CAT 2024 Slot 1 QA Questions with Answers & Solutions

Q47MCQPermutations & Combinations

Consider two sets

A = \{2, 3, 5, 7, 11, 13\}

and

B = \{1, 8, 27\}

. Let

f

be a function from

A

to

B

such that for every element

b

in

B

, there is at least one element

a

in

A

such that

f(a) = b

. Then, the total number of such functions

f

is

665
667
537
540

Answer and solution

Answer: (D) 540

Every element of

B

must be hit, so count the onto functions by inclusion–exclusion. All functions: each of the 6 elements of

A

can go to any of the 3 elements of

B

, giving

3^6 = 729

. Functions missing one given element of

B

map into the other 2:

2^6 = 64

. There are 3 choices of the missed element:

3 \times 64 = 192

. Functions missing two given elements send everything to the third:

1^6 = 1

. There are

\binom{3}{2} = 3

such pairs. Each of these constant functions was subtracted twice above but counted only once in 729, so add the 3 back. Onto functions

= 729 - 192 + 3 = 540

. Option C, 537, is

729 - 192

: it forgets to add back the 3 constant functions. Hence, option D (540).

Q48MCQQuadratic & Polynomial Equations

Let

x

,

y

, and

z

be real numbers satisfying

4(x^2 + y^2 + z^2) = a,

4(x - y - z) = 3 + a

Then

a

equals

3
$1\frac{1}{3}$
4
1

Answer and solution

Answer: (A) 3

Substitute

a = 4(x^2 + y^2 + z^2)

from the first equation into the second:

4(x - y - z) = 3 + 4x^2 + 4y^2 + 4z^2

4x^2 - 4x + 4y^2 + 4y + 4z^2 + 4z + 3 = 0

Write

3

as

1 + 1 + 1

and complete the squares:

(2x - 1)^2 + (2y + 1)^2 + (2z + 1)^2 = 0

A sum of squares of real numbers is 0 only if each square is 0, so

x = \frac{1}{2}

,

y = -\frac{1}{2}

,

z = -\frac{1}{2}

. Then

a = 4\left(\frac{1}{4} + \frac{1}{4} + \frac{1}{4}\right) = 3

. The second equation agrees:

4\left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2}\right) = 6 = 3 + a

. Since

x

,

y

,

z

are forced,

a

has no other value; option C, 4, would need

x^2 + y^2 + z^2 = 1

, but it is

\frac{3}{4}

. Hence, option A (3).

Q49TITAQuadratic & Polynomial Equations

If the equations

x^2 + mx + 9 = 0

,

x^2 + nx + 17 = 0

and

x^2 + (m+n)x + 35 = 0

have a common negative root, then the value of

(2m + 3n)

is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 38

Let

\alpha

be the common negative root. Subtract the first equation from the third:

(m+n)\alpha + 35 - m\alpha - 9 = 0

, so

n\alpha = -26

. Subtract the second equation from the third:

(m+n)\alpha + 35 - n\alpha - 17 = 0

, so

m\alpha = -18

. Put

m\alpha = -18

into the first equation:

\alpha^2 - 18 + 9 = 0

, so

\alpha^2 = 9

. The root is negative, so

\alpha = -3

. Then

m = \frac{-18}{-3} = 6

and

n = \frac{-26}{-3} = \frac{26}{3}

. Check in the second equation:

9 + \frac{26}{3}(-3) + 17 = 9 - 26 + 17 = 0

. So

2m + 3n = 12 + 26 = 38

. The answer is 38.

Q50MCQSequences & Series

Suppose

x_1, x_2, x_3, \ldots, x_{100}

are in arithmetic progression such that

x_5 = -4

and

2x_6 + 2x_9 = x_{11} + x_{13}

. Then,

x_{100}

equals

-194
-196
204
206

Answer and solution

Answer: (A) -194

Let the first term be

a

and the common difference

d

, so

x_n = a + (n-1)d

.

2x_6 + 2x_9 = x_{11} + x_{13}

gives

2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d)

. So

4a + 26d = 2a + 22d

, which gives

2a = -4d

, i.e.

a = -2d

.

x_5 = a + 4d = -4

gives

-2d + 4d = -4

, so

d = -2

and

a = 4

.

x_{100} = a + 99d = 4 - 198 = -194

. Option B,

-196

, comes from using

a + 100d

instead of

a + 99d

. Hence, option A (-194).

Q51TITATime & Work

Renu would take 15 days working 4 hours per day to complete a certain task whereas Seema would take 8 days working 5 hours per day to complete the same task. They decide to work together to complete this task. Seema agrees to work for double the number of hours per day as Renu, while Renu agrees to work for double the number of days as Seema. If Renu works 2 hours per day, then the number of days Seema will work, is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 6

Renu alone needs

15 \times 4 = 60

hours, so she does

\frac{1}{60}

of the task per hour. Seema alone needs

8 \times 5 = 40

hours, so she does

\frac{1}{40}

per hour. Let Seema work

d

days. Renu works 2 hours a day for

2d

days, that is

4d

hours. Seema works

2 \times 2 = 4

hours a day for

d

days, also

4d

hours. Together they complete the task:

\frac{4d}{60} + \frac{4d}{40} = 1

\frac{d}{15} + \frac{d}{10} = 1

, so

\frac{2d + 3d}{30} = 1

and

d = 6

. Seema works 6 days, and Renu 12 days. The answer is 6.

Q52MCQRemainders

When

10^{100}

is divided by 7, the remainder is

3
4
1
6

Answer and solution

Answer: (B) 4

Since

10 = 7 + 3

,

10^{100}

leaves the same remainder as

3^{100}

when divided by 7. Powers of 3 modulo 7:

3^1 \equiv 3

,

3^2 \equiv 2

,

3^3 \equiv 6

,

3^4 \equiv 4

,

3^5 \equiv 5

,

3^6 \equiv 1

. The remainders repeat every 6 powers.

100 = 6 \times 16 + 4

, so

3^{100} = (3^6)^{16} \times 3^4 \equiv 1^{16} \times 4 = 4 \pmod{7}

. So

10^{100}

leaves remainder 4. Option C, 1, would need the exponent to be a multiple of 6, and 100 is not. Hence, option B (4).

Q53MCQIndices & Surds

The sum of all real values of

k

for which

\left(\dfrac{1}{8}\right)^k \times \left(\dfrac{1}{32768}\right)^{\frac{1}{3}} = \dfrac{1}{8} \times \left(\dfrac{1}{32768}\right)^{\frac{1}{k}}

, is

2/3
4/3
-2/3
-4/3

Answer and solution

Answer: (C) -2/3

32768 = 2^{15} = 8^5

, so write every term as a power of

8

. Left side:

\left(\dfrac{1}{8}\right)^k \times \left(\dfrac{1}{32768}\right)^{1/3} = 8^{-k} \times 8^{-5/3} = 8^{-k - 5/3}

. Right side:

\dfrac{1}{8} \times \left(\dfrac{1}{32768}\right)^{1/k} = 8^{-1} \times 8^{-5/k} = 8^{-1 - 5/k}

. Equating exponents:

k + \dfrac{5}{3} = 1 + \dfrac{5}{k}

. Multiplying by

3k

(with

k \ne 0

):

3k^2 + 5k = 3k + 15

, so

3k^2 + 2k - 15 = 0

. Its discriminant is

2^2 + 4 \times 3 \times 15 = 184 > 0

, so both roots are real, and they are non-zero since their product is

-5

. Their sum is

-\dfrac{2}{3}

. Option A (

\dfrac{2}{3}

) has the wrong sign: the sum of the roots of

3k^2 + 2k - 15 = 0

is

-\dfrac{2}{3}

. Hence, option C (-2/3).

Q54TITAFunctions & Graphs

For any natural number

n

, let

a_n

be the largest integer not exceeding

\sqrt{n}

. Then the value of

a_1 + a_2 + \cdots + a_{50}

is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 217

a_n = \lfloor \sqrt{n} \rfloor

stays the same between consecutive perfect squares:

a_n = k

for

k^2 \le n \le (k+1)^2 - 1

, which is

2k + 1

values of

n

. So, up to

n = 50

:

a_n = 1

for

n = 1

to

3

(3 terms),

a_n = 2

for

n = 4

to

8

(5 terms),

a_n = 3

for

n = 9

to

15

(7 terms),

a_n = 4

for

n = 16

to

24

(9 terms),

a_n = 5

for

n = 25

to

35

(11 terms),

a_n = 6

for

n = 36

to

48

(13 terms), and

a_n = 7

for

n = 49, 50

(2 terms). Check:

3 + 5 + 7 + 9 + 11 + 13 + 2 = 50

terms. Sum

= 1 \times 3 + 2 \times 5 + 3 \times 7 + 4 \times 9 + 5 \times 11 + 6 \times 13 + 7 \times 2

= 3 + 10 + 21 + 36 + 55 + 78 + 14 = 217

. The answer is 217.

Q55MCQPercentages

In September, the incomes of Kamal, Amal and Vimal are in the ratio

8 : 6 : 5

. They rent a house together, and Kamal pays 15%, Amal pays 12% and Vimal pays 18% of their respective incomes to cover the total house rent in that month. In October, the house rent remains unchanged while their incomes increase by 10%, 12% and 15%, respectively. In October, the percentage of their total income that will be paid as house rent, is nearest to

15.18
13.26
14.84
12.75

Answer and solution

Answer: (B) 13.26

Take the September incomes as

80x

,

60x

and

50x

(ratio

8 : 6 : 5

). Rent paid: Kamal

15\%

of

80x = 12x

, Amal

12\%

of

60x = 7.2x

, Vimal

18\%

of

50x = 9x

. Total rent

= 28.2x

. October incomes: Kamal

1.10 \times 80x = 88x

, Amal

1.12 \times 60x = 67.2x

, Vimal

1.15 \times 50x = 57.5x

. Total

= 212.7x

. The rent is unchanged, so the share of income paid as rent is

\frac{28.2x}{212.7x} \approx 0.1326

, about

13.26\%

. Option C,

14.84

, is September's share,

\frac{28.2x}{190x} \approx 14.84\%

; it ignores the rise in incomes. Hence, option B (13.26).

Q56TITAPermutations & Combinations

The sum of all four-digit numbers that can be formed with the distinct non-zero digits

a

,

b

,

c

, and

d

, with each digit appearing exactly once in every number, is

153310 + n

, where

n

is a single digit natural number. Then, the value of

(a + b + c + d + n)

is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 31

The 4 distinct digits form

4! = 24

numbers. In each place, each digit appears

3! = 6

times, so the digits in any one place add up to

6(a + b + c + d)

. Sum of all the numbers

= 6(a + b + c + d)(1000 + 100 + 10 + 1) = 6666(a + b + c + d)

. This equals

153310 + n

with

n

from 1 to 9, so a multiple of 6666 must lie between 153311 and 153319.

6666 \times 23 = 153318

lies in that range, while

6666 \times 22 = 146652

and

6666 \times 24 = 159984

do not. So

a + b + c + d = 23

and

n = 8

. (A sum of 23 is possible with distinct non-zero digits, e.g. 9, 8, 5, 1.)

a + b + c + d + n = 23 + 8 = 31

. The answer is 31.

Q57TITATriangles & Lines

ABCD

is a rectangle with sides

AB = 56

cm and

BC = 45

cm, and

E

is the midpoint of side

CD

. Then, the length, in cm, of radius of incircle of

\triangle ADE

is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 10

E

is the midpoint of

CD

, and

CD = AB = 56

cm, so

DE = 28

cm. Also

AD = BC = 45

cm.

Solution figure for question 57, CAT 2024 Slot 1

Triangle

ADE

has a right angle at

D

, a corner of the rectangle, with legs

AD = 45

and

DE = 28

. Hypotenuse:

AE^2 = 45^2 + 28^2 = 2025 + 784 = 2809

, so

AE = 53

. For a right triangle with legs

p

,

q

and hypotenuse

h

, the inradius is

r = \frac{p + q - h}{2}

.

r = \frac{45 + 28 - 53}{2} = \frac{20}{2} = 10

. Check with area over semi-perimeter: area

= \frac{1}{2} \times 45 \times 28 = 630

, semi-perimeter

= \frac{45 + 28 + 53}{2} = 63

, and

\frac{630}{63} = 10

. The answer is 10.

Q58MCQCoordinate Geometry

In the

XY

-plane, the area, in sq. units, of the region defined by the inequalities

y \geq x + 4

and

-4 \leq x^2 + y^2 + 4(x - y) \leq 0

is

$2\pi$
$4\pi$
$\pi$
$3\pi$

Answer and solution

Answer: (A) $2\pi$

Complete the squares in the middle expression:

x^2 + y^2 + 4x - 4y = (x + 2)^2 + (y - 2)^2 - 8

. So

-4 \le (x + 2)^2 + (y - 2)^2 - 8 \le 0

means

4 \le (x + 2)^2 + (y - 2)^2 \le 8

. This is the ring between two circles centred at

(-2, 2)

with radii

2

and

2\sqrt{2}

.

Solution figure for question 58, CAT 2024 Slot 1

The line

y = x + 4

passes through the centre, since

2 = -2 + 4

. So it cuts the ring into two equal halves, and

y \ge x + 4

keeps the half above the line. Ring area

= \pi(2\sqrt{2})^2 - \pi(2)^2 = 8\pi - 4\pi = 4\pi

. Required area

= \frac{1}{2} \times 4\pi = 2\pi

. Option B,

4\pi

, is the whole ring; it ignores the condition

y \ge x + 4

. Hence, option A (

2\pi

).

Q59TITALogarithms

If

x

is a positive real number such that

4\log_{10} x + 4\log_{100} x + 8\log_{1000} x = 13

, then the greatest integer not exceeding

x

, is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 31

Change every log to base 10, using

\log_{10^k} x = \frac{1}{k}\log_{10} x

:

\log_{100} x = \frac{1}{2}\log_{10} x

and

\log_{1000} x = \frac{1}{3}\log_{10} x

. The equation becomes

4\log_{10} x + 2\log_{10} x + \frac{8}{3}\log_{10} x = 13

. So

\frac{26}{3}\log_{10} x = 13

, which gives

\log_{10} x = \frac{3}{2}

. Then

x = 10^{3/2} = \sqrt{1000} = 10\sqrt{10}

. Since

31^2 = 961 < 1000 < 1024 = 32^2

, we get

31 < x < 32

(

x \approx 31.62

). The greatest integer not exceeding

x

is 31. The answer is 31.

Q60MCQProfit, Loss & Discount

The selling price of a product is fixed to ensure 40% profit. If the product had cost 40% less and had been sold for 5 rupees less, then the resulting profit would have been 50%. The original selling price, in rupees, of the product is

15
14
10
20

Answer and solution

Answer: (B) 14

Let the original cost price be

c

. A

40\%

profit makes the selling price

1.4c

. The new cost price is

40\%

less,

0.6c

, and the new selling price is

1.4c - 5

. This gives a

50\%

profit, so

1.4c - 5 = 1.5 \times 0.6c = 0.9c

0.5c = 5

, so

c = 10

. Original selling price

= 1.4 \times 10 = 14

rupees. Check: new cost

6

, new price

14 - 5 = 9

, profit

\frac{3}{6} = 50\%

. Option C, 10, is the original cost price, not the selling price. Hence, option B (14).

Q61MCQAverages, Mixtures & Alligations

A glass is filled with milk. Two-thirds of its content is poured out and replaced with water. If this process of pouring out two-thirds the content and replacing with water is repeated three more times, then the final ratio of milk to water in the glass, is

$1 : 27$
$1 : 80$
$1 : 81$
$1 : 26$

Answer and solution

Answer: (B) $1 : 80$

Each time two-thirds of the contents is poured out, one-third of the milk present stays; the water only refills the glass. This happens once and then three more times, 4 times in all. Milk left

= \left(1 - \frac{2}{3}\right)^4 = \left(\frac{1}{3}\right)^4 = \frac{1}{81}

of the glass. Water

= 1 - \frac{1}{81} = \frac{80}{81}

of the glass. Milk : water

= \frac{1}{81} : \frac{80}{81} = 1 : 80

. Option C,

1 : 81

, compares the milk with the whole glass, not with the water. Hence, option B (

1 : 80

).

Q62TITARatios, Proportions & Partnership

A fruit seller has a total of 187 fruits consisting of apples, mangoes and oranges. The number of apples and mangoes are in the ratio

5 : 2

. After she sells 75 apples, 26 mangoes and half of the oranges, the ratio of number of unsold apples to number of unsold oranges becomes

3 : 2

. The total number of unsold fruits is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 66

Let the apples be

5x

and the mangoes

2x

. Then the oranges are

187 - 7x

. After the sales,

5x - 75

apples and

\frac{187 - 7x}{2}

oranges remain, in the ratio

3 : 2

:

\frac{5x - 75}{(187 - 7x)/2} = \frac{3}{2}

4(5x - 75) = 3(187 - 7x)

20x - 300 = 561 - 21x

, so

41x = 861

and

x = 21

. So there were 105 apples, 42 mangoes and

187 - 147 = 40

oranges. Unsold: apples

105 - 75 = 30

, mangoes

42 - 26 = 16

, oranges

40 \div 2 = 20

. Check:

30 : 20 = 3 : 2

. Total unsold

= 30 + 16 + 20 = 66

. The answer is 66.

Q63MCQTime, Speed & Distance

Two places

A

and

B

are 45 kms apart and connected by a straight road. Anil goes from

A

to

B

while Sunil goes from

B

to

A

. Starting at the same time, they cross each other in exactly 1 hour 30 minutes. If Anil reaches

B

exactly 1 hour 15 minutes after Sunil reaches

A

, the speed of Anil, in km per hour, is

18
16
14
12

Answer and solution

Answer: (D) 12

They meet after 90 minutes. Let Sunil need

t

more minutes to reach

A

. Anil reaches

B

75 minutes after Sunil reaches

A

, so Anil needs

t + 75

more minutes. After the meeting, each covers the stretch the other covered in 90 minutes. If their speeds are

v_A

and

v_S

, Anil takes

90\frac{v_S}{v_A}

minutes and Sunil

90\frac{v_A}{v_S}

, so the product of these times is

90^2

:

t(t + 75) = 8100

t^2 + 75t - 8100 = 0

, i.e.

(t - 60)(t + 135) = 0

, so

t = 60

. Anil needs

60 + 75 = 135

minutes after the meeting, so his total time is

90 + 135 = 225

minutes

= 3.75

hours. Anil's speed

= \frac{45}{3.75} = 12

km/h. Option A, 18, is Sunil's speed: his total time is

90 + 60 = 150

minutes, and

\frac{45}{2.5} = 18

km/h. Hence, option D (12).

Q64TITAAverages, Mixtures & Alligations

There are four numbers such that average of first two numbers is 1 more than the first number, average of first three numbers is 2 more than average of first two numbers, and average of first four numbers is 3 more than average of first three numbers. Then, the difference between the largest and the smallest numbers, is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 15

Call the numbers

a

,

b

,

c

,

d

in the given order.

\frac{a + b}{2} = a + 1

gives

b = a + 2

.

\frac{a + b + c}{3} = \frac{a + b}{2} + 2

. Multiply by 6:

2(a + b + c) = 3(a + b) + 12

, so

2c = a + b + 12 = 2a + 14

and

c = a + 7

.

\frac{a + b + c + d}{4} = \frac{a + b + c}{3} + 3

. Multiply by 12:

3(a + b + c + d) = 4(a + b + c) + 36

, so

3d = a + b + c + 36 = 3a + 45

and

d = a + 15

. The numbers are

a

,

a + 2

,

a + 7

,

a + 15

. Whatever

a

is, the smallest is

a

and the largest is

a + 15

. Their difference is 15. The answer is 15.

Q65MCQSimple & Compound Interest

An amount of Rs 10000 is deposited in bank

A

for a certain number of years at a simple interest of 5% per annum. On maturity, the total amount received is deposited in bank

B

for another 5 years at a simple interest of 6% per annum. If the interests received from bank

A

and bank

B

are in the ratio

10 : 13

, then the investment period, in years, in bank

A

is

4
5
3
6

Answer and solution

Answer: (D) 6

Let the money stay in bank

A

for

t

years. Interest from

A

:

10000 \times 0.05 \times t = 500t

. Amount at maturity:

10000 + 500t

. Interest from

B

:

(10000 + 500t) \times 0.06 \times 5 = 0.3(10000 + 500t) = 3000 + 150t

. The ratio is

10 : 13

:

\frac{500t}{3000 + 150t} = \frac{10}{13}

6500t = 30000 + 1500t

, so

5000t = 30000

and

t = 6

. Check: the interests are 3000 and

0.3 \times 13000 = 3900

, in the ratio

10 : 13

. Option B, 5, gives 2500 and

0.3 \times 12500 = 3750

, a ratio of

2 : 3

, not

10 : 13

. Hence, option D (6).

Q66MCQLinear Equations

A shop wants to sell a certain quantity (in kg) of grains. It sells half the quantity and an additional 3 kg of these grains to the first customer. Then, it sells half of the remaining quantity and an additional 3 kg of these grains to the second customer. Finally, when the shop sells half of the remaining quantity and an additional 3 kg of these grains to the third customer, there are no grains left. The initial quantity, in kg, of grains is

50
36
42
18

Answer and solution

Answer: (C) 42

Work backwards from the end. If the shop has

R

kg before a customer, it sells

\dfrac{R}{2} + 3

kg and keeps

\dfrac{R}{2} - 3

kg. After the third customer nothing is left:

\dfrac{R_2}{2} - 3 = 0

, so

R_2 = 6

kg before the third customer. Before the second customer:

\dfrac{R_1}{2} - 3 = 6

, so

R_1 = 18

kg. At the start:

\dfrac{N}{2} - 3 = 18

, so

N = 42

kg. Check: from

42

the shop sells

21 + 3 = 24

and keeps

18

; then sells

9 + 3 = 12

and keeps

6

; then sells

3 + 3 = 6

and keeps

0

. Option D (18) is the stock before the second customer, not the initial quantity. Option B (36) fails the check: it leaves

15

, then

4.5

, and the third customer would need

5.25

kg. Hence, option C (42).

Q67MCQIndices & Surds

If

(a + b\sqrt{n})

is the positive square root of

(29 - 12\sqrt{5})

, where

a

and

b

are integers, and

n

is a natural number, then the maximum possible value of

(a + b + n)

is

18
22
4
6

Answer and solution

Answer: (A) 18

Try to write

29 - 12\sqrt{5}

as a perfect square:

(3 - 2\sqrt{5})^2 = 9 + 20 - 12\sqrt{5} = 29 - 12\sqrt{5}

. Since

2\sqrt{5} \approx 4.47 > 3

, the positive square root is

2\sqrt{5} - 3

. So

a + b\sqrt{n} = -3 + 2\sqrt{5}

. The rational parts must match, so

a = -3

and

b\sqrt{n} = 2\sqrt{5} = \sqrt{20}

. With

b

an integer and

n

natural, this allows

b = 2, n = 5

, giving

a + b + n = 4

, or

b = 1, n = 20

, giving

a + b + n = 18

. The maximum is 18. The option 4 is the trap of stopping at the first form. Hence, option A (18).

Q68MCQMensuration

The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is

$1125\pi$
$750\pi$
$1125\pi\sqrt{2}$
$750\pi\sqrt{2}$

Answer and solution

Answer: (C) $1125\pi\sqrt{2}$

Let the box have edges

l

,

b

,

h

. Edges:

4(l + b + h) = 144

, so

l + b + h = 36

. Surface area:

2(lb + bh + hl) = 846

. So

l^2 + b^2 + h^2 = (l + b + h)^2 - 2(lb + bh + hl) = 1296 - 846 = 450

.

Solution figure for question 68, CAT 2024 Slot 1

The box is inscribed in the sphere, so its space diagonal is a diameter:

\sqrt{l^2 + b^2 + h^2} = 2R

. Then

4R^2 = 450

,

R^2 = \frac{225}{2}

and

R = \frac{15}{\sqrt{2}}

. Volume

= \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \times \frac{3375}{2\sqrt{2}} = \frac{2250\pi}{\sqrt{2}} = 1125\pi\sqrt{2}

. Option A,

1125\pi

, drops the factor

\sqrt{2}

; it stays because

R = \frac{15}{\sqrt{2}}

is irrational, so

R^3

carries it. Hence, option C (

1125\pi\sqrt{2}

).

CAT 2024 Slot 1 — QA questions with answers

Quantitative Ability

Q47MCQPermutations & Combinations

Q48MCQQuadratic & Polynomial Equations

Q49TITAQuadratic & Polynomial Equations

Q50MCQSequences & Series

Q51TITATime & Work

Q52MCQRemainders

Q53MCQIndices & Surds

Q54TITAFunctions & Graphs

Q55MCQPercentages

Q56TITAPermutations & Combinations

Q57TITATriangles & Lines

Q58MCQCoordinate Geometry

Q59TITALogarithms

Q60MCQProfit, Loss & Discount

Q61MCQAverages, Mixtures & Alligations

Q62TITARatios, Proportions & Partnership

Q63MCQTime, Speed & Distance

Q64TITAAverages, Mixtures & Alligations

Q65MCQSimple & Compound Interest

Q66MCQLinear Equations

Q67MCQIndices & Surds

Q68MCQMensuration