Q61. An arithmetic paper is done by 7500 persons of whom $20\%$ are girls and the rest boys. If $5\%$ of the boys and $40\%$ of the girls fail, what percentage of the whole number of candidates passed?
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$\Rightarrow$ Total no. of students = 7500
$\therefore$ No. of girls = $\dfrac{20}{100} \times 7500 = \dfrac{75,000}{5} = 1500$
$\Rightarrow$ Percent of boys in school = $100 - 20 = 80\%$
$\therefore$ No. of boys = $\dfrac{80}{100} \times 7500 = 80 \times 75 = 6000$
$\Rightarrow$ Percent of girls failed = $40\%$
$\therefore$ No. of girls failed = $\dfrac{40}{100} \times 1500 = 40 \times 15 = 600$
$\therefore$ No. of girls passed = $1500 - 600 = 900$
$\Rightarrow$ Percent of boys failed = $5\%$
$\therefore$ No. of boys failed = $\dfrac{5}{100} \times 6000 = \dfrac{6000}{20} = \dfrac{600}{2} = 300$
$\therefore$ No. of boys passed = $6000 - 300 = 5700$
$\Rightarrow$ Total no. of students passed = $5700 + 900 = 6600$
$\therefore$ Total pass percentage = $\dfrac{6600}{7500} \times 100$
$\Rightarrow \dfrac{6600}{75} = \dfrac{1320}{15} = \dfrac{264}{3} = 88\%$ Ans
$\therefore$ No. of girls = $\dfrac{20}{100} \times 7500 = \dfrac{75,000}{5} = 1500$
$\Rightarrow$ Percent of boys in school = $100 - 20 = 80\%$
$\therefore$ No. of boys = $\dfrac{80}{100} \times 7500 = 80 \times 75 = 6000$
$\Rightarrow$ Percent of girls failed = $40\%$
$\therefore$ No. of girls failed = $\dfrac{40}{100} \times 1500 = 40 \times 15 = 600$
$\therefore$ No. of girls passed = $1500 - 600 = 900$
$\Rightarrow$ Percent of boys failed = $5\%$
$\therefore$ No. of boys failed = $\dfrac{5}{100} \times 6000 = \dfrac{6000}{20} = \dfrac{600}{2} = 300$
$\therefore$ No. of boys passed = $6000 - 300 = 5700$
$\Rightarrow$ Total no. of students passed = $5700 + 900 = 6600$
$\therefore$ Total pass percentage = $\dfrac{6600}{7500} \times 100$
$\Rightarrow \dfrac{6600}{75} = \dfrac{1320}{15} = \dfrac{264}{3} = 88\%$ Ans
Q62. The population of a town is 24,000. The no. of males was increased by $6\%$ and of females by $9\%$ that raised the population to 25,620. Find no. of females in town.
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$\Rightarrow$ Population before increase = 24,000
$\Rightarrow$ Population after increase = 25,620
$\therefore$ Increase in population = $25,620 - 24,000 = 1620$
$\Rightarrow$ Let male population be = x
$\therefore$ Female population = $(24,000 - x)$
$\Rightarrow$ Increase in male population = $\dfrac{6}{100} \times x = \dfrac{6x}{100}$
$\Rightarrow$ Increase in female population = $\dfrac{9}{100} \times (24,000 - x) = \dfrac{9(24,000)}{100} - \dfrac{9x}{100}$
So, we have the equation:
$\Rightarrow \dfrac{6x}{100} + \dfrac{2,16,000}{100} - \dfrac{9x}{100} = 1620$
$\Rightarrow \dfrac{6x}{100} - \dfrac{9x}{100} = 1620 - \dfrac{2,16,000}{100}$
$\Rightarrow \dfrac{6x - 9x}{100} = \dfrac{1,62,000 - 2,16,000}{100}$
$\Rightarrow -\dfrac{3x}{100} = -\dfrac{18,000}{100}$
$\Rightarrow x = \dfrac{18,000}{100} \times \dfrac{100}{3}$
$\therefore x = 6000$
$\Rightarrow$ So, No. of males = 6000
$\Rightarrow$ And, No. of females = $24,000 - 6000 = 18,000$ Ans
$\Rightarrow$ Population after increase = 25,620
$\therefore$ Increase in population = $25,620 - 24,000 = 1620$
$\Rightarrow$ Let male population be = x
$\therefore$ Female population = $(24,000 - x)$
$\Rightarrow$ Increase in male population = $\dfrac{6}{100} \times x = \dfrac{6x}{100}$
$\Rightarrow$ Increase in female population = $\dfrac{9}{100} \times (24,000 - x) = \dfrac{9(24,000)}{100} - \dfrac{9x}{100}$
So, we have the equation:
$\Rightarrow \dfrac{6x}{100} + \dfrac{2,16,000}{100} - \dfrac{9x}{100} = 1620$
$\Rightarrow \dfrac{6x}{100} - \dfrac{9x}{100} = 1620 - \dfrac{2,16,000}{100}$
$\Rightarrow \dfrac{6x - 9x}{100} = \dfrac{1,62,000 - 2,16,000}{100}$
$\Rightarrow -\dfrac{3x}{100} = -\dfrac{18,000}{100}$
$\Rightarrow x = \dfrac{18,000}{100} \times \dfrac{100}{3}$
$\therefore x = 6000$
$\Rightarrow$ So, No. of males = 6000
$\Rightarrow$ And, No. of females = $24,000 - 6000 = 18,000$ Ans
Q63. A village lost $12\%$ of its goat in a flood and $5\%$ of the remainder died from disease. If the number left now is 8360, what was original no. of goats before the flood?
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$\Rightarrow$ Let total no. of goats be = $x$
$\Rightarrow$ Total no. of goats lost in flood = $x - \dfrac{12}{100} \times x$
$\Rightarrow x - \dfrac{12x}{100} = \dfrac{100x - 12x}{100} = \dfrac{88x}{100}$
$\Rightarrow$ No. of goats died of disease = $\dfrac{88x}{100} - \dfrac{12}{100} \times \dfrac{88x}{100}$
$\Rightarrow \dfrac{88x}{100} - \dfrac{440x}{10000} = 0.88x - 0.0440x = 0.8360x$
So we have the equation:
$0.8360x = 8360$
$x = \dfrac{8360}{0.8360} = \dfrac{8360 \times 10,000}{8360} = 10,000$ Ans
$\Rightarrow$ Total no. of goats lost in flood = $x - \dfrac{12}{100} \times x$
$\Rightarrow x - \dfrac{12x}{100} = \dfrac{100x - 12x}{100} = \dfrac{88x}{100}$
$\Rightarrow$ No. of goats died of disease = $\dfrac{88x}{100} - \dfrac{12}{100} \times \dfrac{88x}{100}$
$\Rightarrow \dfrac{88x}{100} - \dfrac{440x}{10000} = 0.88x - 0.0440x = 0.8360x$
So we have the equation:
$0.8360x = 8360$
$x = \dfrac{8360}{0.8360} = \dfrac{8360 \times 10,000}{8360} = 10,000$ Ans
Q64. If A's income is $25\%$ more than B's, how much percent B's income is less than A's?
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$\Rightarrow$ Let B's income be = Rs.100
$\therefore$ A's income = $100 + \dfrac{25}{100} \times 100 = 100 + 25 = 125$
$\Rightarrow$ B's income is less by = $125 - 100 = 25$
$\therefore$ Percentage less = $\dfrac{25}{125} \times 100 = \dfrac{100}{5} = 20\%$ Ans
$\therefore$ A's income = $100 + \dfrac{25}{100} \times 100 = 100 + 25 = 125$
$\Rightarrow$ B's income is less by = $125 - 100 = 25$
$\therefore$ Percentage less = $\dfrac{25}{125} \times 100 = \dfrac{100}{5} = 20\%$ Ans
Q65. A's income is $20\%$ higher than that of B's. Find how much percent B's income is lower than A's?
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$\Rightarrow$ Let B's income be = Rs.100
$\therefore$ A's income = $100 + \dfrac{20}{100} \times 100 = 100 + 20 = 120$
$\Rightarrow$ B's income is less by = $120 - 100 = 20$
$\therefore$ Percentage less = $\dfrac{20}{120} \times 100 = \dfrac{100}{6} = 16\dfrac{2}{3}\%$ Ans
$\therefore$ A's income = $100 + \dfrac{20}{100} \times 100 = 100 + 20 = 120$
$\Rightarrow$ B's income is less by = $120 - 100 = 20$
$\therefore$ Percentage less = $\dfrac{20}{120} \times 100 = \dfrac{100}{6} = 16\dfrac{2}{3}\%$ Ans
Q66. A's income is $25\%$ less than that of B. How much percent B's income is more than A's?
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$\Rightarrow$ Let B's income be = Rs.100
$\therefore$ A's income = $100 - \dfrac{25}{100} \times 100 = 100 -25 = 75$
$\Rightarrow$ B's income is more by = $100 - 75 = 25$
$\therefore$ Percentage more = $\dfrac{25}{75} \times 100 = \dfrac{100}{3} = 33\dfrac{1}{3}\%$ Ans
$\therefore$ A's income = $100 - \dfrac{25}{100} \times 100 = 100 -25 = 75$
$\Rightarrow$ B's income is more by = $100 - 75 = 25$
$\therefore$ Percentage more = $\dfrac{25}{75} \times 100 = \dfrac{100}{3} = 33\dfrac{1}{3}\%$ Ans
Q67. A man spent $12.5\%$ of his money and after spending $75\%$ of the remainder he has Rs.875 left. Calculate the money he had at first?
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$\Rightarrow$ Let money at first be = Rs.100
$\Rightarrow$ Amount of money spent = $100 - \dfrac{125}{100 \times 10} \times 100$
$\Rightarrow 100 - \dfrac{125}{10} = 100 - \dfrac{25}{2} = \dfrac{200 - 25}{10} = \dfrac{175}{2}$
$\Rightarrow$ Remaining amount after spening $75\%$ of remainder = $\dfrac{175}{2} - \dfrac{75}{100} \times \dfrac{175}{2}$
$\Rightarrow \dfrac{175}{2} - \dfrac{3}{4} \times \dfrac{175}{2} = \dfrac{175}{2} - \dfrac{525}{8}$
$\Rightarrow \dfrac{700 - 525}{8} = \text{ Rs.}\dfrac{175}{8}$
$\Rightarrow$ When money left is Rs.$\dfrac{175}{8}$, then original money = Rs.100
$\Rightarrow$ When money left is Rs.1, then original money = $\dfrac{100}{\dfrac{175}{8}}$
$\therefore$ When money left is Rs.875, then original money = $\dfrac{100 \times 8}{175} \times 875$
$\Rightarrow 100 \times 8 \times 5 = 100 \times 40 = \text{ Rs.}4000$ Ans
$\Rightarrow$ Amount of money spent = $100 - \dfrac{125}{100 \times 10} \times 100$
$\Rightarrow 100 - \dfrac{125}{10} = 100 - \dfrac{25}{2} = \dfrac{200 - 25}{10} = \dfrac{175}{2}$
$\Rightarrow$ Remaining amount after spening $75\%$ of remainder = $\dfrac{175}{2} - \dfrac{75}{100} \times \dfrac{175}{2}$
$\Rightarrow \dfrac{175}{2} - \dfrac{3}{4} \times \dfrac{175}{2} = \dfrac{175}{2} - \dfrac{525}{8}$
$\Rightarrow \dfrac{700 - 525}{8} = \text{ Rs.}\dfrac{175}{8}$
$\Rightarrow$ When money left is Rs.$\dfrac{175}{8}$, then original money = Rs.100
$\Rightarrow$ When money left is Rs.1, then original money = $\dfrac{100}{\dfrac{175}{8}}$
$\therefore$ When money left is Rs.875, then original money = $\dfrac{100 \times 8}{175} \times 875$
$\Rightarrow 100 \times 8 \times 5 = 100 \times 40 = \text{ Rs.}4000$ Ans
Q68. A man looses $20.5\%$ of his money and after spending $80\%$ of the remainder, he is left with Rs.159. How much money did he had at first?
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$\Rightarrow$ Let money at first be = Rs.100
$\Rightarrow$ Amount of money lost = $100 - \dfrac{205}{100 \times 10} \times 100$
$\Rightarrow 100 - \dfrac{205}{10} = 100 - \dfrac{41}{2} = \dfrac{200 - 41}{2} = \dfrac{159}{2}$
$\Rightarrow$ Remaining amount after spening $80\%$ of remainder = $\dfrac{159}{2} - \dfrac{80}{100} \times \dfrac{159}{2}$
$\Rightarrow \dfrac{159}{2} - \dfrac{8}{10} \times \dfrac{159}{2} = \dfrac{159}{2} - \dfrac{4}{5} \times \dfrac{159}{2} = \dfrac{159}{2} - \dfrac{636}{10}$
$\Rightarrow \dfrac{795 - 636}{10} = \text{ Rs.}\dfrac{159}{10}$
$\Rightarrow$ When money left is Rs.$\dfrac{159}{10}$, then original money = Rs.100
$\Rightarrow$ When money left is Rs.1, then original money = $\dfrac{100}{\dfrac{159}{10}}$
$\therefore$ When money left is Rs.159, then original money = $\dfrac{100 \times 10}{159} \times 159$
$\Rightarrow 100 \times 10 = \text{ Rs.}1000$ Ans
$\Rightarrow$ Amount of money lost = $100 - \dfrac{205}{100 \times 10} \times 100$
$\Rightarrow 100 - \dfrac{205}{10} = 100 - \dfrac{41}{2} = \dfrac{200 - 41}{2} = \dfrac{159}{2}$
$\Rightarrow$ Remaining amount after spening $80\%$ of remainder = $\dfrac{159}{2} - \dfrac{80}{100} \times \dfrac{159}{2}$
$\Rightarrow \dfrac{159}{2} - \dfrac{8}{10} \times \dfrac{159}{2} = \dfrac{159}{2} - \dfrac{4}{5} \times \dfrac{159}{2} = \dfrac{159}{2} - \dfrac{636}{10}$
$\Rightarrow \dfrac{795 - 636}{10} = \text{ Rs.}\dfrac{159}{10}$
$\Rightarrow$ When money left is Rs.$\dfrac{159}{10}$, then original money = Rs.100
$\Rightarrow$ When money left is Rs.1, then original money = $\dfrac{100}{\dfrac{159}{10}}$
$\therefore$ When money left is Rs.159, then original money = $\dfrac{100 \times 10}{159} \times 159$
$\Rightarrow 100 \times 10 = \text{ Rs.}1000$ Ans
Q69. The price of a certain commodity is reduced by by $20\%$. By what percentage its consumption be increased in order to have the same expenditure?
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$\Rightarrow$ Let price of commodity be = Rs.100
$\therefore$ Reduced price = $100 - \dfrac{20}{100} \times 100 = 100 - 20 = \text{ Rs.}80$
$\Rightarrow$ Required increase in consumption = $100 - 20 = 80$
$\therefore$ Percentage increase = $\dfrac{20}{80} \times 100 = \dfrac{100}{4} = 25\%$ Ans
$\therefore$ Reduced price = $100 - \dfrac{20}{100} \times 100 = 100 - 20 = \text{ Rs.}80$
$\Rightarrow$ Required increase in consumption = $100 - 20 = 80$
$\therefore$ Percentage increase = $\dfrac{20}{80} \times 100 = \dfrac{100}{4} = 25\%$ Ans
Q70. The price of sugar is increased by $25\%$. What percentage of consumption is to be decreased so that there would be no increase in the expenditure?
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$\Rightarrow$ Let price of sugar be = Rs.100
$\therefore$ Increased price = $100 - \dfrac{25}{100} \times 100 = 100 + 25 = \text{ Rs.}125$
$\Rightarrow$ Required increase in consumption = $125 - 100 = 25$
$\therefore$ Percentage increase = $\dfrac{25}{125} \times 100 = \dfrac{100}{5} = 20\%$ Ans
$\therefore$ Increased price = $100 - \dfrac{25}{100} \times 100 = 100 + 25 = \text{ Rs.}125$
$\Rightarrow$ Required increase in consumption = $125 - 100 = 25$
$\therefore$ Percentage increase = $\dfrac{25}{125} \times 100 = \dfrac{100}{5} = 20\%$ Ans
Q71. A reduction of $10\%$ in the price of sugar would enable a man to buy 2 kgs of sugar more for Rs.125. Find reduced price per kg.
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$\Rightarrow$ Reduced price of 2 kg sugar = $\dfrac{10}{100} \times 125 = \dfrac{125}{10} = \text{ Rs.}\dfrac{25}{2}$
$\therefore$ Reduced price of 1 kg sugar = $\dfrac{\dfrac{25}{2}}{2} = \dfrac{25}{2 \times 2} = \dfrac{25}{4} = \text{ Rs.}6.25$ Ans
$\therefore$ Reduced price of 1 kg sugar = $\dfrac{\dfrac{25}{2}}{2} = \dfrac{25}{2 \times 2} = \dfrac{25}{4} = \text{ Rs.}6.25$ Ans
Q72. A reduction of $10\%$ in the price of sugar would enable a man to buy 16 kgs of sugar more for Rs.400. Find reduced price per kg. Also, find the original price of sugar per kg.
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$\Rightarrow$ Reduced price of 16 kgs sugar = $\dfrac{10}{100} \times 400 = \text{ Rs.}40$
$\therefore$ Reduced price of 1 kg sugar = $\dfrac{40}{16} = \dfrac{5}{2} = \text{ Rs.}2.5$ Ans
Now, $90\%$ of original price of sugar = $\dfrac{2.5}{90} \times 100$
$\Rightarrow \dfrac{25}{90 \times 10} \times 100 = \dfrac{25}{900} \times 100 = = \dfrac{25}{9} = \text{ Rs.}2\dfrac{7}{9} \text{ OR } 2.77$ Ans
$\therefore$ Reduced price of 1 kg sugar = $\dfrac{40}{16} = \dfrac{5}{2} = \text{ Rs.}2.5$ Ans
Now, $90\%$ of original price of sugar = $\dfrac{2.5}{90} \times 100$
$\Rightarrow \dfrac{25}{90 \times 10} \times 100 = \dfrac{25}{900} \times 100 = = \dfrac{25}{9} = \text{ Rs.}2\dfrac{7}{9} \text{ OR } 2.77$ Ans
Q73. The cost of manufacturing an article is divided between materials and labours in the ratio 8:7. If the cost of materials be increased by $40\%$ and the cost of labour be reduced by $20\%$, find the resulting change, as percent, in the total cost.
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$\Rightarrow$ Cost ratio of materials and labour = $8:7$
$\Rightarrow$ Let cost of materials be Rs.8 and cost of labours be Rs.7
$\therefore$ Total cost of article = $8 + 7 = \text{ Rs.}15$
$\Rightarrow$ New cost of materials = $8 + \dfrac{40}{100} \times 8$
$\Rightarrow 8 + \dfrac{40}{25} \times 2 = 8 + \dfrac{80}{25} = \dfrac{200 + 80}{25} = \dfrac{280}{25}= \text{ Rs.}11.20$
$\Rightarrow$ New cost of labours = $7 - \dfrac{20}{100} \times 7$
$\Rightarrow 7 - \dfrac{1}{5} \times 7 = 7 - \dfrac{7}{5} = \dfrac{35 - 7}{5} = \dfrac{28}{5}= \text{ Rs.}5.60$
$\therefore$ Total new cost of article = $11.20 + 5.60 = \text{ Rs.}16.80$
$\Rightarrow$ Increase in cost of article = $16.80 - 15.00 = \text{ Rs.}1.80$
$\therefore$ Increase in percentage = $\dfrac{1.80}{15} \times 100$
$\Rightarrow \dfrac{180}{15 \times 100} \times 100 = \dfrac{180}{15} = \dfrac{60}{5} = 12\%$ Ans
$\Rightarrow$ Let cost of materials be Rs.8 and cost of labours be Rs.7
$\therefore$ Total cost of article = $8 + 7 = \text{ Rs.}15$
$\Rightarrow$ New cost of materials = $8 + \dfrac{40}{100} \times 8$
$\Rightarrow 8 + \dfrac{40}{25} \times 2 = 8 + \dfrac{80}{25} = \dfrac{200 + 80}{25} = \dfrac{280}{25}= \text{ Rs.}11.20$
$\Rightarrow$ New cost of labours = $7 - \dfrac{20}{100} \times 7$
$\Rightarrow 7 - \dfrac{1}{5} \times 7 = 7 - \dfrac{7}{5} = \dfrac{35 - 7}{5} = \dfrac{28}{5}= \text{ Rs.}5.60$
$\therefore$ Total new cost of article = $11.20 + 5.60 = \text{ Rs.}16.80$
$\Rightarrow$ Increase in cost of article = $16.80 - 15.00 = \text{ Rs.}1.80$
$\therefore$ Increase in percentage = $\dfrac{1.80}{15} \times 100$
$\Rightarrow \dfrac{180}{15 \times 100} \times 100 = \dfrac{180}{15} = \dfrac{60}{5} = 12\%$ Ans
Q74. The cost of manufacturing an article is made up of 3 items, viz materials, labours and overheads. In 1983, the cost of these were in the ratio 6:5:4. In 1984, the cost of materials rose by $20\%$, of labours rose $12\%$ and of overheads rose by $30\%$. Find percentage increase in the cost of articles.
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$\Rightarrow$ Given ratio = 6:5:4
$\Rightarrow$ Let costs of materials, labours and overheads be Rs.6, Rs.5 and Rs.4
$\therefore$ Total cost of article = $6 + 5 + 4 = \text{ Rs.}15$
$\Rightarrow$ New cost of materials = $6 + \dfrac{20}{100} \times 6$
$\Rightarrow 6 + \dfrac{6}{5} = \dfrac{30 + 6}{5} = \dfrac{36}{5} = \text{ Rs.}7.20$
$\Rightarrow$ New cost of labours = $5 + \dfrac{12}{100} \times 5$
$\Rightarrow 5 + \dfrac{12}{20} = 5 + \dfrac{3}{5} = \dfrac{25 + 3}{5} = \dfrac{28}{5} = \text{ Rs.}5.60$
$\Rightarrow$ New cost of overheads = $4 + \dfrac{30}{100} \times 4$
$\Rightarrow 4 + \dfrac{30}{25} = 4 + \dfrac{6}{5} = \dfrac{20 + 6}{5} = \dfrac{26}{5} = \text{ Rs.}5.20$
$\therefore$ New total cost of article = $7.20 + 5.60 + 5.20 = \text{ Rs.}18$
$\Rightarrow$ Increase in cost of article = $18 - 15 = \text{ Rs.}3$
$\therefore$ Increase in percentage = $\dfrac{3}{15} \times 100$
$\Rightarrow \dfrac{100}{5} = 20\%$ Ans
$\Rightarrow$ Let costs of materials, labours and overheads be Rs.6, Rs.5 and Rs.4
$\therefore$ Total cost of article = $6 + 5 + 4 = \text{ Rs.}15$
$\Rightarrow$ New cost of materials = $6 + \dfrac{20}{100} \times 6$
$\Rightarrow 6 + \dfrac{6}{5} = \dfrac{30 + 6}{5} = \dfrac{36}{5} = \text{ Rs.}7.20$
$\Rightarrow$ New cost of labours = $5 + \dfrac{12}{100} \times 5$
$\Rightarrow 5 + \dfrac{12}{20} = 5 + \dfrac{3}{5} = \dfrac{25 + 3}{5} = \dfrac{28}{5} = \text{ Rs.}5.60$
$\Rightarrow$ New cost of overheads = $4 + \dfrac{30}{100} \times 4$
$\Rightarrow 4 + \dfrac{30}{25} = 4 + \dfrac{6}{5} = \dfrac{20 + 6}{5} = \dfrac{26}{5} = \text{ Rs.}5.20$
$\therefore$ New total cost of article = $7.20 + 5.60 + 5.20 = \text{ Rs.}18$
$\Rightarrow$ Increase in cost of article = $18 - 15 = \text{ Rs.}3$
$\therefore$ Increase in percentage = $\dfrac{3}{15} \times 100$
$\Rightarrow \dfrac{100}{5} = 20\%$ Ans
Q75.The tax on a commodity is diminished by $15\%$ and its consumption is increased by $10\%$. Find:
(i) Decrease percent in revenue derived from it.
(ii) With what increase percent in its consumption would the revenue remain the same.
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(i) Let tax be = Rs.100
$\Rightarrow$ And consumption be = Rs.100
$\because$ Revenue = Tax $\times$ Consumption
$\therefore$ Revenue derived originally = $100 \times 100 = \text{ Rs.}10,000$
$\Rightarrow$ Reduced Tax = $100 - \dfrac{15}{100} \times 100 = 100 - 15 = \text{ Rs.}85$
$\Rightarrow$ Increased Consumption = $100 + \dfrac{10}{100} \times 100 = 100 + 10 = \text{ Rs.}110$
$\therefore$ New Revenue = $110 \times 85 = \text{ Rs.}9350$
$\Rightarrow$ Decrease in revenue = $10,000 - 9350 = \text{ Rs.}650$
$\therefore$ Decrease percentage = $\dfrac{9350}{10,000} \times 100$
$\Rightarrow \dfrac{650}{100} = \dfrac{65}{10} = \dfrac{13}{2} = 6\dfrac{1}{2}\%$ Ans
(ii) New tax is = $85\% = \dfrac{85}{100} = \dfrac{17}{20}$
$\because$ New tax is $\dfrac{17}{20}$ of original tax, the revenue would remain the same if new consumption becomes $\dfrac{20}{17}$ of original consumption.
$\therefore$ Required increase = $(\dfrac{20}{17} - 1) \times 100$
$\Rightarrow \dfrac{3}{17} \times 100 = \dfrac{300}{17} = 17\dfrac{11}{17}$ Ans
$\Rightarrow$ And consumption be = Rs.100
$\because$ Revenue = Tax $\times$ Consumption
$\therefore$ Revenue derived originally = $100 \times 100 = \text{ Rs.}10,000$
$\Rightarrow$ Reduced Tax = $100 - \dfrac{15}{100} \times 100 = 100 - 15 = \text{ Rs.}85$
$\Rightarrow$ Increased Consumption = $100 + \dfrac{10}{100} \times 100 = 100 + 10 = \text{ Rs.}110$
$\therefore$ New Revenue = $110 \times 85 = \text{ Rs.}9350$
$\Rightarrow$ Decrease in revenue = $10,000 - 9350 = \text{ Rs.}650$
$\therefore$ Decrease percentage = $\dfrac{9350}{10,000} \times 100$
$\Rightarrow \dfrac{650}{100} = \dfrac{65}{10} = \dfrac{13}{2} = 6\dfrac{1}{2}\%$ Ans
(ii) New tax is = $85\% = \dfrac{85}{100} = \dfrac{17}{20}$
$\because$ New tax is $\dfrac{17}{20}$ of original tax, the revenue would remain the same if new consumption becomes $\dfrac{20}{17}$ of original consumption.
$\therefore$ Required increase = $(\dfrac{20}{17} - 1) \times 100$
$\Rightarrow \dfrac{3}{17} \times 100 = \dfrac{300}{17} = 17\dfrac{11}{17}$ Ans