# Profit and Loss Test Paper 5

21) The marked price of a radio is Rs 480. The owner allows a discount of 10% and still earns the profit of 8%. If no discount is allowed, find his gain percent?

1. 25%
2. 12%
3. 15%
4. 20%

Explanation:

Marked price= 480
The shopkeeper allows a discount of 10%, so, the Selling price will be= 480 * 90/100 = 432.

And the profit is 8%.

Apply formula:

C.P = (100/ (100 + p %))* S.P

C.P = (100/108) * 432
C.P = 400

If the discount is not allowed:
Let Profit= P%

S.P = Marked price = 480, and C.P = 400
Profit = S.P - C.P

Now, apply formula:

P% = (Profit/C.P) * 100

So, gain % = ((480-400)/ 400) * 100
= 80/4 = 20%

22) 5% more is gained by selling a cow for Rs 1010 than by selling it for Rs 1000. What is the C.P of the cow?

1. Rs 100
2. Rs 150
3. Rs 200
4. Rs 250

Explanation:

ATQ, the difference between profit % = 5
The difference between S.P = 10

Apply formula:

C.P = (100* Difference between S.P)/ (% difference in profit)

Hence, the C.P = (100 * 10)/ 5 = 200.

The difference between profit = 5%
Let the C.P = x
And S.P2= 1010, S.P1 = 1000

i.e., P2 % - P1% = 5.................................................... (i)

Apply formula of percent

P% = (Profit/C.P) * 100

And profit = S.P - C.P.

Now according to equation (i)

((S.P 2- x)/ x)*100 - ((S.P 1- x)/ x)*100 = 5

((1010- x)/ x)*100 - ((1000- x)/ x)*100 = 5

Or, take (100/x) as common from the whole equation.

i.e., (100/x) [(1010-x) - (1000-x)] = 5

(100/x) [1010 - x - 1000 + x] = 5

Or, (100/x) * 10 = 5

Or, 5x = 1000
Hence, x = 200, or C.P = Rs 200.

23) A dishonest milkman sells milk at cost price, but he mixes water and earns 16(2/3) % profit. Find the ratio of mixture and milk in the mixture.

1. 7: 5
2. 7: 6
3. 7: 7
4. 7: 8

Explanation:

Solution 1:

Let C.P of 1-litre milk = 10
Brought milk = x liters
Then C.P = 10x
Let water mixed in milk = y liter
Then quantity of mixture = (x+y) liter

Mixture is sold on C.P
Then S.P = 10(x + y)
= 10x + 10y

The profit = S.P- C.P = (10x + 10y) - 10x
= 10y
Profit % = (10y/10x)*100 = 16(2/3) %

Or, y/x = (50/3)*(1/100)
Or, y/x = (50/300)
Or, y/x = (1/6)
So, mixture = x+y = 6+1 = 7
Then, (mixture/milk) = 7/6 or 7:6

Apply formula:

Solution 2:

16(2/3) % profit = (50/3)*(1/100) = 1/6
i.e., +1>water
6> milk
That means if the total mixture is 6+1 = 7
Then milk = 6
Mixture: milk = 7: 6

24) Ayaan bought 30 kg of rice at the rate of Rs 9.50/ kg and 40 kg rice at the rate of Rs 8.50/kg and mixed them. He sold the mixture at the rate of Rs. 8.90/kg. Find his total profit or loss in the transaction.

1. Rs 2 loss
2. Rs 5 profit
3. Rs 2 profit
4. Rs 7 loss

Explanation:

The total cost price = 30* 9.5 + 40* 8.5 = 625
The total selling price = 8.9(30+40) = 623

cost price> selling price = Loss
Hence, the total loss in this transaction = 625-623= Rs 2

25) If a retail seller buys 30 pencils at the marked price of 27 pens, find the profit percent.

1. 11(1/9)%
2. 9(1/11)%
3. 10%
4. 20%

Explanation:

ATQ, if the retailer seller purchase 27 pens, he have gained 3 pens extra
i.e., profit = 3 pens, and C.P = 27
Apply profit % formula:

P% = (profit/C.P) * 100

P% = (3/27) * 100 = 100/9 So, the Profit % = 11(1/9) %.

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Profit & Loss Concepts   