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profit and loss questions for bank exams

profit and loss mcq for bank exams, profit and loss mcq for banking, profit and loss mcq for banking and insurance,

11. The profit earned after selling a pair of shoes for ₹ 2,033 is the same as loss incurred after selling the same pair of shoes for ₹ 1,063. What is the cost of the shoes ?

(a) ₹ 1,650

(b) ₹ 1,548

(c) ₹ 1,532

(d) Cannot be determined

(e) None of these


Solution: (b)
Let the CP of the shoes be ₹ x.
Therefore, 2033 – x = x – 163
\(\displaystyle \Rightarrow \) 2x = 2033 + 1063 = 3096
\(\displaystyle \Rightarrow \) x = \(\displaystyle \frac{{3096}}{2}=1548\)

12. Gauri went to the stationery and bought things worth ₹25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?

(a) ₹15

(b) ₹15.70

(c) ₹19.70

(d) ₹20

(e) None of these


Solution: (c)
Let the amount taxable purchases be Rs. x.
Then, 6% of x = \(\displaystyle \frac{{30}}{{100}}\)
\(\displaystyle \Rightarrow \) x = \(\displaystyle (\frac{{30}}{{100}}\times \frac{{100}}{6})=5\)
Cost of tax-free items = ₹ [25 – (5 + 0.30)] = ₹19.70

13. Naresh purchased a TV set for ₹11,250 after getting discount of 10% on the labelled price. He spent ₹150 on transport and ₹800 on installation. At what price should it be sold so that the profit earned would be 15% if no discount was offered?

(a) ₹12,937.50

(b) ₹14,030

(c) ₹ 13,450

(d) ₹15,467.50

(e) None of these


Solution: (d)
Cost price of TV when discount is not offered
\(\displaystyle 11250\times \frac{{100}}{{90}}=12500\)
Total cost of TV after transport and installation
= 12500 + 800 + 150 = 13450
To earn 15% profit, he must sell at
\(\displaystyle 13450\times \frac{{115}}{{100}}=15467.50\)

14. A person sold an article from ₹3600 and got a profit of 20%. Had he sold the article for  ₹ 3150, how much profit would he have got?

(a) 4%

(b) 5%

(c) 6%

(d) 10%

(e) None of these


Solution: (b)
Let the cost price of the article be ₹ x
After 20% profit
\(\displaystyle \Rightarrow \) x = \(\displaystyle \frac{{120x}}{{100}}=3600\)
x = 3000
Now, profit percentage, when the article is sold for ₹ 3150
\(\displaystyle \Rightarrow \) \(\displaystyle \frac{{3150-3000}}{{3000}}\times 100-\frac{{150}}{{3000}}\times 100=5\%\)

15. A refrigerator and a camera were sold for Rs. 12000 each. The refrigerator was sold at a loss of 20% of the cost and the camera at a gain of 20% of the cost. The entire transaction results in which one of the following?

(a) No loss or gain

(b) Loss of ₹ 1000

(c) Gain of ₹ 1000

(d) Loss of ₹ 2000

(e) None of these


Solution: (b)
\(\displaystyle X+Y+\frac{{XY}}{{100}}=+20-20-\frac{{20\times 20}}{{100}}=-4\%\)
Total selling price of a refrigerator and a camera
= 12000 + 12000 = ₹ 24000
Now, loss is 4%
\(\displaystyle CP\times \frac{{96}}{{100}}=24000\)
CP = ₹ 25000
Loss amount = (25000 – 24000) = ₹ 1000

16. A milkman bought 15 kg of milk and mixed 3 kg of water in it. If the price per kg of the mixture becomes ₹ 22, what is cost price of the milk per kg?

(a) ₹ 28.00

(b) ₹ 26.40

(c) ₹ 24.00

(d) ₹ 22.00

(e) None of these

Solution : (b)

Let cost price of milk ₹ x per kg.

Price of 15kg of milk = ₹ 15x.

Now, mix 3kg of water, therefore quantity of mixture

= (15 + 3) kg = 18 kg

So, price of mixture is ₹22 per kg

According to question.

15x = 22 × 18

x= \(\displaystyle \frac{{22\times 18}}{{15}}=\frac{{132}}{5}=26.40\)

 Alternate Method :

 Let CP of milk be ₹ x per kg.

By Alligation method

A milkman bought 15 kg of milk and mixed 3 kg of water in it. If the price per kg of the mixture becomes ₹ 22, what is cost price of the milk per kg?

\(\displaystyle \Rightarrow \)\(\displaystyle \frac{{22}}{{x-22}}=\frac{{15}}{3}\)

\(\displaystyle \frac{{22}}{{x-22}}=5\)

\(\displaystyle \Rightarrow \) 22 = 5x – 110

\(\displaystyle \Rightarrow \) 22 = 132

Therefore, x = ₹ 26.40

17. The price of an article is ₹ 25. After two successive cuts by the same percentage, the price becomes ₹ 20.25. If each time the cut was x%, then

(a) x = 9

(b) x =10

(c) x =11

(d) x = 11.5

(e) None of these


Solution: (b)

According to the question,

\(\displaystyle \Rightarrow \) \(\displaystyle 25\times (\frac{{100-x}}{{100}})(\frac{{100-x}}{{100}})=20.25\)

\(\displaystyle \Rightarrow \)\(\displaystyle {{(100-x)}^{2}}=\frac{{202500}}{{25}}\)

\(\displaystyle \Rightarrow \)\(\displaystyle {{(100-x)}^{2}}=8100\)

\(\displaystyle \Rightarrow \) 100 – x = 90

Therefore,  x =10

Alternate method

We know that, Successive discount is given by = \(\displaystyle \left( {a+b-\frac{{ab}}{{100}}} \right)\%\)

Here a and b are same. Lets say \(\displaystyle x\)

Then,

Successive discount= \(\displaystyle \left( {x+x-\frac{{x\times x}}{{100}}} \right)\%\)

Given

Price of article = 25

After two successive discount price= 20.25

Then, Profit= 25 – 20.25 = 4.75

Profit % = \(\displaystyle \frac{{SP-CP}}{{CP}}\times 100\)$ = \(\displaystyle \frac{{4.75}}{{25}}\times 100\) = 19%

\(\displaystyle \Rightarrow \left( {x+x-\frac{{x\times x}}{{100}}} \right)=19\)

\(\displaystyle \Rightarrow \left( {2x-\frac{{{{x}^{2}}}}{{100}}} \right)=19\)

\(\displaystyle \Rightarrow \left( {200x-{{x}^{2}}} \right)=190\)

\(\displaystyle \Rightarrow {{x}^{2}}-200x+1900=0\)

\(\displaystyle \Rightarrow {{x}^{2}}-10x-190x+1900=0\)

\(\displaystyle \Rightarrow x=10or190\)

190 is not possible so \(\displaystyle x\) =10     

We can also solve the equation by elimination method from options.

\(\displaystyle x\) =10 is the only option which satisfies the condition of the equation \(\displaystyle \left( {x+x-\frac{{x\times x}}{{100}}} \right)=19\)

18. A dealer marked the price of an item 40% above the cost price. Once he gave successive discounts of 20% and 25% to a particular customer. As a result, he incurred a loss of  ₹ 448. At what price did he sell the item to the mentioned customer?

(a) ₹ 2416

(b) ₹ 2268

(c) ₹ 2352

(d) ₹ 2152

(e) ₹ 2578


Solution: (c)
Let the cost price of the item be 100.
A dealer marked the price of an item 40% above the cost price. Once he gave successive discounts of 20% and 25% to a particular customer. As a result, he incurred a loss of ₹ 448. At what price did he sell the item to the mentioned customer?

\(\displaystyle \Rightarrow \)Loss = 16% and Loss = ₹ 448
\(\displaystyle \Rightarrow \)CP = \(\displaystyle \frac{{448\times 100}}{{16}}=2800\)
\(\displaystyle \Rightarrow \)SP = \(\displaystyle \frac{{2800\times 84}}{{100}}=2352\)

Alternate method

Let the cost price be Rs. 100

Then, CP = 100 means MP = 140

20% discount = 112

25% discount (SP = Rs. 84)

If Rs. 16 loss, cost price Rs. 100

If Rs. 448 loss, cost price = \(\displaystyle \frac{{100}}{{16}}\times 448\)

\(\displaystyle \Rightarrow \) CP = 28 \(\displaystyle \times \) 100 = Rs. 2800

Selling Price =  \(\displaystyle \frac{{2800\times 84}}{{100}}\) = Rs. 2352

19. Deepak found that he had made a loss of 10% while selling his smartphone. He also found that had he sold it for Rs.50 more, he would have made a profit of 5%. The initial loss was what percentage of the profit earned, had he sold the smartphone for a 5% profit ?

(a) 100%

(b) 200%

(c) 75%

(d) 85%

(e) None of the Above


Solution: (b)
Profit = 5%
5% of CP = ₹ 50
CP = ₹ 1000
Now, Loss% = 10%
Loss = ₹ 100
Required % = \(\displaystyle \frac{{100}}{{50}}\times 100=200\%\)

20. On selling an article for ₹ 651, there is a loss of 7%. The cost price of that article is

(a) 744

(b) 751

(c) 793

(d) 700

(e) 750


Solution: (d)
Let the C.P. of article be ‘x’
\(\displaystyle (100-7)\%x=651\)
x= \(\displaystyle \frac{{651}}{{93}}\times 100=700\)
Alternate method:
C.P. = \(\displaystyle SP(\frac{{100}}{{100-loss\%}})\)
CP=\(\displaystyle 651(\frac{{100}}{{100-7}})\)
CP=\(\displaystyle \frac{{651\times 100}}{{93}}\)
C.P = Rs. 700