invested in Scheme Q . If the total interest earned from both schemes is ₹2,625 and the amount invested in Scheme P is ₹2,100, what is the simple interest rate,

r

A.14%
B.13%
C.10%
D.16%

Solution

Simple interest — rate The amount put in Scheme P is ₹2,100 and it is

40 %

more than the amount put in Scheme Q. So

2100 = Q \times \frac{140}{100} \Rightarrow Q = \frac{2100 \times 100}{140} = 1500

, i.e. ₹1,500 goes into Scheme Q.

Interest from Scheme P

= \frac{2100 \times r \times 6}{100} = 126 r

Interest from Scheme Q

= \frac{1500 \times (r + 3) \times 7}{100} = 105 (r + 3)

Total interest: 126r+105(r+3)=2625

231 r + 315 = 2625 \Rightarrow 231 r = 2310 \Rightarrow r = 10

Hence the simple interest rate is

10 %

, so option (c) is correct.

What is the value of

x^{2} + y^{2} + z^{2}

, if

x + y + z = 16

and

x y + y z + z x = 80

A.94
B.96
C.86
D.256

Solution

Algebraic identity Identity:

{(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x)

Rearranging it,

x^{2} + y^{2} + z^{2} = {(x + y + z)}^{2} - 2 (x y + y z + z x)

Substituting x+y+z=16 and xy+yz+zx=80:

x^{2} + y^{2} + z^{2} = 16^{2} - 2 \times 80 = 256 - 160 = 96

Option (d) 256 is the trap — it is only

{(x + y + z)}^{2}

, with 2(xy+yz+zx) not yet subtracted. Hence the required value is 96, so option (b) is correct.

Which of the following numbers can possibly be a perfect square ?

A.364952
B.694828
C.268923
D.430336

Solution

Perfect squares — unit digit The square of any whole number ends only in one of the digits 0, 1, 4, 5, 6 or 9. So a number whose last digit is 2, 3, 7 or 8 can never be a perfect square.

364952 ends in 2, 694828 ends in 8 and 268923 ends in 3, so all three are ruled out at sight. 430336 ends in 6, so it survives the test — and indeed

656^{2} = 430336

. Hence option (d) is correct.

A shopkeeper mixes 6 liters of a milk solution containing

30 %

fat with 8 liters of another milk solution containing

50 %

fat. What is the fat percentage in the final mixture?

A. $41.42 %$
B.32.50%
C.54.41%
D.65.00%

Solution

Mixture — percentage Fat in the first solution

= 6 \times \frac{30}{100} = 1.8

litres Fat in the second solution

= 8 \times \frac{50}{100} = 4

litres Total fat =1.8+4=5.8 litres and total volume =6+8=14 litres

Fat percentage

= \frac{5.8}{14} \times 100 = \frac{580}{14} \approx 41.42 %

Hence option (a) is correct.

If the radius of a sphere is increased by 10 percent, by what percentage will its volume increase?

A. $33.1 %$
B. $41.1 %$
C. $39.2 %$
D.50.0%

Solution

Volume of a sphere Volume of a sphere is

V = \frac{4}{3} π r^{3}

, so V varies as

r^{3}

. Take the original radius as r=10; a

10 %

rise makes it 11.

\frac{V_{2}}{V_{1}} = {(\frac{11}{10})}^{3} = \frac{1331}{1000}

Increase

= \frac{1331 - 1000}{1000} \times 100 = 33.1 %

Hence option (a) is correct.

A 12 cm -radius solid metal sphere is melted and reshaped into smaller spheres, each with a 3 cm radius. From the original sphere, how many smaller spheres can be created?

A.64
B.68
C.62
D.65

Solution

Melting and recasting spheres Melting does not change the metal, so the total volume of the small spheres equals the volume of the big sphere.

Number of spheres

= \frac{\frac{4}{3} π R^{3}}{\frac{4}{3} π r^{3}} = {(\frac{R}{r})}^{3}

= {(\frac{12}{3})}^{3} = 4^{3} = 64

Hence 64 small spheres can be made, so option (a) is correct.

The centroid of a triangle is the center of mass. If the coordinates of the vertices are

A (5, 4), B (7, 9)

, and

C (6, 8)

what is the centroid of the triangle?

A. $(6, 7)$
B. $(7, 6)$
C. $(5, 7)$
D. $(6, 8)$

Solution

Centroid of a triangle Formula: for vertices

(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

the centroid is

(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})

x-coordinate

= \frac{5 + 7 + 6}{3} = \frac{18}{3} = 6

y-coordinate

= \frac{4 + 9 + 8}{3} = \frac{21}{3} = 7

So the centroid is (6,7), and option (a) is correct. Option (d) (6,8) is simply the vertex C, not the centroid.

64 \times 8^{4} = 8^{y}

, what is

y

Solution

Laws of exponents Bring every term to the same base 8: here

64 = 8 \times 8 = 8^{2}

. So

64 \times 8^{4} = 8^{2} \times 8^{4}

Rule:

a^{m} \times a^{n} = a^{m + n}

, hence

8^{2} \times 8^{4} = 8^{2 + 4} = 8^{6}

. Comparing

8^{6} = 8^{y}

with equal bases gives y=6. Hence option (b) is correct.

Two similar triangles have areas in the ratio of

25 : 64

. What is the ratio of their corresponding altitudes?

A.625 : 4096
B.25 : 64
C.5 : 64
D. $5 : 8$

Solution

Similar triangles — area ratio Property: in two similar triangles the ratio of the areas is the square of the ratio of any pair of corresponding sides, medians or altitudes. If

A_{1}, A_{2}

are the areas and

h_{1}, h_{2}

the corresponding altitudes, then

\frac{A_{1}}{A_{2}} = {(\frac{h_{1}}{h_{2}})}^{2}

{(\frac{h_{1}}{h_{2}})}^{2} = \frac{25}{64}

\frac{h_{1}}{h_{2}} = \sqrt{\frac{25}{64}} = \frac{5}{8}

So the corresponding altitudes are in the ratio 5:8, and option (d) is correct. Option (a) squares the area ratio and option (b) repeats it — both forget the square root.

x + y = \sqrt{23}

and

x - y = \sqrt{3}

, find

\frac{(x^{2} + y^{2})}{(x y)}

A. $\frac{26}{5}$
B. $\frac{20}{7}$
C. $\frac{13}{5}$
D. $\frac{17}{5}$

Solution

Algebraic identities Square the first equation:

{(x + y)}^{2} = {(\sqrt{23})}^{2}

x^{2} + y^{2} + 2 x y = 23

… (i) Square the second equation:

{(x - y)}^{2} = {(\sqrt{3})}^{2}

x^{2} + y^{2} - 2 x y = 3

… (ii) Adding (i) and (ii):

2 (x^{2} + y^{2}) = 26 \Rightarrow x^{2} + y^{2} = 13

Subtracting (ii) from (i):

4 x y = 20 \Rightarrow x y = 5

\frac{x^{2} + y^{2}}{x y} = \frac{13}{5}

Hence option (c) is correct.

Simplify:

\frac{(\cos (y) + \sin (y))^{2}}{1 + \cos (y) \sin (y)}

A. $\frac{6 + 5 \cos (y) \sin (y)}{1 + \cos (y) \sin (y)}$
B. $\frac{7 + 8 \cos (y) \sin (y)}{3 + \cos (y) \sin (y)}$
C. $\frac{9 + 2 \cos (y) \sin (y)}{7 + \cos (y) \sin (y)}$
D. $\frac{1 + 2 \cos (y) \sin (y)}{1 + \cos (y) \sin (y)}$

Solution

Trigonometric identity Expand the numerator:

{(\cos y + \sin y)}^{2} = \cos^{2} y + \sin^{2} y + 2 \cos y \sin y

Using the identity

\sin^{2} y + \cos^{2} y = 1

, the numerator becomes

1 + 2 \cos y \sin y

. The denominator is left untouched, so the expression

= \frac{1 + 2 \cos y \sin y}{1 + \cos y \sin y}

This is exactly option (d). It does not cancel down to 1, because the numerator carries

2 \cos y \sin y

while the denominator carries only

\cos y \sin y

\cos A = \frac{12}{13}

, find

\cot A

A. $\frac{13}{7}$
B. $\frac{12}{5}$
C. $\frac{15}{8}$
D. $\frac{13}{9}$

Solution

Trigonometric ratios

\cos A = \frac{12}{13}

tells us that the side adjacent to A is 12 units and the hypotenuse is 13 units. By the Pythagoras theorem the opposite side

= \sqrt{13^{2} - 12^{2}} = \sqrt{169 - 144} = \sqrt{25} = 5

\cot A

is adjacent ÷ opposite, so

\cot A = \frac{12}{5}

Hence option (b) is correct.

In a right triangle PQR , right-angled at Q , if

\cot P = \sqrt{3}

, what is the value of 2 sin P cos P ?

A. $\frac{\sqrt{3}}{2}$
B. $\frac{5}{\sqrt{3}}$
C. $\frac{\sqrt{3}}{8}$
D. $\frac{2}{\sqrt{3}}$

Solution

Trigonometric ratios

\cot P = \sqrt{3}

and

\cot 30^{\circ} = \sqrt{3}

, so

∠ P = 30^{\circ}

Then

\sin P = \sin 30^{\circ} = \frac{1}{2}

and

\cos P = \cos 30^{\circ} = \frac{\sqrt{3}}{2}

2 \sin P \cos P = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}

The same value follows in one step from the double-angle form

2 \sin P \cos P = \sin 2 P = \sin 60^{\circ} = \frac{\sqrt{3}}{2}

Hence option (a) is correct.

The equation of a line is

6 x - 3 y =

12. What is the slope of a line perpendicular to this line?

A. $\frac{1}{2}$
B. $\frac{- 1}{2}$
C. $\frac{2}{3}$
D.2

Solution

Slope of perpendicular lines Put the line in the slope–intercept form y=mx+c.

6 x - 3 y = 12 \Rightarrow 3 y = 6 x - 12 \Rightarrow y = 2 x - 4

So the slope of the given line is

m_{1} = 2

. For two perpendicular lines,

m_{1} \times m_{2} = - 1

2 \times m_{2} = - 1 \Rightarrow m_{2} = - \frac{1}{2}

Hence the required slope is

- \frac{1}{2}

, so option (b) is correct. Option (d) 2 is the slope of the given line itself, not of the perpendicular.

Finished reading? Now test yourself.

The same 25 questions, but with a timer and the answers hidden — 15 minutes.

Take the test

Full contents list