Q46. If $15\%$ of the oranges were bad and the remaining 1360 were good, find the original number of oranges.
Show Answer
$\Rightarrow$ Let total no. of oranges be = $x$
$\therefore x - \dfrac{15}{100} \times x = 1360$
$\Rightarrow x - \dfrac{3x}{20} = 1360$
$\Rightarrow \dfrac{20x - 3x}{20} = 1360$
$\Rightarrow \dfrac{17x}{20} = 1360$
$\Rightarrow x = 1360 \times \dfrac{20}{17} = 80 \times 20 = 1600$ Ans
$\therefore x - \dfrac{15}{100} \times x = 1360$
$\Rightarrow x - \dfrac{3x}{20} = 1360$
$\Rightarrow \dfrac{20x - 3x}{20} = 1360$
$\Rightarrow \dfrac{17x}{20} = 1360$
$\Rightarrow x = 1360 \times \dfrac{20}{17} = 80 \times 20 = 1600$ Ans
Q47. The population of a village is decreased by $23\%$ and now is 1386. What was the original population of the village? Also, find the decrease in its population.
Show Answer
$\Rightarrow$ Let total population be = $x$
$\therefore x - \dfrac{23}{100} \times x = 1386$
$\Rightarrow x - \dfrac{23x}{100} = 1386$
$\Rightarrow \dfrac{100x - 23x}{100} = 1386$
$\Rightarrow \dfrac{77x}{100} = 1386$
$\Rightarrow x = 1360 \times \dfrac{100}{77} = 18 \times 100 = 1800$ Ans
Also, decrease in population = $1800 - 1386 = 414$ Ans
$\therefore x - \dfrac{23}{100} \times x = 1386$
$\Rightarrow x - \dfrac{23x}{100} = 1386$
$\Rightarrow \dfrac{100x - 23x}{100} = 1386$
$\Rightarrow \dfrac{77x}{100} = 1386$
$\Rightarrow x = 1360 \times \dfrac{100}{77} = 18 \times 100 = 1800$ Ans
Also, decrease in population = $1800 - 1386 = 414$ Ans
Q48. A man, with an income of Rs.1020 per month, spends $85\%$ of it. What does he save in a year.
Show Answer
$\Rightarrow$ Savings per month:
$\Rightarrow 1020 - \dfrac{85}{100} \times 1020 = 1020 - \dfrac{85}{10} \times 102 = 1020 - \dfrac{17}{2} \times 102$
$\Rightarrow 1020 - (17 \times 51) = 1020 - 867 = \text{ Rs.}153$
$\therefore$ Yearly savings = $153 \times 12 = \text{ Rs.}1836$ Ans
$\Rightarrow 1020 - \dfrac{85}{100} \times 1020 = 1020 - \dfrac{85}{10} \times 102 = 1020 - \dfrac{17}{2} \times 102$
$\Rightarrow 1020 - (17 \times 51) = 1020 - 867 = \text{ Rs.}153$
$\therefore$ Yearly savings = $153 \times 12 = \text{ Rs.}1836$ Ans
Q49. A factory produced 3000 articles in 1982. It increased by $20\%$ in 1983 and then by $15\%$ in 1984. Calculate the total produce in 1984.
Show Answer
$\Rightarrow$ Production in 1982 = 3000
$\Rightarrow$ Production in 1983 = $3000 + \dfrac{20}{100} \times 3000$
$\Rightarrow 3000 + (20 \times 30) = 3000 + 600 = 3600$
$\Rightarrow$ Production in 1984 = $3600 + \dfrac{15}{100} \times 3600$
$\Rightarrow 3600 + (15 \times 36) = 3600 + 540 = 4140$ Ans
$\Rightarrow$ Production in 1983 = $3000 + \dfrac{20}{100} \times 3000$
$\Rightarrow 3000 + (20 \times 30) = 3000 + 600 = 3600$
$\Rightarrow$ Production in 1984 = $3600 + \dfrac{15}{100} \times 3600$
$\Rightarrow 3600 + (15 \times 36) = 3600 + 540 = 4140$ Ans
Q50. A manufacturer estimates that upon inspection $12\%$ of the articles, he produces, will be rejected. He accepts an order to supply 22,000 articles at Rs.12 each. Calculate the least number of articles he manufacture to ensure the completion of the order. Also, he estimates the profit on his outlay including the manufacturing of rejected articles to be $20\%$. Find the cost of manufacturing of each article.
Show Answer
$\Rightarrow$ Let total no. of articles to be produced be = $x$
$\Rightarrow$ Required percentage of production = $\dfrac{100 - 12}{100} = \dfrac{88}{100} = \dfrac{22}{25}$
As such, we have: $\dfrac{22}{25} \text{ of } x = 22,000$
$\Rightarrow \dfrac{22x}{25} = 22,000$
$\Rightarrow x = \dfrac{22,000 \times 25}{22} = 1000 \times 25 = 25,000$ Ans
Now, S.P. of total supply = $22,000 \times 12 = \text{ Rs.}2,64,000$
$\therefore$ C.P. of total supply = $2,64,000 - \dfrac{20}{100} \times 2,64,000$
$\Rightarrow 2,64,000 - \dfrac{1}{5} \times 2,64,000 = 2,64,000 - 52,800 = \text{ Rs.}2,11,800$
As such, C.P. per article = $\dfrac{2,11,800}{22,000} = \dfrac{2112}{220} = \dfrac{1056}{110} = \dfrac{528}{55} = \text{ Rs.}9.6$ Ans
$\Rightarrow$ Required percentage of production = $\dfrac{100 - 12}{100} = \dfrac{88}{100} = \dfrac{22}{25}$
As such, we have: $\dfrac{22}{25} \text{ of } x = 22,000$
$\Rightarrow \dfrac{22x}{25} = 22,000$
$\Rightarrow x = \dfrac{22,000 \times 25}{22} = 1000 \times 25 = 25,000$ Ans
Now, S.P. of total supply = $22,000 \times 12 = \text{ Rs.}2,64,000$
$\therefore$ C.P. of total supply = $2,64,000 - \dfrac{20}{100} \times 2,64,000$
$\Rightarrow 2,64,000 - \dfrac{1}{5} \times 2,64,000 = 2,64,000 - 52,800 = \text{ Rs.}2,11,800$
As such, C.P. per article = $\dfrac{2,11,800}{22,000} = \dfrac{2112}{220} = \dfrac{1056}{110} = \dfrac{528}{55} = \text{ Rs.}9.6$ Ans
Q51. A manufacturer estimates that on inspection $12\%$ of the articles he produces will be rejected. He accepts an order to supply 22000 articles at Rs. 7.50 each. Calculate the least number of articles he must manufacture to ensure the completion of the order. He estimates the profit on his outlay including the manufacturing of rejected articles to be $12\%$. Find the cost of manufacturing of each article.
Show Answer
$\Rightarrow$ Let total no. of articles to be produced be = $x$
$\Rightarrow$ Required percentage of production = $\dfrac{100 - 12}{100} = \dfrac{88}{100} = \dfrac{22}{25}$
As such, we have: $\dfrac{22}{25} \text{ of } x = 22,000$
$\Rightarrow \dfrac{22x}{25} = 22,000$
$\Rightarrow x = \dfrac{22,000 \times 25}{22} = 1000 \times 25 = 25,000$ Ans
Now, S.P. of total supply = $22,000 \times 7.50 = \text{ Rs.}2,64,000$
$\Rightarrow 22,000 \times \dfrac{750}{100} = 220 \times 750 = \text{ Rs.}1,65,000$
$\therefore$ C.P. of total supply = $1,65,000 - \dfrac{12}{100} \times 1,65,000$
$\Rightarrow 1,65,000 - 12 \times 1650 = 2,64,000 - 19,800 = \text{ Rs.}1,45,200$
As such, C.P. per article = $\dfrac{1,45,200}{22,000} = \dfrac{1452}{220} = \dfrac{726}{110} = \dfrac{363}{55} = \text{ Rs.}6.6$ Ans
$\Rightarrow$ Required percentage of production = $\dfrac{100 - 12}{100} = \dfrac{88}{100} = \dfrac{22}{25}$
As such, we have: $\dfrac{22}{25} \text{ of } x = 22,000$
$\Rightarrow \dfrac{22x}{25} = 22,000$
$\Rightarrow x = \dfrac{22,000 \times 25}{22} = 1000 \times 25 = 25,000$ Ans
Now, S.P. of total supply = $22,000 \times 7.50 = \text{ Rs.}2,64,000$
$\Rightarrow 22,000 \times \dfrac{750}{100} = 220 \times 750 = \text{ Rs.}1,65,000$
$\therefore$ C.P. of total supply = $1,65,000 - \dfrac{12}{100} \times 1,65,000$
$\Rightarrow 1,65,000 - 12 \times 1650 = 2,64,000 - 19,800 = \text{ Rs.}1,45,200$
As such, C.P. per article = $\dfrac{1,45,200}{22,000} = \dfrac{1452}{220} = \dfrac{726}{110} = \dfrac{363}{55} = \text{ Rs.}6.6$ Ans
Q52. A manufacturer estimates that on inspection $12\%$ of his articles will be rejected. He accepts an order to supply 4400 articles at Rs.15 each. He estimates the profit on his outlay including manufacturing of rejected articles to be $20\%$. Find the cost of manufacturing each article. Also, calculate the least number of articles he must manufacture to ensure the completion of the order.
Show Answer
$\Rightarrow$ S.P. of total supply = $4400 \times 15 = \text{ Rs.}66,000$
$\therefore$ C.P. of total supply = $66,000 - \dfrac{20}{100} \times 66,000$
$\Rightarrow 66,000 - 13,200 = \text{ Rs.}52,800$
As such, C.P. per article = $\dfrac{52,800}{4400} = \dfrac{528}{44} = \text{ Rs.}12$ Ans
Now, let total no. of articles to be produced be = $x$
$\Rightarrow$ Required percentage of production = $\dfrac{100 - 12}{100} = \dfrac{88}{100} = \dfrac{22}{25}$
As such, we have: $\dfrac{22}{25} \text{ of } x = 4400$
$\Rightarrow \dfrac{22x}{25} = 4400$
$\Rightarrow x = \dfrac{4400 \times 25}{22} = 200 \times 25 = 5,000$ Ans
$\therefore$ C.P. of total supply = $66,000 - \dfrac{20}{100} \times 66,000$
$\Rightarrow 66,000 - 13,200 = \text{ Rs.}52,800$
As such, C.P. per article = $\dfrac{52,800}{4400} = \dfrac{528}{44} = \text{ Rs.}12$ Ans
Now, let total no. of articles to be produced be = $x$
$\Rightarrow$ Required percentage of production = $\dfrac{100 - 12}{100} = \dfrac{88}{100} = \dfrac{22}{25}$
As such, we have: $\dfrac{22}{25} \text{ of } x = 4400$
$\Rightarrow \dfrac{22x}{25} = 4400$
$\Rightarrow x = \dfrac{4400 \times 25}{22} = 200 \times 25 = 5,000$ Ans
Q53. During a certain week, the number of the unemployed persons diminished by 36060, this being $15\%$ of what it had been. Find the total number of unemployed persons after this decrease.
Show Answer
$\Rightarrow$ Let total no. of people unemployed be = $x$
$\Rightarrow$ Given, decrease in no. of people unemployed be = 36,060
$\Rightarrow$ Decrease percent = $15\%$
$\therefore$ Total no. of people unemployed = $\dfrac{15}{100} \text{ of } x = 36,060$
$\Rightarrow \dfrac{3x}{20} = 36,060$
$\Rightarrow x = 36,060 \times \dfrac{20}{3} = 12,020 \times 20 = 2,40,400$
$\Rightarrow$ Total no. of unemployed after decrease = $2,40,400 - 36060 = 2,04,340$ Ans
$\Rightarrow$ Given, decrease in no. of people unemployed be = 36,060
$\Rightarrow$ Decrease percent = $15\%$
$\therefore$ Total no. of people unemployed = $\dfrac{15}{100} \text{ of } x = 36,060$
$\Rightarrow \dfrac{3x}{20} = 36,060$
$\Rightarrow x = 36,060 \times \dfrac{20}{3} = 12,020 \times 20 = 2,40,400$
$\Rightarrow$ Total no. of unemployed after decrease = $2,40,400 - 36060 = 2,04,340$ Ans
Q54. A man saves $30\%$ of his monthly salary. If on account of dearness of things, he is to increase his monthly expenses by $30\%$, he is able to save Rs.81 only. Find his monthly income.
Show Answer
$\Rightarrow$ Let monthly salary be = $\text{ Rs.}x$
$\therefore$ Monthly savings = $\dfrac{30}{100} \times x$
$\Rightarrow \dfrac{30x}{100} = \dfrac{6x}{20} = \dfrac{3x}{10}$
$\Rightarrow$ Monthly expenses = $x - \dfrac{3x}{10} = \dfrac{10x - 3x}{10} = \dfrac{7x}{10}$
$\Rightarrow$ Increase in monthly expenses = $\dfrac{7x}{10} + 30\% \text{ of } \dfrac{7x}{10}$
$\Rightarrow \dfrac{7x}{10} + \dfrac{3x}{10} \times \dfrac{7x}{10} = \dfrac{7x}{10} + \dfrac{21}{100} = \dfrac{70x + 21x}{100} = \dfrac{91x}{100}$
$\Rightarrow$ New monthly savings = $\dfrac{91x}{100}$
$\therefore$ New savings = $x - \dfrac{91x}{100} = \dfrac{100x - 91x}{100} = \dfrac{9x}{100}$
$\Rightarrow$ Given savings = Rs.81
So we have:
$\Rightarrow \dfrac{9x}{100} = 81$
$\Rightarrow x = 81 \times \dfrac{100}{9} = 9 \times 100 = \text{ Rs.}900$ Ans
$\therefore$ Monthly savings = $\dfrac{30}{100} \times x$
$\Rightarrow \dfrac{30x}{100} = \dfrac{6x}{20} = \dfrac{3x}{10}$
$\Rightarrow$ Monthly expenses = $x - \dfrac{3x}{10} = \dfrac{10x - 3x}{10} = \dfrac{7x}{10}$
$\Rightarrow$ Increase in monthly expenses = $\dfrac{7x}{10} + 30\% \text{ of } \dfrac{7x}{10}$
$\Rightarrow \dfrac{7x}{10} + \dfrac{3x}{10} \times \dfrac{7x}{10} = \dfrac{7x}{10} + \dfrac{21}{100} = \dfrac{70x + 21x}{100} = \dfrac{91x}{100}$
$\Rightarrow$ New monthly savings = $\dfrac{91x}{100}$
$\therefore$ New savings = $x - \dfrac{91x}{100} = \dfrac{100x - 91x}{100} = \dfrac{9x}{100}$
$\Rightarrow$ Given savings = Rs.81
So we have:
$\Rightarrow \dfrac{9x}{100} = 81$
$\Rightarrow x = 81 \times \dfrac{100}{9} = 9 \times 100 = \text{ Rs.}900$ Ans
Q55. In an election contested by two candidates, $10\%$ of the voters did not cast their votes. The successful candidate won by a majority of 300 votes, securing $48\%$ of the total votes. How many votes were cast for each candidate?
Show Answer
$\Rightarrow$ Votes secured by winning candidate = $48\%$
$\therefore$ Votes secured by loosing candidate = $100 - 48 = 52\%$
$\Rightarrow$ Votes not cast = $52 - 10 = 42\%$
$\Rightarrow$ Majority percent = $48 - 42 = 6\%$
$\Rightarrow$ Given, majority of votes = 3600
$\Rightarrow$ Let total votes polled be = $x$
So we have: $\dfrac{6}{100} \text{ of } x = 3600$
$\Rightarrow \dfrac{3x}{50} = 3600$
$\Rightarrow x = 3600 \times \dfrac{50}{3} = 1200 \times 50 = 60,000$
$\therefore$ Votes secured by winning candidate = $\dfrac{48}{100} \times 60,000$
$\Rightarrow 48 \times 600 = 28,800$ Ans
$\therefore$ Votes secured by loosing candidate = $\dfrac{42}{100} \times 60,000$
$\Rightarrow 42 \times 600 = 25,200$ Ans
$\therefore$ Votes secured by loosing candidate = $100 - 48 = 52\%$
$\Rightarrow$ Votes not cast = $52 - 10 = 42\%$
$\Rightarrow$ Majority percent = $48 - 42 = 6\%$
$\Rightarrow$ Given, majority of votes = 3600
$\Rightarrow$ Let total votes polled be = $x$
So we have: $\dfrac{6}{100} \text{ of } x = 3600$
$\Rightarrow \dfrac{3x}{50} = 3600$
$\Rightarrow x = 3600 \times \dfrac{50}{3} = 1200 \times 50 = 60,000$
$\therefore$ Votes secured by winning candidate = $\dfrac{48}{100} \times 60,000$
$\Rightarrow 48 \times 600 = 28,800$ Ans
$\therefore$ Votes secured by loosing candidate = $\dfrac{42}{100} \times 60,000$
$\Rightarrow 42 \times 600 = 25,200$ Ans
Q56. In an election, contested by two candidates only, the candidate who gets $34\%$ of the votes polled was defeated by a majority of 1600 votes. Find the total number of votes polled and the votes secured by each candidate.
Show Answer
$\Rightarrow$ Votes secured by loosing candidate = $34\%$
$\therefore$ Votes secured by winning candidate = $100 - 34 = 66\%$
$\Rightarrow$ Majority percent = $66 - 34 = 32\%$
$\Rightarrow$ Given, majority of votes = 1600
$\Rightarrow$ Let total votes polled be = $x$
So we have: $\dfrac{32}{100} \text{ of } x = 1600$
$\Rightarrow \dfrac{8x}{25} = 1600$
$\Rightarrow x = 1600 \times \dfrac{25}{8} = 200 \times 25 = 5,000$ Ans
$\therefore$ Votes secured by winning candidate = $\dfrac{66}{100} \times 5,000$
$\Rightarrow 66 \times 50 = 3300$ Ans
$\therefore$ Votes secured by loosing candidate = $5000 - 3300 = 1700$ Ans
$\therefore$ Votes secured by winning candidate = $100 - 34 = 66\%$
$\Rightarrow$ Majority percent = $66 - 34 = 32\%$
$\Rightarrow$ Given, majority of votes = 1600
$\Rightarrow$ Let total votes polled be = $x$
So we have: $\dfrac{32}{100} \text{ of } x = 1600$
$\Rightarrow \dfrac{8x}{25} = 1600$
$\Rightarrow x = 1600 \times \dfrac{25}{8} = 200 \times 25 = 5,000$ Ans
$\therefore$ Votes secured by winning candidate = $\dfrac{66}{100} \times 5,000$
$\Rightarrow 66 \times 50 = 3300$ Ans
$\therefore$ Votes secured by loosing candidate = $5000 - 3300 = 1700$ Ans
Q57. In an election, $8\%$ of the votes cast were declared invalid. The candidate who got $50\%$ of the total votes won by 240 votes. Find the total number of votes cast, if the winning candidate had no invalid votes.
Show Answer
$\Rightarrow$ Votes secured by winning candidate = $50\%$
$\therefore$ Votes secured by loosing candidate = $100 - 50 = 50\%$
$\Rightarrow$ Invalid votes = $8\%$
$\therefore$ Valid votes = $50 - 8 = 42\%$
$\Rightarrow$ Majority percent = $50 - 42 = 8\%$
$\Rightarrow$ Given, majority of votes = 240
$\Rightarrow$ Let total votes polled be = $x$
So we have: $\dfrac{8}{100} \text{ of } x = 240$
$\Rightarrow \dfrac{2x}{25} = 240$
$\Rightarrow x = 240 \times \dfrac{25}{2} = 120 \times 25 = 3,000$ Ans
$\therefore$ Votes secured by loosing candidate = $100 - 50 = 50\%$
$\Rightarrow$ Invalid votes = $8\%$
$\therefore$ Valid votes = $50 - 8 = 42\%$
$\Rightarrow$ Majority percent = $50 - 42 = 8\%$
$\Rightarrow$ Given, majority of votes = 240
$\Rightarrow$ Let total votes polled be = $x$
So we have: $\dfrac{8}{100} \text{ of } x = 240$
$\Rightarrow \dfrac{2x}{25} = 240$
$\Rightarrow x = 240 \times \dfrac{25}{2} = 120 \times 25 = 3,000$ Ans
Q58. An architects' fees are worked out at $10\%$ of the first Rs.25,000 of cost and $2\dfrac{1}{2}\%$ on the remainder of costs.
(a) Find Architects' fees on given building costs:
(i) Rs.60,650
(ii) Rs.80,460
(iii) Rs.89,540
(b) Express to one decimal place, architects' total fee on the three buildings as a percentage of the total building costs.
Show Answer
(a)(i)
$\Rightarrow$ Given cost = Rs.60,650
$\Rightarrow$ $10\%$ on first Rs.25,000 = $\dfrac{10}{100} \times 25,000 = 10 \times 250 = \text{ Rs.}2500$
$\Rightarrow$ Remainder cost = $60,650 - 25,000 = \text{ Rs.}35,650$
$\Rightarrow 2\dfrac{1}{2}\%$ on remainder cost = $2\dfrac{1}{2} \times 35,650$
$\Rightarrow \dfrac{5}{2 \times 100} \times 35,650 = \dfrac{5}{20} \times 3565 = \dfrac{1}{4} \times 3565 = \text{ Rs.}891.25$
$\therefore$ Architects' fees = $2500 + 891.25 = \text{ Rs.}3391.25$ Ans
(ii) Given cost = Rs.80,460
$\Rightarrow$ $10\%$ on first Rs.25,000 = $\dfrac{10}{100} \times 25,000 = 10 \times 250 = \text{ Rs.}2500$
$\Rightarrow$ Remainder cost = $80,460 - 25,000 = \text{ Rs.}55,460$
$\Rightarrow 2\dfrac{1}{2}\%$ on remainder cost = $2\dfrac{1}{2} \times 55,460$
$\Rightarrow \dfrac{5}{2 \times 100} \times 55,460 = \dfrac{5}{20} \times 5546 = \dfrac{1}{4} \times 5546 = \text{ Rs.}1386.50$
$\therefore$ Architects' fees = $2500 + 1386.50 = \text{ Rs.}3886.50$ Ans
(iii) Given cost = Rs.89,540
$\Rightarrow$ $10\%$ on first Rs.25,000 = $\dfrac{10}{100} \times 25,000 = 10 \times 250 = \text{ Rs.}2500$
$\Rightarrow$ Remainder cost = $89,540 - 25,000 = \text{ Rs.}64,540$
$\Rightarrow 2\dfrac{1}{2}\%$ on remainder cost = $2\dfrac{1}{2} \times 64,540$
$\Rightarrow \dfrac{5}{2 \times 100} \times 64,540 = \dfrac{5}{20} \times 6454 = \dfrac{1}{4} \times 6454 = \text{ Rs.}1613.50$
$\therefore$ Architects' fees = $2500 + 1613.50 = \text{ Rs.}4116.50$ Ans
(b) Total cost of three buildings = $60,650 + 80,460 + 89,540 = \text{ Rs.}2,30,650$
$\Rightarrow$ $10\%$ on first Rs.25,000 = $\dfrac{10}{100} \times 25,000 = 10 \times 250 = Rs.2500$
$\Rightarrow$ Remainder cost = $2,30,650 - 25,000 = Rs.2,06,650$
$\Rightarrow 2\dfrac{1}{2}\%$ on remainder cost = $2\dfrac{1}{2} \times 2,05,650$
$\Rightarrow \dfrac{5}{2 \times 100} \times 20,565 = \dfrac{5}{20} \times 20,565 = \dfrac{1}{4} \times 20,565 = \text{ Rs.}5141.25$
$\therefore$ Architects' fees = $5141.25 + 2500 = \text{ Rs.}7641.25$ Ans
$\Rightarrow$ Given cost = Rs.60,650
$\Rightarrow$ $10\%$ on first Rs.25,000 = $\dfrac{10}{100} \times 25,000 = 10 \times 250 = \text{ Rs.}2500$
$\Rightarrow$ Remainder cost = $60,650 - 25,000 = \text{ Rs.}35,650$
$\Rightarrow 2\dfrac{1}{2}\%$ on remainder cost = $2\dfrac{1}{2} \times 35,650$
$\Rightarrow \dfrac{5}{2 \times 100} \times 35,650 = \dfrac{5}{20} \times 3565 = \dfrac{1}{4} \times 3565 = \text{ Rs.}891.25$
$\therefore$ Architects' fees = $2500 + 891.25 = \text{ Rs.}3391.25$ Ans
(ii) Given cost = Rs.80,460
$\Rightarrow$ $10\%$ on first Rs.25,000 = $\dfrac{10}{100} \times 25,000 = 10 \times 250 = \text{ Rs.}2500$
$\Rightarrow$ Remainder cost = $80,460 - 25,000 = \text{ Rs.}55,460$
$\Rightarrow 2\dfrac{1}{2}\%$ on remainder cost = $2\dfrac{1}{2} \times 55,460$
$\Rightarrow \dfrac{5}{2 \times 100} \times 55,460 = \dfrac{5}{20} \times 5546 = \dfrac{1}{4} \times 5546 = \text{ Rs.}1386.50$
$\therefore$ Architects' fees = $2500 + 1386.50 = \text{ Rs.}3886.50$ Ans
(iii) Given cost = Rs.89,540
$\Rightarrow$ $10\%$ on first Rs.25,000 = $\dfrac{10}{100} \times 25,000 = 10 \times 250 = \text{ Rs.}2500$
$\Rightarrow$ Remainder cost = $89,540 - 25,000 = \text{ Rs.}64,540$
$\Rightarrow 2\dfrac{1}{2}\%$ on remainder cost = $2\dfrac{1}{2} \times 64,540$
$\Rightarrow \dfrac{5}{2 \times 100} \times 64,540 = \dfrac{5}{20} \times 6454 = \dfrac{1}{4} \times 6454 = \text{ Rs.}1613.50$
$\therefore$ Architects' fees = $2500 + 1613.50 = \text{ Rs.}4116.50$ Ans
(b) Total cost of three buildings = $60,650 + 80,460 + 89,540 = \text{ Rs.}2,30,650$
$\Rightarrow$ $10\%$ on first Rs.25,000 = $\dfrac{10}{100} \times 25,000 = 10 \times 250 = Rs.2500$
$\Rightarrow$ Remainder cost = $2,30,650 - 25,000 = Rs.2,06,650$
$\Rightarrow 2\dfrac{1}{2}\%$ on remainder cost = $2\dfrac{1}{2} \times 2,05,650$
$\Rightarrow \dfrac{5}{2 \times 100} \times 20,565 = \dfrac{5}{20} \times 20,565 = \dfrac{1}{4} \times 20,565 = \text{ Rs.}5141.25$
$\therefore$ Architects' fees = $5141.25 + 2500 = \text{ Rs.}7641.25$ Ans
Q59. 65 pupils from school A entered for an examination and $80\%$ of them passed. School B entered 10 more pupils than school A, but the number of pupils who passed from school the number of pupils who passed from school B was 4 less than the number of pupils who passed from school A. Find the percentage of passes in school B.
Show Answer
$\Rightarrow$ No. of pupil from school A = 65
$\Rightarrow$ Pass percentage from school A = $80\%$
$\therefore$ No. of pupil passed from school A = $\dfrac{80}{100} \times 65$
$\Rightarrow \dfrac{8}{10} \times 65 = \dfrac{4}{5} \times 65 = 4 \times 13 = 52$
$\Rightarrow$ No. of pupil from school A = $65 + 10 = 75$
$\Rightarrow$ No. of pupil passed from school A = $52 - 4 = 48$
$\therefore$ Pass percentage from school A = $\dfrac{48}{75} \times 100$
$\Rightarrow \dfrac{48}{15} \times 20 = \dfrac{48}{3} \times 4 = 16 \times 4 = 64\%$ Ans
$\Rightarrow$ Pass percentage from school A = $80\%$
$\therefore$ No. of pupil passed from school A = $\dfrac{80}{100} \times 65$
$\Rightarrow \dfrac{8}{10} \times 65 = \dfrac{4}{5} \times 65 = 4 \times 13 = 52$
$\Rightarrow$ No. of pupil from school A = $65 + 10 = 75$
$\Rightarrow$ No. of pupil passed from school A = $52 - 4 = 48$
$\therefore$ Pass percentage from school A = $\dfrac{48}{75} \times 100$
$\Rightarrow \dfrac{48}{15} \times 20 = \dfrac{48}{3} \times 4 = 16 \times 4 = 64\%$ Ans
Q60. In a survey it was found that 224 children representing $28\%$ of the total number of children in a school came to school by buses, 510 walked and the rest used their bicycles. Calculate:
(i) The total number of children in school.
(ii) The percentage of children who used bicycles.
Show Answer
(i) Let total no. of students be = $x$
$\Rightarrow$ No. of students using the bus = 224
$\Rightarrow$ Given, $28\%$ = 224
$\therefore$ Total no. of students = $\dfrac{28}{100} \times x = 224$
$\Rightarrow x = 224 \times \dfrac{100}{28} = 8 \times 100 = 800$ Ans
(ii) No. of students walked = $510$
$\Rightarrow$ No. of students used bus and walked = $224 + 510 = 734$
$\Rightarrow$ No. of students using bicycle = $800 - 734 = 66$
$\therefore$ Percent of students using bicycle = $\dfrac{66}{800} \times 100$
$\Rightarrow \dfrac{66}{8} = \dfrac{33}{4} = 8.25\%$ Ans
$\Rightarrow$ No. of students using the bus = 224
$\Rightarrow$ Given, $28\%$ = 224
$\therefore$ Total no. of students = $\dfrac{28}{100} \times x = 224$
$\Rightarrow x = 224 \times \dfrac{100}{28} = 8 \times 100 = 800$ Ans
(ii) No. of students walked = $510$
$\Rightarrow$ No. of students used bus and walked = $224 + 510 = 734$
$\Rightarrow$ No. of students using bicycle = $800 - 734 = 66$
$\therefore$ Percent of students using bicycle = $\dfrac{66}{800} \times 100$
$\Rightarrow \dfrac{66}{8} = \dfrac{33}{4} = 8.25\%$ Ans