CMAT 2023 Slot 2 Questions Paper

Sum is 24: The only possibility is (6,6,6,6) -> 1 way

Sum is 23: The only possibility is (6,6,6,5) -> $$\frac{4!}{3!}$$ = 4 ways

Sum is 22: There are two possibilities, i.e. (6,6,6,4) and (6,6,5,5)

(6,6,6,4) -> $$\frac{4!}{3!}$$ = 4 ways

(6,6,5,5) -> $$\frac{4!}{2!2!}$$ = 6 ways

Total number of ways satisfying the given condition = 1 + 4 + 4 + 6 = 15

Total number of possibilities = (6)(6)(6)(6)

Required probability = $$\frac{15}{1296}$$ = $$\frac{5}{432}$$

The answer is option A.

Speed of train = 18 km/hr = 5 m/s

Let the length of train be 'P'

It is given,

$$\frac{P}{5}=9$$

P = 45m

Let the length of platform be 'Q

It is given,

$$\frac{P+Q}{5}=25$$

P+Q = 125

Q = 125 - 45 = 80m

(P, Q) = (45, 80)

The answer is option A.

It is given that P, Q and R together can complete the work in 6 hrs.

P, Q and R together worked for 2 hours, this implies they completed 1/3rd of the work.

R left and remaining work is finished by P and Q in 7 hours.

Therefore, P and Q completed 2/3rd of the work in 7 hours.

This implies P and Q together can complete the work in 21/2 hours.

Let the total work be LCM(6, 21/2), i.e. 42 units.

P, Q and R together can complete 7 units in one hour.

P and Q together can complete 4 units in one hour.

This implies, R can complete 3 units in one hour.

Time taken by R alone to complete the work = $$\frac{42}{3}$$ = 14 hours

The answer is option C.

If a is the length of equilateral triangle, height is $$\frac{\sqrt{3}a}{2}$$

Inradius = $$\frac{\sqrt{\ 3}a}{2}\left(\frac{1}{3}\right)=\frac{a}{2\sqrt{3}}$$

It is given, R = $$\frac{a}{2\sqrt{3}}$$

Area of triangle = Inradius * semiperimeter = $$R\left(\frac{3a}{2}\right)$$ = $$\frac{3R}{2}\left(2\sqrt{3}R\right)=3\sqrt{3}R^2$$

The answer is option D.

Let $$n^2+96=k^2$$

$$k^2-n^2=96$$

$$\left(k+n\right)\left(k-n\right)=96$$

$$\left(k+n\right)\left(k-n\right)=2\times48,\ 4\times24,\ 6\times16,\ 8\times12,\ 12\times8,\ 16\times6,\ 24\times4,\ 48\times2$$

n can take 8 values, i.e. -23, -10, -5, -2, 2, 5, 10 and 23.

The answer is option C.

Statement 1 

We need to pack 6 indistinguishable books into 4 indistinguishable boxes 

The ways we can pack the books are

6 ,0,0,0 (only one way as 0,6,0,0 or 0,0,6,0 won’t be different as boxes are identical)
5, 1,0,0
4, 2,0,0
4, 1, 1,0
3, 3,0,0
3, 2, 1,0
3, 1, 1, 1
2, 2, 2,0
2, 2, 1, 1.

so total 9 ways

Hence statement 1 is correct. 

Statement 2 

To calculate the minimum number of students required, we need to consider the worst case scenario i.e. the birthdays of students are evenly spread across all the 12 months. We'll have to assume that there are atleast 5 birthday is each month, so the total number of students required = 5* 12 = 60. To assure that there is 1 month in which we have 6 birthdays, we'll have to add 1 to this number, hence the minimum number of students required = 60 + 1 = 61

Statement I:

It is given that the ratio of son's age to father's age is 1:5

Let the son's age be x and father's age be 5x.

(x)(5x) = 245

x = 7

Therefore, son's age is 7 years and father's age is 35 years.

After 9 years, son's age is 16 years and father's age is 44 years.

Ratio = 16 : 44 = 4 : 11

Statement II:

Initially - 70L milk and 30L water

After selling 1/5th of the solution - 56L milk and 24L water

Adding water - 56L milk and 44L water

water:milk = 44:56 = 11:14

The answer is option B.

It is given,

$$x-\frac{1}{x}=y$$, $$y-\frac{1}{y}=z$$ and $$z-\frac{1}{z}=x$$

Adding all the given equations, we get

$$x\ +\ y\ +\ z\ -\ \frac{1}{x}-\frac{1}{y}-\frac{1}{z}=y+z+x$$

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$

Therefore, equation A is correct.

$$x-\frac{1}{x}=y$$, $$y-\frac{1}{y}=z$$ and $$z-\frac{1}{z}=x$$

Squaring on both the sides and adding three resultant equations, we get

$$x^2+\frac{1}{x^2}-2+y^2+\frac{1}{y^2}-2+z^2+\frac{1}{z^2}-2=y^2+z^2+x^2$$

$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=6$$

Therefore, equation B is incorrect.

The answer is option B.

It is given,

$$5^a=9^b=2045$$

$$5^a=2045$$

Applying log on both the sides, we get

$$\log_55^a=\log_52045$$

$$a=\log_52045$$

Similarly, we get $$b=\log_92045$$

$$\dfrac{ab}{a+b}=\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}$$

$$\dfrac{1}{a}=\log_{2045}5$$

$$\dfrac{1}{b}=\log_{2045}9$$

$$\dfrac{1}{a}+\dfrac{1}{b}=\log_{2045}45$$

$$\dfrac{ab}{a+b}=\log_{45}2045\ =\ 2$$

The answer is option C.

8 = $$\left(\sqrt{59}-\sqrt{51}\right)\left(\ \sqrt{\ 59}+\sqrt{51}\right)$$

8 = $$\left(\sqrt{37}-\sqrt{29}\right)\left(\ \sqrt{37}+\sqrt{29}\right)$$

We know that, $$\sqrt{59}+\sqrt{51}$$ is greater than $$\sqrt{37}+\sqrt{29}$$

This implies, $$\sqrt{37}-\sqrt{29}$$ is greater than $$\sqrt{59}-\sqrt{51}$$ as the product is constant.

Similarly, $$\sqrt{59}-\sqrt{51}$$ is greater than $$\sqrt{79}-\sqrt{71}$$ and $$\sqrt{79}-\sqrt{71}$$ is greater than $$\sqrt{87}-\sqrt{79}$$.

The correct order is B > A > D > C

The answer is option D.