CAT 2022 Question Paper Quant Slot 2

Let the total investment me 15x and the no. of years required be T years

$$\frac{\left(3x\times6\times T\right)}{100}+\frac{\left(5x\times10\times T\right)}{100}+\frac{\left(7x\times1\times T\right)}{100}\ge15x$$

or, $$\frac{75xT}{100}\ge15x$$

or, $$T\ge20$$

So minimum value of T is 20 years

$$a_1+a_2+.....+a_N=300N$$

$$6a_1+a_2+.....+a_N=400N$$

$$5a_1=100N$$

$$a_1=20N$$

As the given sequence of numbers is non-decreasing sequence, N can take values from 2 to 15.

N is not equal to 1, if N = 1, then average of N numbers is 300 wouldn't satisfy.

Therefore, N can take values from 2 to 15, i.e. 14 values.

Case 1: When $$x^2-3x-10=0$$ and $$x^2-10\ne\ 0$$

$$x^2-3x-10=0\ $$or, $$(x-5)(x+2) = 0$$

or, x= 5 or -2

Case 2: $$x^2-10=1$$

$$x^2-11=0$$

No integer solutions

Case 3: $$x^2-10=-1\ and\ x^2-3x-10\ is\ even$$

$$x^2-9=0$$

or, (x+3)(x-3)=0

or, x= -3 and 3

for x= -3 and +3 $$x^2-3x-10$$ is even

In total 4 values of x satisfy the equations

Savings target in a year = 550*12 = Rs 6600

Saving in first 9 months = 9(4000-3500) = Rs 4500

Saving for remaining 3 months should be 6600-4500, i.e. Rs 2100

Savings for each month in last 3 months = $$\frac{2100}{3}$$ = Rs 700

It is given, monthly expenses in last 3 months = Rs 3700

This implies, his monthly earnings from 10th month should be 3700+700, i.e. Rs 4400

The answer is option A.

It is given, Angle BAE = 45 degrees

This implies AE = BE

Let AE = BE = x 

In right-angled triangle ABD, it is given $$\angle ABC=\theta\ $$

$$\sin\theta=\frac{AD}{AB}\ $$

$$\sin\theta=\frac{AD}{x\sqrt{\ 2}}\ $$

$$\sqrt{\ 2}\sin\theta=\frac{AD}{BE}\ $$

The answer is option D.

$$f(x) \geq 0$$for all real numbers $$x$$, so D<=0

Since f(2)=0 therefore x=2 is a root of f(x)

Since the discriminant of f(x) is less than equal to 0 and 2 is a root so we can conclude that D=0

Therefore f(x) = $$a\left(x-2\right)^2$$

f(4)=6

or, 6 = $$a\left(x-2\right)^2$$

a= 3/2

$$f\left(-2\right)=\ -\frac{3}{2}\left(-4\right)^2=24$$

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

Step 1: Half the content of the first container is transferred to the second container

Step 2: Half of the mixture of second container is transferred back to the first container

Step 3:  Half the content of the first container is transferred back to the second container

Sugar syrup : Milk in second container = 62.5 : 75 = 5 : 6

The answer is option D.

Sum of n terms in an A.P = $$\frac{n}{2}\left(2a+\left(n-1\right)d\right)$$

$$A_n=\frac{n}{2}\left(6+\left(n-1\right)4\right)=n\left(2n+1\right)$$

$$\Sigma\ A_n=\Sigma\ n\left(2n+1\right)=2\Sigma\ n^2+\Sigma\ n=\ \frac{\ 2n\left(n+1\right)\left(2n+1\right)}{6}+\ \frac{\ n\left(n+1\right)}{2}$$

Substituting n = 25, we get

$$\frac{1}{25} \sum_{n=1}^{25} A_{n}$$ = $$\frac{1}{25}\left(\ \frac{\ 2\left(25\right)\left(25+1\right)\left(50+1\right)}{6}+\ \frac{\ 25\left(25+1\right)}{2}\right)$$

$$\frac{1}{25} \sum_{n=1}^{25} A_{n}$$ = $$\ 26\left(17\right)+13$$ = 455

The answer is option A. 

Let $$\ \log_2n=y$$

$$\ \ \frac{\ 4-y}{3-\frac{y}{2}}<0$$

$$\ \ \left(4-y\right)\left(3-\frac{y}{2}\right)<0$$

$$\ \ \left(4-y\right)\left(6-y\right)<0$$

$$\ \ \left(y-4\right)\left(y-6\right)<0$$

$$4 < y < 6$$

$$4<\log_2n<6$$

$$2^4<n<2^6$$

$$16<n<64$$

n can take values from 17 to 63(inclusive).

The number of n values possible = 47

a + 2b = 6

From the above equation, we can say that maximum value b can take is 3 and minimum value b can take is 0.

a + b + b = 6

a + b = 6 - b

a + b is maximum when b is minimum, i.e. b = 0

Maximum value of a + b = 6 - 0 = 6

a + b is minimum when b is maximum, i.e. b = 3

Minimum value of a + b = 6 - 3 = 3

Average = $$\ \frac{\ 6+3}{2}$$ = 4.5

The answer is option D.

The average marks for all the students is 38.

Sum = 5*38 = 190

To find the minimum marks scored by Amit, we need to maximise the score of remaining students.

Maximum scores sum of remaining students = 50 + 49 + 48 + 32 = 179

Minimum possible score of Amit = 190 - 179 = 11

It is given, Amit scored least. This implies maximum possible score of Amit is 31.

Difference = 31 - 11 = 20

The answer is option D.

Area of triangle ABD is twice the area of triangle ACD

$$\angle\ ADB=\theta\ $$

$$\frac{1}{2}\left(AD\right)\left(BD\right)\sin\theta\ =2\left(\frac{1}{2}\left(AD\left(CD\right)\sin\left(180-\theta\ \right)\right)\right)$$

$$BD\ =2CD$$

Therefore, BD = 2 and CD = 1

$$\angle\ ABC=\angle\ ACB=60^{\circ\ }$$

Applying cosine rule in triangle ADC, we get

$$\cos\angle\ ACD=\ \frac{\ AC^2+CD^2-AD^2}{2\left(AC\right)\left(CD\right)}$$

$$\frac{1}{2}=\ \frac{\ 9+1-AD^2}{6}$$

$$AD^2=7$$

$$AD=\sqrt{\ 7}$$

The answer is option C.