Topic Included: | Formulas, Definitions & Exmaples. |

Main Topic: | Quantitative Aptitude. |

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

Questions for practice: | 10 Questions & Answers with Solutions. |

If there is discount on an item, \(x \ \%\) and \(y \ \%\) successively, then the overall discount on the item will be $$ \left[x + y - \frac{xy}{100}\right] \ \% $$

If three or more successive discounts are given, then first of all we will determine single discount for first two successive discounts and after that we will find single discount for next successive discounts (discovered discount and next one discount) and so on. [see example (2)].

**Example (1):** If there are discounts on an item, \(10 \ \%\) and \(20 \ \%\) successively, then find the overall discount on the item?

**Solution:**Given values, \(x = 10\), \(y = 20\), then according to formula of discount given above, the overall discount will be, $$ \left[x + y - \frac{xy}{100}\right] \ \% $$ $$ \left[10 + 20 - \frac{10 \times 20}{100}\right] $$ $$ \left[30 - \frac{200}{100}\right] \ \% $$ $$ \left[30 - 2\right] \ \% = 28 \ \% $$

**Example (2):** Find the single Discount which is equivalent to three successive discounts \(5 \ \%\), \(10 \ \%\) and \(20 \ \%\)?

**Solution: (1st Method)**first of all we will find single discount for first two successive discounts, \(x = 5\), \(y = 10\), then according to formula of discount given above, the single discount will be, $$ \left[x + y - \frac{xy}{100}\right] \ \% $$ $$ \left[5 + 10 - \frac{5 \times 10}{100}\right] $$ $$ \left[15 - \frac{50}{100}\right] \ \% $$ $$ \left[15 - 0.5\right] \ \% = 14.5 \ \% $$ now we will find single discount for next successive discounts [discovered discount \((14.5 \ \%)\) and next one discount \((20 \ \%)\) given in the question],

\(x = 14.5\), \(y = 20\) then, $$ \left[x + y - \frac{xy}{100}\right] \ \% $$ $$ \left[14.5 + 20 - \frac{14.5 \times 20}{100}\right] $$ $$ \left[34.5 - \frac{290}{100}\right] \ \% $$ $$ \left[34.5 - 2.9\right] \ \% = 31.6 \ \% $$

**Solution: (2nd Method)** Let marked price = \(Rs.100\)

then selling price will be \(95 \ \%\) of \(90 \ \%\) of \(80 \ \%\) of \(Rs.100\), $$ = \frac{95}{100} \times \frac{90}{100} \times \frac{80}{100} \times 100 $$ $$ = Rs.68.4 $$ Hence single discount, $$ = (100 - 68.4) \ \% = 31.6 \ \% $$

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