Class 8 Percentage 12.2-3

Q18. Aman’s income is 20% less than that of Anil. How much percent is Anil’s income more than Aman’s income?\begin{array}{l} \text {Sol. Let’s assume Anil income be Rs. x } \\ \Rightarrow \quad \text {Aman’s income } = x – (20\% \text { of } x) = \frac {4}{5}x \\\Rightarrow \quad \text {Difference between Anil’s and Aman’s income } = x – \frac {4}{5}x =\frac {1}{5}x \\\text {Lets assume Anil’s incoem is z% more than of Aman.} \\\Rightarrow \quad \text {z% of Aman’s income } = \frac {1}{5}x \\\Rightarrow \quad \frac {z}{100} \times \frac {4}{5} x = \frac {1}{5}x \\\Rightarrow \quad z = \frac {1}{5}x \times \frac {5}{4x} \times 100\\\Rightarrow \quad z = 25\% \therefore \text {Anil’s income is 25% more than of Aman’s income.} \\ \end{array}Q19. The value of a machine depreciates every year by 5%. If the present value of the machine be Rs 100000, what will be its value after 2 years?\begin{array}{l} \text {Sol. Given that the value of the machine depreciates by 5% every year.} \\ \text {Present value of the machine Rs. }=100000 \\ \therefore \text {Value of the machine after Ist year Rs. } = 100000 – (5\% \text { of } 100000) = 100000 -5000 = 95000 \\\Rightarrow \quad \text {Value of the machine after 2nd year } =95000 – (5\% \text { of } 95000) = 95000 -4750 = 90250 \\ \therefore \text {After two years, the value of the machine will be Rs.90250 } \\ \end{array}Q20. The population of a town increases by 10 % annually. If the present population is 60000, what will be its population after 2 years?\begin{array}{l} \text {Sol. Given that the population of a town increased by 10% every year.} \\ \text {Present population of the town }=60000 \\ \therefore \text {Population of the town after Ist year } = 60000 + (10\% \text { of } 60000) = 60000 +6000 = 66000 \\\Rightarrow \quad \text {Population of the town after 2nd year } =66000 + (10\% \text { of } 66000) = 66000 + 6600 = 72600 \\ \therefore \text {After two years, the population of the town will be 72600 } \\ \end{array}Q21. The population of a town increases by 10% annually. If the present population is 22000, find its population a year ago.\begin{array}{l} \text {Sol. Lets assume the population of the town be x one year ago.} \\ \text {Its is given that populations increases by 10% every year.} \\\Rightarrow \quad x + 10\% \text { of } x = 22000 \\\Rightarrow \quad \frac {110}{100} x = 22000 \\\Rightarrow \quad x = \frac {100}{110} \times 22000 \\\Rightarrow \quad x = 20000 \\\therefore \text {One year ago, population of the town was 20000. } \\ \end{array}Q22. Ankit was given an increment of 10% on his salary. His new salary is Rs 3575. What was his salary before increment?\begin{array}{l} \text {Sol. Lets assume that salary before increent was Rs. x} \\ \text {Its is given that an increment of 10% was given to Ankit.} \\\Rightarrow \quad x + 10\% \text { of } x = 3575 \\\Rightarrow \quad \frac {110}{100} x = 3575 \\\Rightarrow \quad x = \frac {100}{110} \times 3575 \\\Rightarrow \quad x = 3250 \\\therefore \text {Salary before increment was Rs.3250 } \\ \end{array}Q23. In the new budget, the price of petrol rose by 10%. By how much percent must one reduce the consumption so that the expenditure does not increase?\begin{array}{l} \text {Sol. Lets assume x be the original petrol price and y be the petrol consumption.} \\\Rightarrow \quad \text {Expenditure } = \text { Price } \times \text { Consumption } =xy \\\text {Petrol price is incresed by 10%. } \\ \text {Therefore incremented price } = x + 10\% \text { of } x = \frac{110}{100} x \\\text {Lets assume that new consumption be y1 } \\\Rightarrow \quad \text {Expenditure } = \text { Incremented Price } \times \text { new Consumption } =\frac{110}{100} x \times y1 \\\text {As we know the Expenditure remain same, therefore } \\\Rightarrow \quad \frac{110}{100} x \times y1 = xy \\\Rightarrow \quad y1 = \frac{100}{110} \times y \\\text {Reduction in consumption } = y – y1 = y – (\frac{100}{110} \times y) = \frac{10}{110} \times y \\\text {Lets assume that overall consumption be reduced by z%, therefore } \\ z\% \text { of } y = \frac{10}{110} \times y \\ \Rightarrow \quad z = \frac{10}{110} \times 100 \\\Rightarrow \quad z = 9 \frac{1}{11} \\\therefore \text {To keep the expenditure same, we need to reduce the consumtion by } 9 \frac{1}{11} \% \\\text {Note : We can also increase the formula to find the increase or decrease directly.} \\ \text {(i) If the price of a commodity is increased by x%, then reduction in consumption } \\ \text { for keeping the expenditure same is} \\ [\frac {x}{100+x} \times 100] \% \\ \\\text {(ii) If the price of a commodity is decreased by x%, then increase in consumption } \\ \text { for keeping the expenditure same is} \\ [\frac {x}{100-x} \times 100 ]\% \\\end{array}Q24. Mohan’s income is Rs 15500 per month. He saves 11% of his income. If his income increases by 10%, then he reduces his saving by 1%, how much does he save now?\begin{array}{l} \text {Sol. Given that Mohan’s monthly income is Rs. 15500} \\ \text {Mohan savings }= 11\% \text { of } 15500 \\ =\frac {11}{100} \times 15500 = 1705 \\\text {When monthly income increased by 10%, the new monthly income } \\ =15500 + 10\% \text { of } 15500 =15500+1550 = 17050 \\\text {When savings reduced by 1%, new savings } =10\% \text { of } 17050 = 1705 \\\therefore \text {Current savings is Rs. 1705, which is same old saving.} \\ \end{array}Q25. Shikha’s income is 60% more than that of Shalu. What percent is Shalu’s income less than Shikha’s?\begin{array}{l} \text {Sol. Let’s assume Shalu’s income be Rs x. } \\ \text {Shikha’s Income } =x + 60\% \text { of } x \\ =\frac {8}{5} \times x \\ \text {Difference between Shikha’s and Shalu’s income } \\ =\frac {8}{5} \times x-x=\frac {3}{5} \times x \\\text {Percentage of Shalu’s income less than shikha’s } \\z \% \text { of } \frac {8}{5} \times x = \frac {3}{5} \times x \\ \Rightarrow \quad z = \frac {3}{5} \times x \times \frac {5}{8x} \times 100\\ \Rightarrow \quad z = \frac {75}{2} = 37 \frac {1}{2} \\ \therefore \text {Shalu’s income is } 37 \frac {1}{2} \% \text {less than Shikha’s income.} \end{array}Q26. Rs 3500 is to be shared among three people so that the first person gets 50% of the second, who in turn gets 50% of the third. How much will each of them get?\begin{array}{l} \text {Sol. Let’s assume that third person got Rs. x} \\\text {Second person got 50% of third } = 50\% \text { of } x = \frac {1}{2} x \\\text {Third person got 50% of second } = 50\% \text { of } \frac {1}{2} x = \frac {1}{4} x \\\text {According to given question, } \\ x + \frac {1}{2} x + \frac {1}{4} x = 3500 \\ \Rightarrow \quad \frac {4+2+1}{4} x = 3500 \\\Rightarrow \quad x = \frac {3500 \times 4}{7} \\\Rightarrow \quad x = 2000 \\\therefore \text {Money recieved by third person is Rs. 2000} \\ \text {Money recieved by second person is Rs. } \frac {1}{2} \times 2000 = 1000 \\ \text {Money recieved by first person is Rs. } \frac {1}{4} \times 2000 = 500 \\\end{array}Q27. After a 20% hike, the cost of Chinese Vase is Rs 2000. What was the original price of the object?\begin{array}{l} \text {Sol. Let’s assume x be the original cost of Chinese Vase. } \\ \text {After 20% hike, cost of Chinese Vase } = x +20\% \text { of } x = \frac {6}{5}x \\ \text {According to question, } \\\frac {6}{5}x = 2000 \\ \Rightarrow \quad x = \frac {5}{6} \times 2000 \\ \Rightarrow \quad x =1666.67 \\ \therefore \text {Original cost of Chinese Vase is Rs. 1666.67 } \end{array}