CAT 2022 Slot 2 Question Paper

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.