1. If the numerator of certain fraction is increased by 200% and the denominator is increased by 150% the new fraction thus formed is \(\displaystyle \frac{9}{{10}}\) . What is the original fraction?
(a)\(\displaystyle \frac{3}{4}\)
(b)\(\displaystyle \frac{1}{4}\)
(c)\(\displaystyle \frac{3}{5}\)
(d)\(\displaystyle \frac{2}{5}\)
(e)None of these
2. If the numerator of a certain fractions increased by 100% and the denominator is increased by 200%; the new fraction thus formed is \(\displaystyle \frac{4}{{21}}\). What is the original fraction?
(a) \(\displaystyle \frac{2}{7}\)
(b) \(\displaystyle \frac{3}{7}\)
(c) \(\displaystyle \frac{2}{5}\)
(d) \(\displaystyle \frac{4}{7}\)
(e) None of these
Solution: (a) Let the original fraction be = \(\displaystyle \frac{x}{y}\) \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{x\times 200}}{{y\times 300}}=\frac{4}{{21}} $ \(\displaystyle \Rightarrow \)latex \displaystyle \frac{x}{y}=\frac{4}{{21}}\times \frac{3}{2}=\frac{2}{7}$
3. The difference between 55% of a number and 14% of the same number is 8610. What is 85% of that number?
(a) 15850
(b) 17020
(c) 17850
(d) 18450
(e) None of these
Solution: (c) Let the number be x. Now, according to the question, (55 – 14) % of x = 8610 \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{x\times 41}}{{100}}=8610$ \(\displaystyle \Rightarrow \)latex \displaystyle x=\frac{{8610\times 100}}{{41}}=2100$ \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{21000\times 85}}{{100}}$=17850
4. Animesh got 102 marks in Hindi, 118 marks in Science, 104 marks in Sanskrit, 114 marks in Maths and 96 marks in English. The maximum marks of each subject are 120. How much overall percentage of marks did Animesh get?
(a) 82
(b) 89
(c) 77
(d) 71
(e) None of these
Solution: (b) Total marks obtained by Animesh = 102 + 118 + 104 + 114 + 96 = 534 Total maximum marks = 120 × 5 = 600 Therefore, required percentage= \(\displaystyle \frac{{534}}{{600}}\times 100\) =89
5. Madhur got 101 marks in Hindi, 100 marks in Science, 96 marks in Sanskrit, 108 marks in Maths and 78 marks in English. If the maximum marks of each subject is equal and if Madhur scored 84 per cent marks in all the subjects together, what is the maximum marks of each subject?
(a) 110
(b) 120
(c) 115
(d) 100
(e) None of these
Solution: (c) Total marks = 101 + 100 + 96 + 108 + 78 = 483 Since it is 84% of the total maximum marks, Let total maximum marks be x \(\displaystyle \frac{{84}}{{100}}x=483\) x=\(\displaystyle \frac{{483\times 100}}{{84}}=575\) Maximum marks of each subject = \(\displaystyle \frac{{575}}{5}\) = 115 Alternate method, If each paper is of “x” marks, then total marks=5x Given, total marks = 101 + 100 + 96 + 108 + 78 = 483 Mathur scored 84% marks in total, (which is 484 marks) 483 marks —— 5x marks 84%————–100% Cross-multiply \(\displaystyle \begin{array}{l}5x\times 84=483\times 100\\5x=\frac{{483\times 100}}{{84}}=575\\x=115\end{array}\)
6. In an examination, the maximum aggregate marks that a student can get is 1040. In order to pass the exam, a student is required to get 676 marks out of the aggregate marks. Mina got 624 marks. By what per cent did Mina fail in the exam?
(a) 5%
(b) 8%
(c) 7%
(d) Cannot be determined
(e) None of these
Solution: (a) Minal failed by (676 – 624) = 52 marks % marks =\(\displaystyle \frac{{52}}{{1040}}\times 100=5\%\)
7. In a town three newspapers A, B and C are published. 42% of the people in that town read A, 68% read B, 51% read C, 30% read A and B, 28% read B and C, 36% A and C and 18% do not read any paper. Find the % of population of town that reads all the three.
(a) 15%
(b) 25%
(c) 20%
(d) 35%
(e) None of these
Solution: (a) \(\displaystyle n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)\) \(\displaystyle \Rightarrow \)100 – 18 = 42 + 68 + 51 – 30 – 28 – 36 + x \(\displaystyle \Rightarrow \)x = 15 (or) Elaborated method, \(\displaystyle \begin{array}{l}\begin{array}{*{20}{l}} \begin{array}{l}Let\text{ }the\text{ }no.\text{ }of\text{ }persons\text{ }in\text{ }the\text{ }city\text{ }be\text{ }100x\\So,\text{ }n\left( A \right)\text{ }=\text{ }42x\text{ },\text{ }\left[ {n\left( A \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ A}} \right]\end{array} \\ {n\left( B \right)\text{ }=\text{ }68x,\text{ }\left[ {n\left( B \right)\text{ }=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ B}} \right]} \\ {n\left( C \right)\text{ }=\text{ }51x\text{ }\left[ {n\left( C \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ C}} \right]} \end{array}\\n\left( {A\cap B} \right)=30x\left[ {n\left( {A\cap B} \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ A and B}} \right]\\n\left( {B\cap C} \right)=28x\left[ {n\left( {B\cap C} \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ B and C}} \right]\\n\left( {A\cap C} \right)=36x\left[ {n\left( {A\cap C} \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ A and C}} \right]\\n(A\cap B\cap C)=people\text{ }that\text{ }read\text{ }newspaper\text{ A, B and C}\\n(A\cup B\cup C)=82x[people\text{ who donot }read\text{ any }newspaper\text{ A, B and C }\!\!]\!\!\text{ }\\\text{(100-18=82)}\\\Rightarrow n\left( {A\cup B\cup C} \right)=n(A)+n(b)+n(c)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\Putting\text{ }the\text{ }above\text{ }values,\text{ }we\text{ }have\\82x=42x+68x+51x-30x-28x-36x+n(A\cap B\cap C)\\\Rightarrow n(A\cap B\cap C)=82x-67x=15x\\Therefore\text{ }the\text{ }number\text{ }of\text{ }people\text{ }reading\text{ }all\text{ }the\text{ }three\text{ }newspapers\text{ }are\text{ }15x\text{ }peopleout\text{ }of\text{ }100x\text{ }which\text{ }is\text{ }15\%\text{ }of\text{ }the\text{ }total\text{ }people\end{array}\)
8. Ramola’s monthly income is three times Ravina’s monthly income. Ravina’s monthly income is fifteen percent more than Ruchira’s monthly income. Ruchira’s monthly income is ₹ 32,000. What is Ramola’s annual income?
9. If the numerator of a fraction is increased by 300% and the denominator is increased by 200%, the resultant fraction is \(\displaystyle \frac{4}{15}\). What is the original fraction?
(a) \(\displaystyle \frac{3}{5}\)
(b) \(\displaystyle \frac{4}{5}\)
(c) \(\displaystyle \frac{2}{5}\)
(d) \(\displaystyle \frac{1}{5}\)
(e) None of these
Solution: (d) Let the original fraction be \(\displaystyle \frac{x}{y}\) According to the question Therefore, \(\displaystyle \frac{{x\times 400}}{{y\times 300}}=\frac{4}{{15}}\) \(\displaystyle \Rightarrow \)latex \displaystyle \frac{x}{y}=\frac{4}{{15}}\times \frac{3}{4}=\frac{1}{5}$
10. A student was awarded certain marks in an examination. However, after re-evaluation, his marks were reduced by 40% of the marks that were originally awarded to him so that the new score now became 96. How many marks did the student lose after re-evaluation?
(a) 58
(b) 68
(c) 63
(d) 56
(e) 64
Solution: (e) Let initial marks of student = x After re-evaluation marks reduced by 40% of x New score = 60% of x = 96 = \(\displaystyle \frac{{60}}{{100}}\times x=96\) \(\displaystyle x=\frac{{96\times 100}}{{60}}\) x=160 Marks lose = 160 – 96 = 64
11. Mathew scored 42 marks in Biology, 51 marks in Chemistry, 58 marks Mathematics, 35 marks in Physics and 48 marks in English. The maximum marks a student can score in each subject are 60. How much overall percentage did Mathew get in this exam?
(a) 76
(b) 82
(c) 68
(d) 78
(e) None of these
Solution: (d) Total maximum marks of 5 subjects = 60 × 5 = 300 Total marks of Mathew = 42 + 51 + 58 + 35 + 48 = 234 % of Marks = \(\displaystyle \frac{{234}}{{300}}\times 100=78\%\)
12. The respective ratio of salaries of A and B is 8 : 7. If the salary of B increases by 20% and the salary of A increases by 21%, the new ratio becomes 96 : 77 respectively. What is A’s salary?
a) ₹ 22560
(b) ₹21600
(c) ₹ 20640
(d) ₹ 23040
(e) Cannot be determined
Solution: (e) \(\displaystyle \frac{{A'ssalary}}{{B'ssalary}}=\frac{8}{7}\) A’s salary= \(\displaystyle \frac{{8x}}{{15}}\) B’s salary=\(\displaystyle \frac{{7x}}{{15}}\) Now, A’s salary= \(\displaystyle \frac{{8x}}{{15}}+\frac{{8x}}{{15}}\times \frac{{21}}{{100}}=\frac{{8x+1.68x}}{{15}}=\frac{{9.68x}}{{15}}\) Now B’s salary= \(\displaystyle \frac{{7x}}{{15}}+\frac{{7x}}{{15}}\times \frac{{20}}{{100}}=\frac{{7x+1.4x}}{{15}}=\frac{{8.4x}}{{15}}\) Therefore, \(\displaystyle \frac{{9.68x}}{{15}}\times \frac{{15}}{{8.4x}}=\frac{{96}}{{77}}\) \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{9.68}}{{8.4}}=\frac{{96}}{{77}}$ Here x is cancelled. So, salary of A can’t be determined.
13. A merchant bought some goods worth Rs. 6000 and sold half of them at 12% profit. At what profit percent should he sell the remaining goods to make and overall profit of 18%?
(a) 24
(b) 28
(c) 18
(d) 20
(e) 26
Solution: (a) Profit on all the goods = 18% of 6000 = ₹ 1080 Profit on half of the goods = 12% of 3000 = ₹ 360 Therefore, the profit on remaining half of the objects = 1080 – 360 = ₹ 720 Hence, required profit percentage=\(\displaystyle \frac{{720}}{{3000}}\times 100\%=24\%\)
14. Prema decided to donate 15% of her salary to an orphanage. On the day of donation she changed her mind and donated ₹ 1,896 which was 80% of what she had decided earlier. How much is Prerna’s salary?
(a) ₹18,500
(b) ₹ 10,250
(c) ₹15,800
(d) Cannot be determined
(e) None of these
Solution: (c) Let Prerna’s salary be ₹ x According to the question, 80% of 15% of x = 1896 \(\displaystyle \Rightarrow \)latex \displaystyle x\times \frac{{15}}{{100}}\times \frac{4}{5}=1896$ Therefore, \(\displaystyle x=\frac{{1896\times 5\times 100}}{{15\times 4}}=₹15800\)
15. If the numerator of a fraction is increased by 600% and the denominator is increased by 200%, the resulting fraction is \(\displaystyle 2\frac{4}{5}\) What was the original fraction?
(a) \(\displaystyle \frac{4}{7}\)
(b) \(\displaystyle \frac{13}{12}\)
(c) \(\displaystyle \frac{11}{12}\)
(d) \(\displaystyle \frac{6}{5}\)
(e) None of these
Solution: (d) Let the original fraction is \(\displaystyle \frac{a}{b}\) Numerator is increased by 600%, \(\displaystyle a\to a+\frac{{600}}{{100}}\times a=7a\) Denominator is increased by 200%, \(\displaystyle b\to b+\frac{{200}}{{100}}\times b=3b\) According to the question \(\displaystyle \frac{{7a}}{{3b}}=\frac{{14}}{5}\) Therefore, \(\displaystyle \frac{a}{b}=\frac{{14\times 3}}{{5\times 7}}=\frac{{42}}{{35}}\) or \(\displaystyle \frac{a}{b}=\frac{6}{5}\)
16. If the numerator of a fraction is increased by 20% and the denominator is increased by 25%, the fraction obtained is \(\displaystyle \frac{3}{5}\). What was the original fraction?
(a) 5/7
(b) 4/7
(c) 3/8
(d) Cannot be determined
(e) None of these
Solution: (e) Let fraction be \(\displaystyle \frac{x}{y}\) According to the question \(\displaystyle \frac{{x\times 120\%}}{{y\times 125\%}}=\frac{3}{5}\) \(\displaystyle \Rightarrow \)latex \displaystyle \frac{x}{y}=\frac{3}{5}\times \frac{{125}}{{120}}=\frac{5}{8}$
17. Mr Alok spends 50% of his monthly income on household items and out of the remaining he spends 50% on transport, 25% on entertainment, 10% on sports and the remaining amount of ₹900 is saved. What is Mr Alok’s monthly income?
(a) ₹ 6000
(b) ₹9000
(c) ₹12000
(d) Cannot be determined
(e) None of these
Solution: (c) Let total monthly income of Mr. Alok be ₹ x. According to question, Therefore, \(\displaystyle x\times \frac{{50}}{{100}}\times \frac{{15}}{{100}}=900\) x = ₹ 12000 Hence, monthly income of Mr. Alok = ₹12000
18. The salaries of A,B,C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be the new ratio of their salaries?
(a) 3 : 3 : 10
(b) 10 : 11 : 20
(c) 23 : 33 : 60
(d) Cannot be determined
(e) None of these
Solution: (c) Let A = 2k, B = 3k and C = 5k. A’s new salary = \(\displaystyle \frac{{115}}{{100}}\times 2k=\frac{{23}}{{10}}k\) B’s new salary = \(\displaystyle \frac{{110}}{{100}}\times 3k=\frac{{33}}{{10}}k\) C’s new salary = \(\displaystyle \frac{{120}}{{100}}\times 5k=6k\) Therefore, New ratio = \(\displaystyle \frac{{23k}}{{10}}:\frac{{33k}}{{10}}:6k=23:33:60\)
19. Barath spends 25 per cent of his salary on house rent, 5 percent on food, 15 percent on travel, 10 percent on clothes and the remaining amount of ₹ 27,000 is saved. What is Barath’s income?
(a) ₹60,000
(b) ₹60,500
(c) ₹60,700
(d) ₹70,500
(e) None of these
Solution: (a) Saving percentage = (100 – 55) % = 45% If the income of Barath be ₹ x, then, \(\displaystyle \frac{{45\times x}}{{100}}=27000\) \(\displaystyle \Rightarrow \)x = \(\displaystyle x=\frac{{27000\times 100}}{{45}}=60000\)
20. Five-ninths of number is equal to twenty five percent of the second number. The second number is equal to one-fourth of the third number. The value of the third number is 2960. What is 30 percent of the first number?
(a) 99.9
(b) 88.8
(c) 66.6
(d) Cannot be determined
(e) None of these
Solution: (a) Second number = \(\displaystyle \frac{1}{4}\times 2960=740\) Let the first number be x. \(\displaystyle \frac{5}{9}x=\frac{{25}}{{100}}\times 740\) x= \(\displaystyle x=\frac{9}{5}\times \frac{1}{4}\times 740=333\) 30% of 1st number =\(\displaystyle \frac{{30}}{{100}}\times 333=99.9\)
percentage questions with solutions for bank exams
21. A manufacture undertakes to supply 2000 pieces of a particular component at ₹25 per piece. According to his estimates, even if 5% fail to pass the quality tests, then he will make a profit of 25%. However, as it turned out, 50% of the components were rejected. What is the loss to the manufacture?
(a) ₹ 12,000
(b) ₹ 13,000
(c) ₹ 14,000
(d) ₹ 15,000
(e) None of these
Solution: (b) 5% of 2000 = 100 2000 – 100 =1900 If he sells 1900 he will get 25% profit cost per piece ₹25 \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{25\times 1900\times 100}}{{125}}=38000$ CP if 50% rejected, only 1000 pieces sold so 1000 × 25 = 25000 = SP Loss = CP – SP = 38000 – 25000 = 13000
22. In an examination, 40% of the candidates wrote their answers in Hindi and the others in English. The average marks of the candidates written in Hindi is 74 and the average marks of the candidates written in English is 77. What is the average marks of all the candidates?
(a) 75.5
(b) 75.8
(c) 76.0
(d) 76.8
(e) None of these
Solution: (b) Let total number of candidates = 100 Therefore, total marks of 40 candidates = 40 × 74 Total marks of 60 candidates = 60 × 77 Therefore, required average marks = \(\displaystyle \frac{{40\times 74+60\times 77}}{{100}}\) = \(\displaystyle \frac{{2960+4620}}{{100}}=\frac{{7580}}{{100}}=75.80\)
23. Ram spends 50% of his monthly income on household items, 20% of his monthly income on buying clothes, 5% of his monthly income on medicines and saves remaining ₹ 11,250. What is Ram’s monthly income?
(a) ₹ 38,200
(b) ₹ 34,000
(c) ₹ 41,600
(d) ₹ 45,000
(e) None of these
Solution: (d) Let total income of Ram be x. Then (100 – 50 – 20 – 5)% of x = 11250 x = 45000.
24. If the numerator of a fraction is increased by 350% and the denominator of the fraction is increased by 300% the resultant fraction is \(\displaystyle \frac{9}{22}\). What is the original fraction?
(a) \(\displaystyle \frac{3}{4}\)
(b) \(\displaystyle \frac{5}{12}\)
(c) \(\displaystyle \frac{7}{9}\)
(d) \(\displaystyle \frac{4}{11}\)
(e) None of these
Solution: (d) Let the original fraction is \(\displaystyle \frac{a}{b}\) According to question, \(\displaystyle \frac{{a+\frac{{350}}{{100}}\times a}}{{b+\frac{{300}}{{100}}\times b}}=\frac{9}{{22}}\) \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{4.5a}}{{4b}}=\frac{9}{{22}}$ \(\displaystyle \Rightarrow \)latex \displaystyle \frac{a}{b}=\frac{9}{{22}}\times \frac{4}{{4.5}}=\frac{4}{{11}}$
25. If the numerator of a fraction is increased by 300% and the denominator is increased by 500%, the resultant fraction is \(\displaystyle \frac{5}{12}\). What was the original fraction?
(a) \(\displaystyle \frac{8}{5}\)
(b) \(\displaystyle \frac{5}{{11}}\)
(c) \(\displaystyle \frac{{12}}{5}\)
(d) \(\displaystyle \frac{5}{7}\)
(e) None of these
Solution: (e) Let the original fraction be \(\displaystyle \frac{x}{y}\) According to the question, \(\displaystyle \frac{{x\times 400}}{{y\times 600}}=\frac{5}{{12}}\) \(\displaystyle \Rightarrow \)latex \displaystyle \frac{x}{y}=\frac{5}{{12}}\times \frac{6}{4}=\frac{5}{8}$
26. Bina’s monthly income is 90% of Anita’s monthly income. The total of both their monthly incomes is Mr. Sen’s monthly income. Mr. Sen’s annual income is 7,75,200. What is Bina’s monthly income?
(a) ₹ 34,000
(b) ₹ 36,000
(c) ₹ 30,600
(d) ₹ 30,000
(e) None of these
Solution: (c) Sen's monthly income = \(\displaystyle \frac{{775200}}{{12}}=64600\) Let the monthly income of Anita be Rs. x. Therefore, Bina's monthly income = \(\displaystyle \frac{{90\times x}}{{100}}=0.9x\) Now, according to the question, x+0.9x=64600 \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{64600}}{{1.9}}=34000$ Bina's monthly income = 34000 × 0.9 = ₹ 30600
27. A has double the money of B and B has 50% more money than C. If average money of all the three persons is 12000, how much money A have?
(a) \(\displaystyle \frac{{211000}}{{11}}\)
(b) \(\displaystyle \frac{{315000}}{{11}}\)
(c) \(\displaystyle \frac{{216000}}{{11}}\)
(d) \(\displaystyle \frac{{316000}}{{11}}\)
(e) None of these
Solution: (c) Let the money of C be x. According to the question, Total money of B = x + x + 50% =\(\displaystyle x+\frac{{50x}}{{100}}=\frac{{3x}}{2}\) Total money of A = \(\displaystyle 2\times \frac{{3x}}{2}=3x\) Average money of three persons = 12000 Therefore, total money to thee 12000 × 3 \(\displaystyle 3x+\frac{{3x}}{2}+x=12000\times 3\) \(\displaystyle \frac{{6x+3x+2x}}{2}=36000\) 3x=\(\displaystyle 3x=\frac{{3\times 7200}}{{11}}\) Therefore, x \(\displaystyle \frac{{36000\times 2}}{{11}}=\frac{{72000}}{{11}}\) Now, money of A =\(\displaystyle 3x=3\times \frac{{72000}}{{11}}=\frac{{216000}}{{11}}\)
28. Fresh grapes contain 80% water, while dry grapes contain 10% water. If the weight of dry grapes is 500 kg, then what is its total weight when it is fresh?
(a) 2350 kg
(b) 2085 kg
(c) 2255 kg
(d) 2250 kg
(e) None of these
Solution: (d) Let the weight of fresh grapes be x. Quantity of water in it = \(\displaystyle \frac{{80}}{{100}}\times x=\frac{{4x}}{5}\) Quantity of pulp in it = \(\displaystyle x-\frac{{4x}}{5}=\frac{x}{5}\) Quantity of water in 500 kg dry grapes = \(\displaystyle \frac{{10}}{{100}}\times 500=50kg\) Therefore, quantity of pulp in it = (500 – 50) = 450 kg \(\displaystyle \frac{x}{5}=450\) x = 2250 kg
29. Number of students in 4th and 5th class is in the ratio 6 : 11. 40% in class 4 are girls and 48% in class 5 are girls. What percentage of students in both the classes are boys?
(a) 62.5%
(b) 54.8%
(c) 52.6%
(d) 55.8%
(e) 53.5%
Solution: (b) Boys in class 4 = \(\displaystyle \frac{{60}}{{100}}\times 6x=\frac{{360x}}{{100}}\) Boys in class 5 = \(\displaystyle \frac{{52}}{{100}}\times 11x=\frac{{572x}}{{100}}\) So total boys = \(\displaystyle \frac{{360x}}{{100}}+\frac{{572x}}{{100}}=\frac{{932x}}{{100}}=9.32x\) % of boys = \(\displaystyle \frac{{9.32x}}{{17x}}\times 100=54.8\%\)
30. In a school the number of boys and girls are in the ratio of 4:7. If the number of boys are increased by 25% and the number of girls are increased by 15%. What will be the new ratio of number of boys to that of girls?
(a) 100:131
(b) 100:151
(c) 100:161
(d) 100:181
(e) None of these
Solution: (c) Boys = 4x and girls = 7x Ratio =\(\displaystyle 4x\times \frac{{125}}{{100}}:7x\times \frac{{115}}{{100}}=100:161\) Alterative method, Let x be the constant ratio. Boys : Girls = 4 : 7 Boys : Girls = 4x : 7x The number of boys is increased by 25%. 25% of 4x = 0.25 x 4x = x Number of boys now = 4x + x = 5x The number of girls is increased by 15%. 15% of 7x = 0.15 x 7x = 1.05x Number of girls now = 7x + 1.05x = 8.05x New Ratio Boys : Girls = 5x : 8.05x Boys : Girls = 100(5x) : 100(8.05x) Boys : Girls =500x : 805x Boys : Girls =100x : 161x Boys : Girls =100 : 161
31. Sujata invests 7% i.e. ₹ 2170 of her monthly salary in mutual funds. Later she invests 18% of her monthly salary in recurring deposits also; she invests 6% of her salary on NSC’s. What is the total annual amount invested by Sujata?
(a) ₹ 1,25,320
(b) ₹ 1,13,520
(c ) ₹ 1,35,120
(d) ₹ 1,15,320
(e) None of these
Solution: (d) Let her monthly salary be ₹ x. According to the question \(\displaystyle \frac{7}{{100}}\times x=2170\) \(\displaystyle \Rightarrow \)x= \(\displaystyle x=\frac{{2170\times 100}}{7}=31000\) Total monthly investment = (18 + 6 + 7)% of 31000 \(\displaystyle \frac{{31}}{{100}}\times 31000=9610\) Total annual investment = 12 × 9610 = ₹ 115320
32. If tax on a commodity is reduced by 10%, total revenue remains unchanged. What is the percentage increase in its consumption?
(a) \(\displaystyle 11\frac{1}{9}\%\)
(b) 20%
(c) 10%
(d) \(\displaystyle 9\frac{1}{{11}}\%\)
(e) None of these
Solution: (a) Percentage increase in the consumption =\(\displaystyle \frac{{10}}{{100-10}}\times 100=\frac{{100}}{9}=11\frac{1}{9}\%\)
33. Five-ninths of a number is equal to 25% of the second number. The second number is equal to one-fourth of the third number. The value of the third number is 2960. What is 30% of the first number?
(a) 88.8
(b) 99.9
(c) 66.6
(d) Can’t be determined
(e) None of these
Solution: (b) Second number = \(\displaystyle \frac{1}{4}\times 2960=740\) Let the first number be x. \(\displaystyle \frac{5}{9}x=\frac{{25}}{{100}}\times 740\) \(\displaystyle x=\frac{9}{5}\times \frac{1}{4}\times 740=333\) 30% of 1st number = \(\displaystyle \frac{{30}}{{100}}\times 333=99.9\)
34. If 8% of x is the same as 4% of y, then 20% of x is the same as:
(a) 10% of y
(b) 16% of y
(c) 80% of y
(d) 50% of y
(e) 70% of y
Solution: (a) \(\displaystyle \frac{{8x}}{{100}}=\frac{{4y}}{{100}}\) \(\displaystyle \Rightarrow \)y = 2x Therefore, 20% of x = 10% of y.
35. In a school 40% of the students play football and 50% play cricket. If 18% of the students neither play football nor cricket, the percentage of the students playing both is:
(a) 40%
(b) 32%
(c) 22%
(d) 8%
(e) 20%
Solution: (d) Since 18% of the students neither play football nor cricket. It means 82% of the students either play football or cricket or both. Using set theory \(\displaystyle n(A\cup B)=n(A)+n(B)-n(A\cap B)\) \(\displaystyle \Rightarrow \)82 = 40 + 50 –\(\displaystyle n(A\cap B)\) \(\displaystyle \Rightarrow \)latex \displaystyle n(A\cap B)$ =90-82=8 Therefore, 8% students play both games.
36. If 120% of a is equal to 80% of b, then \(\displaystyle \frac{{b+a}}{{b-a}}\) is equal to
37. The population of a village is 25,000. One fifth are females and the rest are males. 5% of males and 40% of females are uneducated. What percentage on the whole are educated?
38. The sum of the numbers of boys and girls in a school is 150. If the number of boys is x, the number of girls becomes x% of the total number of students. The number of boys is :
39. Due to 25% fall in the rate of eggs, one can buy 2 dozen eggs more than before by investing Rs.162. Then the original rate per dozen of the eggs is
(a) Rs. 22
(b) Rs. 24
(c) Rs. 27
(d) Rs. 30
(e) Rs.32
Solution: (c) Let Initial price of eggs = Rs. X per dozen New price = Rs. 3x/4 per dozen According to the question, \(\displaystyle \frac{{162}}{{\frac{{3x}}{4}}}-\frac{{162}}{x}=2\) \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{162\times 4}}{{3x}}-\frac{{162}}{x}=2$ \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{216}}{x}-\frac{{162}}{x}=2$ \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{54}}{x}=2$ \(\displaystyle \Rightarrow \)2x = 54 \(\displaystyle \Rightarrow \) x = Rs. 27 per dozen
40. A supply of juice lasts for 35 days. If its use is increased by 40% the number of days would the same amount of juice lasts, is
(a) 25 days
(b) 30 days
(c) 24 days
(d) 27 days
(e) 29 days
Solution: (a) Required time = \(\displaystyle \frac{{35\times 100}}{{140}}\) =25 days
41. When 75 is added to 75% of a number, the answer is the number. Find 40% of that number.
(a) 100
(b) 80
(c) 120
(d) 160
(e) 180
Solution: (c) If the number be x, then \(\displaystyle x\times \frac{{75}}{{100}}+75=x\) \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{3x}}{4}+75=x$ \(\displaystyle \Rightarrow \)latex \displaystyle x-\frac{{3x}}{4}=75$ \(\displaystyle \Rightarrow \)latex \displaystyle \frac{x}{4}=75$ \(\displaystyle \Rightarrow \)latex \displaystyle x=4\times 75=300$ Therefore, 40% of 300 = \(\displaystyle \frac{{300\times 40}}{{100}}=120\)
42. In an office, 40% of the staff is female. 70% of the female staff and 50% of the male staff are married. The percentage of the unmarried staff in the office is
(a) 64
(b) 60
(c) 54
(d) 42
(e) 45
Solution: (d) Total staff strength in the office = 100 (let) Females = 40 Males = 60 Married females = \(\displaystyle \frac{{40\times 70}}{{100}}=28\) Unmarried females = 40 – 28 = 12 Unmarried males = 30 Therefore, unmarried staff = 30 + 12 = 42 i.e. 42%
43. A boy found the answer for the question “subtract the sum of \(\displaystyle \frac{1}{4}\) and \(\displaystyle \frac{1}{5}\) from unity and express the answer in decimals” as 0.45. The percentage of error in his answer was
44. Two numbers are less than a third number by 30% and 37% respectively. The per cent by which the second number is less than the first is
(a) 10%
(b) 7%
(c) 4%
(d) 3%
(e) 5%
Solution: (a) Third number = 100 First number = 70 Second number = 63 Therefore, required percentage = \(\displaystyle \frac{7}{{70}}\times 100=10\)
45. 28% members of a certain group are married. What is the respective ratio between the number of married members to the number of unmarried members ?
46. 52% students from a college participated in a survey. What is the respective ratio between the number of students who did not participate in the survey to the number of students who participated?
(a) 11 : 13
(b) 12 : 13
(c) 12: 17
(d) Cannot be determined
(e) None of these
Solution: (b) Required ratio = 48 : 52 = 12 : 13
47. In an examination it is required to get 55% of the aggregate marks to pass. A student gets 520 marks and is declared failed by 5% marks. What are the maximum aggregate marks a student can get?
(a) 960
(b) 1250
(c) 1040
(d) Cannot be determined
(e) None of these
Solution: (c) Let maximum aggregate marks be x. Student gets 520 marks and is declared failed by 5% marks it means student get 50% marks 50% of x = 520 \(\displaystyle \frac{x}{2}\) = 520 x = 1040
48. The product of 5% of a positive number and 2% of the same number is 211.6. What is half of that number?
(a) 230
(b) 460
(c) 920
(d) 115
(e) None of these
Solution: (a) Let the number be x. Then, according to the question, \(\displaystyle \frac{{5x}}{{100}}\times \frac{{2x}}{{100}}=211.6\) or \(\displaystyle {{x}^{2}}=\frac{{211.6\times 100\times 100}}{{5\times 2}}=211600\) x = + 460 Therefore, half of eight number = 230
49. In an examination, the maximum aggregate marks a 1020. In order to pass the exam a student is required to obtain 663 marks out of the aggregate marks. Shreya obtained 612 marks. By what percent did Shreya fail the exam?
50. The product of 5% of a positive number and 3% of the same number is 504.6. What is half of that number?
(a) 290
(b) 340
(c) 680
(d) 580
(e) None of these
Solution: (d) Let the positive number be x Then, \(\displaystyle \frac{{5x}}{{3x}}\times \frac{{3x}}{{100}}=504.6\) \(\displaystyle \frac{{15{{x}^{2}}}}{{10000}}=504.6\) or, \(\displaystyle {{x}^{2}}=\frac{{504.6\times 10000}}{{15}}\) Therefore, x = 580
51. The sum of 15% of a positive number and 20% of the same number is 126. What is one-third of that number?
(a) 360
(b) 1080
(c) 120
(d) 40
(e) None of these
Solution: (c) (15% + 20%) of x = 126 \(\displaystyle x=\frac{{126\times 100}}{{35}}=360\) Therefore, required answer = \(\displaystyle 360\times \frac{1}{3}=120\)
52. Nandita scored 80% marks in five subjects together viz Hindi, Science, Maths, English and Sanskrit, where in the maximum marks of each subject were 105. How many marks did Nandita score in Science if she scored 89 marks in Hindi, 92 marks in Sanskrit, 98 marks in Maths and 81 marks in English?
(a) 60
(b) 75
(c) 65
(d) 70
(e) None of these
Solution: (a) (Total marks scored by Nandita =\(\displaystyle 525\times \frac{{80}}{{100}}=420\) Let Score in Science be x 89 + 92 + 98 + 81 + x = 420 360 + x = 420 \(\displaystyle \Rightarrow \)x = 60
53. Niraj incurred a loss of 55 percent on selling an article for ₹ 9,549. What was the cost price of the article?
(a) ₹ 27,700
(b) ₹ 25,600
(c) ₹ 21,220
(d) ₹ 29,000
(e) None of these
Solution: (c) Let cost Price of article be x \(\displaystyle x-\frac{{55}}{{100}}x=9549\) \(\displaystyle \frac{{45x}}{{100}}=9549\) x=\(\displaystyle \frac{{9549\times 100}}{{45}}=21220\)
54. In order to pass in an exam, a student is required to get 780 marks out of the aggregate marks. Sonu got 728 marks and was declared failed by 5 percent. What are the maximum aggregate marks a student can get in the examination?
(a) 1040
(b) 1100
(c) 1000
(d) Cannot be determined
(e) None of these
Solution: (a) 5% of maximum aggregate marks = 780 – 728 = 52 Let maximum aggregate marks be x 5% of x = 52 Maximum aggregate marks = \(\displaystyle \frac{{52}}{5}\times 100=1040\)
55. The sum of 15% of a positive number and 10% of the same number is 70. What is twice of that number?
(a) 440
(b) 280
(c) 560
(d) 140
(e) None of these
Solution: (c) Let the positive no. be x. According to question. 15% of x + 10% of x = 70 \(\displaystyle x\times \frac{{15}}{{100}}+\frac{{x\times 10}}{{100}}=70\) \(\displaystyle \frac{{15x}}{{100}}+\frac{{10x}}{{100}}=70\) \(\displaystyle \frac{{24x}}{{100}}=70\) x= \(\displaystyle \frac{{70\times 100}}{{25}}=280\) Therefore, Double of the given number = 280 × 2 = 560
56. In order to pass in an exam a student is required to get 975 marks out of the aggregate marks. Priya got 870 marks and was declared failed by 7 percent. What are the maximum aggregate marks a student can get in the examination?
(a) 1500
(b) 1000
(c) 1200
(d) Cannot be determined
(e) None of these
Solution: (a) Minimum marks to pass = 975 Priya failed by 975 – 870 = 105 marks Maximum marks = \(\displaystyle \frac{{105}}{7}\times 100=1500\)
57. A salesgirl’s terms were changed from a flat commission of 5% on all her sales to a fixed salary of ₹ 1000 plus 2.5% commission on all sales exceeding ₹4000. If her remuneration as per the new scheme was ₹600 more than that by the previous scheme, her total sales was
58. The product of one-third of a number and 150% of another number is what percent of the product of the original numbers?
(a) 80%
(b) 50%
(c) 75%
(d) 120%
(e) None of these
Solution: (b) Let the original numbers be x and y and their product be xy Product of 1/3rd of x and 150% of y = \(\displaystyle \frac{x}{3}\times \frac{3}{2}y=\frac{{xy}}{2}\) Required answer = \(\displaystyle \frac{{xy}}{{2\times xy}}\times 100=50\%\) (or) if xy is original number, then \(\displaystyle \frac{{xy}}{2}\) is half or 50% of xy
59. Akbar salary increases every year by 10% in June. If there is no other increase or reduction in the salary and his salary in June 2022 was ₹22,385, what was his salary in June 2020?
(a) ₹18,650
(b) ₹18,000
(c) ₹19,250
(d) ₹18,500
(e) None of these
Solution: (d) Salary in June 2022 = 22385 Using compound interest \(\displaystyle A=P\mathop{{\left( {1+\frac{r}{{100}}} \right)}}^{n}\) \(\displaystyle 22385=P\mathop{{\left( {1+\frac{{10}}{{100}}} \right)}}^{2}\) \(\displaystyle 22385=P\mathop{{\left( {1+0.1} \right)}}^{2}\) \(\displaystyle 22385=P\mathop{{\left( {1.1} \right)}}^{2}\) \(\displaystyle P=\frac{{22385}}{{\mathop{{\left( {1.1} \right)}}^{2}}}\) \(\displaystyle P=18500\) Hence his salary in 2020 is ₹18,500 (or) Alternate direct method, Salary in June 2020 = \(\displaystyle \frac{{22385}}{{1.1\times 1.1}}=18500\)
60. An HR Company employees 4800 people, out of which 45 percent are males and 60 percent of the males are either 25 years or older. How many males are employed in that HR Company who are younger than 25 years?
(a) 2640
(b) 2160
(c) 1296
(d) 864
(e) None of these
Solution: (d) Total Number of employees = 4800 45% of these 4800 are male Therefore, total male = \(\displaystyle \frac{{45}}{{100}}\times 4800=2160\) 60% of these 2160 are 25years or older Therefore, total male who are 25 years or older=\(\displaystyle \frac{{60}}{{100}}\times 2160=1296\) The number of males less that 25 years=2160 – 1296 = 864 (or) Direct alternate method, Required number = \(\displaystyle 4800\times \frac{{45}}{{100}}\times \frac{{40}}{{100}}=864\)
mcq on percentage questions and answers for rbi assistant
61. In an Entrance Examination Ritu scored 56 percent marks, Smita scored 92 percent marks and Rina scored 634 marks. The maximum marks of the examination are 875. What are the average marks scored by all the three girls together?
(a) 1929
(b) 815
(c) 690
(d) 643
(e) None of these.
Solution: (d) Marks scored by Ritu = \(\displaystyle 875\times \frac{{56}}{{100}}=490\) Marks scored by Smita = \(\displaystyle 875\times \frac{{92}}{{100}}=805\) Therefore, Average marks scored by all the three together =\(\displaystyle \frac{{490+805+634}}{3}=\frac{{1929}}{3}=643\)
62. The sum of 55% of a number and 40% of the same number is 180.5. What is 80% of that number?
(a) 134
(b) 152
(c) 148
(d) 166
(e) None of these
Solution: (b) Let the number be x. Now (55 + 40)% of x = 180.5 \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{x\times 95}}{{100}}=180.5$ x= \(\displaystyle \frac{{180.5\times 100}}{{95}}=190\) Now 80% of 190 = \(\displaystyle \frac{{190\times 80}}{{100}}=152\)
63. In an examination it is required to get 65% of the aggregate marks to pass, A student gets 847 marks and is declared failed by 10% marks. What are the maximumaggregate marks a student can get?
(a) 1450
(b) 1640
(c) 1500
(d) Cannot be determined
(e) None of these
Solution: (e) Let maximum marks = x Student got 55% x = 847 Therefore, x = \(\displaystyle x=\frac{{847\times 100}}{{55}}=1540\)
64. Last year there were 610 boys in a school. The number decreased by 20 percent this year. How many girls are there in the school if the number of girls is 175 percent of the total number of boys in the school this year?
(a) 854
(b) 848
(c) 798
(d) 782
(e) None of these
Solution: (a) No. of boys, last year = 610 20% of 610 = 122 No. of boys, current year = 610 – 122 = 488 No. of girls = 175% of 488 = \(\displaystyle \frac{{175\times 488}}{{100}}=854girls\)
65. 855 candidates applied for a job, out of which 80% of the candidates were rejected. How many candidates were selected for the job?
(a) 684
(b) 151
(c) 676
(d) 179
(e) None of these
Solution: (e) No. of candidates selected for job = 20% of 855 \(\displaystyle \frac{{20\times 855}}{{100}}=171\)
65. What should come in place of the question mark so that it satisfies equality of the equation — 32% of 750 < ?
(a) 23% of 600
(b) 46% of 207
(c) 98% of 250
(d) 75% of 320
(e) None of these
Solution: (c) 32% of 750=\(\displaystyle \frac{{32\times 750}}{{100}}=240\) 23% of 600 =\(\displaystyle \frac{{23\times 600}}{{100}}=138\) 46% of 207 =\(\displaystyle \frac{{46\times 207}}{{100}}=95.22\) 98% of 250 =\(\displaystyle \frac{{98\times 250}}{{100}}=245\)
66. Sum of three consecutive numbers is 2262. What is 41 % of the highest number?
(a) 301.51
(b) 303.14
(c) 308.73
(d) 306.35
(e) 309.55
Solution: (e) Let the numbers are x, x + 1, x + 2 Sum of three consecutive numbers = 2262 x + x + 1 + x + 2 = 2262 3x + 3 = 2262 3x = 2259 x = 753 Number are 753, 754, 755 Therefore, 41% of 755 = 309.55
67. In an examination, Raman scored 25 marks less than Rohit. Rohit scored 45 more marks than Sonia. Rohan scored 75 marks which is 10 more than Sonia. Ravi’s score is 50 less than, maximum marks of the test. What approximate percentage of marks did Ravi score in the examination, if he gets 34 marks more than Raman?
(a) 90
(b) 70
(c) 80
(d) 60
(e) 85
Solution: (b) Rohan’s marks = 75 Sonia’s marks = 65 Rohit’s marks = 65 + 45 = 110 Raman’s marks = 110 – 25 = 85 Ravi got marks = 85 + 34 = 119 Total maximum marks = 119 + 50 + 169 Percentage of Ravi’s marks= \(\displaystyle \frac{{119}}{{169}}\times 100\%=70.4\%=70\%\)
68. 10% of the inhabitants of a village having died of cholera, a panic set in, during which 25% of the remaining inhabitants left the village. The population is then reduced to 4050. Find the number of original inhabitants.
(a) 5000
(b) 6000
(c) 7000
(d) 8000
(e) None of these
Solution: (b) Let the total number of original inhabitants be x. Then, (100 – 25)% of (100 –10)% of x = 4050 \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{75}}{{100}}\times \frac{{90}}{{100}}\times x=4050$ \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{27}}{{40}}x=4050$ \(\displaystyle \Rightarrow \)latex \displaystyle x=\frac{{4050\times 40}}{{27}}=6000$ Number of original inhabitants = 6000.
69. ‘A’ sells a good to ‘B’ at a profit of 20 % and ‘B’ sells it to C at profit of 25 %. If ‘C’ pays ₹ 225 for it, what was cost price for ‘A’ ?
(a) 150
(b) 120
(c) 200
(d) 110
(e) None of these
Solution: (a) During both the transaction there are profits. So our calculating figures would be 120, 125 and 100. A’s cost is certainly less than C’s selling price Therefore, Required price = \(\displaystyle 225\times \frac{{100}}{{120}}\times \frac{{100}}{{125}}=150\)
70. Naresh’s monthly income is 30% more than that of Raghu. Raghu’s monthly income is 20% less than that of Vishal. If the difference between the monthly incomes of Naresh and Vishal is ₹ 800, what is the monthly income of Raghu?
(a) ₹ 16,000
(b) ₹ 20,000
(c) ₹ 12,000
(d) Data inadequate
(e) None of these
Solution: (a) N = R + 30% of R = 1.3 R R = V – 20% of V = 80% of V = 0.8 V Therefore, N = 1.3 × 0.8V = 1.04 V Now, N – V = 1.04 V – V = 0.04 V = ₹800 (given) Therefore, V = ₹ 20000 Hence, R = 0.8 × 20000 = ₹16000
71. Groundnut oil is now being sold at ₹ 27 per kg. During last month its cost was ₹ 24 per kg. Find by how much % a family should reduce its consumption, so as to keep the expenditure same.
(a) \(\displaystyle 11\frac{1}{9}\%\)
(b) \(\displaystyle 11\frac{1}{{11}}\%\)
(c) \(\displaystyle 11\frac{9}{{10}}\%\)
(d) \(\displaystyle 9\frac{1}{{10}}\%\)
(e) None of these
Solution: (a) % change in rate =(27-24)/24×100=100/8% For fixed expenditure, % change in consumption =\(\displaystyle \frac{{\%changeinrate}}{{100+\%changeinrate}}\times 100\) =\(\displaystyle \frac{{100/8}}{{100[1+\frac{1}{8}]}}\times 100=\frac{{100}}{9}\%=11\frac{1}{9}\%\)
72. If 50% of a certain number is equal to \(\displaystyle \frac{3}{4}\) th of another number, what is the ratio between the numbers ?
(a) 3 : 2
(b) 2 : 5
(c) 5 : 2
(d) 3 : 4
(e) 4 : 3
Solution: (a) First number = x Second number = y Therefore, \(\displaystyle x\times \frac{{50}}{{100}}=y\times \frac{3}{4}\) \(\displaystyle \Rightarrow \)latex \displaystyle \frac{x}{2}=y\times \frac{3}{4}$ \(\displaystyle \Rightarrow \)latex \displaystyle \frac{x}{y}=\frac{3}{4}\times 2=\frac{3}{2}$
73. A petrol pump owner mixed leaded and unleaded petrol in such a way that the mixture contains 10% unleaded petrol. What quantity of leaded petrol should be added to 1 litre mixture so that the percentage of unleaded petrol becomes 5%?
(a) 900 ml
(b) 1000 ml
(c) 1800 ml
(d) 1900 ml
(e) None of these
Solution: (b) In 1 litre quantity of unlead petrol = 100 ml (given 10%) Let x ml leaded petrol be added, then 5% of (1000 + x) = 100 ml or, 5(1000 + x) = 100 × 100 \(\displaystyle \Rightarrow \)latex \displaystyle x=\frac{{5000}}{5}=1000ml$
74. A man losses 20% of his money. After spending 25% of the remaining, he has ₹ 480 left. What is the amount of money he originally had?
(a) ₹600
(b) ₹ 720
(c) ₹ 800
(d) ₹ 840
(e) None of these
Solution: (c) Let man has originally ₹x After 20% loss = \(\displaystyle \frac{{x\times 80}}{{100}}=\frac{{8x}}{{10}}\) After spending 25% = \(\displaystyle \frac{{8x}}{{10}}\times \frac{{75}}{{100}}=\frac{{8x}}{{10}}\times \frac{3}{4}\) \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{8x}}{{10}}\times \frac{3}{4}=480$ Therefore, \(\displaystyle \frac{{480\times 4\times 10}}{{8\times 3}}=800\)
75. A person could save 10% of his income. But 2 years later, when his income increased by 20%, he could save the same amount only as before. By how much percentage has his expenditure increased?
(a) \(\displaystyle 22\frac{2}{9}\%\)
(b) \(\displaystyle 23\frac{1}{3}\%\)
(c) \(\displaystyle 24\frac{2}{9}\%\)
(d) \(\displaystyle 25\frac{2}{9}\%\)
(e) None of these
Solution: (a) Let income be ₹ 100 Expenditure amount = \(\displaystyle 100\times \frac{{90}}{{100}}\) Now, income increased by 20% = \(\displaystyle 100\times \frac{{120}}{{100}}\) Expenditure amount = (120 – 10) = ₹110 Increase in expenditure = 110 – 90 = ₹ 20 Increase in % of expenditure = \(\displaystyle \frac{{20}}{{90}}\times 100\) = \(\displaystyle \frac{{200}}{9}=22\frac{2}{9}\%\)
76. Sujata scored 2240 marks in an examination that is 128 marks more than the minimum passing percentage of 64%. What is the percentage of marks obtained by Meena if she scores 907 marks less than Sujata?
(a) 35
(b) 40
(c) 45
(d) 36
(e) 48
Solution: (b) If total maximum marks be x, then, \(\displaystyle \frac{{x\times 64}}{{100}}2240-128=2112\) \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{2112\times 100}}{{64}}=3300$ Marks obtained by Meena = 2240 – 907 = 1333 Required percentage = \(\displaystyle \frac{{1333}}{{3300}}\times 100=40\)
77. Ms. Pooja invests 13% of her monthly salary, i.e.,₹ 8554 in Mediclaim Policies, Later she invests 23% of her monthly salary on Child. Education Policies; also she invests another 8% of her monthly salary on Mutual Funds. What is the total annual amount invested by Ms. Pooja?
(a) ₹ 28952
(b) ₹ 43428
(c) ₹ 347424
(d) ₹ 173712
(e) None of these
Solution: (c) Let Ms. Pooja monthly salary = ₹ x According to the question, 13% of the x = ₹8554 \(\displaystyle \Rightarrow \)latex \displaystyle x=\frac{{8554\times 100}}{{13}}$ = Rs. 65800 Total monthly investment in percentage= 13 + 23 + 8 = 44 Therefore, Total monthly investment = 44% of ₹65800 =\(\displaystyle \frac{{44\times 65800}}{{100}}\) = ₹ 28952 Therefore, total annual investments= (12 × 28952) = 347424
78. In a class of 240 students, each student got sweets that are 15% of the total number of students. How many sweets were there?
(a) 3000
(b) 3125
(c) 8640
(d) Cannot be determined
(e) None of these
Solution: (c) Number of sweets received by each student= 15% of 240 =\(\displaystyle \frac{{15\times 240}}{{100}}=36\) Therefore, total number of sweets = 240 × 36 = 8640
79. What is the value of three fourth of sixty percent of 480?
(a) 216
(b) 218
(c) 212
(d) 214
(e) None of these
Solution: (a) Required Value = \(\displaystyle 480\times \frac{{60}}{{100}}\times \frac{3}{4}=216\)
80. It is required to get 40% marks to pass an exam. A candidate scored 200 marks and failed by 8 marks. What were the maximum marks of that exam?
(a) 530
(b) 540
(c) 502
(d) Cannot be determined
(e) None of these
Solution: (e) Maximum marks = \(\displaystyle \frac{{100\times 208}}{{40}}=520\)
81. In an examination out of 480 students, 85% of the girls and 70% of the boys have passed. How many boys appeared in the examination, if total pass percentage was 75%?
(a) 37
(b) 340
(c) 320
(d) 360
(e) None of these
Solution: (c) Total number of students = 480 Percentage of total students passed 75% of total students =\(\displaystyle \frac{{75\times 480}}{{100}}=360\) students Now, using the condition from the question, Let the number of boys be x. Then, 70% of x + 85% of (480 – x) = 360 \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{75\times x}}{{100}}+\frac{{85\times (480-x}}{{100}}=360$ \(\displaystyle \Rightarrow \)70x – 85x + 40800 = 36000 \(\displaystyle \Rightarrow \)40800 – 36000 = 85x – 70x \(\displaystyle \Rightarrow \)x = \(\displaystyle \frac{{4800}}{{15}}=320\) Therefore, there are 320 boys who appeared for the examination.
82. When the price of rice is increased by 25 percent, a family reduces its consumption such that the expenditure is only 10 percent more than before. If 40 kg of rice is consumed by family before, then find the new consumption of family.
(a) 35.2
(b) 35.2
(c) 36.2
(d) 37.2
(e) None of these
Solution: (b) Suppose initialy price per kg of rice is 100 then their expenditure is 4000. Now their expenditure is only increased by only 10% i.e – 4400. Increased price of rice = 125. So new consumption = \(\displaystyle \frac{{4400}}{{125}}=35.2\)
83. If 80% of A = 50% of B and B =x% of A, then the value of x is :
(a) 400
(b) 300
(c) 160
(d) 150
(e) 320
Solution: (c) According to question, \(\displaystyle A\times \frac{{80}}{{100}}=B\times \frac{{50}}{{100}}\) Therefore, B=\(\displaystyle B=\frac{{A\times 80}}{{100}}1.6A\) B = 160% of A x = 160
84. If x is 80% of y, what percent of y is x ?
(a) 75%
(b) 80%
(c) 100%
(d) 125%
(e) 120%
Solution: (d) According to question, y= \(\displaystyle \frac{{100\times 100}}{{80}}\times x\) y = 125% of x
85. If 15% of (A + B) = 25% of (A – B), then what per cent of B is equal to A?
(a) 10%
(b) 60%
(c) 200%
(d) 400%
(e) 450%
Solution: (d) 15% of (A + B)= 25% of (A – B) \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{15}}{{100}}(A+B)=\frac{{25}}{{100}}(A-B)$ \(\displaystyle \Rightarrow \)15 (A + B) = 25 (A – B) \(\displaystyle \Rightarrow \)15 A + 15 B = 25A – 25 B \(\displaystyle \Rightarrow \)10 A = 40 B \(\displaystyle \Rightarrow \)A = 4 B Now, let x% of B is equal to A Therefore, \(\displaystyle \frac{X}{{100}}\times B=A\to \frac{X}{{100}}\times B=4B\) x =400%
86. What is 20% of 25% of 300?
(a) 15
(b) 25
(c) 45
(d) 60
(e) 150
Solution: (a) 20% of 25% of 300 =\(\displaystyle \frac{{20}}{{100}}\times \frac{{25}}{{100}}\times 300\) =\(\displaystyle \frac{1}{5}\times \frac{1}{4}\times 300=15\)
87. The time duration of 1 hour 45 minutes is what percent of a day?
89. Two numbers are respectively 20% and 50% of a third number. What per cent is the first number of the second?
(a) 10%
(b) 20%
(c) 30%
(d) 40%
(e) 50%
Solution: (d) Let the third number be x, According to the question; First number = \(\displaystyle \frac{{20}}{{100}}\times x=\frac{x}{5}\) Second number = \(\displaystyle \frac{{50}}{{100}}\times x=\frac{x}{2}\) Therefore, required percentage = \(\displaystyle \frac{{\frac{x}{5}\times 100}}{{\frac{x}{2}}}=\frac{x}{5}\times \frac{2}{x}\times 100=40\%\)
90. Two numbers are respectively 25% and 20% less than a third number. What percent is the first number of the second ?
(a) 5%
(b) 75%
(c) 80%
(d) 93.75%
(e) 95%
Solution: (d) If two numbers are respectively x% and y% less than the third number, first number as a percentage of second is \(\displaystyle \frac{{100-x}}{{100-y}}\times 100\%\) Therefore, required percentage = \(\displaystyle \frac{{100-25}}{{100-20}}\times 100\%\) =\(\displaystyle \frac{{75}}{{80}}\times 100\%=93.75\%\)
Solution: (a) Let the number be x. According to the question, \(\displaystyle x\times \frac{{18}}{{100}}=75\times \frac{{12}}{{100}}\) \(\displaystyle \Rightarrow \)latex \displaystyle x=\frac{{75\times 12}}{{18}}=50$
92. If X is 20% less than Y, then find the values of \(\displaystyle \frac{{Y-X}}{Y}\) and \(\displaystyle \frac{X}{{X-Y}}\) .
(a) \(\displaystyle \frac{1}{5},-4\)
(b) \(\displaystyle 5,-\frac{1}{4}\)
(c) \(\displaystyle \frac{2}{5},-\frac{5}{2}\)
(d) \(\displaystyle \frac{3}{5},-\frac{5}{3}\)
(e) \(\displaystyle \frac{4}{5},-\frac{4}{3}\)
Solution: (a) X is 20% less than Y. If Y = 100, X = 80 \(\displaystyle \frac{{Y-X}}{Y}=\frac{{100-80}}{{100}}\) =\(\displaystyle \frac{{20}}{{100}}=\frac{1}{5}\) \(\displaystyle \frac{X}{{X-Y}}=\frac{{80}}{{80-100}}\) =\(\displaystyle \frac{{80}}{{-20}}=-4\)
96. 50% of a number when added to 50 is equal to the number. The number is
(a) 50
(b) 75
(c) 100
(d) 150
(e) 200
Solution: (c) Let the number be x. According to the question, \(\displaystyle \frac{{x\times 50}}{{100}}+50=x\) \(\displaystyle \frac{x}{2}+50=x\) \(\displaystyle x-\frac{x}{2}=50\) \(\displaystyle \frac{x}{2}=50\) \(\displaystyle \Rightarrow \)x=100
97. 51% of a whole number is 714. 25% of that number is
(a) 350
(b) 450
(c) 550
(d) 250
(e) 650
Solution: (a) Let the whole number be x. According to the question, 51% of x = 714 \(\displaystyle \Rightarrow \)latex \displaystyle \frac{{x\times 51}}{{100}}=714$ \(\displaystyle \Rightarrow \)X=\(\displaystyle \frac{{714\times 100}}{{51}}=1400\) \(\displaystyle \Rightarrow \)25% of 1400= \(\displaystyle \frac{{1400\times 25}}{{100}}=350\)
98. There are 950 employees in an organization, out of which 28% got promoted. How many employees got promoted?
(a) 226
(b) 256
(c) 266
(d) 216
(e) None of these
Solution: (c) Number of promoted employees = \(\displaystyle \frac{{950\times 28}}{{100}}=266\)
99. Arjun kapoor and Anil Kapoor appear for a test. For each correct answer is awarded 1 mark and for each wrong answer 1/2 mark is deducted. Arjun kapoor answers some questions and gets 10% of his answers wrong. He secures a score of 85% which is 6 marks more than the pass marks. Anil Kapoor also answers some questions and gets 20% of his answers wrong. He gets a score of 70% which is 3 marks less than the pass mark. No marks are awarded or deducted for the unanswered questions. What is the pass mark?
(a) 64
(b) 51
(c) 45
(d) 25
(e) None of these
Solution: (c) Let Arjun kapoor attempt x questions, he gets 10% of the answers wrong. Arjun kapoor’s score = \(\displaystyle 0.9x-(0.1x)\times \frac{1}{2}=0.85x\) \(\displaystyle 0.85x=0.85z\), where z is the total number of marks as well as total number of marks possible. So, x = z \(\displaystyle \Rightarrow x=100\%ofz\) Similarly let Anil Kapoor attempt y questions Anil kapoor’s score = \(\displaystyle 0.8y-(0.2y)\times \frac{1}{2}=0.7y\) \(\displaystyle 0.7y=0.7z\) \(\displaystyle \Rightarrow y=100\%ofz\) Now, \(\displaystyle 0.85z=P+6\), where P is pass mark …….. (i) Also, \(\displaystyle 0.7z=P-3\)…….. (ii) From (i) and (ii), we get \(\displaystyle 0.15z=9\) \(\displaystyle \Rightarrow z=60\) Putting the value of z in (ii), we get \(\displaystyle 0.7\times 60=P-3\) \(\displaystyle P=42+3=45\)
100. The production of a company has ups and downs every year. The production increases for two consecutive years consistently by 15% and in the third year it decreases by 20%. Again in the next two years it increases by 25% each year and decreases by 10% in the third year. If we start counting from the year 2014 approximately what will be the effect on the production of the Company in 2018?
(a) 22
(b) 32
(c) 30
(d) 20
(e) None of these
Solution: (b) Suppose the production of the company in the year 2014 be x. Then production of the company in year 2018 \(\displaystyle =x\times \frac{{115}}{{100}}\times \frac{{115}}{{100}}\times \frac{{80}}{{100}}\times \frac{{125}}{{100}}=1.32x\) Therefore, Increase % in the production in year 2018 \(\displaystyle =\frac{{(1.32x-x)}}{x}=32\%\)
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