[Download PDF] CAT 2022 Slot 2 Question Paper Paper with Solutions

Instructions

For the following questions answer them individually

Question 51

Let r and c be real numbers. If r and -r are roots of $$5x^{3} + cx^{2} - 10x + 9 = 0$$, then c equals

Video Solution

Solution

Let the roots of the given equation $$5x^{3} + cx^{2} - 10x + 9 = 0$$ be r, -r and p

r - r + p = $$-\frac{c}{5}$$

p = $$-\frac{c}{5}$$ ...... (1)

$$-r^2-pr+pr=-2$$

$$r^2=2$$ ...... (2)

$$-r^2p=-\frac{9}{5}$$

$$p=\frac{9}{10}$$ ...... (3)

Substituting p in (1), we get

$$\frac{9}{10}=-\frac{c}{5}$$

$$-\frac{9}{2}=c$$

The answer is option A.

Question 52

Two ships meet mid-ocean, and then, one ship goes south and the other ship goes west, both travelling at constant speeds. Two hours later, they are 60 km apart. If the speed of one of the ships is 6 km per hour more than the other one, then the speed, in km per hour, of the slower ship is

Video Solution

Solution

Let the speeds of two ships be 'x' and 'x+6' km per hour

Distance covered in 2 hours will be 2x and 2x+12

It is given,

$$\left(2x\right)^2+\left(2x+12\right)^2=60^2$$

$$\left(x\right)^2+\left(x+6\right)^2=30^2$$

$$2x^2+12x+36=900$$

$$x^2+6x+18=450$$

$$x^2+6x-432=0$$

Solving, we get x = 18

The speed of slower ship is 18 kmph

The answer is option C.

Question 53

Suppose for all integers x, there are two functions f and g such that $$f(x) + f (x - 1) - 1 = 0$$ and $$g(x ) = x^{2}$$. If $$f\left(x^{2} - x \right) = 5$$, then the value of the sum f(g(5)) + g(f(5)) is

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Video Solution

Solution

Given,

$$f\left(x\right)+f\left(x-1\right)=1$$ ...... (1)

$$f\left(x^2-x\right)=5$$ ...... (2)

$$g\left(x\right)=x^2$$

Substituting x = 1 in (1) and (2), we get

f(0) = 5

f(1) + f(0) = 1

f(1) = 1 - 5 = -4

f(2) + f(1) = 1

f(2) = 1 + 4 = 5

f(n) = 5 if n is even and f(n) = -4 if n is odd

f(g(5)) + g(f(5)) = f(25) + g(-4) = -4 + 16 = 12

Question 54

In an examination, there were 75 questions. 3 marks were awarded for each correct answer, 1 mark was deducted for each wrong answer and 1 mark was awarded for each unattempted question. Rayan scored a total of 97 marks in the examination. If the number of unattempted questions was higher than the number of attempted questions, then the maximum number of correct answers that Rayan could have given in the examination is

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Video Solution

Solution

Let the number of questions attempted be x+y out of which x are correct and y are incorrect and the number of questions unattempted be z.

It is given,

x + y + z = 75 ...... (1)

3x - y + z = 97 ...... (2)

(2)-(1) -> x - y = 11

(1)+(2) -> 2x + z = 86

z > x + y

z > 75 - z

z > 37.5

Minimum possible value of z is 38

2x + 38 = 86

2x = 48

x = 24

The maximum number of correct answers is 24.

Question 55

Regular polygons A and B have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then the number of sides of B equals

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Video Solution

Solution

Let the number of sides of polygons A and B be n and 2n, respectively.

$$\ \dfrac{\ \ \dfrac{\ \left(n-2\right)180}{n}}{\ \dfrac{\ \left(2n-2\right)180}{2n}}=\dfrac{3}{4}$$

$$\ \dfrac{\ \ \ n-2}{\ \ n-1}=\dfrac{3}{4}$$

$$\ \ \ \ 4n-8=3n-3$$

$$n=5$$

The number of sides of polygon B is 2*5, i.e. 10.

Question 56

In an election, there were four candidates and 80% of the registered voters casted their votes. One of the candidates received 30% of the casted votes while the other three candidates received the remaining casted votes in the proportion 1 : 2 : 3. If the winner of the election received 2512 votes more than the candidate with the second highest votes, then the number of registered voters was

Video Solution

Solution

Let the number of registered votes be 100x

The number of votes casted = 80x

Votes received by one of the candidates = $$\frac{30}{100}\times80x$$ = 24x

Remaining votes = 80x - 24x = 56x

Votes received by other three candidates is $$\frac{56x}{6},\frac{2\times56x}{6},\ \frac{\ 3\times56x}{6}$$

It is given,

28x - 24x = 2512

4x = 2512

x = 628

The number of registered votes = 100x = 62800

The answer is option D.

Question 57

On day one, there are 100 particles in a laboratory experiment. On day n, where $$n\ge2$$, one out of every n articles produces another particle. If the total number of particles in the laboratory experiment increases to 1000 on day m, then m equals

Video Solution

Solution

Given, the number of particles on day 1 = 100

On day 2, one out of every 2 articles produces another particle.

The number of particles on day 2 will be $$\frac{100}{2}$$, i.e. 50 particles

On day 3, one out of every 3 articles produces another particle.

The number of particles on day 3 will be $$\frac{100+50}{3}$$, i.e. 50 particles

On day 4, one out of every 4 articles produces another particle.

The number of particles on day 4 will be $$\frac{100+50+50}{4}$$, i.e. 50 particles

Series will be 100, 50, 50, 50,....

It is given,

100 + (m-1)50 = 1000

m = 19

The answer is option A.

Question 58

The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is

Video Solution

Solution

Case 1: 4-digit numbers

Given digits - 0, 1, 2, 3, 4, 5

_, _, _, _

As the numbers should be greater than 2000, first digit can be 2, 3, 4 and 5.

For remaining digits, we need to arrange 3 digits from the remaining 5 digits, i.e. 5*4*3 = 60 ways

Total number of possible 4-digit numbers = 4*60 = 240

Case 2: 5-digit numbers

_, _, _, _, _

First digit cannot be zero.

Therefore, total number of cases = 5*5*4*3*2 = 600

Case 3: 6-digit numbers

_, _, _, _, _, _

First digit cannot be zero.

Therefore, total number of cases = 5*5*4*3*2*1 = 600

Total number of integers possible = 600 + 600 + 240 = 1440

The answer is option A.

Question 59

For some natural number n, assume that (15,000)! is divisible by (n!)!. The largest possible value of n is

Video Solution

Solution

To find the largest possible value of n, we need to find the value of n such that n! is less than 15000.

7! = 5040

8! = 40320 > 15000

This implies 15000! is not divisible by 40320!

Therefore, maximum value n can take is 7.

The answer is option B.

Question 60

Working alone, the times taken by Anu, Tanu and Manu to complete any job are in the ratio 5 : 8 : 10. They accept a job which they can finish in 4 days if they all work together for 8 hours per day. However, Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day. Then, the number of hours that Manu will take to complete the remaining job working alone is

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Solution

Let the time taken by Anu, Tanu and Manu be 5x, 8x and 10x hours.

Total work = LCM(5x, 8x, 10x) = 40x

Anu can complete 8 units in one hour

Tanu can complete 5 units in one hour

Manu can complete 4 units in one hour

It is given, three of them together can complete in 32 hours.

32(8 + 5 + 4) = 40x

x = $$\frac{68}{5}$$

It is given,

Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day, i.e. 36 + 4 = 40 hours

40(8 + 5) + y(4) = 40x

4y = 24

y = 6

Manu alone will complete the remaining work in 6 hours.

CAT Content

CAT 2022 Slot 2 Question Paper

Let r and c be real numbers. If r and -r are roots of $$5x^{3} + cx^{2} - 10x + 9 = 0$$, then c equals

Two ships meet mid-ocean, and then, one ship goes south and the other ship goes west, both travelling at constant speeds. Two hours later, they are 60 km apart. If the speed of one of the ships is 6 km per hour more than the other one, then the speed, in km per hour, of the slower ship is

Suppose for all integers x, there are two functions f and g such that $$f(x) + f (x - 1) - 1 = 0$$ and $$g(x ) = x^{2}$$. If $$f\left(x^{2} - x \right) = 5$$, then the value of the sum f(g(5)) + g(f(5)) is

Regular polygons A and B have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then the number of sides of B equals

On day one, there are 100 particles in a laboratory experiment. On day n, where $$n\ge2$$, one out of every n articles produces another particle. If the total number of particles in the laboratory experiment increases to 1000 on day m, then m equals

The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is

For some natural number n, assume that (15,000)! is divisible by (n!)!. The largest possible value of n is

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CAT Content

CAT 2022 Slot 2 Question Paper

Let r and c be real numbers. If r and -r are roots of $$5x^{3} + cx^{2} - 10x + 9 = 0$$, then c equals

Two ships meet mid-ocean, and then, one ship goes south and the other ship goes west, both travelling at constant speeds. Two hours later, they are 60 km apart. If the speed of one of the ships is 6 km per hour more than the other one, then the speed, in km per hour, of the slower ship is

Suppose for all integers x, there are two functions f and g such that $$f(x) + f (x - 1) - 1 = 0$$ and $$g(x ) = x^{2}$$. If $$f\left(x^{2} - x \right) = 5$$, then the value of the sum f(g(5)) + g(f(5)) is

Regular polygons A and B have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then the number of sides of B equals

On day one, there are 100 particles in a laboratory experiment. On day n, where $$n\ge2$$, one out of every n articles produces another particle. If the total number of particles in the laboratory experiment increases to 1000 on day m, then m equals

The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is

For some natural number n, assume that (15,000)! is divisible by (n!)!. The largest possible value of n is

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