CAT 2020 Question Paper | CAT QA
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Anil, Sunil and Ravi start from the same point at the same time.

Anil and Sunil will first meet each other at the starting point

Hence, option 1.

Assume that ‘x’ litres of solution A is mixed with ‘3x’ litres of solution B initially.

Later solution A isadded to double the quantity of total liquid, i.e., (x + 3x) = 4x litres more solution A is added.

Thusthe resultant solution has ‘5x’ litres of solution A and ‘3x’ litres of solution B.

Let the percentage of alcohol in solution B be ‘y’.

By the given condition,

∴ 192 = 100 + y

∴ y = 92%

 Hence, option 3.

14a = 36b = 84c ⇒ log 14a = log 36b = log 84c

 ⇒ a × log 14 = b × log 36 = c × log 84 = k (assume)

Answer: 3.

4²⁰¹⁷= 2⁴⁰³⁴

We know that ab = 2⁴⁰³⁴

As a, b are positive integers, a, b ∈{2⁰, 2¹, 2²,…, 2⁴⁰³², 2⁴⁰³³, 2⁴⁰³⁴}

If a = b, then a = b = 2²⁰¹⁷

As a ≤ b, a can take following values: {2⁰, 2¹, 2²,…, 2²⁰¹⁶, 2²⁰¹⁷} and b can take {2²⁰¹7, 2²⁰¹⁸, 2²⁰¹⁹, 2²⁰²⁰, …, 2⁴⁰³⁴}

Thus, there are 2018 pairs (a, b) of positive integers are there such that a ≤ b and ab = 4²⁰¹⁷.

Hence, option 1.

2 < x < 10 and 14 < y < 23 ⇒ 3 ≤ x ≤ 9 and 15 ≤ y ≤ 22

As N = x + y, 3 + 15 ≤ x + y ≤ 9 + 22 ⇒ 18 ≤ N ≤ 31

As N > 25, N = 26, 27, 28, 29, 30, 31

Thus, there are 6 distinct values are possible for N.

Answer: 6.

x² + mx + 2n = 0 and x² + 2nx + m = 0 have real roots.

Therefore m² + 4 × 1× (2n) ≥ 0 and (2n)² + 4 × 1 × m ≥ 0

i.e., m² ≥ 8n and n² ≥ m

n² ≥ m ⇒ n⁴ ≥ m²

As m² ≥ 8n, n⁴ ≥ 8n ⇒ n³ ≥ 8 ⇒ n ≥ 2

m² ≥ 8n ≥ 8 × 2 ⇒ m ≥ 4 ....(m is a positive integer)

∴ the smallest possible value of m + n = 4 + 2 = 6

 Hence, option 1.

Consider y = |x – 2| + 4

Case (i) x ≤ 2

For x = 2, y = 4 and for x = 0, y = 6.

So, y = |x – 2| + 4 is a line that passes through (2, 4) and (0, 6).

Case (ii) x ≥ 2

For x = 2, y = 4 and for x = 3, y = 5.

So, y = |x – 2| + 4 is a line that passes through (2, 4) and (3, 5).

As we need to find area enclosed by x = 2, y = |x – 2| + 4, the x-axis and the y-axis, we need notconsider case (ii) above.

The required area is area of the trapezium ABCO = 1/2 × (AO + BC) × OC = 1/2 × (6 + 4) × 2 = 10 sq. units

 Hence, option 1.

Let Tom's age be 'x' years.

 Therefore Dick's age = (3x) years and Harry's age = (6x) years

 By the given condition,

Solving this, we get x =3

Harry's age = 6x = 18 years

Answer: 18.

The diagonals of parallelogram bisect each other.

 Let A(x, y) be the point of intersection of the diagonals of the parallelogram.

Hence, option 4.

Let the time taken by Vimla to reach office at 40 km/hr be ‘t’ hours.

∴ Distance = 40t

Given: She is late by 6 minutes if she drives at 35 km/hr.

Therefore time taken by her at 35 km/hr

Therefore 40t = 35(t + 0.1) ⇒ 5t = 3.5 ⇒ t = 0.7 hours

∴ Distance = 28 km

Given: One day, she covers two-thirds of her distance to office in one-thirds of her usual time to reach office, and then stops for 8 minutes.

Therefore she has to cover the remaining distance

= 28 km/hr

Hence, option 2.

Note that every option has term ( A + B).

∴ ( A + B ) = log_a30 – log_a(5/3) = loga[30/(5/3)] = log_a18

= log_a(3² × 2) = 2log_a3 + log_a2

As log₂a = 1/3 ⇒  log_a2 = 3, (A + B) = 2log_a3 + 3

∴  2log_a3 = A + B – 3

Hence, option 3.

The two line will have unique solution if the lines are not parallel to each other. In other words, the slopes of the two lines are different.

Hence, option 3.

Let the cost price, CP, be Rs. 'x' per kg.

∴ the marked price = Rs. (1.2x)

The man sold 5 kg sugar at Rs. (1.2x) per kg.

The rate at which the man sold 15 kg sugar (i.e., at 10% discount) = 0.9 × (1.2x) = Rs. (1.08x)

He sells remaining sugar i.e., 35 – 5 – 15 – 3 = 12 kg by raising

By the given condition,

∴ p ≈ 25%

Hence, option 2.

f(10) = f(5 + 5) = f(5) f(5) = 4 × 4 = 16

f(20) = f(10 + 10) = f(10) f(10) = 16 × 16 = 256

f(10) = f(20 – 10) = f(20 + (–10)) f(20) × f(–10) ⇒ 16 = 256 × f(–10)

Hence, option 2.

n(A) = n(Set of natural numbers less then or equal to 120 and are divisible by 2) = 120/2 = 60

n(B) = n(Set of natural numbers less then or equal to 120 and are divisible by 5) = 120/5 = 24

n(C) = n(Set of natural numbers less then or equal to 120 and are divisible by 7) = 120/7 = 17

n(A â‹‚ B) = 120/10 = 12, n(A â‹‚ C) = 120/14 = 8, n(C â‹‚ B) = 120/35 = 3

n(A â‹‚ B â‹‚ C) = 120/70 = 1

Using this information, we can draw the folloing diagram.

∴ n(A ⋃ B ⋃ C) = 79

Among the given integers, there are 120 – 79 = 41 integers divisible by none 2, 5 and 7.

Hence option 1.

As n is even natural number, n = 4

⇒ 20 > m > 8 ⇒ m can take following values: 9, 10, ..., 19

⇒ m can take following values : 5, 6, 7, 8, 9

From (i) and (ii), m = 9 

Therefore, m – 2n = 9 – 2(4) = 1

Hence, option 2.

x_m+1 = x_m – (m + 1)

x₁₊₁ = x₁ – (1 + 1) ⇒ x₂ = x₁ – (1 + 1) = x₁ + 2 = –1 –2 = –3

x₃ = x₂ – (2 + 1) = x₂ −3 = −1 −2 −3

x₄ = x₃ − (3 + 1) = x₃ −4 = −1 −2 −3 –4

.... and so on.

Hence, option 3.

Construction: DE ⊥ AB.

 As AB is parallel to DC, BC is perpendicular to DC, DE = BC = 4 cm. Also, EB = 5 cm

Consider âˆ†AED. m∠AED = 90° and ED = 4 cm

m∠BAD = m∠ADE = 45° ⇒ AE = ED 4 cm

Therefore, AB = 4 + 5 = 9 cm

Answer: 28.

The set {100, 101, 102, ..., 999} has 999 - 99 = 900 integers.

The number of integers having all distinct digits:

The hundreds digit can be selected using 9 non-zero digits in 9 ways.

The tens place can be selected using remaining 9 digits in 9 ways.

the units place can be selected using remaining 8 digits in 8 ways.

∴ The number of integers having all distinct digits = 9 × 9 × 8 = 648

∴ The integers in the set {100, 101, 102, ..., 999} have at least one digit repeated = 900 - 648 = 252

Answer: 252

Total runs scored in (n + 2) innings = Total runs scored in first n innings + 38 +15

∴ 29 (n + 2) = Total runs scored in first n innings + 53 = 30n + 53

∴ n = 5

Total runs scored  in first  5 innings = 30 × 5 = 150

To minimize the value of ‘x’, he must have scored 37 runs in maximum number of innings.

150 = x + 4 × 37

∴ x = 2

Hence, option 2.

Answer: 24.

N = 1.5 x 2 = 3

Answer: 16000.

140 workers completed 1.5 km working for 60 days.

Therefore, 140 × 60 = 8400 man-days are required to construct 1.5 km road.

To construct remaining part of the road i.e., 4.5 km road, man-days required = 8400 × 3

The days remaining to complete the construction in time = 200 – 60 = 140

∴ the number of persons should be working for these 140 days

The contractor already has 180 persons. Therefore, he need to employ 180 – 140 = 40 persons inorder to finish the work exactly on time.

Answer: 40.

Let the maximum possible marks be (100x).

Bishnu and Asha and Geeta scored (52x), (64x) and (84x) respectively.

Let the marks obtained by Ramesh be (y).

By the given condition,

52x + 23 = y

64x – y = 34

64x – 52x = 34 + 23 ⇒ 12x = 57

The marks obtained by Geeta = 84x = 7 × 12x = 7 × 57 = 399

Hence, option 3.

A(0, 0), B(4, 0) and C(3, 9) are the vertices of the triangle.

Note that A and B are on X axis. Therefore AB = 4 units

Now height of ΔABC = distance between X axis and C(3, 9) = 9 units

A(ΔABC) = ½ × 4 × 9 = 18 sq. units

Hence, option 2.

The distance between the two trains at 10:30am = 90 – 60 = 30 km

The time required for the trains to meet each other (after 10:30 am)

Therefore, the trains will meet 11:00 am.

Hence, option 4.