Questions and Answers Type: | MCQ (Multiple Choice Questions). |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Profit and Loss Aptitude Questions and Answers. |

Number of Questions: | 10 Questions with Solutions. |

- The difference between a discount of \(50 \ \%\) on \(Rs.1000\) and two successive discounts of \(20 \ \%\) and \(30 \ \%\) on the same amount will be?
- \(Rs.50\)
- \(Rs.55\)
- \(Rs.60\)
- \(Rs.65\)

Answer: (c) \(Rs.60\)

Solution: \(50 \ \%\) discount on \(Rs.1000\), $$ = \frac{50}{100} \times 1000 = Rs.500 $$ overall discount of two successive discounts \(20 \ \%\) and \(30 \ \%\), $$ = x + y - \frac{xy}{100} $$ $$ = 20 + 30 - \frac{20 \times 30}{100} $$ $$ = 44 \ \% $$ now amount of \(44 \ \%\) discount on \(Rs.1000\), $$ = \frac{44}{100} \times 1000 = Rs.440 $$ then $$ Difference = 500 - 440 $$ $$ = Rs.60 $$

Solution: \(50 \ \%\) discount on \(Rs.1000\), $$ = \frac{50}{100} \times 1000 = Rs.500 $$ overall discount of two successive discounts \(20 \ \%\) and \(30 \ \%\), $$ = x + y - \frac{xy}{100} $$ $$ = 20 + 30 - \frac{20 \times 30}{100} $$ $$ = 44 \ \% $$ now amount of \(44 \ \%\) discount on \(Rs.1000\), $$ = \frac{44}{100} \times 1000 = Rs.440 $$ then $$ Difference = 500 - 440 $$ $$ = Rs.60 $$

- find the single discount equivalent to discounts \(20 \ \%\) and \(25 \ \%\), successively?
- \(35 \ \%\)
- \(40 \ \%\)
- \(42 \ \%\)
- \(45 \ \%\)

Answer: (b) \(40 \ \%\)

Solution: Given, \(x = 20 \ \%\)

\(y = 25 \ \%\)

then overall discount, $$ = x + y - \frac{xy}{100} $$ $$ = 20 + 25 - \frac{20 \times 25}{100} $$ $$ = 40 \ \% $$

Solution: Given, \(x = 20 \ \%\)

\(y = 25 \ \%\)

then overall discount, $$ = x + y - \frac{xy}{100} $$ $$ = 20 + 25 - \frac{20 \times 25}{100} $$ $$ = 40 \ \% $$

- find the single discount equivalent to discounts \(10 \ \%\), \(20 \ \%\) and \(25 \ \%\), successively?
- \(46 \ \%\)
- \(48 \ \%\)
- \(50 \ \%\)
- \(52 \ \%\)

Answer: (a) \(46 \ \%\)

Solution: Let marked price = \(Rs.100\)

then selling price will be \(90 \ \%\) of \(80 \ \%\) of \(75 \ \%\) of \(Rs.100\), $$ = \frac{90}{100} \times \frac{80}{100} \times \frac{75}{100} \times 100 $$ $$ = Rs.54 $$ Hence single discount, $$ = (100 - 54) \ \% = 46 \ \% $$

Solution: Let marked price = \(Rs.100\)

then selling price will be \(90 \ \%\) of \(80 \ \%\) of \(75 \ \%\) of \(Rs.100\), $$ = \frac{90}{100} \times \frac{80}{100} \times \frac{75}{100} \times 100 $$ $$ = Rs.54 $$ Hence single discount, $$ = (100 - 54) \ \% = 46 \ \% $$

- A chair is offered for \(Rs.600\) with \(10 \ \%\) and \(20 \ \%\) discounts successively. If in addition, a discount of \(25 \ \%\) is offered on card payment, then find the price of the chair after discounts?
- \(Rs.325\)
- \(Rs.324\)
- \(Rs.322\)
- \(Rs.320\)

Answer: (b) \(Rs.324\)

Solution: Price of the chair after discounts will be \(90 \ \%\) of \(80 \ \%\) of \(75 \ \%\) of \(Rs.600\), $$ = \frac{90}{100} \times \frac{80}{100} \times \frac{75}{100} \times 600 $$ $$ = Rs.324 $$

Solution: Price of the chair after discounts will be \(90 \ \%\) of \(80 \ \%\) of \(75 \ \%\) of \(Rs.600\), $$ = \frac{90}{100} \times \frac{80}{100} \times \frac{75}{100} \times 600 $$ $$ = Rs.324 $$

- find the single discount equivalent to discounts \(5 \ \%\) \(10 \ \%\), \(20 \ \%\) and \(25 \ \%\), successively?
- \(45.6 \ \%\)
- \(46.8 \ \%\)
- \(48.7 \ \%\)
- \(49.6 \ \%\)

Answer: (c) \(48.7 \ \%\)

Solution: Let marked price = \(Rs.100\)

then selling price will be \(95 \ \%\) of \(90 \ \%\) of \(80 \ \%\) of \(75 \ \%\) of \(Rs.100\), $$ = \frac{95}{100} \times \frac{90}{100} \times \frac{80}{100} \times \frac{75}{100} \times 100 $$ $$ = Rs.51.3 $$ Hence single discount, $$ = (100 - 51.3) \ \% = 48.7 \ \% $$

Solution: Let marked price = \(Rs.100\)

then selling price will be \(95 \ \%\) of \(90 \ \%\) of \(80 \ \%\) of \(75 \ \%\) of \(Rs.100\), $$ = \frac{95}{100} \times \frac{90}{100} \times \frac{80}{100} \times \frac{75}{100} \times 100 $$ $$ = Rs.51.3 $$ Hence single discount, $$ = (100 - 51.3) \ \% = 48.7 \ \% $$

- A shopkeeper sold two mobiles for \(Rs.7500\) each, gaining \(10 \ \%\) on first mobile and losing \(5 \ \%\) on second mobile. Find his net profit or loss amount?
- \(Rs.287.09\)
- \(Rs.282.03\)
- \(Rs.288.06\)
- \(Rs.280.05\)

Answer: (a) \(Rs.287.09\)

Solution: selling price of two mobiles = \(Rs.15000\)

cost price of two mobiles, $$ = \frac{100}{110} \times 7500 + \frac{100}{95} \times 7500 $$ $$ = 14712.91 \ Rs. $$ then net profit = \(15000 - 14712.91\) = \(287.09 \ Rs.\)

Solution: selling price of two mobiles = \(Rs.15000\)

cost price of two mobiles, $$ = \frac{100}{110} \times 7500 + \frac{100}{95} \times 7500 $$ $$ = 14712.91 \ Rs. $$ then net profit = \(15000 - 14712.91\) = \(287.09 \ Rs.\)

- A seller sold an article at \(5 \ \%\) profit. If he had sold the article at \(10 \ \%\) profit, he would have recieved \(Rs.500\) more. Find the selling price of the article?
- \(Rs.10,000\)
- \(Rs.10,500\)
- \(Rs.10,800\)
- \(Rs.10,900\)

Answer: (b) \(Rs.10,500\)

Solution: Let cost price of the article = \(Rs.K\), then $$ k + k \times \frac{10}{100} = k + k \times \frac{5}{100} + 500 $$ $$ \frac{5 \ k}{100} = 500 $$ $$ k = 10,000 \ Rs. $$ Hence, selling price of the article, $$ = 10,000 + 10,000 \times \frac{5}{100} $$ $$ = Rs.10,500 $$

Solution: Let cost price of the article = \(Rs.K\), then $$ k + k \times \frac{10}{100} = k + k \times \frac{5}{100} + 500 $$ $$ \frac{5 \ k}{100} = 500 $$ $$ k = 10,000 \ Rs. $$ Hence, selling price of the article, $$ = 10,000 + 10,000 \times \frac{5}{100} $$ $$ = Rs.10,500 $$

- A man sold two chairs for \(Rs.1500\) each, gaining \(25 \ \%\) on first chair and losing \(25 \ \%\) on second chair. Find his net profit or loss amount?
- \(Rs.200\)
- \(Rs.220\)
- \(Rs.250\)
- \(Rs.260\)

Answer: (a) \(Rs.200\)

Solution: selling price of two chairs = \(Rs.3000\)

cost price of two chairs, $$ = \frac{100}{125} \times 1500 + \frac{100}{75} \times 1500 $$ $$ = 3200 \ Rs. $$ then net loss = \(3200 - 3000\) = \(200 \ Rs.\)

Solution: selling price of two chairs = \(Rs.3000\)

cost price of two chairs, $$ = \frac{100}{125} \times 1500 + \frac{100}{75} \times 1500 $$ $$ = 3200 \ Rs. $$ then net loss = \(3200 - 3000\) = \(200 \ Rs.\)

- x sells a laptop to y at a profit of \(25 \ \%\) and y sells the laptop to z at a profit of \(10 \ \%\). If z pays \(Rs.20,000\), then what did x pay for laptop?
- \(Rs.12545.45\)
- \(Rs.14545.45\)
- \(Rs.15545.55\)
- \(Rs.16545.55\)

Answer: (b) \(Rs.14545.45\)

Solution: the amount x paid for the laptop is \(110 \ \%\) of \(125 \ \%\) of x, $$ \frac{110}{100} \times \frac{125}{100} \times x = 20,000 $$ $$ \frac{55 \ x}{40} = 20,000 $$ $$ x = 14545.45 \ Rs. $$

Solution: the amount x paid for the laptop is \(110 \ \%\) of \(125 \ \%\) of x, $$ \frac{110}{100} \times \frac{125}{100} \times x = 20,000 $$ $$ \frac{55 \ x}{40} = 20,000 $$ $$ x = 14545.45 \ Rs. $$

- A girl purchased a book at \(Rs.250\) and sold it at \(Rs.150\). Find the loss percent of the girl?
- \(25 \ \%\)
- \(30 \ \%\)
- \(35 \ \%\)
- \(40 \ \%\)

Answer: (d) \(40 \ \%\)

Solution: Given, cost price (CP) = \(Rs.250\)

selling price (SP) = \(Rs.150\), then $$ Loss = CP - SP $$ $$ = 250 - 150 = Rs.100 $$ Hence the loss percent, $$ Loss \ \% = \frac{Loss}{CP} \times 100 $$ $$ = \frac{100}{250} \times 100 $$ $$ = 40 \ \% $$

Solution: Given, cost price (CP) = \(Rs.250\)

selling price (SP) = \(Rs.150\), then $$ Loss = CP - SP $$ $$ = 250 - 150 = Rs.100 $$ Hence the loss percent, $$ Loss \ \% = \frac{Loss}{CP} \times 100 $$ $$ = \frac{100}{250} \times 100 $$ $$ = 40 \ \% $$

Lec 1: Introduction
Exercise-1
Lec 2: Profit and Loss Case-1
Exercise-2
Lec 3: Profit and Loss Case-2
Exercise-3
Lec 4: Profit and Loss Case-3
Exercise-4
Lec 5: Case of Discount
Exercise-5