CAT 2021 Question Paper Slot 2

Let Total matches played be n and in initial n-10 matches his goals be x 
so we get $$\frac{\left(x+1\right)}{n}=0.15$$
we get x+1 =0.15n                (1)
From condition (2) we get :
$$\frac{\left(x+2\right)}{n}=0.2$$
we get x+2 = 0.2n                  (2)
Subtracting (1) and (2)
we get 1 =0.05n
n =20
So initially he played n-10 =10 matches

Considering the three kinds of tea are A, B, and C.

The price of kind A = Rs 800 per kg.

The price of kind B = Rs 500 per kg.

The price of kind C = Rs 300 per kg.

They were mixed in the ratio of 2 : 3: 5.

1/6 of the total mixture is sold for Rs 700 per kg.

Assuming the ratio of mixture to A = 12kg, B = 18kg, C =30 kg.

The total cost price is 800*12+500*18+300*30 = Rs 27600.

Selling 1/6 which is 10kg for Rs 700/kg the revenue earned is Rs 7000.

In order to have an overall profit of 50 percent on Rs 27600.

Thes selling price of the 60 kg is Rs 27600*1.5 = Rs 41400.

Hence he must sell the remaining 50 kg mixture for Rs 41400 - Rs 7000 = 34400.

Hence the price per kg is Rs 34400/50 = Rs 688

We have :
$$x^2-xy-x\ =22\ \ \ \ \ \ \left(1\right)$$ 
And $$y^2-xy+y\ =34\ \ \ \ \ \ \left(2\right)$$         
Adding (1) and (2)
we get $$x^2-2xy+y^2-x+y\ =56$$
we get $$\left(x-y\right)^2-\ \left(x-y\right)\ =56$$
Let (x-y) =t
we get $$t^2-t=56$$
$$t^2-t-56=0$$
(t-8)(t+7) =0
so t=8
so x-y =8

We have :

Now area of ADE = $$\frac{1}{2}\times\ AD\times\ AE\times\ \sin A$$
=$$\frac{1}{2}\times\ 2x\times\ 2y\times\ \sin A=8$$
we get xy sinA =4
Now Area of triangle ABC = $$\frac{1}{2}AB\times\ AC\times\ \sin A$$
we get $$\frac{1}{2}\times3x\ \times5y\ \sin A=\frac{15}{2}xy\ \sin A=\ \frac{15}{2}\times\ 4$$
we get Area of ABC = 30

Let the amount invested by Anil Bobby and Chintu be x, y, and z.

Considering x+y+z = 100*p.

Given Anil's share was 70 percent = 70*p.

As per the information provided :

His share of profit decreases by ₹ 420 if the overall profit goes down from 18% to 15%.

Since the profits are distributed in the ratio of their investments :

With a 3% decrease in the profits the value of profit earned by A decreased by Rs 420 which was 70 percent of the total invested.

Hence for all three of them would be combinedly losing $$\left(420\right)\cdot\left(\frac{10}{7}\right)\ =\ 600$$

Hence 3 percent profit was equivalent to  Rs 600.

The initial investment is equivalent to Rs 20000.

This is the total amount invested.

Chintu's profit share increased by Rs 80 when the profit percentage increased by 2 %. A 2 percent increase in profit is equivalent to Rs 20000*2/100 = Rs 400.

Of which Rs 80 is earned by Chintu which is 20% of the total Rs 400.

Hence he invested 20% of the total amount.

Bobby invested the other 10 percent.

10 percent of Rs 20000 = Rs 2000

Let A fill the tank at x liters/hour and B drain it at y liters/hour 
Now as per Condition 1 : 
We get Volume filled till 10pm = 8x-7y       (1) .
Here A operates for 8 hours and B operates for 7 hours .
As per condition 2
We get Volume filled till 6pm = 4x-2y        (2)
Here A operates for 4 hours and B operates for 2 hours .
Now equating (1) and (2)
we get 8x-7y =4x-2y
so we get 4x =5y 
y =4x/5
So volume of tank = $$8x-7\times\ \frac{4x}{5}=\frac{12x}{5}$$
So time taken by A alone to fill the tank = $$\frac{\frac{12x}{5}}{x}=\frac{12}{5}hrs\ $$
= 144 minutes