CAT 2020 QA Slot 3 | Previous Year CAT Paper

Answer: 252

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Here we have to find all 3-digit numbers which have at least one digit repeated.

This can also be calculated by subtracting all 3-digit number which do not have any repeated digits from total 3-digit numbers.

Total 3-digit numbers = 9 × 10 × 10 = 900

3-digit numbers without repetition,

​​​​​​​

∴ The number 3-digit numbers without repetition = 9 × 9 × 8 = 648

Therefore, the number of 3-digit numbers which have at least one digit repeated 900 – 648 = 252.

Hence, 252.

Answer: Option B

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Train A starts at 9 a.m. whereas Train B starts at 10:30 a.m.

Distance travelled by A till 10:30 a.m. = 40 × 1.5 = 60 kms.

At 10:30 a.m. remaining distance between A and B = 90 – 60 = 30 kms.

Time taken for both of them to meet now = 30/(40 + 20) = 0.5 hours i.e., 30 mins.

∴ Both trains meet half after 10:30 a.m. i.e., at 11 a.m.

Hence, option (b).

Concept:

Two People Moving Between Two Points1

Answer: Option D

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Average score in n innings is 30 and in n + 2 innings is 29. Score in last 2 innings is 38 and 15

∴ 30n + 38 + 15 = 29(n + 2)

⇒ n = 5

∴ Total runs score in first n innings = 5 × 30 = 150

Also, he scored less than 38 runs in each of these 5 innings.

To calculate the lowest possible score in an innings we try to maximize the score of other 4 innings. Maximum runs score in each of the 4 innings can be 37.

∴ x + 4 × 37 = 150

⇒ x = 2

Hence, option (d).

Answer: 28

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

The following diagram can be drawn using the information given.

​​​​​​​

Since, ∆ADE is an isosceles right triangle hence, DE = AE = 4 cm

⇒ AB = BE + AE = 5 + 4 = 9 cm

∴ Area of trapezium = $\frac{1}{2}\times (CD+AB)\times BC$ = $\frac{1}{2}\times 14\times 4$ = 28 cm².

Hence, 28.

Answer: 40

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

In 60 days 1.5 kms out of 6 kms is built.

∴ In 60 days 1.5/6 = 1/4^th work is done.

To complete the whole work, it would take 4 × 60 = 240 days, i.e., 180 more days.

Days remaining now is 200 – 60 = 140 days.

∴ To complete the remaining work in 140 days, contractor will have to hire more people. Assuming he hires x more men.

The work which 140 men would take 180 more days, now need to be completed by 140 + x men in 140 days.

∴ 180 × 140 = 140 × (140 + x)

⇒ x = 40

∴ Contractor hires 40 more people.

Hence, 40.

Answer: Option C

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Given, x_m = x_m+1 + (m + 1)

⇒ x_m+1 = x_m - (m + 1)

Put m = 1 ⇒ x₂ = -1 – 2
Put m = 2 ⇒ x₃ = -1 - 2 – 3
Put m = 3 ⇒ x₄ = - 1 – 2 - 3 – 4

And so on.

∴ x₁₀₀ = - 1 – 2 – 3 -4 … - 100

⇒ x₁₀₀ = - (100 × 101)/2 = - 5050

Hence, option (c).

Answer: 3

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Assuming, k = 14^a = 36^b = 84^c

⇒ 14 = k^1/a

⇒ 36 = 6² = k^1/b or 6 = k^1/2b 

⇒ 84 = k^1/c

We know, 84 = 6 × 14

⇒ k^1/c = k^1/2b × k^1/a

⇒ k^{1/c – 1/a} = k^1/2b

⇒ $\frac{1}{c}-\frac{1}{a}=\frac{1}{2b}$

⇒ $2\mathrm{b}\left(\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{a}}\right)$ = 1

Hence, $6\mathrm{b}\left(\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{a}}\right)$ = 3

Hence, 3.

Answer: Option D

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

We know, 0.2 <  m/20 <0.5 ⇒ 4 < m < 10

Values m can take = {5, 6, 7, 8, 9}

We know, 0.2 < n/11 <0.5 ⇒ 2.2 < n < 5.5

Values n can take = {3, 4, 5}

Since n is even, the only possible value of n is 4.

Also, 0.2 < $\frac{\mathrm{m}}{20},\frac{\mathrm{n}}{\mathrm{m}},\frac{\mathrm{n}}{11}$ < 0.5 ⇒ m/5 < n < m/2

Since n = 4 ⇒ 20 > m > 8

The only possible value of m is 9.

∴ m – 2n = 9 – 2 × 4 = 1.

Hence, option (d).

Answer: Option B

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Given, f(x + y) = f(x) × f(y)

Substitute y = 0

⇒ f(x + 0) = f(x) × f(0)

f(x) = f(x) × f(0)

[f(x) cannot be ‘0’ since it is given that f(5) = 4]

∵ f(0) = 1

f(5) = 4 (Given)

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

f(10) = 16

We know that f(0) = 1

f(0) = f(10 - 10) = f(10 + (-10)) = f(10) × f(-10) = 1

16 × f(-10) = 1

f(-10) = 1/16 = 0.0625

f(10) - f(-10) = 16 - 0.0625 = 15.9375.

Hence, option (b).

Answer: Option D

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

One of the diagonals of the parallelogram connects the points (2, 1) and (-3, -4).

Line x + 9y + c = 0, represents the other diagonal of the parallelogram.

We know, diagonals of a parallelogram bisect each other. 

∴ x + 9y + c = 0 should pass through the midpoint of the line joining (2, 1) and (-3, -4).

The midpoint of the line joining (2, 1) and (-3, -4) = $\left(\frac{2- 3}{2},\frac{1 - 4}{2}\right)$ = (-0.5, -1.5)

⇒ Point (-0.5 , -1.5) lies on x + 9y + c = 0.

⇒ -0.5 + 9 × (-1.5) + c = 0

⇒ -0.5 - 13.5 + c = 0

⇒ c = 14.

Hence, option (d).

Answer: Option B

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Let the distance from home to office be d kms.

∴ $\frac{d}{35}-\frac{d}{40}=\frac{6}{60}$

⇒ d = 28 kms.

∴ Normal time taken = $\frac{28}{40}$ = 0.7 hour = 42 minutes.

Now, she covers 2/3^rd of this distance in 1/3rd time i.e., in 14 minutes.

She now takes rest for 8 minutes.

∴ Time elapsed = 14 + 8 = 22 minutes.

Time remaining = 20 minutes.

In 20 minutes she needs to travel 1/3^rd of the distance i.e., 28/3 kms.

∴ Time taken = $\frac{28/3}{20/60}$ = 28 kmph.

Hence, option (b).

Answer: Option A

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Let the cost price of sugar be Rs. 100/kg and marked price will be = Rs. 120/kg

∴ Total cost for the man = 35 × 100 = Rs. 3500

He earns 15% profit on this, hence total selling price = 3500 × 1.15 = Rs. 4025

Total Selling price for 5 kg sugar = 5 × 120 = 600

Total Selling price for 15 kg sugar = 15 × 120 × 0.9 = 1620

Total Selling price for 3 kg sugar = 3 × 0 = 0

Total Selling price for remaining 12 kg sugar = 12 × 120(1 + p%)

∴ 4025 = 600 + 1620 + 0 + 1440(1 + p%)

⇒ 1805 = 1440(1 + p%)

⇒ 1 + p% ≈ 1.25

⇒ p = 25%

Hence, option (a).

Answer: Option D

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

First, we have the quadratic equation, x² + mx + 2n = 0
Since its roots are real, that means D ≥ 0.
⇒ m² – 8n ≥ 0  
⇒ m ≥ 2√(2n)   …(1)

Also, we have the quadratic equation, x² + 2nx + m = 0
Since its roots are real, that means D ≥ 0.
⇒ (2n)² – 4m ≥ 0
⇒ 4n² – 4m ≥ 0    
⇒ n² ≥ m         …(2)

From (1) and (2)
⇒ n² ≥ m ≥ 2√(2n)
⇒ n^3/2 ≥ 2^3/2
⇒ n ≥ 2

Hence, least possible value of n = 2.
⇒ m ≥ 2√(2n)
m will be least when n is least
⇒ m ≥ 2√(4)
⇒ m ≥ 4
Hence, least possible value of m = 4.

So, the smallest possible values of m and n are 4 and 2 respectively.
Hence, smallest value of m + n = 4 + 2 = 6.

Hence, option (d).

Answer: Option D

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Speeds of Anil, Sunil and Ravi, which are 15 kmph, 10 kmph and 8 kmph respectively.

Length of the track is 3 km.

Time taken by Anil to complete one round = 3/15 × 60 = 12 minutes.

Time taken by Sunil to complete one round = 3/10 × 60 = 18 minutes.

Both Anil and Sunil will reach at the starting points after LCM[12, 18] = 36 minutes.

The distance covered by Ravi in 36 minutes (= 0.6 hours) = 8 × 0.6 = 4.8 km.

Hence, option (d).

Concept:

Circular Race

Answer: 6

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Given, 2 < x < 10

x can take any of the values from {3, 4, 5, 6, 7, 8, 9}

Also, 14 < y < 23

y can take any of the values from {15, 16, 17, 18, 19, 20, 21, 22}

The highest value N (i.e., x + y) can take = 9 + 22 = 31 (when x = 9; y = 22)

The lowest value N (i.e., x + y) can take = 3 + 15 = 18 (when x = 3; y = 15)

But, N = x + y > 25. Hence the different values of x + y are {31, 30, 29, 28, 27, 26}.

Hence, x + y, and thereby N can take 6 distinct values.

Hence, 6.

Answer: Option A

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Let the maximum marks be 100x.

Marks scored by Bishnu = 52x and Asha = 64x

Bishnu scored 23 less, Asha scored 34 more than the marks obtained by Ramesh

∴ Asha – Bishnu = 34 + 23 = 57

⇒ 64x – 52x = 57

⇒ 12x = 57

Now, Geeta scored 84%, her score = 84x = 7 × 12x 

= 7 × 57 = 399.

Hence, option (a).

Answer: 18

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Let Tom's age to be 'x' years.

∴ Dick's age = 3x.

∴ Harry's age = 6x.

Given, Dick's age is 1 year less than the average age of all three,

∴ 3x = (x + 3x + 6x)/3 – 1

⇒ 9x = 10x – 3

⇒ x = 3

∴ Harry’s age = 6x = 6 × 3 = 18.

Hence, 18.

Answer: Option B

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

The diagram can be drawn as follows:

​​​​​​​

Height of the given triangle = 9 and base = 4

∴ Area of the triangle = $\frac{1}{2}\times 9\times 4$ = 18

We know area of a triangle = abc/4R

(where a, b, and c are the sides of the triangle and R is the circumradius)

a = AB = 4 units
b = BC = $\sqrt{9^2+1^2}=\sqrt{82}$ units
c = AC = $\sqrt{3^2+9^2}=\sqrt{90}$ units

The sides of the triangle are 4, √82 and √90

⇒ 18 = $\frac{4\times \sqrt{82}\times \sqrt{90}}{4R}$

⇒ R = $\frac{\sqrt{205}}{3}$

∴ Area of the circumcircle = πR² = $\frac{205\mathrm{\pi }}{9}$

Hence, option (b).

Answer: 24

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Given, $\frac{2\times 4\times 8\times 16}{\left({{\log }_{2}4)}^2\right({{\log }_{4}8)}^3({{\log }_{8}16)}^4}$

= $\frac{2\times 2^2\times 2^3\times 2^4}{\left({{\log }_2{2}^2)}^2\right({{\log }_{2^2}{2}^3)}^3({{\log }_{2^3}{2}^4)}^4}$

= $\frac{2^{10}}{{\left(2\right)}^2{\left(\frac{3}{2}\right)}^3{\left(\frac{4}{3}\right)}^4}$

= $\frac{2^{10}\times 2^3\times 3^4}{2^2\times 3^3\times 2^8}$

= 2³ × 3 = 24 

Hence, 24.

Answer: 16000

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Rate of interest is 10% per annum. Hence, interest for half a year is 5%.

The Principal, P is invested for 1.5 years or 3 terms of half years.

Therefore, P (1 + 5%)³ = 18522.

⇒ P × 1.05³ = 18522

⇒ P × ${\left(\frac{105}{100}\right)}^3$= 18522

⇒ P × ${\left(\frac{21}{20}\right)}^3$ = 2 × 213

⇒ P = 2 × 20³ = 16000.

Hence, 16000.

Answer: Option C

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Since all the 4 options have A + B, lets add A and B.

∴ A + B = log_a30 + (-log_a(5/3)) 

= log_a30 – log_a(5/3) = log_a(30 × 3/5) 

= log_a(18) = log_a(2 × 3²)

⇒ A + B = log_a(2) + 2log_a(3)

⇒ A + B = 3 + 2log_a(3) [∵ log₂a = 1/3 ∴ log_a2 = 3]

⇒ log_a(3) = (A + B - 3)/2

⇒ log₃(a) = 2/(A + B - 3)

Hence, option (c).

Answer: Option A

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Given, A and B are mixed in the ratio 1 : 3

Let 10 liters of A and 30 liters of B is mixed initially i.e., total 40 liters solution.

Now volume is double by adding 40 liters of A.

∴ In the solution A = 10 + 40 = 50 liters and B = 30 liters

Concentration of alcohol in this 80-liter solution = 72%

Let the alcohol concentration in B = b%

∴ 80 × 72% = 50 × 60% + 30 × b%

⇒ 576 = 300 + 30b

⇒ b = 92%

Hence, option (a). 

Answer: Option D

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

Given, a × b = 4²⁰¹⁷ = 2⁴⁰³⁴

Here, a and b and integers and will be some power of 2 since RHS is 2⁴⁰³⁴.

Let’s say a = 2^x and b = 2^y. [x and y are non-negative integers]

Also, a ≤ b, Hence, x ≤ y.

Now, 2^x × 2^y = 2⁴⁰³⁴

Since x ≤ y, x ≤ 4034/2 = 2017

The values x can take is 0, 1, 2, …. till 2017.

Hence, x can take a total of 2018 values.

Hence, option (d).

Answer: Option B

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

We can plot x = 2 and y = |x - 2| + 4

​​​​​​​

The required area is sum of the triangle (orange) + rectangle (green).

Area of triangle (orange) = 1/2 × 2 × 2 = 2

Area of rectangle = 2 × 4 = 8

∴ Total required area = 2 + 8 = 10 sq. units.

Hence, option (b).

Answer: Option B

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

There are total 120 numbers given.

Integers not divisible by 2, 5 and 7 = 120 – (numbers that are divisible by at least one of 2, 5 and 7)

We know, n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C).

n(x) = number of multiples of x.

∴ n(2 ∪ 5 ∪ 7) = n(2) + n(5) + n(7) - n(2 ∩ 5) - n(5 ∩ 7) - n(7 ∩ 2) + n(2 ∩ 5 ∩ 7)

⇒ n(2 ∪ 5 ∪ 7) = n(2) + n(5) + n(7) - n(10) - n(35) - n(14) + n(70)

⇒ n(2 ∪ 5 ∪ 7) = 60 + 24 + 17 - 12 - 3 - 8 + 1 = 79.

Now, Integers not divisible by 2, 5 and 7 = 120 – (numbers that are divisible by at least one of 2, 5 and 7)

= 120 – 79 = 41

∴ Out of 120 numbers given, 41 numbers are not divisible by any of 2, 5 and 7.

Hence, option (b).

Answer: Option C

Video Explanation

Join our Telegram Group for CAT/MBA Preparation.

Explanation :

For 2 equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ to have a unique solution, the condition is: $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$.

∴ $\frac{k}{4}=\frac{1}{k}$

⇒ k² = 4

⇒ k = ±2 or |k| ≠ 2.

Hence, option (c).