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CAT 2024 Question Paper Slot 1 | All Questions

Previous Year Questions

01.

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

A.
750π
B.
1125π√2
C.
1125π
D.
750π√2

Option B is the correct answer.

Video Explanation

Explanatory Answer

Given: A closed rectangular box

Step 1: Surface area of the box
2(lb + lh + bh) = 846

Step 2: Sum of the lengths of all edges
4(l + b + h) = 144
=> l + b + h = 144 / 4 = 36 …(1)

Step 3: Square equation (1)
(l + b + h)^2 = l^2 + b^2 + h^2 + 2(lb + lh + bh) = 36^2 = 1296
l^2 + b^2 + h^2 = 1296 − 846 = 450

Step 4: Diameter of the sphere (space diagonal of the box)
d = √(l^2 + b^2 + h^2) = √450 = 15√2 = 2r

Step 5: Radius of the sphere
r = (15√2) / 2 = 15/√2

Step 6: Volume of the sphere
V = (4/3) π r^3 = (4/3) π (15√2 / 2)^3
V = (4/3) π * 1125√2 / 8
V = 1125 π √2 / 6
V = 1125 π √2

02.

If (a + b√n) is the positive square root of (29 − 12√5),
where a and b are integers, and n is a natural number,

Then, find the maximum possible value of (a + b + n).

A.
18
B.
22
C.
4
D.
6

Option A is the correct answer.

Video Explanation

Explanatory Answer

Given: √(29 − 12√5) = a + b√n

Step 1: Square both sides
(a + b√n)^2 = 29 − 12√5
a^2 + n b^2 + 2ab√n = 29 − 12√5

Step 2: Equate rational and irrational parts
Rational part: a^2 + n b^2 = 29
Irrational part: 2ab√n = −12√5
=> ab√n = −6√5
=> n = 5, ab = −6

Step 3: Solve a^2 + 5b^2 = 29 with ab = −6
Possible values: (a = 3, b = −2) or (a = −3, b = 2)

Step 4: Identify positive and negative roots
Positive root: −3 + 2√5
Negative root: 3 − 2√5

Step 5: To maximize a + b + n, modify the expression:
Take a = −3, b = 1, n = 20

Step 6: Compute maximum possible value
a + b + n = −3 + 1 + 20 = 18

Answer: Maximum possible value of (a + b + n) = 18

03.

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

A.
50
B.
36
C.
42
D.
18

Option C is the correct answer.

Video Explanation

Explanatory Answer

Let the initial quantity of grains be x kg.

Step 1: First customer
Grains sold = (1/2)x + 3
Remaining grains = x - [(1/2)x + 3] = (1/2)x - 3

Step 2: Second customer
Grains sold = (1/2)((1/2)x - 3) + 3 = (1/4)x - 3/2 + 3 = (1/4)x + 3/2
Remaining grains = ((1/2)x - 3) - ((1/4)x + 3/2) = (1/4)x - 9/2

Step 3: Third customer
Grains sold = (1/2)((1/4)x - 9/2) + 3 = (1/8)x - 9/4 + 3 = (1/8)x + 3/4

Remaining grains = ((1/4)x - 9/2) - ((1/8)x + 3/4) = 0

Step 4: Solve for x
(1/4)x - 9/2 - 1/8 x - 3/4 = 0
(1/8)x - 21/4 = 0
(1/8)x = 21/4
x = (21/4) * 8 = 42

Answer: Initial quantity of grains = 42 kg

04.

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

A.
4
B.
5
C.
3
D.
6

Option D is the correct answer.

Video Explanation

Explanatory Answer

Interest from bank A = 10000*n*5 / 100 = 500n ( n is the Investment period, in years, in Bank A)
Interest from bank B = (10000*6*5/ 100) + (500n*6*5/100) = 3000 + 150n

Ratio of interest from Bank A and Bank B= 500n/ 3000 + 150n = 10/13
6500n = 30000 + 1500n
5000n = 30000
N = 30000/5000 = 6

Investment period, in years, in Bank A = 6

05.

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

Answer : 15

Video Explanation

Explanatory Answer

Average of first two numbers is 1 more than first number.
Lets take first number = a
Second number = a+2

Average of first three numbers is 2 more than the average of first two numbers which is a+3.
Therefore 3rd number is a+7

Average of first four numbers is 3 more than average of first three numbers which is a+6
Therefore the 4th number is a+15

The difference between the largest and smallest numbers is a+15 - a = 15

06.

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

A.
18
B.
16
C.
14
D.
12

Option D is the correct answer.

Video Explanation

Explanatory Answer

Ratio of the distances, $90 / T = (T + 75) / 90$

T 2 + 75 T = 8100 T 2 + 75 T - 8100 = 0 (T + 135) (T - 60) = 0 T = 60 mins

So, Anil takes $90 + 60 + 75 = 225 m i n s = 3 h r 45 m i n = 3 (3 / 4) = 15 / 4 h r$
Speed of Anil in Km/hr $= 45 / (15 / 4) = 45 * 4 / 15 = 12 k m / h r$

07.

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

Answer : 66

Video Explanation

Explanatory Answer

atio of number of apples and mangoes is $5 : 2$
Let the Number of apples $= 5 x$
Number of mangoes $= 2 x$
Number of oranges $= 2 y$

7 x + 2 y = 187 - - - - - - - - (1)

She sells 75 apples, 26 mangoes and half of the oranges.
Number of unsold apples= $5 x - 75$
Number of unsold mangoes $= 2 x - 26$
Number of unsold oranges $= y$
Ratio of unsold apples to unsold oranges=3:2

5 x - 75 : y = 3 : 2 10 x - 150 = 3 y

10 x - 3 y = 150 - - - - - - (2)

Multiplying equation(1) with 3 , we get $21 x + 6 y = 561$
Multiplying equation(2) with 2 , we get $20 x - 6 y = 300$
Adding the above two equations, $41 x = 861$

x = 861 / 41 = 21

Substituting $x = 21$ in equation 1 ,

7 (21) + 2 y = 187

2 y = 40 y = 20

Total number of unsold fruits $= 5 x - 75 + 2 x - 26 + y = 105 - 75 + 42 - 26 + 20 = 66$

08.

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

A.
1 : 27
B.
1 : 80
C.
1 : 81
D.
1 : 26

Option B is the correct answer.

Video Explanation

Explanatory Answer

Final concentration of milk $=$ Initial concentration * $(1 - (x / v))^{n}$
$x =$ volume replaced
$v =$ total volume
$n =$ number of iterations
Let initial quantity of milk $= y$ litres
Final concentration $= y (1 - 2 / 3)^{4} = y (1 / 3)^{4} = 1 / 81 y$
Volume of milk $= 1 / 81 y$
Volume of water $= y - (1 / 81 y) = 80 / 81 y$
Final ratio of milk to water in the glass $= 1 : 80$

09.

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

A.
15
B.
14
C.
10
D.
20

Option B is the correct answer.

Video Explanation

Explanatory Answer

Let CP=xSP=1.4x New CP=0.6x New SP=1.4x−5 New Profit %=50%1.4x−5=0.6x∗(3/2)1.4x−5=0.9x0.5x=5x=10 New SP=1.4(10)=14

10.

If x is a positive real number such that:
4 log₁₀(x) + 4 log₁₀₀(x) + 8 log₁₀₀₀(x) = 13

Then, find the greatest integer not exceeding x.

Answer : 31

Video Explanation

Explanatory Answer

4 (log x / log 10) + 4 (log x / 2 log 10) + 8 (log x / 3 log 10) = 13 Simplifying: 4 log x + 2 log x + (8/3) log x = 13 (4 + 2 + 8/3) log x = 13 (26/3) log x = 13 log x = 13 * 3 / 26 log x = 3/2 x = 10^(3/2) x = \sqrt1000 \approx 31.62 The greatest integer not exceeding x = 31

Previous year papers

2024

CAT 2024 Slot 1

CAT 2024 Slot 2

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2023

CAT 2023 Slot 1

CAT 2023 Slot 2

CAT 2023 Slot 3

2022

CAT 2022 Slot 1

CAT 2022 Slot 2

CAT 2022 Slot 3

2021

CAT 2021 Slot 1

CAT 2021 Slot 2

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2020

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2019

CAT 2019 Slot 1

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2018

CAT 2018 Slot 1

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