Crack IBPS Exams 2017 – Topic wise Discussion on Quantitative Aptitude Day2 (Profit and Loss: PART2):
Dear Readers, We all knew a famous saying of the Legend Aristotle, “Well begun is half done” and we also knew that this statement is 100% true. To Crack the Bank Exams you need to know where to start and how to start. Now this is the absolute correct time to kick start the preparation for upcoming IBPS Exams 2017. To help you in this aspect and to be a part of your preparation here we, IBPS Guide Team providing Topic wise Discussion on Quantitative Aptitude, this session will be conducted regularly on daily basis. This will provide a complete overview on the topics along with exercise questions. Kindly Make use of it.
PROFIT AND LOSS: PART2
Type 6:
1). An article was purchased for Rs. 78,350/. Its price was marked up by 30%. It was sold at a discount of 20% on the marked up price. What was the profit percent on the cost price?
Solution:
Cost Price = Rs.78350
Marked Price = 78350 x 130/100 = Rs.101855
Selling Price = 101855 x 80/100 = Rs.81484
Profit = 81484 – 78350 = 3134
Therefore Profit% = 3134/78350 x 100 = 4%
Type 7:
2). A shopkeeper purchases 12 balloons for Rs.10 and sells them at 10 balloons for Rs.12.Thus, he earns a profit of
Solution:
Number of Balloons purchased and sold be = LCM (12,10) = 60
CP = 60/12 x 10 = 50; SP = 60/10 x 12 = 72
Profit = (72 – 50)/50 x 100 = 22/50 x 100 = 44%
Type 8:
3). The cost price of an article is Rs. 800. After allowing a discount of 10%, a gain of 12.5% was made. Then the marked price of the article is
Solution:
CP = Rs 800
Gain% = 12.5%
SP = 800 + 12.5% of 800 = 800 + 25/200 x 800 = 800 + 100 = Rs 900
MP = SP + Discount
MP = 900 + 10% of MP
MP – MP/10 = 900
9MP/10 = 900
MP = (900 x 10)/9 = Rs 1000
Type 9:
4). If the selling price of an article is doubled, then its loss per cent is converted into equal profit per cent. The loss per cent on the article is
Solution:
Let CP = C, SP = S, Loss% = x %
=> x = (C – S)/C x 100 ————–(1)
When SP is doubled, loss% becomes profit%.
x = (2S – C)/C x 100 ————–(2)
From eqns (1) and (2)
2S – C = C – S
3S = 2C => S = 2/3 C
Substituting b = 2/3 a in eqn (1)
x = (C – 2/3 C)/C x 100= (C/3)/C x 100 = 100/3 = 33 1/3%
Type 10:
5). A man bought an article listed at Rs. 1500 with a discount of 20% offered on the list price. What additional discount must be offered to man to bring the net price to Rs. 1,104?
Solution:
SP of the article = MP – Discount = 1500 – 20% of 1500 = 1500 – 300 = Rs 1200
Therefore, additional discount = (1200 – 1104)/1200 x 100 = (96/1200) x 100 = 8%
More Types on Profit and Loss Problems will be discussed in the next Session, Kindly follow us daily.
Dear Aspirants, Here below we have given exercise questions on Profit and Loss based on the above types, solve these questions by yourself and comment your answers below. Correct Answers with explanation will be updated in the end of the day.
Exercise Questions:
1). Giri purchased 100 shirts at Rs. 450 per piece. While selling he offered 10%discount on the labeled price and earned a profit of 20%. What was the labeled price of each shirt?
a) Rs. 540
b) Rs. 650
c) Rs. 590
d) Rs. 600
e) None of these
2). If the cost price of 15 Pencils is equal to the selling price of 12 pencils, find gain percent.
a) 20
b) 25
c) 18
d) 21
e) None of these
3). A dealer purchased a printing machine for Rs. 7660. He allows a discount of 12% on its marked price and still gains 10%. Find the marked price of the machine.
a) 9575
b) 8765
c) 10985
d) 8995
e) None of these
4). A fruit seller buys oranges at 2 for a rupee and sells them at 5 for three rupees. His gain percent is
a) 10%
b) 15%
c) 20%
d) 25%
e) 30%
5). The labeled price of a Juicer is Rs.620. If it is sold at 15% discount and the dealers earn a 20% profit, what is the cost price?
a) Rs.439.16
b) Rs.423.16
c) Rs.412.16
d) Rs.436.11
e) None of these
6). A person sold an toy for Rs. 136 and got 15% loss, had he sold it for Rs. N, he would have got a profit of 15%. Which one of the following is correct?
a) 190< N< 200
b) 180< N< 190
c) 170< N< 180
d) 160< N< 170
e) None of these
7). A person bought an old Bike at Rs.7200 and spent 15% of it on repairing. If he wants to make a profit of Rs.648 then what percentage should be added to purchase price?
a) 15%
b) 20%
c) 24%
d) 25%
e) None of these
8). A shopkeeper gave an additional 40% discount on the reduced price after giving 25% standard concession on that item, if a person bought that item for Rs.1260, what is the original price of the item?
a) Rs.2400
b) Rs.2800
c) Rs.3200
d) Rs.2000
e) None of these
9). 20 mangoes were bought for a rupee. Approximately how many mangoes must be sold for a rupee to gain a profit of 30%?
a) 12
b) 10
c) 15
d) 20
e) 25
10). A seller wants to earn 12% profit on an item after giving 20% discount to the customer. By what percentage should he increase his marked price to arrive at the label price?
a) 24%
b) 32%
c) 42%
d) 16%
e) None of these
Answers with Explanation:
1). Answer: d)
Let the labeled price of each shirt be Rs. x.
According to the question,
90% of x = 120 × 450 / 100
90 / 100 × x = 120 × 450 / 100
x = 120 × 450 / 90
x = Rs. 600
2). Answer: b)
Gain percent = (15 – 12) / 12 × 100 = 25%
3). Answer: a)
Cost price of the machine = Rs. 7660, Gain% = 10%.
Therefore, selling price = [{(100 + gain%)/100} × CP]
= Rs. [{(100 + 10)/100} × 7660]
= Rs. [(110/100) × 7660]
= Rs. 8426.
Let the marked price be Rs. x.
Then, the discount = 12% of x
= {x × (12/100)}
= 3x/25
Therefore, SP = (Marked Price) – (discount)
= (x – 3x/25)
= 22x/25.
But, the SP = Rs. 8426.
Therefore, 22x/25 = 8426
⇒ x = (8426 × 25/22)
⇒ x = 9575.
Hence, the marked price of the Printing machine is Rs. 9575.
Therefore, selling price = [{(100 + gain%)/100} × CP]
= Rs. [{(100 + 10)/100} × 7660]
= Rs. [(110/100) × 7660]
= Rs. 8426.
Let the marked price be Rs. x.
Then, the discount = 12% of x
= {x × (12/100)}
= 3x/25
Therefore, SP = (Marked Price) – (discount)
= (x – 3x/25)
= 22x/25.
But, the SP = Rs. 8426.
Therefore, 22x/25 = 8426
⇒ x = (8426 × 25/22)
⇒ x = 9575.
Hence, the marked price of the Printing machine is Rs. 9575.
4). Answer: c)
The cost price of 1 lemon =Rs.0 .50
Selling price of 1 lemon = Rs.0.60;
Selling price of 1 lemon = Rs.0.60;
Therefore, profit = Rs.0.10
:. Gain per cent = 0.10 × 100/0.50 = 20%.
:. Gain per cent = 0.10 × 100/0.50 = 20%.
5). Answer: a)
The selling price = Rs.620 – 15% of 620 = 620 – 93 = Rs.527
Rs.527 is 120% of the cost price;
Therefore, cost price = 527 × 100/120 = Rs.439.16.
6). Answer: b)
Cost price = (selling price x 100)/(100 – loss%)
= (136 x 100)/(100 – 15)
= (136 x 100)/85
= Rs. 160
Selling price (N) = 160 x (100 + 15)/100 = (160 x 115)/100
= Rs. 184
∴ Option (b) is correct because [180< N< 190].
= (136 x 100)/(100 – 15)
= (136 x 100)/85
= Rs. 160
Selling price (N) = 160 x (100 + 15)/100 = (160 x 115)/100
= Rs. 184
∴ Option (b) is correct because [180< N< 190].
7). Answer: c)
CP = Rs.7200
CP + repair cost + profit expected = Rs.7200 + 1080 +648 = Rs.8928.
The amount to be added to CP = Rs.8928 – Rs.7200 = Rs.1728
The percentage of amount to be added to CP = 1728 x 100/7200 = 24%
CP + repair cost + profit expected = Rs.7200 + 1080 +648 = Rs.8928.
The amount to be added to CP = Rs.8928 – Rs.7200 = Rs.1728
The percentage of amount to be added to CP = 1728 x 100/7200 = 24%
8). Answer: b)
Let the original price be ‘x’
The price after 1st concession of 25% = x – 25x/100 = x – x/4 = 3x/4
The Price after additional discount 40% = 3x/4 – 3x/4 X 40/100 = 30x – 12x/40
= 9x/20, i.e. 9x/20 = Rs.1260,
:. x = 1260 x 20 /9 = Rs.2800.
9). Answer: c)
Given 20 mangoes cost 100 paisa.
:. 1 mango’s cost = 5 paisa.
For a getting a profit of 30% , the SP per mango should be equal to 6.5 paise.
:. For 1 Rupee, the No. of mangoes to be sold = 100/6.5
= 15 nos. approximately.
10). Answer: e)
Let x = marked price and the CP be Rs.100.
Initial SP = Rs.112.
To give 20% discount = x – 20x/100 = Rs.112.
= x – x/5 = Rs.112, 4x = 112 x 5
x = Rs.140.
Here marked profit = Rs.40.
Percent profit = 40%
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