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Profit And Loss - Aptitude MCQ Questions and Solutions with Explanations

Profit And Loss MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Profit And Loss with easy and logical explanations. Profit And Loss MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-7 section-3

Each section contains 100 MCQs on Profit and Loss .
61
To gain 10% on selling the mixture of milk and water at the cost price of pure milk, the quantity of water to be mixed with 50 kg of pure milk is = ?
Discuss
Answer & Solution
Answer: Option B
Solution:
Let cost price of 1 kg = Rs. 1
Cost price of 50 kg = Rs. 50
$$\eqalign{ & \Rightarrow {\text{Profit}} = \frac{{10}}{{100}} \times {\text{50}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. 5}} \cr & \therefore {\text{Quantity to added}} \cr & = \frac{5}{1} = {\text{5 kg}} \cr} $$
62
A milkman bought 70 litres of milk for Rs. 630 and added 5 litres of water. If sells it at Rs. 9.00 per litres, his profit percentage is = ?
Discuss
Answer & Solution
Answer: Option D
Solution:
According to the question,
Cost price of 70 litres of milk = Rs. 630
Added 5 litres of water
Now, the solution becomes = 75 litres
Cost price of water = Rs. 0
∴ Selling price of 1 litre milk = Rs. 9
Selling price of 75 litres milk = 9 × 75 = Rs. 675
Profit = Selling price - Cost price
= 675 - 630
= 45
$$\eqalign{ & {\text{Profit}} = \frac{{\operatorname{Profit} }}{{{\text{Cost price}}}} \times 100 \cr & = \frac{{45}}{{630}} \times 100 \cr & = \frac{{50}}{7} \cr & = 7\frac{1}{7}\% \cr} $$
63
A farmer sold a cow and an ox for Rs. 800 and got a profit of 20% on the cow and 25% on the ox. If he sells the cow and the ox for Rs. 820 he gets a profit of 25% on the cow and 20% on the ox. The individual cost price of the cow and the ox is :
Discuss
Answer & Solution
Answer: Option A
Solution:
Let the C.P. of the cow be Rs. x and that of the ox be Rs. y
Then,
= 120% of x + 125% of y = 800
$$ \Rightarrow \frac{{6x}}{5} + \frac{{5y}}{4} = 800$$
⇒ 24x + 25y = 16000 .......(i)
And
125% of x + 120% of y = 820
$$ \Rightarrow \frac{{5x}}{4} + \frac{{6y}}{5} = 820$$
⇒ 25x + 24y = 16400 ......(ii)
Adding (i) and (ii), we get :
49x + 49y = 32400
$$ \Rightarrow x + y = \frac{{32400}}{{49}}......(iii)$$
Substracting (i) from (ii), we get :
x - y = 400 ........(iv)
Adding (iii) and (iv), we get :
$$\eqalign{ & 2x = \frac{{32400}}{{49}} + 400 \cr & \,\,\,\,\,\,\,\,\, = \frac{{52000}}{{49}} \cr & \Rightarrow x = \frac{{26000}}{{49}} \approx 530.60 \cr & {\text{Putting }}x = \frac{{26000}}{{49}}{\text{ in }}(iii), \cr & {\text{we get}}: \cr & y = \frac{{32400}}{{49}} - \frac{{26000}}{{49}} \cr & \,\,\,\,\,\, = \frac{{6400}}{{49}} \approx 130.60 \cr} $$
64
The C.P. of two watches taken together is Rs. 840. If by selling one at a profit of 16% and the other at a loss of 12% , there is no loss or gain in the whole transaction, then the C.P. of the two watches are respectively.
Discuss
Answer & Solution
Answer: Option A
Solution:
C.P. = Cost Price.
Let the C.P. of the watches be Rs. x and Rs. (840 - x)
∴ (116% of x) + [88% of (840 - x)] = 840
⇒ 116x + 73920 - 88x = 84000
⇒ 28x = 10080
⇒ x = 360
∴ Their cost prices are Rs. 360 and Rs. 480
65
A space research company wants to sell its two products A and B. If the product A is sold at 20% loss and the product B at 30% gain, the company will not lose anything. If the product A is sold at 15% loss and the product B at 15% gain, the company will lose Rs. 6 million in the deal. What is the cost of product B ?
Discuss
Answer & Solution
Answer: Option A
Solution:
Let the cost of product A be Rs. x and B be Rs. y
According to first sell
    0.80x + 1.3y = x + y
⇒ 0.2x = 0.3y
⇒ x = $$\frac{3\text{y}}{2}$$ . . . . . . (i)
According to second sell
    0.85x + 1.15y = (x + y) - 6
⇒ 0.15y = 0.15x - 6 (Multiply by 100)
⇒ 15y = 15x - 600 (Putting the value of x)
⇒ 15y = 15 × $$\frac{3\text{y}}{2}$$ - 600
⇒ 15y = $$\frac{45\text{y}}{2}$$ - 600 (Multiply by 2 both side)
⇒ 30y = 45y - 1200
⇒ 15y = 1200
⇒ y = 80 million
66
A milkman makes 20% profit by selling milk mixed with water at Rs. 9 per litre, if the cost price of 1 litre pure milk is Rs. 10, then the ratio of milk and water in the said mixture is = ?
Discuss
Answer & Solution
Answer: Option A
Solution:
According to question,
S.P. at 20% profit = Rs. 9/litres
Cost of mixture $$ = \frac{9}{{120}} \times 100 = {\text{Rs}}{\text{.}}\,7.5/{\text{litre}}$$
Now, let the ratio of milk and water in the mixture = x : y
$$\eqalign{ & \Rightarrow \frac{{\left( {10 \times x} \right) + \left( {0 \times y} \right)}}{{x + y}} = 7.5 \cr & \Rightarrow 10x = 7.5\left( {x + y} \right) \cr & \Rightarrow x:y = 3:1 \cr} $$
67
A shopkeeper bought 15 kg of rice at the rate of Rs. 29 per kg and 25 kg of rice at the rate of Rs. 20 per kg. He sold the mixture of both types of rice at the rate of Rs. 27 per kg. His profit in this transaction is = ?
Discuss
Answer & Solution
Answer: Option D
Solution:
According to the question,
Cost price of mixture of rice
= 15 × 29 + 25 × 20
= 435 + 500
= Rs. 935
Selling price of 1 kg mixture of rice = Rs. 27
Selling price of 40 kg mixture of rice
= 27 × 40 = Rs. 1080.
Profit = Selling price - Cost price
= 1080 - 935
= Rs. 145
68
On selling 17 balls at Rs. 720 there is a loss equal to the cost price of 5 balls. The cost price of a ball is = ?
Discuss
Answer & Solution
Answer: Option C
Solution:
According to the question,
Let cost price of 1 ball is = Rs. x
∴ Cost price - Selling price = Loss
17x - 720 = 5x
12x = 720
⇒ x = 60
∴ Cost price of 1 ball is Rs. 60
69
A small and medium enterprise imports two components A and B from Taiwan and China respectively and assembles them with other components to form a toy. Components A contributes to 10% of production cost while components B contributes to 20% of production cost. Usually the company sells this toy at 20% above the production cost. Due to increase in the raw material and labour cost in both the countries ,component A became 20% costlier and components B became 40% costlier. Owing to these reasons the company increased its selling price by 15%. Considering that cost of other components does not change. What will be the profit percentage if the toy is sold at the new price?
Discuss
Answer & Solution
Answer: Option B
Solution:
Let the original cost of the toy be Rs. 100.
Then,
Original cost of component A
= 10% of Rs. 100
= Rs. 10
Original cost of component B
= 20% of Rs. 100
= Rs. 20
Original S.P. of the toy
= 120% of Rs. 100
= Rs. 120
New cost of component A
= 120% of Rs. 10
= Rs. 12
New cost of component B
= 140% of Rs. 20
= Rs. 28
New price of the toy
= Rs. [100 + (12 + 28) - (10 + 20)]
= Rs. 110
New S.P. of the toy
= 115% of Rs. 120
= Rs. 138
Profit = Rs. ( 138 - 110) = Rs. 28
$$\eqalign{ & \therefore {\text{Profit }}\% = \left( {\frac{{28}}{{110}} \times 100} \right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 25.45\% \approx 25.5\% \cr} $$
70
A firm of readymade garments makes both men's and women's shirts. Its average profit is 6% of the sales. Its profit in men's shirt average 8% of the sales and women's shirts comprise 60% of the output. The average profit per sales rupee in women's shirts is
Discuss
Answer & Solution
Answer: Option B
Solution:
Let the total sales be Rs. x and let the profit in women's shirts average y% of the sales.
Then,
= 8% of 40% of x + y% of 60% of x = 6% of x
$$\eqalign{ & \Rightarrow \frac{8}{{100}} \times \frac{{40}}{{100}} \times x + \frac{y}{{100}} \times \frac{{60}}{{100}} \times x = \frac{6}{{100}} \times x \cr & \Rightarrow \frac{{32}}{{10}} + \frac{{6y}}{{10}} = 6 \cr & \Rightarrow \frac{{6y}}{{10}} = \frac{{28}}{{10}} \cr & \Rightarrow y = \frac{{28}}{6} = 4.66 \cr} $$
∴ Average profit per sales rupee in women's shirt
$$\eqalign{ & = 4.66\% {\text{ of Rs}}.1 \cr & = {\text{Rs}}.\left( {\frac{{4.66}}{{100}}} \right) \cr & = {\text{Rs}}{\text{. }}0.0466 \cr} $$