Q16. In an objective tpe paper of 150 questions, John got $80\%$ correct answers and Mohan got $64\%$ correct answers. Find:
(i) How many correct answers did each get?
(ii) What percentage is Mohan's correct answers to John's correct answers.
Show Answer
(i) Given, total no. of questions = 150
$\Rightarrow$ Percentage of Mohan's correct answers = $80\%$
$\therefore$ no. of correct answers by Mohan = $\dfrac{80}{100} \times 150 = 8 \times 15 = 120$ Ans
$\Rightarrow$ Percentage of John's correct answers = $64\%$
$\therefore$ no. of correct answers by John = $\dfrac{64}{100} \times 150$
$\Rightarrow \dfrac{64}{10} \times 15 = \dfrac{64}{2} \times 3 = 32 \times 3 = 96$ Ans
(ii) Percentage of Mohan's correct answer to that of John's = $\dfrac{96}{120} \times 100$
$\Rightarrow \dfrac{96}{12} \times 10 = 8 \times 10 = 80\%$ Ans
$\Rightarrow$ Percentage of Mohan's correct answers = $80\%$
$\therefore$ no. of correct answers by Mohan = $\dfrac{80}{100} \times 150 = 8 \times 15 = 120$ Ans
$\Rightarrow$ Percentage of John's correct answers = $64\%$
$\therefore$ no. of correct answers by John = $\dfrac{64}{100} \times 150$
$\Rightarrow \dfrac{64}{10} \times 15 = \dfrac{64}{2} \times 3 = 32 \times 3 = 96$ Ans
(ii) Percentage of Mohan's correct answer to that of John's = $\dfrac{96}{120} \times 100$
$\Rightarrow \dfrac{96}{12} \times 10 = 8 \times 10 = 80\%$ Ans
Q17. A man bought a certain number of oranges, out of which $13\%$ were rotten. He gave $75\%$ of the remaining in charity and still has 522 left. Find how many had he bought?
Show Answer
$\Rightarrow$ Let total no. of oranges be = 100
$\Rightarrow$ Percentage of rotten oranges = $13\%$
$\therefore$ Remaining no. of oranges = $100 - \dfrac{13}{100} \text{ of } 100 = 100 - 13 = 87$
$\Rightarrow$ Percentage of oranges given to charity = $13\%$
$\therefore$ Remaining no. of oranges = $87 - \dfrac{75}{100} \text{ of } 87$
$\Rightarrow 87 - \dfrac{3}{4} \times 87 = 87 - \dfrac{261}{4} = \dfrac{348 - 261}{4} = \dfrac{87}{4}$
$\Rightarrow$ Now, when remaining no. of oranges is $\dfrac{87}{4}$, then original no. = 100
$\Rightarrow$ And, when remaining no. of oranges is 1, then original no. = $\dfrac{4}{87} \times 100$
$\therefore$ when remaining no. of oranges is 522, then original no. = $\dfrac{4 \times 100 \times 522}{87}$
$\Rightarrow 4 \times 100 \times 6 = 100 \times 24 = 2400$ Ans
$\Rightarrow$ Percentage of rotten oranges = $13\%$
$\therefore$ Remaining no. of oranges = $100 - \dfrac{13}{100} \text{ of } 100 = 100 - 13 = 87$
$\Rightarrow$ Percentage of oranges given to charity = $13\%$
$\therefore$ Remaining no. of oranges = $87 - \dfrac{75}{100} \text{ of } 87$
$\Rightarrow 87 - \dfrac{3}{4} \times 87 = 87 - \dfrac{261}{4} = \dfrac{348 - 261}{4} = \dfrac{87}{4}$
$\Rightarrow$ Now, when remaining no. of oranges is $\dfrac{87}{4}$, then original no. = 100
$\Rightarrow$ And, when remaining no. of oranges is 1, then original no. = $\dfrac{4}{87} \times 100$
$\therefore$ when remaining no. of oranges is 522, then original no. = $\dfrac{4 \times 100 \times 522}{87}$
$\Rightarrow 4 \times 100 \times 6 = 100 \times 24 = 2400$ Ans
Q18. $5\%$ pupil in town died due to some diseases and $3\%$ of the remaining have left the town. If 2,76,450 pupil are still in town, find original number of pupil in the town.
Show Answer
$\Rightarrow$ Let original no. of people be = 100
$\therefore$ No. of people died of disease = $100 - \dfrac{5}{100} \times 100 = 100 - 5 = 95$
$\therefore$ No. of people left the town = $95 - \dfrac{3}{100} \times 95$
$\Rightarrow 95 - \dfrac{3}{100} \times 95 = 95 - \dfrac{3}{20} \times 19 = 95 - \dfrac{57}{20}$
$\Rightarrow 95 - \dfrac{57}{20} = \dfrac{1900 - 57}{20} = \dfrac{1843}{20}$
$\Rightarrow$ Now, when remaining no. of oranges is $\dfrac{1843}{20}$, then original no. = 100
$\Rightarrow$ And, when remaining no. of oranges is 1, then original no. = $\dfrac{20}{1843} \times 100$
$\therefore$ when remaining no. of oranges is 2,76,450, then original no. = $\dfrac{20 \times 100 \times 2,76,450}{1843}$
$\Rightarrow 20 \times 100 \times 150 = 100 \times 3000 = 3,00,000$ Ans
$\therefore$ No. of people died of disease = $100 - \dfrac{5}{100} \times 100 = 100 - 5 = 95$
$\therefore$ No. of people left the town = $95 - \dfrac{3}{100} \times 95$
$\Rightarrow 95 - \dfrac{3}{100} \times 95 = 95 - \dfrac{3}{20} \times 19 = 95 - \dfrac{57}{20}$
$\Rightarrow 95 - \dfrac{57}{20} = \dfrac{1900 - 57}{20} = \dfrac{1843}{20}$
$\Rightarrow$ Now, when remaining no. of oranges is $\dfrac{1843}{20}$, then original no. = 100
$\Rightarrow$ And, when remaining no. of oranges is 1, then original no. = $\dfrac{20}{1843} \times 100$
$\therefore$ when remaining no. of oranges is 2,76,450, then original no. = $\dfrac{20 \times 100 \times 2,76,450}{1843}$
$\Rightarrow 20 \times 100 \times 150 = 100 \times 3000 = 3,00,000$ Ans
Q19. In a combined test in English and Physics, $36\%$ candidates failed in English, $28\%$ failed in Physics and $12\%$ in both, find:
(i) Percentage of passed candidates.
(ii) Total no. of candidates apeared, if 208 candidates have failed.
Show Answer
$\Rightarrow$ Percentage of candidate failed only in English = $36 - 12 = 24\%$
$\Rightarrow$ Percentage of candidate failed only in Physics = $28 - 12 = 16\%$
(i) Total percentage of candidates failed = $24 + 16 + 12 = 52\%$
$\therefore$ Percentage of candidates passed = $100 - 52 = 48\%$ Ans
(ii) Given, total no. of candidates failed = 208
$\Rightarrow$ Let total no. of candidates be = $x$
$\Rightarrow$ We have, $52\%$ of total candidates = 208
$\therefore \dfrac{52}{100} \text{ of } x = 208$
$\Rightarrow x = \dfrac{100}{52} \times 208 = 100 \times 4 = 400$ Ans
$\Rightarrow$ Percentage of candidate failed only in Physics = $28 - 12 = 16\%$
(i) Total percentage of candidates failed = $24 + 16 + 12 = 52\%$
$\therefore$ Percentage of candidates passed = $100 - 52 = 48\%$ Ans
(ii) Given, total no. of candidates failed = 208
$\Rightarrow$ Let total no. of candidates be = $x$
$\Rightarrow$ We have, $52\%$ of total candidates = 208
$\therefore \dfrac{52}{100} \text{ of } x = 208$
$\Rightarrow x = \dfrac{100}{52} \times 208 = 100 \times 4 = 400$ Ans
Q20. In a combined test in Maths and Chemistry, $84\%$ candidates passed in Maths, $76\%$ in Chemistry and $8\%$ failed in both. Find:
(i) The percentage of failed candidates.
(ii) If 340 candidates passed in the test, then how many candidates appeared?
Show Answer
$\Rightarrow$ Percentage of candidate passed in Maths = $84\%$
$\therefore$ Percentage of candidate failed in Maths = $100 - 84 = 16\%$
$\Rightarrow$ Percentage of candidate passed only in Maths = $16 - 8 = 8\%$
$\Rightarrow$ Percentage of candidate passed in Physics = $76\%$
$\therefore$ Percentage of candidate failed in Physics = $100 - 76 = 24\%$
$\Rightarrow$ Percentage of candidate passed only in Physics = $24 - 8 = 16\%$
(i) Total percentage of failed candidates = $8 + 16 + 8 = 32\%$ Ans
(ii) Total percentage of passed candidates = $100 - 32 = 68\%$
$\Rightarrow$ Given, total no. of candidates passed = 340
$\Rightarrow$ Let total no. of candidates appeared be = $x$
$\therefore \dfrac{68}{100} \text{ of } x = 340$
$\Rightarrow x = \dfrac{100}{68} \times 340 = 100 \times 5 = 500$ Ans
$\therefore$ Percentage of candidate failed in Maths = $100 - 84 = 16\%$
$\Rightarrow$ Percentage of candidate passed only in Maths = $16 - 8 = 8\%$
$\Rightarrow$ Percentage of candidate passed in Physics = $76\%$
$\therefore$ Percentage of candidate failed in Physics = $100 - 76 = 24\%$
$\Rightarrow$ Percentage of candidate passed only in Physics = $24 - 8 = 16\%$
(i) Total percentage of failed candidates = $8 + 16 + 8 = 32\%$ Ans
(ii) Total percentage of passed candidates = $100 - 32 = 68\%$
$\Rightarrow$ Given, total no. of candidates passed = 340
$\Rightarrow$ Let total no. of candidates appeared be = $x$
$\therefore \dfrac{68}{100} \text{ of } x = 340$
$\Rightarrow x = \dfrac{100}{68} \times 340 = 100 \times 5 = 500$ Ans
Q21. A's income is $25\%$ more than B's. Find B's income is how much percent less than A's.
Show Answer
$\Rightarrow$ Let B's income be = Rs.100
$\therefore$ A's income = $100 + \dfrac{25}{100} \times 100 = 100 + 25 = 125$
$\Rightarrow$ And B's is less by = $125 - 100 =$ Rs.25
$\therefore \dfrac{25}{125} \times 100 = \dfrac{100}{5} = 20\%$ Ans
Note - Include larger value as base value to calculate how much less of percent, and use the lower value as base value to find how much more of percentage.
$\therefore$ A's income = $100 + \dfrac{25}{100} \times 100 = 100 + 25 = 125$
$\Rightarrow$ And B's is less by = $125 - 100 =$ Rs.25
$\therefore \dfrac{25}{125} \times 100 = \dfrac{100}{5} = 20\%$ Ans
Note - Include larger value as base value to calculate how much less of percent, and use the lower value as base value to find how much more of percentage.
Q22. Mona is $20\%$ younger than Neetu. How much percentage is Neetu older than Mona?
Show Answer
$\Rightarrow$ Let Neetu's age be = $x$
$\therefore$ Mona's age = $x - \dfrac{20}{100} \times x$
$\Rightarrow x - \dfrac{x}{5} = \dfrac{5x - x}{5} = \dfrac{4x}{5}$
$\Rightarrow$ Age difference = $x - \dfrac{4x}{5} = \dfrac{5x - 4x}{5} = \dfrac{x}{5}$
$\therefore \dfrac{\text{Age difference}}{\text{Age of Mona}} \times 100$
$\Rightarrow \dfrac{\dfrac{x}{5}}{\dfrac{4x}{5}} \times 100 = \dfrac{x \times 5}{4x \times 5} \times 100 = \dfrac{100x}{4x} = 25\%$ Ans
$\therefore$ Mona's age = $x - \dfrac{20}{100} \times x$
$\Rightarrow x - \dfrac{x}{5} = \dfrac{5x - x}{5} = \dfrac{4x}{5}$
$\Rightarrow$ Age difference = $x - \dfrac{4x}{5} = \dfrac{5x - 4x}{5} = \dfrac{x}{5}$
$\therefore \dfrac{\text{Age difference}}{\text{Age of Mona}} \times 100$
$\Rightarrow \dfrac{\dfrac{x}{5}}{\dfrac{4x}{5}} \times 100 = \dfrac{x \times 5}{4x \times 5} \times 100 = \dfrac{100x}{4x} = 25\%$ Ans
Q23. If the price of sugar is increased by $25\%$ today.By what percentage should it be decreased tommorow to bring the price price back to the original?
Show Answer
$\Rightarrow$ Let original price of sugar be = Rs.100
$\therefore$ Today's price = $100 + \dfrac{25}{100} \times 100 = 100 + 25 = 125$
$\Rightarrow$ Difference = $125 - 100 = 25$
$\therefore \dfrac{25}{125} \times 100 = \dfrac{100}{5} = 20\%$ Ans
Note - Include larger value as base value to calculate how much less of percent, and use the lower value as base value to find how much more of percentage.
$\therefore$ Today's price = $100 + \dfrac{25}{100} \times 100 = 100 + 25 = 125$
$\Rightarrow$ Difference = $125 - 100 = 25$
$\therefore \dfrac{25}{125} \times 100 = \dfrac{100}{5} = 20\%$ Ans
Note - Include larger value as base value to calculate how much less of percent, and use the lower value as base value to find how much more of percentage.
Q24. A number increased by $15\%$ becomes 391. Find the number.
Show Answer
$\Rightarrow$ Let the number be = $x$
$\therefore x + \dfrac{15}{100} \times x = 391$
$\Rightarrow x + \dfrac{3x}{20} = 391$
$\Rightarrow \dfrac{20x + 3x}{20} = 391$
$\Rightarrow \dfrac{23x}{20} = 391$
$\Rightarrow x = \dfrac{391 \times 20}{23} = 17 \times 20 = 340$ Ans
$\therefore x + \dfrac{15}{100} \times x = 391$
$\Rightarrow x + \dfrac{3x}{20} = 391$
$\Rightarrow \dfrac{20x + 3x}{20} = 391$
$\Rightarrow \dfrac{23x}{20} = 391$
$\Rightarrow x = \dfrac{391 \times 20}{23} = 17 \times 20 = 340$ Ans
Q25. A number decreased by $23\%$ becomes 539. Find the number.
Show Answer
$\Rightarrow$ Let the number be = $x$
$\therefore x - \dfrac{23}{100} \times x = 539$
$\Rightarrow x - \dfrac{23x}{100} = 539$
$\Rightarrow \dfrac{100x - 23x}{100} = 539$
$\Rightarrow \dfrac{77x}{100} = 539$
$\Rightarrow x = \dfrac{539 \times 100}{77} = 7 \times 100 = 700$ Ans
$\therefore x - \dfrac{23}{100} \times x = 539$
$\Rightarrow x - \dfrac{23x}{100} = 539$
$\Rightarrow \dfrac{100x - 23x}{100} = 539$
$\Rightarrow \dfrac{77x}{100} = 539$
$\Rightarrow x = \dfrac{539 \times 100}{77} = 7 \times 100 = 700$ Ans
Q26. Two numbers are respectively $20\%$ and $50\%$ more than a third number. What percent is the second of the first?
Show Answer
$\Rightarrow$ Let the third number be = 100
$\therefore$ First number = $100 + \dfrac{20}{100} \times 100 = 100 + 20 = 120$
$\Rightarrow$ Second number = $100 + \dfrac{50}{100} \times 100 = 100 + 50 = 150$
$\therefore$ Required percentage = $\dfrac{150}{120} \times 100$
$\Rightarrow \dfrac{15}{12} \times 100 = \dfrac{5}{4} \times 100 = 5 \times 25 = 125\%$ Ans
$\therefore$ First number = $100 + \dfrac{20}{100} \times 100 = 100 + 20 = 120$
$\Rightarrow$ Second number = $100 + \dfrac{50}{100} \times 100 = 100 + 50 = 150$
$\therefore$ Required percentage = $\dfrac{150}{120} \times 100$
$\Rightarrow \dfrac{15}{12} \times 100 = \dfrac{5}{4} \times 100 = 5 \times 25 = 125\%$ Ans
Q27. Two numbers are respectively $20\%$ and $50\%$ of a third number. What percent is the second of the first?
Show Answer
$\Rightarrow$ Let the third number be = 100
$\therefore$ First number = $\dfrac{20}{100} \times 100 = 20$
$\Rightarrow$ Second number = $\dfrac{50}{100} \times 100 = 50$
$\therefore$ Required percentage = $\dfrac{50}{20} \times 100$
$\Rightarrow \dfrac{50}{20} \times 100 = 50 \times 5 = 250\%$ Ans
$\therefore$ First number = $\dfrac{20}{100} \times 100 = 20$
$\Rightarrow$ Second number = $\dfrac{50}{100} \times 100 = 50$
$\therefore$ Required percentage = $\dfrac{50}{20} \times 100$
$\Rightarrow \dfrac{50}{20} \times 100 = 50 \times 5 = 250\%$ Ans
Q28. Two numbers are respectively $30\%$ and $40\%$ less than a third number. What percent is the second of the first?
Show Answer
$\Rightarrow$ Let the third number be = 100
$\therefore$ First number = $100 - \dfrac{30}{100} \times 100 = 100 - 30 = 70$
$\Rightarrow$ Second number = $100 - \dfrac{40}{100} \times 100 = 100 - 40 = 60$
$\therefore$ Required percentage = $\dfrac{60}{70} \times 100$
$\Rightarrow \dfrac{6}{7} \times 100 = \dfrac{600}{7} = 85\dfrac{5}{7}\%$ Ans
$\therefore$ First number = $100 - \dfrac{30}{100} \times 100 = 100 - 30 = 70$
$\Rightarrow$ Second number = $100 - \dfrac{40}{100} \times 100 = 100 - 40 = 60$
$\therefore$ Required percentage = $\dfrac{60}{70} \times 100$
$\Rightarrow \dfrac{6}{7} \times 100 = \dfrac{600}{7} = 85\dfrac{5}{7}\%$ Ans
Q29. A bag contains 8 red balls, 11 blue balls and 6 green balls. Find the percentage of blue balls in the bag.
Show Answer
$\Rightarrow$ Total no. of balls = $8 + 11 + 6 = 25$
$\Rightarrow$ No. of blue balls = 11
$\therefore$ Percentage of blue balls = $\dfrac{11}{25} \times 100 = 11 \times 4 = 44\%$ Ans
$\Rightarrow$ No. of blue balls = 11
$\therefore$ Percentage of blue balls = $\dfrac{11}{25} \times 100 = 11 \times 4 = 44\%$ Ans
Q30. Mohan gets Rs.1350 from Geeta and Rs. 650 from Rohit. Out of the total money that Mohan gets from Geeta and Rohit, what percent does he get from Rohit?
Show Answer
$\Rightarrow$ Total money recieved = $1350 + 650 = 2000$
$\Rightarrow$ Rohit gave = Rs.650
$\therefore$ Rohit's percentage = $\dfrac{650}{2000} \times 100$
$\Rightarrow \dfrac{650}{20} = \dfrac{65}{2} = 32.5\%$ Ans
$\Rightarrow$ Rohit gave = Rs.650
$\therefore$ Rohit's percentage = $\dfrac{650}{2000} \times 100$
$\Rightarrow \dfrac{650}{20} = \dfrac{65}{2} = 32.5\%$ Ans