CAT 2022 Question Paper | CAT QA
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Let ‘a’ be the number of students who join the class. Let ‘b’ kg be the average weight of the original number of students.

Let there be ‘m’ students in the class initially. 

By the given condition;

∴ mb + 0.6m + ab + 0.6a = mb + ab + 3a

∴ 0.6m + 0.6a = 3a

∴ 0.6m = 2.4a

∴ m : a = 4 : 1 

 Hence, option 4.

a² + ab + a = 14 and b² + ab + b = 28
 
∴ a² + b² + 2ab + a + b = 42 ⇒ (a + b)² + (a + b) = 42 ⇒ (a + b)(a + b + 1) = 42 ⇒ a + b = 6 and a + b + 1 = 7
 
a² + ab + a = 14 ⇒ a(a + b) + a = 14 ⇒ 6a + a = 14 ⇒ a = 2 and hence b = 4
 
∴ 2a + b = 8
 
Hence, option 3

Let x, y and z be the number of students who like exactly one, exactly two and all the three drinks respectively. Let n students do not like any drink.

∴ x + y + z + n = 100

∴ 100 – n = 73 + 80 + 52 – y – 2z ⇒ y + 2z – n = 105 ⇒ y – n = 105 – 2z

Thus, (n – y) has to be odd ⇒ Either n or y is even and the other is odd.

At the most 52 students may like all the three drinks. Note that z ≠ 0.

For z = 1, 2 and 3, (y – n) = 103, 101, 99. But this is not possible as which is not possible as y, n < 100

If z = 4, y – n = 98, which is not possible as y cannot be more than 96.

For z = 5, y – n = 95. In this case, n = 0 and y = 95.

Thus the minimum possible value of z can be 5.

Therefore, the required difference = 52 – 5 = 47. 

Hence, option 2

Let there be (3x) persons standing ahead of Pinky.

Therefore, (5x) persons are behind her.

Total number of persons in the queue = 3x + 1 + 5x = 8x + 1

8x + 1 < 300

∴ x < 37.5 ⇒ x ≤ 37

The maximum number of persons ahead of Pinky ≤ 3 × 37 = 111

Therefore, the required answer is 111.

Answer: 111

For n < 20, < 1

For n = 20, = 1

= 1

For 20 < n < 44, < 2

= 1 + (1 × 24) = 25

Therefore, the required answer is 44.

Answer: 44

Assume that 4 children get (2a), (2b), (2c) and (2d) balloons.

∴ 2a + 2b + 2c + 2d = 20

∴ a + b + c + d = 10

As a, b, c and d ≠ 0, the required number of ways = ^{(10 – 1)}C_{(4 – 1)} = ⁹C₃ = 84

Therefore, the required answer is 84.

Answer: 84

Assume that the two trains meet each other after ‘t’ minutes.

Let the two trains meet each other at P (X-P-Y). Assume that train A and B travel with speeds a and b respectively.

To travel XP, train A takes t minutes and train B takes 9 minutes. To travel PY, train A takes (10 – t) minutes and train B takes t minutes

∴ = =

∴ t² = 90 – 9t

Solving this, we get t = 6 or –15

The value of t cannot be negative. So t = 6

Therefore the total time taken by train B = 6 + 9 = 15 minutes

Hence, option 2.

Let the three integers be a, b and c such that

a + b + c = 3 × 13 = 39
 
a + b + c + n = 4 × odd number

∴ 39 + n = even number ⇒ n = odd
 
For n = 1 or 3, the average of a, b, c and n will be 10 and 10.5 respectively.
 
For n = 5, the average of a, b, c and n will be 11.
 
Therefore, the minimum possible value of n = 5.

Hence, option 3.

Let there be (5x) males and (4x) females. 
 
The ratio of number of literate males to literate females is 2 : 3
 
3600 males are literate ⇒ 5400 females are literate.
 
∴ the number of illiterate males = (5x – 3600) and the number of illiterate females = (4x – 5400) 
 
∴ (5x – 3600) : (4x – 5400) = 4 : 3
 
Solving this, we get x = 10800 
 
Thus, the total number of females in the village = 3 × 10800 = 43200 
 
Therefore, the required answer is 43200. 
 

Answer: 43200

Let (2x) be the length and (2y) be the breadth of the rectangle.
 
Therefore, area of the rectangle = 4xy 
 
Also, R² = x² + y² 
 
It is given that P = 2(2x + 2y) = 4(x + y)
 
∴ P² = 16(x² + y² + 2xy)
 
∴ 32xy = P² – 16(x² + y²) = P² – 16R²
 
∴ 4xy =  – 2R² 
 

Hence, option 2.

2 litres of original mixture has 1 litre lemon juice and 1 litre sugar syrup.

2 litres of the original mixture is mixed with 6 litres of sugar syrup.

This 8 litres of new mixture will have 1 litre lemon juice and (1 + 6 =) 7 litres of sugar syrup. 

Therefore the ratio of lemon juice and sugar syrup in the new mixture is 1 : 7.  

Hence, option 3.

f(x) = |x – a| + |x – 100| + |x – a – 50| = (100 – a) + |x – a – 50|
 
For a = 50,  f(x) = (100 – a) + |x – a – 50| = 50 + |x – 100|. In this case the function has maximum value for x = 50. The maximum value = 100.
 
For a = 25,  f(x) = (100 – a) + |x – a – 50| = 75 + |x – 75|. In this case the function has maximum value for x = 25. The maximum value = 125. Hence, rejected.
 
For a = 100,  f(x) = (100 – a) + |x – a – 50| = 0 + |x – 150|. In this case the function has maximum value for x = 100. The maximum value = 50. Hence, rejected.
 
For a = 0,  f(x) = (100 – a) + |x – a – 50| = 100 + |x – 50|. In this case, the maximum value > 100. Hence, rejected.
 
Hence, option 1

S_n = n + 2n², S_{(n – 1)} = (n – 1) + 2(n – 1)² = n – 1 + 2n² – 4n + 2 = 2n² – 3n + 1

 
∴ T_n = S_n – S_{(n – 1)} = 4n – 1
 
As T_n is divisible by 9, n = 7. 

Hence, the correct answer is option 2.

Let the CP of cashews, peanuts and almonds be a, b and c per kg respectively. 

The cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds ⇒ 7a = 30b = 9c

Therefore, a : b : c = 90 : 21 : 70

Cost price of the mixture = 4 × 90k + 14 × 21k + 6 × 70k = ₹ (1074k), for some k.

∴ CP =  = ₹  (44.75k) per kg 

Ankita wants to earn profit of  = ₹ 73 per kg. 

She must have marked the price of the mixture = ₹ (44.75k + 73) per kg 

∴ 4 × (44.75k + 73) + 20 × [0.8 × (44.75k + 73)] – 1074k = 744

∴ 179k + 292 + 716k + 1168 – 1074k = 744

∴ k = 4 

The amount, in rupees, that she had spent in buying almonds = 6 × 70 × 4 = ₹ 1680

Hence, option 1.

3^k + 4^k + 5^k is an even number.
 
For k = 1, 3^k + 4^k + 5^k = 12 and for k = 2, 3^k + 4^k + 5^k = 50. HCF of 12 and 50 is 2. Thus, A = 2
 
4^k + 3(4^k) + 4^{k + 2} = (1 + 3)4^k + 4^{k + 2} = (4 + 16) × 4^k = 20 × 4^k
 
Thus, 20 × 4 = 80 divides this number when k = 1. Also, 20 divides all numbers of the form 4^k + 3(4^k) + 4^{k + 2}. Thus, B = 80
 
Therefore, (A + B) = 82
 
Therefore, the required answer is 82. 

Answer: 82

Let the cost price of juice be (100a) per kg. Therefore, the cost price of syrup is (80a) per kg.
 
∴ 10 × 1.1 × (80a) + 20 × 1.2 × (100a) + 200 × 308.32 = [(110 × 80a) + (120 × 100a)] × 1.64
 
∴ 880a + 2400a + 200 × 308.32 = 34112a
 
∴ 3280a + 200 × 308.32 = 30832a
 
Solving this, we get a = 2.
 
Amal’s cost price for syrup = 80 × 2 = 160
 
Therefore, the required answer is 160. 

Answer: 160

Assume that Alex invested ‘a’ at 15% for 4 years and ‘b’ at 12% for 3 years.
 ∴ a × 0.15 × 4 = b × 0.12 × 3

∴ the required percentage

Hence, option 2.

3 + 8 + x + y = 36 ⇒ x + y = 25
As ∆CPD is a right triangle,

Solving this, we get x = 13 and hence, y = 12.

A(â–¡ABCD) = 1/2 × (3 + 8) × 12 = 66 sq.cm

Therefore, the required answer is 66. 

 Answer: 66

Let (x, y) are the coordinates of D.

As ABCD is a parallelogram,

Sum of X-coordinates of A and C = Sum of X-coordinates of B and D

Sum of Y-coordinates of A and C = Sum of Y-coordinates of B and D

1 + (–2) = 3 + x ⇒ x = –4

1 + 8 = 4 + y ⇒ y = 5

Thus, the coordinates of the vertex D are (–4, 5).

Hence, the correct answer is option 1

b² < 4ac ⇒ the roots are imaginary.

 
Therefore, the graph will not intersect X-axis, i.e., the graph will always be above X-axis or below X-axis.
 
If the graph is above the X-axis, the value of the function will always be positive and hence, for no real value m, f(m) < 0. In this case, S will be an empty set.
 
If the graph is below X-axis, the value of the function will always be negative and hence, for every real value m, f(m) < 0. In this case, S will have all integers. 
 

Hence, the correct answer is option 4

y(x + z) = 19 and z(x + y) = 51

xz – xy = 32 ⇒ x(z – y) = 32 

xy + yz = 19 ⇒ y(x + z) = 19 ⇒ y = 1 or 19. As x, y and z are natural numbers; y cannot be 19. Therefore, y = 1. Therefore, (x + z) = 19 and hence, either x or z is odd and the other is even. 

yz + xz = 51 ⇒ z(x + 1) = 51 ⇒ z = 1, 3, 17 or 51. We know that y and z are odd, x has to be even. As x + z = 19, z ≠ 51. 

y(x + z) = 19 and z(x + y) = 51

xz – xy = 32 ⇒ x(z – y) = 32 ⇒ x(z – 1) = 32. Therefore, z = 3 or 17 and hence, x = 16 or 2. 

Thus, for (x, y, z) = (16, 1, 3) ⇒ xyz = 48

For (x, y, z) = (2, 1, 17) ⇒ xyz = 34 

Thus, the minimum possible value of xyz is 34. 

Therefore, the required answer is 34.

|x + a| is the distance of (–a) from x and |x – 1| is the distance of 1 from x.
 
|x + a| + |x – 1| = 2 is the sum of the distance of x from 1 and (–a) is 2.
 
This is possible if –a = –1 or 3 i.e., a = 1 or –3.
 
Hence, option 4