Q1. Evaluate:
(i) $55\% \text{ of } 160 + 24\% \text{ of } 50 - 36\% \text{ of } 150$
(ii) $9.3\% \text{ of } 500 - 4.8\% \text{ of } 250 - 2.5\% \text{ of } 240$
Show Answer
(i) $55\% \text{ of } 160 + 24\% \text{ of } 50 - 36\% \text{ of } 150$
$\Rightarrow \dfrac{55}{100} \times 160 + \dfrac{24}{100} \times 50 - \dfrac{36}{100} \times 150$
$\Rightarrow \dfrac{55}{10} \times 16 + \dfrac{24}{10} \times 5 - \dfrac{36}{10} \times 15$
$\Rightarrow \dfrac{11}{2} \times 16 + \dfrac{24}{2} - \dfrac{36}{2} \times 3$
$\Rightarrow (11 \times 8) + 12 - (18 \times 3)$
$\Rightarrow 88 + 12 - 54 = 100 - 54 = 46$ Ans
(ii) $9.3\% \text{ of } 500 - 4.8\% \text{ of } 250 - 2.5\% \text{ of } 240$
$\Rightarrow \dfrac{93}{100 \times 10} \times 500 - \dfrac{48}{100 \times 10} \times 250 - \dfrac{25}{100 \times 10} \times 240$
$\Rightarrow \dfrac{93}{10} \times 5 - \dfrac{48}{100} \times 25 - \dfrac{25}{100} \times 24$
$\Rightarrow \dfrac{93}{2} - \dfrac{48}{4} - \dfrac{1}{4} \times 24$
$\Rightarrow 46.5 - 12 - 6 = 34.5 + 6 = 28.5$ Ans
$\Rightarrow \dfrac{55}{100} \times 160 + \dfrac{24}{100} \times 50 - \dfrac{36}{100} \times 150$
$\Rightarrow \dfrac{55}{10} \times 16 + \dfrac{24}{10} \times 5 - \dfrac{36}{10} \times 15$
$\Rightarrow \dfrac{11}{2} \times 16 + \dfrac{24}{2} - \dfrac{36}{2} \times 3$
$\Rightarrow (11 \times 8) + 12 - (18 \times 3)$
$\Rightarrow 88 + 12 - 54 = 100 - 54 = 46$ Ans
(ii) $9.3\% \text{ of } 500 - 4.8\% \text{ of } 250 - 2.5\% \text{ of } 240$
$\Rightarrow \dfrac{93}{100 \times 10} \times 500 - \dfrac{48}{100 \times 10} \times 250 - \dfrac{25}{100 \times 10} \times 240$
$\Rightarrow \dfrac{93}{10} \times 5 - \dfrac{48}{100} \times 25 - \dfrac{25}{100} \times 24$
$\Rightarrow \dfrac{93}{2} - \dfrac{48}{4} - \dfrac{1}{4} \times 24$
$\Rightarrow 46.5 - 12 - 6 = 34.5 + 6 = 28.5$ Ans
Q2. (i) A number is increased from 125 to 150, find the percentage increase.
(ii) A number is decreased from 250 to 200, find percentage decrease.
Show Answer
(i) $\because$ Given, increase = $150 - 125 = 25$
$\therefore$ Required increase percentage = $\dfrac{25}{125} \times 100 = \dfrac{1}{5} \times 100 = 20\%$ Ans
(ii) $\because$ Given, decrease = $250 - 200 = 50$
$\therefore$ Required decrease percentage = $\dfrac{50}{250} \times 100 = \dfrac{1}{5} \times 100 = 20\%$ Ans
$\therefore$ Required increase percentage = $\dfrac{25}{125} \times 100 = \dfrac{1}{5} \times 100 = 20\%$ Ans
(ii) $\because$ Given, decrease = $250 - 200 = 50$
$\therefore$ Required decrease percentage = $\dfrac{50}{250} \times 100 = \dfrac{1}{5} \times 100 = 20\%$ Ans
Q3. Find:
(i) 45 is what percent of 54?
(ii) 2.7 is what percent of 18?
Show Answer
(i) $\dfrac{45}{54} \times 100 = \dfrac{5}{6} \times 100 = \dfrac{5}{3} \times 50 = \dfrac{250}{3} = 83\dfrac{1}{3}\%$ Ans
(ii) $\dfrac{27}{18 \times 10} \times 100 = \dfrac{27}{18} \times 10 = \dfrac{27}{9} \times 5 = 3 \times 5 = 15\%$ Ans
(ii) $\dfrac{27}{18 \times 10} \times 100 = \dfrac{27}{18} \times 10 = \dfrac{27}{9} \times 5 = 3 \times 5 = 15\%$ Ans
Q4. (i) 252 is $35\%$ of a certain number, find the number.
(ii) If $14\%$ of a number is 315, find the number.
Show Answer
(i) $\dfrac{35}{100} \text{ of Required number} = 252$
$\therefore \text{ Required number} = 252 \times \dfrac{100}{35} = 252 \times \dfrac{20}{7} = 36 \times 20 = 720$ Ans
(ii) $\dfrac{14}{100} \text{ of Required number} = 315$
$\therefore \text{ Required number} = 315 \times \dfrac{100}{14} = 315 \times \dfrac{50}{7} = 45 \times 50 = 2250$ Ans
$\therefore \text{ Required number} = 252 \times \dfrac{100}{35} = 252 \times \dfrac{20}{7} = 36 \times 20 = 720$ Ans
(ii) $\dfrac{14}{100} \text{ of Required number} = 315$
$\therefore \text{ Required number} = 315 \times \dfrac{100}{14} = 315 \times \dfrac{50}{7} = 45 \times 50 = 2250$ Ans
Q5. Find percentage change, when a number is changed from:
(i) 80 to 100
(ii) 100 to 80
(iii) 6.25 to 7.50
Show Answer
(i) $\because$ Increase = $100 - 80 = 20$
$\therefore$ Increase percentage = $\dfrac{20}{80} \times 100 = \dfrac{1}{4} \times 100 = 25\%$ Ans
(ii) $\because$ Decrease = $100 - 80 = 20$
$\therefore$ Decrease percentage = $\dfrac{20}{100} \times 100 = \dfrac{1}{5} \times 100 = 20\%$ Ans
(iii) $\because$ Increase = $7.50 - 6.25 = 1.25$
$\therefore$ Increase percentage = $\dfrac{1.25}{6.25} \times 100 = \dfrac{125 \times 100}{625 \times 100} \times 100$
$\Rightarrow \dfrac{125}{625} \times 100 = \dfrac{5}{25} \times 100 = 5 \times 4 = 20\%$ Ans
$\therefore$ Increase percentage = $\dfrac{20}{80} \times 100 = \dfrac{1}{4} \times 100 = 25\%$ Ans
(ii) $\because$ Decrease = $100 - 80 = 20$
$\therefore$ Decrease percentage = $\dfrac{20}{100} \times 100 = \dfrac{1}{5} \times 100 = 20\%$ Ans
(iii) $\because$ Increase = $7.50 - 6.25 = 1.25$
$\therefore$ Increase percentage = $\dfrac{1.25}{6.25} \times 100 = \dfrac{125 \times 100}{625 \times 100} \times 100$
$\Rightarrow \dfrac{125}{625} \times 100 = \dfrac{5}{25} \times 100 = 5 \times 4 = 20\%$ Ans
Q6. An auctioneer charges $8\%$ for selling a house. If the house is sold for Rs.2,30,500. Find charges of the auctioneer.
Show Answer
$\Rightarrow$ Given, percentage charged = $8\%$
$\Rightarrow$ S.P. of house = Rs.$2,30,500$
$\therefore$ Auctioneer's amount = $\dfrac{8}{100} \times 2,30,000 = 8 \times 2305 = \text{Rs.}18,440$ Ans
$\Rightarrow$ S.P. of house = Rs.$2,30,500$
$\therefore$ Auctioneer's amount = $\dfrac{8}{100} \times 2,30,000 = 8 \times 2305 = \text{Rs.}18,440$ Ans
Q7. Out of 800 oranges, 50 are rotten. Find percentage of good oranges.
Show Answer
$\Rightarrow$ Total no. of oranges = $800$
$\Rightarrow$ No. of rotten oranges = $50$
$\therefore$ No. of good oranges = $800 - 50 = 750$
$\Rightarrow$ Percentage of good oranges = $\dfrac{750}{800} \times 100$
$\Rightarrow \dfrac{750}{8} = \dfrac{375}{4} = 93\dfrac{3}{4}\%$ Ans
$\Rightarrow$ No. of rotten oranges = $50$
$\therefore$ No. of good oranges = $800 - 50 = 750$
$\Rightarrow$ Percentage of good oranges = $\dfrac{750}{800} \times 100$
$\Rightarrow \dfrac{750}{8} = \dfrac{375}{4} = 93\dfrac{3}{4}\%$ Ans
Q8. A cistern contains 5 thousand litres of water. if $6\%$ of is leaked. Find how many litres of water is left in the cistern.
Show Answer
$\Rightarrow$ Litres of water in cistern = $5000$
$\Rightarrow$ Percent of water leaked = $6\%$
$\Rightarrow$ Litres of water leaked = $\dfrac{6}{100} \times 5000 = 6 \times 50 = 300$
$\therefore$ Remaining Litres of water = $5000 - 300 = 4700$ Ans
$\Rightarrow$ Percent of water leaked = $6\%$
$\Rightarrow$ Litres of water leaked = $\dfrac{6}{100} \times 5000 = 6 \times 50 = 300$
$\therefore$ Remaining Litres of water = $5000 - 300 = 4700$ Ans
Q9. A man spends $87\%$ of his salary. If he saves Rs.325, find his salary.
Show Answer
$\Rightarrow$ Percentage spent = $87\%$
$\therefore$ Percentage saved = $100 - 87 = 13\%$
$\Rightarrow$ Let salary be = $x$
$\Rightarrow \dfrac{13}{100} \text{ of } x = 325$
$\Rightarrow x = 325 \times \dfrac{100}{13} = 25 \times 100 = \text{Rs.}2500$ Ans
$\therefore$ Percentage saved = $100 - 87 = 13\%$
$\Rightarrow$ Let salary be = $x$
$\Rightarrow \dfrac{13}{100} \text{ of } x = 325$
$\Rightarrow x = 325 \times \dfrac{100}{13} = 25 \times 100 = \text{Rs.}2500$ Ans
Q10. (i) A number 3.625 is wrongly read as 3.265, find percentage error.
(ii) A number is $5.78 \times 10^3$ is wrongly written as $5.87 \times 10^3$, find percentage error.
Show Answer
(i) Error = $3.625 - 3.265 = 0.360$
$\therefore$ Percentage error = $\dfrac{0.360}{3.625} \times 100$
$\Rightarrow \dfrac{360 \times 1000}{3625 \times 1000} \times 100 = \dfrac{360}{3625} \times 100$
$\Rightarrow \dfrac{72}{725} \times 100 = \dfrac{72}{145} \times 20 = \dfrac{72}{29} \times 4 = \dfrac{288}{29} = 9.93\%$ Ans
(ii) $1^{st}$ No. = $5.78 \times 10^3 = \dfrac{578}{100} \times 1000 = 578 \times 10 = 5780$
$\Rightarrow 2^{nd}$ No. = $5.87 \times 10^3 = \dfrac{587}{100} \times 1000 = 587 \times 10 = 5870$
$\Rightarrow$ Error = $5870 - 5780 = 90$
$\therefore$ Percentage error = $\dfrac{90}{5780} \times 100$
$\Rightarrow \dfrac{90}{578} \times 10 = \dfrac{90}{289} \times 5$
$\Rightarrow \dfrac{450}{289} = 1.56\%$ Ans
$\therefore$ Percentage error = $\dfrac{0.360}{3.625} \times 100$
$\Rightarrow \dfrac{360 \times 1000}{3625 \times 1000} \times 100 = \dfrac{360}{3625} \times 100$
$\Rightarrow \dfrac{72}{725} \times 100 = \dfrac{72}{145} \times 20 = \dfrac{72}{29} \times 4 = \dfrac{288}{29} = 9.93\%$ Ans
(ii) $1^{st}$ No. = $5.78 \times 10^3 = \dfrac{578}{100} \times 1000 = 578 \times 10 = 5780$
$\Rightarrow 2^{nd}$ No. = $5.87 \times 10^3 = \dfrac{587}{100} \times 1000 = 587 \times 10 = 5870$
$\Rightarrow$ Error = $5870 - 5780 = 90$
$\therefore$ Percentage error = $\dfrac{90}{5780} \times 100$
$\Rightarrow \dfrac{90}{578} \times 10 = \dfrac{90}{289} \times 5$
$\Rightarrow \dfrac{450}{289} = 1.56\%$ Ans
Q11. In an election, a candidate secured $58\%$ of the votes polled and won the election by 18,336 votes. Find the total number of votes polled and the votes secured by each candidate.
Show Answer
$\Rightarrow$ Vote percentage secured by winning candidate = $58\%$
$\therefore$ Vote percentage secured by losing candidate = $100 - 58 = 42\%$
$\Rightarrow$ Majority percentage = $58 - 42 = 16\%$
$\Rightarrow$ Given, majority of vote = 18,336
$\Rightarrow$ Let total no. of votes be = $x$
$\therefore 16\% \text{ of } x = 18,336$
$\Rightarrow x = \dfrac{100}{16} \times 18,336$
$\Rightarrow x = 100 \times 1146 = 1,14,600$ Ans
$\Rightarrow$ Now, votes secured by winning candidate = $\dfrac{58}{100} \times 1,14,600$
$\Rightarrow 58 \times 1146 = 66,468$ Ans
$\Rightarrow$ And, votes secured by losing candidate = $\dfrac{42}{100} \times 1,14,600$
$\Rightarrow 42 \times 1146 = 48,132$ Ans
$\therefore$ Vote percentage secured by losing candidate = $100 - 58 = 42\%$
$\Rightarrow$ Majority percentage = $58 - 42 = 16\%$
$\Rightarrow$ Given, majority of vote = 18,336
$\Rightarrow$ Let total no. of votes be = $x$
$\therefore 16\% \text{ of } x = 18,336$
$\Rightarrow x = \dfrac{100}{16} \times 18,336$
$\Rightarrow x = 100 \times 1146 = 1,14,600$ Ans
$\Rightarrow$ Now, votes secured by winning candidate = $\dfrac{58}{100} \times 1,14,600$
$\Rightarrow 58 \times 1146 = 66,468$ Ans
$\Rightarrow$ And, votes secured by losing candidate = $\dfrac{42}{100} \times 1,14,600$
$\Rightarrow 42 \times 1146 = 48,132$ Ans
Q12. In an election, a candidate secured $47\%$ of the votes polled and lost the election by 12,336 votes. Find the total votes polled and the votes secured by the winning candidate.
Show Answer
$\Rightarrow$ Vote percentage secured by losing candidate = $47\%$
$\therefore$ Vote percentage secured by winning candidate = $100 - 47 = 53\%$
$\Rightarrow$ Losing percentage = $53 - 47 = 6\%$
$\Rightarrow$ Given, losing candidate lost by = 12,366 votes
$\Rightarrow$ Let total no. of votes be = $x$
$\therefore 6\% \text{ of } x = 12,366$
$\Rightarrow x = \dfrac{100}{6} \times 12,366$
$\Rightarrow x = 100 \times 2061 = 2,06,100$ Ans
$\Rightarrow$ Now, votes secured by winning candidate = $\dfrac{47}{100} \times 2,06,100$
$\Rightarrow 47 \times 2061 = 1,09,233$ Ans
$\therefore$ Vote percentage secured by winning candidate = $100 - 47 = 53\%$
$\Rightarrow$ Losing percentage = $53 - 47 = 6\%$
$\Rightarrow$ Given, losing candidate lost by = 12,366 votes
$\Rightarrow$ Let total no. of votes be = $x$
$\therefore 6\% \text{ of } x = 12,366$
$\Rightarrow x = \dfrac{100}{6} \times 12,366$
$\Rightarrow x = 100 \times 2061 = 2,06,100$ Ans
$\Rightarrow$ Now, votes secured by winning candidate = $\dfrac{47}{100} \times 2,06,100$
$\Rightarrow 47 \times 2061 = 1,09,233$ Ans
Q13. The cost of a scooter depreciates every year by $15\%$ of its value at the begining of the year. If the present cost of the scooter is Rs.8000, find its cost:
(i) After 1 year.
(ii) After 2 year.
Show Answer
(i) Present cost of scooter = Rs.8000
$\Rightarrow$ Depreciation after 1 year = $\dfrac{15}{100} \times 8000 = 15 \times 80 = \text{ Rs.}1200$
$\therefore$ Cost of scooter after 1 year = $8000 - 1200 = \text{ Rs.}6800$ Ans
(ii) Cost at the begining of $2^{nd}$ year = Rs.6800
$\Rightarrow$ Depreciation after 2 years = $\dfrac{15}{100} \times 6800 = 15 \times 68 = \text{ Rs.}1020$
$\therefore$ Cost of scooter after 2 years = $6800 - 1020 = \text{ Rs.}5780$ Ans
$\Rightarrow$ Depreciation after 1 year = $\dfrac{15}{100} \times 8000 = 15 \times 80 = \text{ Rs.}1200$
$\therefore$ Cost of scooter after 1 year = $8000 - 1200 = \text{ Rs.}6800$ Ans
(ii) Cost at the begining of $2^{nd}$ year = Rs.6800
$\Rightarrow$ Depreciation after 2 years = $\dfrac{15}{100} \times 6800 = 15 \times 68 = \text{ Rs.}1020$
$\therefore$ Cost of scooter after 2 years = $6800 - 1020 = \text{ Rs.}5780$ Ans
Q14. In an examination. the pass mark is $40\%$. if a candidate gets 65 marks and fails by 3 marks, find maximum marks.
Show Answer
$\Rightarrow$ Given, pass marks = $40\%$
$\Rightarrow$ Marks secured by the candidate = 65
$\Rightarrow$ Marks obtained less than pass marks = 3
$\therefore$ Pass marks = $65 + 3 = 68$
$\Rightarrow$ Let maximum marks be = $x$
$\therefore 40\%$ of $x$ = 68
$\Rightarrow x = \dfrac{100}{40} \times 68$
$\Rightarrow x = \dfrac{10}{4} \times 68 = 10 \times 17 = 170$ Ans
$\Rightarrow$ Marks secured by the candidate = 65
$\Rightarrow$ Marks obtained less than pass marks = 3
$\therefore$ Pass marks = $65 + 3 = 68$
$\Rightarrow$ Let maximum marks be = $x$
$\therefore 40\%$ of $x$ = 68
$\Rightarrow x = \dfrac{100}{40} \times 68$
$\Rightarrow x = \dfrac{10}{4} \times 68 = 10 \times 17 = 170$ Ans
Q15. In an examination, a candidate secured 125 marks and failed by 15 marks. If the pass percentage was $35\%$, find the maximum marks.
Show Answer
$\Rightarrow$ Given, pass marks = $35\%$
$\Rightarrow$ Marks secured by the candidate = 125
$\Rightarrow$ Marks obtained less than pass marks = 15
$\therefore$ Pass marks = $125 + 15 = 140$
$\Rightarrow$ Let maximum marks be = $x$
$\therefore 35\%$ of $x$ = 140
$\Rightarrow x = \dfrac{100}{35} \times 140$
$\Rightarrow x = \dfrac{20}{7} \times 140 = 20 \times 20 = 400$ Ans
$\Rightarrow$ Marks secured by the candidate = 125
$\Rightarrow$ Marks obtained less than pass marks = 15
$\therefore$ Pass marks = $125 + 15 = 140$
$\Rightarrow$ Let maximum marks be = $x$
$\therefore 35\%$ of $x$ = 140
$\Rightarrow x = \dfrac{100}{35} \times 140$
$\Rightarrow x = \dfrac{20}{7} \times 140 = 20 \times 20 = 400$ Ans