Profit & Loss Questions For Railways Set-2 PDF:
Download Profit & Loss Questions For Railways PDF for Railway exams of RRB ALP & Group-D with detailed solutions. Important Practice questions based on previous year railway exam papers.
Download Profit & Loss Questions For Railways PDF
Question 1: A shopkeeper labelled the price of the articles so as to earn a profit of 30% on the cost price. He then sold the articles by offering a discount of 10% on the labelled price what is the actual per profit earned in the deal ?
a) 18%
b) 15%
c) 20%
d) Cannot be determined
e) None of these
Question 2: Pure milk costs Rs. 16 per litre. After adding water the milkman sells the mixture Rs. 15 per litre and thereby makes a profit of 25%. In what respective ratio does he mix milk with water?
a) 3: 1
b) 4: 3
c) 3: 2
d) 5: 3
e) 4: 1
Question 3: A starts a business with Rs. 2500. After one month from the start of the business, B joined with Rs. 4500 and A withdrew completely after eleven months from the start of the business. If the difference between A’s and B’s respective shares in the annual profit was Rs. 4800, what was the annual profit earned?
a) Rs. 14800
b) Rs. 16800
c) Rs. 14400
d) Rs. 11400
e) Rs. 15600
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Question 4: A trader sells an item to a retailer at 20% discount, but charges 10% on the discounted price, for delivery and packaging. The retailer sells it for Rs. 2046 more, thereby earning a profit of 25%. At what price had the trader marked the item?
a) Rs. 9400
b) Rs. 9000
c) Rs. 8000
d) Rs. 12000
e) Rs. 9300
Question 5: Shri Ramlal purchased a TV set for Rs. 12,500 and spent Rs. 300 on transportation and Rs. 800 on installation. At what price should he sell it so as to earn an overall profit of 15% ?
a) Rs. 14,560
b) Rs. 14,375
c) Rs. 15,460
d) Rs. 15,375
e) None of these
Question 6: An item was bought at Rs. X and sold at Rs. Y, there by earning a profit of 20%. Had the value of X been 15% less and the value of Y been Rs. 76 less, a profit of 30% would have been earned. What was the value of ‘X’
a) Rs. 640
b) Rs.400
c) Rs.600
d) Rs.800
e) Rs.840
Answers & Solutions:
1) Answer (E)
Let the cost price be x.
As per given condition, the labelled price will be 1.3x.
Discounted price will be 0.9*1.3x= 1.17x
Now, Profit = Price at which item was sold – Cost Price = 1.17x – x = .17x = 17% of Cost price.
Hence, profit is 17%.
2) Answer (A)
Selling price is Rs 15 and Profit is 25%
=> Cost Price = $\frac{15}{1.25}$ = Rs 12
=> In every litre of solution, there is Rs 12 worth milk.
But the actual cost of milk is Rs 16.
=> Ratio = 12 : 4 = 3 : 1
3) Answer (B)
Amount invested by A = Rs. 2500 and by B = Rs. 4500
Both invested for 11 months.
Ratio of profit shared by A and B
= $(2500 \times 11) : (4500 \times 11)$
= $5 : 9$
Let total profit earned by A and B respectively = $Rs. 5x$ and $Rs. 9x$
=> $9x – 5x = 4800$
=> $x = \frac{4800}{4} = 1200$
$\therefore$ Total profit = $9x + 5x = 14x$
= $14 \times 1200 = Rs. 16,800$
Important Question & Answers for Railways
4) Answer (E)
Let Marked price of item = $Rs. 100x$
=> Selling price of trader = Cost price of retailer = $100x \times \frac{80}{100} \times \frac{110}{100}$
= $Rs. 88x$
Selling price of retailer = $Rs. (88x + 2046)$
Profit % = $\frac{(88x + 2046) – 88x}{88x} \times 100 = 25$
=> $\frac{2046}{88x} = \frac{25}{100} = \frac{1}{4}$
=> $x = \frac{2046 \times 4}{88} = 93$
$\therefore$$ Marked price = $$100 \times 93 = Rs. 9,300$
5) Answer (E)
Cost price of TV = Rs. 12,500
Amount spent on transportation = Rs. 300 and installation = Rs. 800
Net spent = Rs. (12500 + 300 + 800) = Rs. 13,600
Let selling price = $Rs.x$
Profit % = $\frac{x-13600}{13600} \times 100=15$
=> $x-13600=15 \times 136$
=> $x=2040+13600$
=> $x=Rs.$$ $$15,640$
=> Ans – (E)
6) Answer (D)
C.P. = $Rs. x$
S.P. = $Rs. y$
Profit % = $\frac{y – x}{x} \times 100 = 20$
=> $\frac{y – x}{x} = \frac{20}{100} = \frac{1}{5}$
=> $5y – 5x = x$ => $6x = 5y$
=> $y = \frac{6 x}{5}$ ———–(i)
If, value of X been 15% less and the value of Y been Rs. 76 less
=> $x’ = \frac{85}{100} \times x = \frac{17 x}{20}$
=> $y’ = y – 76$
Profit % = $\frac{y’ – x’}{x’} \times 100 = 30$
=> $\frac{(y – 76) – (\frac{17 x}{20})}{\frac{17 x}{20}} = \frac{30}{100} = \frac{3}{10}$
=> $10 \times [(y – 76) – (\frac{17 x}{20}] = 3 \times \frac{17 x}{20}$
=> $10y – 760 – \frac{170 x}{20} = \frac{51 x}{20}$
=> $10y – \frac{221 x}{20} = 760$
Using, equaiton (i), we get :
=> $(10 \times \frac{6 x}{5}) – \frac{221 x}{20} = 760$
=> $12x – \frac{221 x}{20} = 760$
=> $\frac{19 x}{20} = 760$
=> $x = 760 \times \frac{20}{19}$
=> $x = 40 \times 20 = Rs. 800$
