Q31. The monthly income of a man is Rs.16,000 $15\%$ of it is paid as income-tax and $75\%$ of the remainder is spent on rent, food, clothing etc. How much money is still left with the man?
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$\Rightarrow$ Monthly income = Rs.16,000
$\Rightarrow$ Amount spent on income-tax = $\dfrac{15}{100} \times 16,000$
$\Rightarrow 15 \times 160 = \text{ Rs.}2400$
$\Rightarrow$ Remaining amount = $16,000 - 2400 = 13,600$
$\Rightarrow$ Amount spent on rent, food etc. = $\dfrac{75}{100} \times 13,600$
$\Rightarrow 75 \times 136 = \text{ Rs.}10,200$
$\therefore$ Remaining amount = $13,600 - 10,200 = \text{ Rs.}3,400$ Ans
$\Rightarrow$ Amount spent on income-tax = $\dfrac{15}{100} \times 16,000$
$\Rightarrow 15 \times 160 = \text{ Rs.}2400$
$\Rightarrow$ Remaining amount = $16,000 - 2400 = 13,600$
$\Rightarrow$ Amount spent on rent, food etc. = $\dfrac{75}{100} \times 13,600$
$\Rightarrow 75 \times 136 = \text{ Rs.}10,200$
$\therefore$ Remaining amount = $13,600 - 10,200 = \text{ Rs.}3,400$ Ans
Q32. A number is first increased by $20\%$ and the resulting number is then decreased by $10\%$. Find the overall change in the number as percent.
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$\Rightarrow$ Let the no. be = 100
$\Rightarrow$ No. after $20\%$ increase = $100 + \dfrac{20}{100} \times 100$
$\Rightarrow 100 + 20 = 120$
$\Rightarrow$ No. after $10\%$ decrease = $120 - \dfrac{10}{100} \times 120$
$\Rightarrow 120 - 12 = 108$
$\Rightarrow$ Change in no. = $108 - 100 = 8$
$\therefore$ Percentage change = $\dfrac{8}{100} \times 100 = 8\%$ Ans
$\Rightarrow$ No. after $20\%$ increase = $100 + \dfrac{20}{100} \times 100$
$\Rightarrow 100 + 20 = 120$
$\Rightarrow$ No. after $10\%$ decrease = $120 - \dfrac{10}{100} \times 120$
$\Rightarrow 120 - 12 = 108$
$\Rightarrow$ Change in no. = $108 - 100 = 8$
$\therefore$ Percentage change = $\dfrac{8}{100} \times 100 = 8\%$ Ans
Q33. A number is increased by $10\%$ and the resulting number is again increased by $20\%$. What is the overall percentage increase in the number?
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$\Rightarrow$ Let the number be = 100
$\Rightarrow$ No. after $10\%$ increase = $100 + \dfrac{10}{100} \times 100$
$\Rightarrow 100 + 10 = 110$
$\Rightarrow$ No. after $20\%$ increase = $110 + \dfrac{20}{100} \times 110$
$\Rightarrow 110 + 22 = 132$
$\Rightarrow$ Change in no. = $132 - 100 = 32$
$\therefore$ Percentage change = $\dfrac{32}{100} \times 100 = 32\%$ Ans
$\Rightarrow$ No. after $10\%$ increase = $100 + \dfrac{10}{100} \times 100$
$\Rightarrow 100 + 10 = 110$
$\Rightarrow$ No. after $20\%$ increase = $110 + \dfrac{20}{100} \times 110$
$\Rightarrow 110 + 22 = 132$
$\Rightarrow$ Change in no. = $132 - 100 = 32$
$\therefore$ Percentage change = $\dfrac{32}{100} \times 100 = 32\%$ Ans
Q34. During 2003, the production of a factory decreased by $25\%$. But, during 2004, it increased by $40\%$ of what it was at the begining of 2004. Calculate the resulting change, increase or decrease, in production during these two years.
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$\Rightarrow$ Let production in 2003 be = 100
$\Rightarrow$ Decrease in 2003 = $100 - \dfrac{25}{100} \times 100$
$\Rightarrow 100 - 25 = 75$
$\therefore$ Production in 2004 = 75
$\Rightarrow$ Increase in 2004 = $75 + \dfrac{40}{100} \times 75$
$\Rightarrow 75 + \dfrac{4}{10} \times 75 = 75 + \dfrac{2}{5} \times 75 = 75 + 2 \times 15 = 75 + 30 = 105$
$\Rightarrow$ Change in production = $105 - 100 = 5$
$\therefore$ Percentage change = $\dfrac{5}{100} \times 100 = 5\%$ (increase) Ans
$\Rightarrow$ Decrease in 2003 = $100 - \dfrac{25}{100} \times 100$
$\Rightarrow 100 - 25 = 75$
$\therefore$ Production in 2004 = 75
$\Rightarrow$ Increase in 2004 = $75 + \dfrac{40}{100} \times 75$
$\Rightarrow 75 + \dfrac{4}{10} \times 75 = 75 + \dfrac{2}{5} \times 75 = 75 + 2 \times 15 = 75 + 30 = 105$
$\Rightarrow$ Change in production = $105 - 100 = 5$
$\therefore$ Percentage change = $\dfrac{5}{100} \times 100 = 5\%$ (increase) Ans
Q35. Last year, oranges were available at Rs.24 per dozen, but this year, they are available at Rs.50 per score. Find the percentage change in the price of oranges. Note: 1 score = 20
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For Previous Year:
$\Rightarrow$ Cost of 12 oranges = Rs.24
$\Rightarrow$ Cost of 1 orange = $\dfrac{24}{12} = \text{ Rs.}2$
For Current Year:
$\Rightarrow$ Cost of 20 oranges = Rs.50
$\Rightarrow$ Cost of 1 orange = $\dfrac{50}{20} = \text{ Rs.}2.5$
$\Rightarrow$ Change in cost = $2.5 - 2 = 0.5$
$\therefore$ Percentage change = $\dfrac{0.5}{2} \times 100$
$\Rightarrow \dfrac{5}{2 \times 10} \times 100 = \dfrac{1}{2 \times 2} \times 100 = \dfrac{1}{4} \times 100 = 25\%$ (increase) Ans
$\Rightarrow$ Cost of 12 oranges = Rs.24
$\Rightarrow$ Cost of 1 orange = $\dfrac{24}{12} = \text{ Rs.}2$
For Current Year:
$\Rightarrow$ Cost of 20 oranges = Rs.50
$\Rightarrow$ Cost of 1 orange = $\dfrac{50}{20} = \text{ Rs.}2.5$
$\Rightarrow$ Change in cost = $2.5 - 2 = 0.5$
$\therefore$ Percentage change = $\dfrac{0.5}{2} \times 100$
$\Rightarrow \dfrac{5}{2 \times 10} \times 100 = \dfrac{1}{2 \times 2} \times 100 = \dfrac{1}{4} \times 100 = 25\%$ (increase) Ans
Q36. In an examination, Kavita scored 120 out of 150 in Maths, 136 out of 200 in English and 108 out of 150 in Science. Find the percentage score in each subject and also on the whole (aggregate).
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$\Rightarrow$ Percentage score in Mahs = $\dfrac{120}{150} \times 100$
$\Rightarrow \dfrac{12}{15} \times 100 = \dfrac{12}{3} \times 20 = 4 \times 20 = 80\%$
$\Rightarrow$ Percentage score in English = $\dfrac{136}{200} \times 100$
$\Rightarrow \dfrac{136}{2} = 68\%$
$\Rightarrow$ Percentage score in Science = $\dfrac{108}{150} \times 100$
$\Rightarrow \dfrac{108}{15} \times 10 = \dfrac{108}{3} \times 2 = 36 \times 2 = 72\%$
$\Rightarrow$ Total marks obtained = $120 + 136 + 108 = 364$
$\Rightarrow$ And maximum marks = $150 + 150 + 200 = 500$
$\therefore$ Percentage on aggregate = $\dfrac{364}{500} \times 100$
$\Rightarrow \dfrac{364}{500} \times 100 = \dfrac{364}{5} = 72.8\%$ Ans
$\Rightarrow \dfrac{12}{15} \times 100 = \dfrac{12}{3} \times 20 = 4 \times 20 = 80\%$
$\Rightarrow$ Percentage score in English = $\dfrac{136}{200} \times 100$
$\Rightarrow \dfrac{136}{2} = 68\%$
$\Rightarrow$ Percentage score in Science = $\dfrac{108}{150} \times 100$
$\Rightarrow \dfrac{108}{15} \times 10 = \dfrac{108}{3} \times 2 = 36 \times 2 = 72\%$
$\Rightarrow$ Total marks obtained = $120 + 136 + 108 = 364$
$\Rightarrow$ And maximum marks = $150 + 150 + 200 = 500$
$\therefore$ Percentage on aggregate = $\dfrac{364}{500} \times 100$
$\Rightarrow \dfrac{364}{500} \times 100 = \dfrac{364}{5} = 72.8\%$ Ans
Q37. A is $25\%$ older than B. By what percent is B younger than A?
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$\Rightarrow$ Let B's age be = $x$
$\therefore$ A's age = $x + \dfrac{25}{100} \times x$
$\Rightarrow x + \dfrac{x}{4} = \dfrac{4x + x}{4} = \dfrac{5x}{4}$
$\Rightarrow$ Age difference = $\dfrac{5x}{4} - x = \dfrac{5x - 4x}{4} = \dfrac{x}{4}$
$\therefore \dfrac{\text{Age difference}}{\text{A's Age}} \times 100$
$\Rightarrow \dfrac{\dfrac{x}{4}}{\dfrac{5x}{4}} \times 100 = \dfrac{x \times 4}{5x \times } \times 100 = \dfrac{100x}{5x} = 20\%$ Ans
$\therefore$ A's age = $x + \dfrac{25}{100} \times x$
$\Rightarrow x + \dfrac{x}{4} = \dfrac{4x + x}{4} = \dfrac{5x}{4}$
$\Rightarrow$ Age difference = $\dfrac{5x}{4} - x = \dfrac{5x - 4x}{4} = \dfrac{x}{4}$
$\therefore \dfrac{\text{Age difference}}{\text{A's Age}} \times 100$
$\Rightarrow \dfrac{\dfrac{x}{4}}{\dfrac{5x}{4}} \times 100 = \dfrac{x \times 4}{5x \times } \times 100 = \dfrac{100x}{5x} = 20\%$ Ans
Q38. (i) Increase 180 by $25\%$.
(ii) Decrease 140 by $18\%$.
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(i) $180 + \dfrac{25}{100} \times 180 = 180 + \dfrac{1}{4} \times 180 = 180 + 45 = 225$ Ans
(ii) $140 - \dfrac{18}{100} \times 140 = 140 - \dfrac{18}{10} \times 14 = 140 - \dfrac{18}{5} \times 7 = 140 - \dfrac{126}{5}$
$\Rightarrow \dfrac{700 - 126}{5} = \dfrac{574}{5} = 114.8\%$ Ans
(ii) $140 - \dfrac{18}{100} \times 140 = 140 - \dfrac{18}{10} \times 14 = 140 - \dfrac{18}{5} \times 7 = 140 - \dfrac{126}{5}$
$\Rightarrow \dfrac{700 - 126}{5} = \dfrac{574}{5} = 114.8\%$ Ans
Q39. In an election, three candidates contested and secured 29,200, 58,800 and 72,000 votes. Find the percentage of votes scored by winning candidate.
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$\Rightarrow$ Total votes = $29,200 + 58,800 + 72,000 = 1,60,000$
$\Rightarrow$ Votes secured by winning candidate = 72,000
$\therefore$ Vote percentage of winning candidate = $\dfrac{72,000}{1,60,000} \times 100$
$\Rightarrow \dfrac{72,000}{1,600} = \dfrac{720}{16} = 45\%$ Ans
$\Rightarrow$ Votes secured by winning candidate = 72,000
$\therefore$ Vote percentage of winning candidate = $\dfrac{72,000}{1,60,000} \times 100$
$\Rightarrow \dfrac{72,000}{1,600} = \dfrac{720}{16} = 45\%$ Ans
Q40. (i) A number when increased by $23\%$ becomes 861, find the number.
(ii) A number when decreased by $16\%$ becomes 798, find the number.
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(i) Let no. be = $x$
$\therefore x + \dfrac{23}{100} \times x = 861$
$\Rightarrow x + \dfrac{23x}{100} = 861$
$\Rightarrow \dfrac{100x + 23x}{100} = 861$
$\Rightarrow \dfrac{123x}{100} = 861$
$\Rightarrow x = 861 \times \dfrac{100}{123} = 7 \times 100 = 700$ Ans
(ii) Let no. be = $x$
$\therefore x - \dfrac{16}{100} \times x = 798$
$\Rightarrow x - \dfrac{16x}{100} = 798$
$\Rightarrow \dfrac{100x - 16x}{100} = 798$
$\Rightarrow \dfrac{84x}{100} = 798$
$\Rightarrow x = 798 \times \dfrac{100}{84}$
$\Rightarrow x = 798 \times \dfrac{25}{21} = 38 \times 25 = 950$ Ans
$\therefore x + \dfrac{23}{100} \times x = 861$
$\Rightarrow x + \dfrac{23x}{100} = 861$
$\Rightarrow \dfrac{100x + 23x}{100} = 861$
$\Rightarrow \dfrac{123x}{100} = 861$
$\Rightarrow x = 861 \times \dfrac{100}{123} = 7 \times 100 = 700$ Ans
(ii) Let no. be = $x$
$\therefore x - \dfrac{16}{100} \times x = 798$
$\Rightarrow x - \dfrac{16x}{100} = 798$
$\Rightarrow \dfrac{100x - 16x}{100} = 798$
$\Rightarrow \dfrac{84x}{100} = 798$
$\Rightarrow x = 798 \times \dfrac{100}{84}$
$\Rightarrow x = 798 \times \dfrac{25}{21} = 38 \times 25 = 950$ Ans
Q41. The price of sugar is incresed by $20\%$. By what percent must the consumption of sugar be decreased so that the expenditure on sugar may remain the same?
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$\Rightarrow$ Let price of sugar be = Rs.100
$\Rightarrow$ $20\%$ increase in price = $100 + \dfrac{20}{100} \times 100 = 100 + 20 = 120$
$\Rightarrow$ Required reduction = $120 - 100 = 20$
$\therefore$ Percentage decrease = $\dfrac{20}{120} \times 100$
$\Rightarrow \dfrac{1}{6} \times 100 = \dfrac{50}{3} = 16\dfrac{2}{3}\%$ Ans
$\Rightarrow$ $20\%$ increase in price = $100 + \dfrac{20}{100} \times 100 = 100 + 20 = 120$
$\Rightarrow$ Required reduction = $120 - 100 = 20$
$\therefore$ Percentage decrease = $\dfrac{20}{120} \times 100$
$\Rightarrow \dfrac{1}{6} \times 100 = \dfrac{50}{3} = 16\dfrac{2}{3}\%$ Ans
Q42. (i) Express $\dfrac{4}{7}$ as percentage.
(ii) Express $6.25\%$ as fraction.
(iii) Express $\dfrac{1}{5}$ as percentage of $\dfrac{20}{11}$.
(iv) Express 1.14 as a percentage of 2.5.
(v) Find $40\%$ of 820.
Show Answer
(i) $\dfrac{4}{7} \times 100 = \dfrac{400}{7} = 57\dfrac{1}{7}\%$ Ans
(ii) $6.25\% = \dfrac{625}{100 \times 100} = \dfrac{625}{10000} = \dfrac{25}{400} = \dfrac{1}{16}$ Ans
(iii) $\dfrac{\dfrac{1}{5}}{\dfrac{20}{11}} \times 100 = \dfrac{1 \times 11}{20 \times 5} \times 100 = \dfrac{11}{100} \times 100 = 11\%$ Ans
(iv) $\dfrac{1.14}{2.5} \times 100 = \dfrac{\dfrac{114}{100}}{\dfrac{25}{10}} \times 100 = \dfrac{114 \times 10}{25 \times 100} \times 100$
$\Rightarrow \dfrac{114 \times 10}{25} = \dfrac{114 \times 2}{5} = \dfrac{228}{5} = 45.6\%$ Ans
(v) $\dfrac{40}{100} \times 820 = \dfrac{4}{10} \times 820 = \dfrac{2}{5} \times 820 = 2 \times 164 = 328$ Ans
(ii) $6.25\% = \dfrac{625}{100 \times 100} = \dfrac{625}{10000} = \dfrac{25}{400} = \dfrac{1}{16}$ Ans
(iii) $\dfrac{\dfrac{1}{5}}{\dfrac{20}{11}} \times 100 = \dfrac{1 \times 11}{20 \times 5} \times 100 = \dfrac{11}{100} \times 100 = 11\%$ Ans
(iv) $\dfrac{1.14}{2.5} \times 100 = \dfrac{\dfrac{114}{100}}{\dfrac{25}{10}} \times 100 = \dfrac{114 \times 10}{25 \times 100} \times 100$
$\Rightarrow \dfrac{114 \times 10}{25} = \dfrac{114 \times 2}{5} = \dfrac{228}{5} = 45.6\%$ Ans
(v) $\dfrac{40}{100} \times 820 = \dfrac{4}{10} \times 820 = \dfrac{2}{5} \times 820 = 2 \times 164 = 328$ Ans
Q43. Find the number if:
(i) $28\%$ of the number is 84.
(ii) $12.5\%$ of the number is 128.
Show Answer
(i) Let no. be = $x$
$\therefore \dfrac{28}{100} \times x = 84$
$\Rightarrow \dfrac{28x}{100} = 84$
$\Rightarrow x = 84 \times \dfrac{100}{28} = 3 \times 100 = 300$ Ans
(ii) Let no. be = $x$
$\therefore \dfrac{12.5}{100} \times x = 128$
$\Rightarrow \dfrac{125x}{100 \times 10} = 128$
$\Rightarrow \dfrac{125x}{1000} = 128$
$\Rightarrow \dfrac{5x}{40} = 128$
$\Rightarrow \dfrac{x}{8} = 128$
$\Rightarrow x = 128 \times 8 = 1024$ Ans
$\therefore \dfrac{28}{100} \times x = 84$
$\Rightarrow \dfrac{28x}{100} = 84$
$\Rightarrow x = 84 \times \dfrac{100}{28} = 3 \times 100 = 300$ Ans
(ii) Let no. be = $x$
$\therefore \dfrac{12.5}{100} \times x = 128$
$\Rightarrow \dfrac{125x}{100 \times 10} = 128$
$\Rightarrow \dfrac{125x}{1000} = 128$
$\Rightarrow \dfrac{5x}{40} = 128$
$\Rightarrow \dfrac{x}{8} = 128$
$\Rightarrow x = 128 \times 8 = 1024$ Ans
Q44. What Percent of:
(i) 390 is 19.5
(ii) Rs.2 is 50 paise
Show Answer
(i) $\dfrac{19.5}{390} \times 100 = \dfrac{195}{10 \times 390} \times 100$
$\Rightarrow \dfrac{195}{3900} \times 100 = \dfrac{195}{39} = 5\%$ Ans
(ii) $\dfrac{.50}{2} \times 100 = \dfrac{50}{2 \times 100} \times 100$
$\Rightarrow \dfrac{50}{2} = 25\%$ Ans
$\Rightarrow \dfrac{195}{3900} \times 100 = \dfrac{195}{39} = 5\%$ Ans
(ii) $\dfrac{.50}{2} \times 100 = \dfrac{50}{2 \times 100} \times 100$
$\Rightarrow \dfrac{50}{2} = 25\%$ Ans
Q45. (i) Express Rs. 37.94 as an exact percentage of Rs.271.
(ii) Find the percentage error in taking $2.53 \times 10^5$ as $2.5 \times 10^5$
Show Answer
(i) $\dfrac{37.94}{271} \times 100 = \dfrac{3794}{271 \times 100} \times 100$
$\Rightarrow \dfrac{3794}{271} = 14\%$ Ans
(ii) $\dfrac{2.53 - 2.5}{2.53} \times 100 = \dfrac{.03}{2.53} \times 100$
$\Rightarrow \dfrac{\dfrac{3}{100}}{\dfrac{253}{100}} \times 100 = \dfrac{3 \times 100}{253 \times 100} \times 100 = \dfrac{3}{253} \times 100$
$\Rightarrow \dfrac{3}{253} = 1\dfrac{47}{253}\%$ Ans
$\Rightarrow \dfrac{3794}{271} = 14\%$ Ans
(ii) $\dfrac{2.53 - 2.5}{2.53} \times 100 = \dfrac{.03}{2.53} \times 100$
$\Rightarrow \dfrac{\dfrac{3}{100}}{\dfrac{253}{100}} \times 100 = \dfrac{3 \times 100}{253 \times 100} \times 100 = \dfrac{3}{253} \times 100$
$\Rightarrow \dfrac{3}{253} = 1\dfrac{47}{253}\%$ Ans