CAT 2022 Question Paper Quant Slot 1

In the question, it is given that the ratio of number of literate males to literate females is 2 : 3.

Given, the number of literate males = 3600

The number of literate females = $$\frac{3600}{2}\times3$$ = 5400

Males : Females = 5 : 4

Let the number of males be 5y and the number of females be 4y

Illiterate males = 5y - 3600

Illiterate females = 4y - 5400

It is given,

$$\ \frac{\ 5y-3600}{4y-5400}=\frac{4}{3}$$

15y - 10800 = 16y - 21600

y = 10800

Number of females = 4y = 4*10800 = 43200

Let the original number of students be 'n' whose average weight is 'x'

Let the number of students added be 'm' and the average weight will be x + 3

We need to find the value of n : m

It is given, average weight of students in a class increased by 0.6 after new students are added.

Therefore,

$$\ \frac{\ nx+m\left(x+3\right)}{n+m}=x+0.6$$

$$\ \ nx+mx+3m=mx+nx+0.6n+0.6m$$

$$2.4m=0.6n$$

$$4m=n$$

$$\frac{n}{m}=\frac{4}{1}$$

The answer is option C.

It is given,

$$S_n=2n^2+n$$

$$S_{n-1}=2\left(n-1\right)^2+\left(n-1\right)$$

$$S_{n-1}=2n^2-3n+1$$

$$T_n=S_n-S_{n-1}=2n^2+n-2n^2+3n-1=4n-1$$

$$T_n=4n-1$$

The terms are 3, 7, 11, 15, 19, 23, 27,......

27 is the first term in the series divisible by 9.

27 is the 7th term.

Therefore, the least possible value of n is 7.

The answer is option C.

x>=a, so |x-a| = x-a

x<100, so |x-100| = 100-x

f(x) = (x-a) + (100-x) + |x-a-50| =100

or, |x-a-50| = a

From the graph we can can see that when x=a then

|x-a-50|=a

or, a= 50

Similarly when x=a+100

|x-a-50|=a

or, a= 50

So value of a is 50 when f(x) is 100.

M - First meeting point

Let the speeds of trains A and B be 'a' and 'b', respectively.

$$\frac{x}{a}=\ \frac{\ D-x}{b}$$

It is given,

$$\frac{D}{a}=10$$ and $$\frac{x}{b}=9$$

$$\frac{x}{\frac{D}{10}}=\ \frac{\ D-x}{\frac{x}{9}}$$

$$\frac{10x}{D}=\ \frac{\ 9D-9x}{x}$$

$$10x^2=\ \ 9D^2-9Dx$$

$$10x^2+9Dx-9D^2=\ 0$$

Solving, we get $$x=\frac{3D}{5}$$

$$\frac{x}{b}=9$$

$$\frac{3D}{b\times5}=9$$

$$\frac{D}{b}=15$$ 

The total time taken by train B to travel from station Y to station X is 15 minutes.

The answer is option B

CD = $$\sqrt{\ y^2+25}$$

$$11+y+\sqrt{y^2+25}=36$$

$$\sqrt{y^2+25}=25-y$$

$$y^2+25=25^2+y^2-50y$$

2y = 24

y = 12

Area of trapezium = $$3y+\frac{5y}{2}=\frac{11y}{2}=\frac{11}{2}\left(12\right)=66$$

It is given,

7C = 30P = 9A and Ankita bought 4C, 14P and 6A.

Let 7C = 30P = 9A = 630k

C = 90k, P = 21k, and A = 70k

Cost price of 4C, 14P and 6A = 4(90k)+14(21k)+6(70k) = 1074k

Marked up price = 1074k + 1752

S.P = $$\frac{1}{6}\left(1074k+1752\right)+\left(\frac{4}{5}\right)\left(\frac{5}{6}\right)\left(1074k+1752\right)$$ = $$\frac{5}{6}\left(1074k+1752\right)$$

S.P - C.P = profit

$$1460-\frac{1074k}{6}=744$$

$$\frac{1074k}{6}=716$$

k = 4

Money spent on buying almonds = 420k = 420*4 = Rs 1680

The answer is option A.

A is the HCF of $$3^k+4^k+5^k$$ for different values of k.

For k = 1, value is 12

For k = 2, value is 50

For k = 3, value is 216

HCF is 2. Therefore, A = 2

$$4^k+3\left(4^k\right)+4^{k+2}=4^k\left(1+3\right)+4^{k+2}=4^{k+1}+4^{k+2}=4^{k+1}\left(1+4\right)=5\cdot4^{k+1}$$

HCF of the values is when k = 1, i.e. 5*16 = 80

Therefore, B = 80

A + B = 82

$$b^2 < 4ac$$ means that the discriminant is less than 0. Therefore, f(x)>0 for all x if the coefficient of $$x^2$$ is positive, and f(x)<0 for all x if the coefficient of $$x^2$$ is negative.

We are given that f(m)<0 and m is an integer.

So the set containing values of m will either be empty if the coefficient of $$x^2$$ is positive, or it will be a set of all integers if the coefficient of $$x^2$$ is negative.

Let the number of balloons each child received be 2a, 2b, 2c and 2d

2a + 2b + 2c + 2d = 20

a + b + c + d = 10

Each of them should get more than zero balloons.

Therefore, total number of ways = $$(n-1)_{C_{r-1}}=(10-1)_{C_{4-1}}=9_{C_3}=84$$

a(a + b + 1) = 14 ...... (1)

b(a + b + 1) = 28 ...... (2)

$$\frac{a}{b}=\frac{1}{2}$$

b = 2a

Substituting in (1), we get

a(3a + 1) = 14

$$3a^2+a-14=0$$

$$3a^2-6a+7a-14=0$$

$$3a\left(a-2\right)+7\left(a-2\right)=0$$

Given, a and b are natural numbers.

Therefore, a = 2 and b = 4

2a + b = 2(2) + 4 = 8

The answer is option A.

Total syrup - 110 kg

Total juice - 120 kg

It is given, cost price of syrup is 20% less than the cost price of juice.

Let the cost price of juice per kg be 10CP

Cost price of syrup per kg is 8CP

10kg syrup -> cost price = 80CP

It is given, 10kg syrup is sold at 10% profit. This implies selling price = 1.1*80CP = 88CP

20kg juice -> cost price = 200CP

It is given, 20kg juice is sold at 20% profit. This implies selling price = 1.2*200CP = 240CP

It is given, Mixing the remaining juice and syrup, Amal sells the mixture at ₹ 308.32 per kg

Selling price of the remaining mixture = 308.32*200 = Rs 61664

Total S.P = 61664 + 328CP

Total C.P = 880CP + 1200CP = 2080CP

Overall profit = 64%

$$61664+328CP=\frac{164}{100}\left(2080CP\right)$$

Solving, we get CP = 20

Cost price for syrup per kg = 8CP = 8*20 = Rs 160

$$A=lb$$

$$l^2+b^2=4r^2$$

$$P\ =2\left(l+b\right)$$

$$\frac{P}{2}=l+b$$

Squaring on both the sides, we get

$$\frac{P^2}{4}=l^2+b^2+2lb$$

$$\frac{P^2}{4}=4r^2+2lb$$

$$\frac{P^2}{8}-2r^2=lb$$

The answer is option B.

It is given that average of three numbers is 13.

Sum = 3*13 = 39

It is given, $$\ \frac{\ 39+n}{4}$$ is a odd number.

Minimum value $$\ \frac{\ 39+n}{4}$$ can take such that n is a natural number is 11

$$\ \frac{\ 39+n}{4}=11$$

n = 5

The answer is option C.

Lemon juice : sugar syrup in the mixture is 1:1, i.e. 50% Lemon juice and 50% sugar syrup.

In sugar syrup, 100% is sugar syrup.

These two are mixed in the ratio 1:3.

Lemon juice = $$\ \frac{\ 1\left(50\%\right)}{1+3}$$

Sugar syrup = $$\ \frac{\ 1\left(50\%\right)+3\left(100\%\right)}{1+3}=\frac{350}{4}$$

Required ratio = 50:350 = 1:7

The answer is option D.

In the question, it is given that the equation $$\mid x + a \mid + \mid x - 1 \mid = 2$$ has an infinite number of solutions for any value of x. This is possible when x in |x+a| and x in |x-1| cancels out.

Case 1:

x + a < 0, x - 1 $$\ge\ $$ 0

- a - x + x - 1 = 2

a = -3

Case 2:

x + a $$\ge\ $$ 0 and x - 1 < 0

x + a - x + 1 = 2

a = 1

Largest value of a is 1.

The answer is option C.

Let n, s, d and t be the number of students who likes none of the drinks, exactly one drink, exactly 2 drinks and all three drinks, respectively.

It is given, 

n + s + d + t = 100 ...... (1)

s + 2d + 3t = 73 + 80 + 52

s + 2d + 3t = 205 ...... (2)

(2)-(1), we get

d + 2t - n = 105

Maximum value t can take is 52, i.e. t = 52, d = 1 and n = 0

Minimum value t can take is 5, i.e. t = 5, d = 95 and n = 0 (This also satisfies equation (1))

Difference = 52 - 5 = 47

The answer is option A.

In a parallelogram, two diagonals of parallelogram bisects each other, which concludes that mid-point of both diagonals are the same.

Midpoint of AC = $$\left(\ \frac{\ 1-2}{2},\ \frac{\ 1+8}{2}\right)$$

Let the coordinates of vertex D be (x,y)

$$\left(\ \frac{\ x+3}{2},\ \frac{\ y+4}{2}\right)=\left(\ \frac{\ 1-2}{2},\ \frac{\ 1+8}{2}\right)$$

x = -4 and y = 5

The answer is option D.

It is given, y(x + z) = 19

y cannot be 19. 

If y = 19, x + z = 1 which is not possible when both x and z are natural numbers.

Therefore, y = 1 and x + z = 19

It is given, z(x + y) = 51

z can take values 3 and 17

Case 1:

If z = 3, y = 1 and x = 16

xyz = 3*1*16 = 48

Case 2:

If z = 17, y = 1 and x = 2

xyz = 17*1*2 = 34

Minimum value xyz can take is 34.

Let the savings invested in first part and second part be 'x' and 'y', respectively.

It is given,

$$\ \frac{\ x\times15\times4}{100}=\ \frac{\ y\times12\times3}{100}$$

60x = 36y

5x = 3y

Required percentage = $$\frac{x}{x+y}\times100=\frac{3}{3+5}\times100=37.5\%$$

The answer is option A.

Let the number of persons standing ahead and behind of Pinky be 3a and 5a.

Total number of persons = 3a + 5a + 1(including pinky) = 8a + 1

8a + 1 < 300

8a < 299

a < 37.375

Maximum value a can take is 37.

The maximum possible number of persons standing ahead of Pinky = 3a = 3*37 = 111

It is given,

$$\Sigma_{n=1}^N\ \left[\frac{1}{5}+\frac{n}{25}\right]=25$$

$$\Sigma_{n=1}^N\ \left[\frac{5+n}{25}\right]=25$$

For n = 1 to n = 19, value of function is zero.

For n = 20 to n = 44, value of function will be 1.

44 = 20 + n - 1

n = 25 which is equal to given value.

This implies N = 44