Rankers Hub

Mcq on profit and loss for bank clerk prelims

profit and loss questions for bank po, profit and loss questions for bank po with solution, profit and loss questions for bank clerk,

51. An article was bought for ₹ 5600. Its price was marked up by 12%. Thereafter it was sold at a discount of 5% on the market price. What was the market price of the article?

(a) ₹ 6207/-

(b) ₹ 6242/-

(c) ₹ 6292/-

(d) ₹ 6192/-

(e) ₹ 6272/-


Solution: (e)
Cost price of article = ₹ 5600
Marked price = \(\displaystyle 5600+5600\times \frac{{12}}{{100}}\)
SP = \(\displaystyle 6272+6272\times \frac{5}{{100}}=5958.4=6272\)

52. An article was bought for ₹ 5600. Its price was marked up by 12%. Thereafter it was sold at a discount of 5% on the market price. What was the percent profit on the transaction?

(a) 6.8%

(b) 6.3%

(c) 6.4%

(d) 6.6%

(e) 6.2%


Solution: (c)
SP = 6272 × 95% = 5958.4
Profit = 5958.4 – 5600 = 358.4.
Profit% = \(\displaystyle \frac{{5958.4-5600}}{{5600}}\times 100\)
\(\displaystyle \frac{{358.4}}{{56}}=6.4\%\)

53. An article was bought for ₹ 5600. Its price was marked up by 12%. Thereafter it was sold at a discount of 5% on the market price. What was the amount of discount given?

(a) ₹ 319.6

(b) ₹ 303.6

(c) ₹ 306.3

(d) ₹ 313.6

(e) ₹ 316.9


Solution: (d)
Price of article=\(\displaystyle 5600\times \frac{{12}}{{100}}+5600=672+5600=6272\)
Amount of discount = \(\displaystyle 6272\times \frac{5}{{100}}=313.6\)

54. The cost price of an article is ₹ 1700. If it was sold at a price of ₹ 2006, what was the percentage profit on the transaction?

(a) 18

(b) 12

(c) 10

(d) 15

(e) 20


Solution: (a)
%profit = \(\displaystyle \frac{{2006-1700}}{{1700}}\times 100=\frac{{306}}{{1700}}\times 100=18\%\)

55. Manish brought 25 kg of rice at ₹ 32 per kg and 15 kg of rice at ₹ 36 per kg. what profit did he get when he mixed the two varieties together and sold it at ₹ 40.20 per kg?

(a) 25%

(b) 40%

(c) 30%

(d) 20%

(e) None of these


Solution: (d)
C.P. of 40 kg of mixture
= ₹ \(\displaystyle \left[ {(25\times 32)+(15\times 36)} \right]\)
= ₹ \(\displaystyle (800+540)\)
= ₹ 1340
S.P.of 40 kg of mixture = ₹ \(\displaystyle (4\times 40.2)\)
Profit= ₹ \(\displaystyle (1608-1340)\)= ₹ 268
Profit % = \(\displaystyle \frac{{268}}{{1340}}\times 100=268\)
\(\displaystyle \frac{{268}}{{130}}\times 100=20\%\)

56. A grocer purchased 80 kg of sugar at ₹ 13.50 per kg and mixed it with 120 kg sugar at ₹16 per kg. At what rate should he sell the mixture to gain 16% ?

(a) ₹ 17 per kg

(b) ₹ 17.40 per kg

(c) ₹ 16.5 per kg

(d) ₹ 16 per kg

(e) None of these


Solution: (b)
C.P. of 200 kg of mixture = ₹ \(\displaystyle (80\times 13.50+120\times 16)\)= ₹3000.
S.P. = 116% of ₹ 3000 = \(\displaystyle \frac{{116}}{{100}}\times 3000=3480\)
\(\displaystyle \Rightarrow \)Rate of S.P. of the mixture = ₹ \(\displaystyle \frac{{3480}}{{200}}\)
= ₹ 17.40 per kg.

57. A shopkeeper purchased 200 bulbs for ₹ 10 each. However, 5 bulbs were fused and had to be thrown away. The remaining were sold at ₹ 12 each. What will be the percentage profit ?

(a) 25

(b) 15

(c) 13

(d) 17

(e) None of these


Solution: (d)
Total cost price = \(\displaystyle 200\times 10\) = ₹ 2000
Total selling price = \(\displaystyle 12\times 195\)= ₹ 2340
Therefore, Profit percent = \(\displaystyle \frac{{2340-2000}}{{2000}}\times 100=17\%\)
= ₹ 17.40 per kg.

58. 10% discount and then 20% discount in succession is equivalent to total discount of

(a) 15%

(b) 30%

(c) 24%

(d) 28%

(e) None of these


Solution: (d)
Successive discount can be given by = \(\displaystyle x+y+\frac{{xy}}{{100}}\)
= \(\displaystyle -10-20+\frac{{-10\times -20}}{{100}}=-30+2=28\%\)
Hence, the successive dicount in equal to 28%
Alternate method:
Let the MP was Rs 100
After first discount, price = 100 – 10 = Rs. 90
After second discount, price = 90 – (90 × 20)/100 = 90 – 18 = Rs. 72
Therefore, Total single discount = [(100 – 72)/100] × 100 = 28%

59. Allowing 20% and 15% successive discounts, the selling price of an article becomes ₹3,060; then the marked price will be

(a) ₹4,400

(b) ₹5,000

(c) ₹4,500

(d) ₹4,000

(e) None of these

Solution (c)

S.P. of an article = 20% and 15% successive discount \(\displaystyle \times \)marked price of an article

\(\displaystyle 3060=\frac{{80}}{{100}}\times \frac{{85}}{{100}}\times MP\)

Therefore, Marked Price of an article = \(\displaystyle \frac{{3060\times 100\times 100}}{{80\times 85}}=4500\)

Alternative method

The two successive discounts are 20% and 15%.

S.P = 3060

We know,

\(\displaystyle SP=MP\left( {1-\frac{{d1}}{{100}}} \right)\left( {1-\frac{{d2}}{{100}}} \right)\)

where d1 and d2 are the given discount.

Now, let the M.P be \(\displaystyle x\).

\(\displaystyle \Rightarrow 3060=x\left( {1-\frac{{20}}{{100}}} \right)\left( {1-\frac{{15}}{{100}}} \right)\)

\(\displaystyle \Rightarrow 3060=x\left( {\frac{4}{5}} \right)\left( {\frac{{17}}{{20}}} \right)\)

\(\displaystyle \Rightarrow x=3060\left( {\frac{5}{4}} \right)\left( {\frac{{20}}{{17}}} \right)\)

\(\displaystyle \Rightarrow x=4500\)

One more method

We know that,

Equivalent discount = a + b – (ab/100)

where, a is the first discount. b is the second discount.

Total discount % = \(\displaystyle 20+15-\frac{{20\times 15}}{{100}}=35-3=32\%\)

32% discount on MRP of Rs. 100

So, for 1 part/unit

MRP \(\displaystyle \to \) Discount \(\displaystyle \to \) Selling Price

100 \(\displaystyle \to \) 32 \(\displaystyle \to \) 68

\(\displaystyle \Rightarrow \) 68 parts ——- 3060

Then, 1 part ———–?

? = \(\displaystyle {\frac{{3060}}{{68}}}\) = 45

⇒ then MRP of 100 parts = 45 \(\displaystyle \times \) 100 = Rs. 4500


60. The average weight of 15 oarsmen in a boat is increased by 1.6 kg when one of the crew, who weighs 42 kg is replaced by a new man. Find the weight of the new man (in kg).

(a) 65

(b) 66

(c) 43

(d) 67

(e) None of these


Solution: (b)
Let the average weight of 15 Oarsmen at the start = x kg
Let the new man’s weight = y kg
According to question
15x – 42 = 15 (x + 1.6) – y
15x – 42 = 15x + 24 – y
y = 24 + 42 = 66 kg