Quant For CAT 2022 (Slot - 1) PDF Download

Substituting in (1), we get

Given, a and b are natural numbers.

Therefore, a = 2 and b = 4

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

The answer is option C.

M - First meeting point

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

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

The answer is option C

b² < 4ac means that the discriminant is less than 0. Therefore, f(x) > 0 for all x if the coefficient of x² is positive, and f(x)<0 for all x if the coefficient of x² 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 x² is positive, or it will be a set of all integers if the coefficient of x² is negative.

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.

It is given that average of three numbers is 13.

Sum = 3*13 = 39

It is given,

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.

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

k = 4

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

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,

In the question, it is given that the equation ∣x + a∣ + ∣x − 1∣ = 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 ≥ ≥  0

- a - x + x - 1 = 2

a = -3

Case 2:

x + a ≥ ≥  0 and x - 1 < 0

x + a - x + 1 = 2

a = 1

Largest value of a is 1.

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

x = -4 and y = 5

The answer is option D.

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.

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

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

It is given,

60x = 36y

5x = 3y