Download Profit and Loss Questions for SBI PO Mains, IBPS Mains and RBI Grade B 2021

Directions : Read the following questions carefully and choose the right answer.
1
Tarak Mehta sold a pair of Tshirt and jeans at 25% profit. The profit obtained on Tshirt was 20%. Had he reduced the both cost price of the Tshirt and selling price of the Tshirt by Rs. 200, his profit on the pair would have risen to 30%. Earlier cost price of the jeans was Rs.80 less than half of the selling price of the Tshirt. What price should he mark on a pair of Tshirt and jeans to earn a profit of 50%?
» Explain it
B
Let Cost price of each Tshirt = s
 
Cost price of each Jeans = p
 
Selling price of each Tshirt = h 
 
And, Selling price of each Jeans = a
 
From question 
 
h + a = 1.25 (s + p) …….(i)
 
h = 1.2s …… (ii)
 
(h + a – 200) = 1.3 (s + p – 200) …. (iii)

And, p =  h – 80 ……… (iv)
2

Solving the above system of equation we get, 
 
s = 800, p = 400, h = 960 and, a = 540 
 
Total cost price of a pair of Tshirt and jeans = 800 + 400 = 1200
 
Required marked price = 1200 × 1.5 = Rs.1800
 
Hence, option (B) is correct.
2
Sohan bought an old Honda Bike and spent Rs. 1500 on its repairs. Then Sohan sold it to Rakesh at a prolit of 20%. Rakesh sold it to Raj at a loss of 10%. Raj finally sold it for Rs. 12100 at a profit of 10%. How much did Sohan pay for the old Honda Bike?
» Explain it
C
Cost price of Honda Bike for Raj =  12100  = 11000
1.1

Since selling price of Honda Bike by Rakesh = Rs. 11000

From question Cost price of Honda Bike for Rakesh

 =  11000  = Rs. 12222.22
0.9

Since selling price of Honda Bike by Sohan = Rs. 12222.22

According to question sohan got 20% profit

Since Cost Price of Honda Bike for Sohan

 =  12222.22  = Rs. 10185.2
1.2

This cost price is includes a Rs. 1500 for repairs.

Hence purchase price for Sohan = 10185.2 – 1500 = Rs. 8685

Therefore option (C) is correct.
3
Sourav Ganguly wants to buy a total of 100 sports equipment using exactly a sum of Rs.1000. He can buy ball at Rs.20 per unit, wicket at Rs.5 per unit and bat at Rs.1 per unit. If he has to buy at least one of each equipment and cannot buy any other type of equipment, then in how many distinct ways can he make his purchase?
» Explain it
C
Let the number of ball, wickets and bats purchased be A, B and C, respectively.
 
Thus,
 
20A + 5B + C = 1000 and A + B + C = 100
 
Solving the above two equations by eliminating C, we Get 
 
19A + 4B = 900
 
⇒ B = 225 –  19 A
4
 

Now, as B is the number of wickets and 0 < B < 99,
 
So, putting these limiting values of B in the above equation will provide the value of A as 27 < A < 47.
 
Since A has to be the multiple of 4, so possible values of A are 28, 32, 36, 40 and 44.
 
Now, for A = 28 and 32; A + B > 100, so these values of A can be rejected.
 
For all other values of A, we get the desired solution:
 
A = 36, B = 54, C = 10
 
A = 40, B = 35, C = 25
 
A = 44, B = 16, C = 40
 
Thus, there are three possible solutions.

Hence, option (C) is correct.
4
In St. Peter’s college Agra an exhibition was organised, hand-made crafts are displayed for sale. Some students are assigned the work of selling crafts. The overall profit p depends on the number of students x selling the crafts on that particular day and is given by the equation p = 250x – 5x2. The school manager claims to have made a maximum profit. Find the number of students engaged in selling the crafts and the maximum profit made. 
» Explain it
C
For profit to be maximum, the derivative of p with reference to x must be 0 and hence

d (250x – 5x2)  = 0 = 250 – 10x = 0
dx

So x = 25 
 
Now p for x = 25 is 
 
= 250 (25) – 5 (25)2 = Rs3125 

Hence, option (C) is correct.
5
In MG Road Delhi PVR has 300 seats. The price of each ticket, when the theatre is houseful, is Rs.60. For every Rs.1 increase in the price of the ticket, the number of tickets sold goes down by 2. What is the price of the ticket (in Rs.) for which the theatre owner would earn the maximum possible revenue?
» Explain it
B
Let the price of the ticket be = Rs. (60 + x), where x is greater than zero. 
 
The number of people in the audience would then be (300 – 2 × x). 
 
The revenue of the theatre owner be = (60 + x) × (300 – 2 × x) 
 
= (18000 + 180 × x – 2 × x2
 
This is quadratic expression which achieves a maximum value when x 
 
= –  coefficient of x
2 × coefficient of x2

Quadratic equation has achieved a maximum value when x

= –  b
2a

So, x = –  180  = 45
– 2 × 2

Hence, the price of ticket at maximum revenue = (60 + 45) = Rs.105. 

Therefore, option (B) is correct.