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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
In the production of fans three types of costs are involved. Cost of raw material, expenditure on labour and miscellaneous costs. Ratio of expenditure on these cost in one year is 4 : 3 : 3. In the next year cost of raw material increased by 10%, expenditure on labour increased by 6% and miscellaneous expenditure decreased by 5%. Find the percentage rise in the cost of fan.1). 6.3%2). 8.1%3). 4.3%4). 4% |
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Answer» Three TYPES of cost are cost on RAW material, labour cost and miscellaneous cost Ratio of expenditure on these cost is 4 : 3 : 3 Let the cost of the raw material be 4x. Then increase in cost of raw material = 10% of 4x = 0.4x Let the expenditure on the labour be 3x Then increase in expenditure on the labour = 6% of 3x = 0.18x Let the expenditure on the miscellaneous cost be 3x Then DECREASE in miscellaneous cost be = 5% of 3x = 0.15x Total increase in expenditure = 0.4x + 0.18x – 0.15x = 0.43x Now, the percentage rise $(= \;\frac{{0.43x}}{{10x}} \TIMES 100 = 4.3)$ Thus, the percentage rise in the cost of fan is 4.3% |
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| 2. |
If x% of a is the same as y% of b, then z% of b will be1). \(\frac{{{\rm{yz}}}}{{\rm{x}}}{\rm{\% \;of\;a}}\)2). \(\frac{{zx}}{y}{\rm{\% \;of\;a}}\)3). \(\frac{{xy}}{z}{\rm{\% \;of\;a}}\)4). \(\frac{y}{z}{\rm{\% \;of\;a}}\) |
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Answer» $(\frac{x}{{100}} \times a = \frac{y}{{100}} \times B)$ ⇒ ax = by $(\Rightarrow b = \frac{{ax}}{y})$ HENCE, Z% of b = $(\frac{z}{{100}} \times b = \frac{z}{{100}} \times \frac{{ax}}{y} = \frac{{XZ}}{y}\% \;of\;a)$ |
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| 3. |
How much quantity (in kg) of wheat costing Rs. 84 per kg must be mixed with 81 kg of wheat costing Rs. 60 per kg so that on selling the mixture at Rs. 75.9 per kg, there is a gain of 15%?1). 272). 20.53). 22.754). 24 |
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| 4. |
1). 502). 723). 754). 76 |
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Answer» Let the number of BOYS = 3x and that of girls = 2x Number of boys who do not hold scholarship = 80% of 3x $(= 3x \times \frac{{80}}{{100}} = \frac{{12x}}{5})$ Number of girls who do not hold scholarship $( = 2x \times \frac{{70}}{{100}} = \frac{{14X}}{{10}})$ Number of students who do not hold scholarship $(= \frac{{12x}}{5} + \frac{{14x}}{{10}} = \frac{{24X + 14x}}{{10}} = \frac{{38x}}{{10}}\;)$ Required percentage $( = \frac{{\frac{{38x}}{{10}}}}{{5X}} \times 100 = \frac{{38}}{{10 \times 5}} \times 100 = 76\% )$ |
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| 5. |
The rent of a house increases at the rate of 10% per annum. If a family paid Rs. 6000 at present, find the rent of the house they have to pay after 2 years?1). Rs. 72202). Rs. 72003). Rs. 72604). Rs. 7240 |
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Answer» Rent of house after 2 years = Rent of House at present × 110/100 × 110/100 ∴ Rent of house after 2 years = 6000 × 110/100 × 110/100 = RS. 7260 |
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| 6. |
5% of a = b, then b% of 20 is the same as __________.1). 20% of a/22). 50% of a/203). 50% of a/24). 20% of a/20 |
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Answer» ⇒ 5a/100 = b ⇒ a = 20b ⇒ (20/100) × (a/20) = (20/100) × b ⇒ 20% of a/20 ∴ b% of 20 = 20% of a/20 |
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| 7. |
If two numbers X and Y are 25% and 32% more than the third number, then find the ratio between X and Y.1). 125 ∶ 1412). 125 ∶ 1323). 125 ∶ 1394). 125 ∶ 129 |
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Answer» X = a + 25a/100 = 1.25a Y = a + 32a/100 = 1.32a Required RATIO = (1.25a) : (1.32a) ∴ Answer is 125 ? 132 |
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| 8. |
1). 38%2). 7%3). 28%4). 42% |
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Answer» LET the total PASSENGERS who USE metro train be 100 So, number of passengers who use metro card = 100 × 49/100 = 49 So, number of male passengers who use metro card = 49 × 1/7 = 7 So, number of FEMALES passengers who use metro card = 49 - 7 = 42 So, there are 42% females use metro card. |
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| 9. |
A greengrocer bought some Gooseberries at a certain price. He sold 3/5th of them in the morning at 25% profit. Due to festival on next day, he increased the prices in the evening and started selling them at 40% profit. But, due to increased prices, his sales reduced and 25% of those remaining Gooseberries got wasted. Find his overall profit or loss percent.1). 17% profit2). 21% profit3). 22% loss4). 21% loss |
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Answer» Let the greengrocer buys ‘y’ Gooseberries for Rs. 100 each Cost price of 1 GOOSEBERRY = Rs. 100 ∴ Cost price of y Gooseberries = 100y When 3/5th of total was sold: Selling price of 1 Gooseberry = $(100 \times \left( {1{\rm{\;}} + \frac{{25}}{{100}}} \right))$ = Rs. 125 ∴ Selling price of 3Y/5 Gooseberries = 125 × (3y/5) = 75Y Now, Remaining Gooseberries = $(\left( {1 - \frac{{3{\rm{y}}}}{5}} \right){\rm{}} = \frac{{2{\rm{y}}}}{5})$ Out of these 25% wasted Remaining Gooseberries = $(\frac{{2y}}{5} \times \left( {1 - \frac{{25}}{{100}}} \right) = \frac{{3y}}{{10}})$ These are sold at 40% profit i.e. for Rs. 140 each ∴ Total Selling price = 75y + 140 × 3y/10 = 117y We can see that, profit is made. Profit percentage = $(\frac{{117{\rm{y\;}} - {\rm{\;}}100{\rm{y}}}}{{100{\rm{y}}}} \times 100 = 17{\rm{\% \;}})$ |
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| 10. |
A student multiplied a number by 4/7 instead of 7/4. What is the percentage error in the calculation?1). 206.25 percent2). 67.35 percent3). 33.67 percent4). 103.13 percent |
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| 11. |
1). 2% increment2). No change3). 4% increment4). 4% decrement |
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Answer» For net CHANGE, Net change = A + B + AB/100 Where, A = 20% increment = +20 B = 20% decrement = -20 ∴ Net change = 20 + (-20) + [(20) × (-20)]/100 = 20 - 20 - (400/100) = -4% = 4% DECREASED |
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| 12. |
1). 7682). 5443). 6184). 448 |
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Answer» ? Q is 70% more than 640 ⇒ Q = (100 + 70)% of 640 = 170% of 640 ⇒ Q = 1.7 × 640 = 1088 ⇒ P = (100 - 50)% of Q ⇒ P = 50% of 1088 ⇒ P = 0.5 × 1088 ∴ P = 544 |
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| 13. |
1). 6.25% decrease2). 6.25% increase3). No change4). None |
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Answer» If the wages increase by 25% and then decrease by 25% than the CHANGE is generated as, Change generated = A + B + AB/100 Here A = Increase by 25% B = Decrease by 25% For DECREMENT NEGATIVE sign is used = 25 - 25 - 625/100 = -6.25% Negative sign shows Reduction. |
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| 14. |
Rahul’s salary is 50% more than Rohit’s salary. If Rahul saves Rs 2250 which is 8% of his salary, then what is Rohit’s salary (in Rs)?1). 222502). 244503). 263504). 18750 |
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Answer» Given, 8% of (Rahul’s SALARY) = 2250 ⇒ Rahul’s salary = 2250/0.08 = RS. 28125 Now, Rahul’s salary = (100 + 50)% of (ROHIT’s salary) = 1.5 × (Rohit’s salary) ∴ Rohit’s salary = 28125/1.5 = Rs. 18750 |
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| 15. |
If A is 6 times of B, then by what percentage is B is less than A?1). 64.822). 83.333). 28.564). 85.71 |
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Answer» We have, a = 6B Difference between a and B = 6b - b = 5b ∴ % by which B is LESS than A = (5b/6b) × 100 = 500/6% = 83.33% |
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| 16. |
1). 48.6% increase2). 20% increase3). 33.33% decrease4). 12.5% decrease |
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Answer» Let the income of the MAN be Rs. 100. So, he spends Rs. 65 and saves Rs.35. His income increased by 30%. Thus, his new income = 100 + (30/100) × 100 = Rs. 130 His increased expenditure = 65 + (20/100) × 65 = Rs. 78 New SAVINGS = 130 - 78 =Rs.52 % increase in the savings = [(new savings - OLD savings)/old savings] × 100 = [(52 - 35)/35] × 100 ≈ 48.6% Hence, the savings increased by 48.6%. |
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| 17. |
When the price of apple was reduced by 25% its sale increased by 80%. What was the net effect on the sale?1). 44% increase2). 44% decrease3). 66% increase4). 35% increase |
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Answer» ⇒ Let original price be = p and Sale be s, then Sale receipts = p × s ⇒ Price reduced by 25%, then NEW Price = x - 25% of p ⇒ new Price = 0.75p ⇒ sale INCREASED by 80%, new sale = s + 80% 0f s ⇒ new sale = 1.80s ⇒ New sale receipt = 0.75p × 1.80s = 1.35ps ⇒ So % net effect on sale receipts = (1.35ps - PS) × 100/ps ∴ Net effect = 35% increase |
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| 18. |
If P got 20% marks less than Q, then the marks of Q is how much percent more than P?1). 202). 103). 254). 15 |
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Answer» <P>Let the MARKS OBTAINED by Q be x Marks obtained by P = x - 20/100 × x = 0.8x Percentage marks Q has more than P = (DIFFERENCE in marks)/(Marks obtained by P) × 100 = 25% Percentage marks Q has more than P = (x - 0.8x)/0.8x × 100 = 25% |
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| 19. |
1). 28.572). 16.223). 43.244). 75 |
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Answer» Total boys = 40 + 40 = 80 Total students = 80 + 105 = 185 Boys PERCENTAGE = (80/185) × 100 = 43.24% |
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| 20. |
In an examination, 84% candidates passed in Science and 76% candidates passed in History. If 70% candidates passed in both these subjects, then what percent of candidates failed in both the subject?1). 82). 103). 254). 18 |
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Answer» In an examination 84% candidates passed in Science and 76% candidates passed in History. 70% candidates passed in both these SUBJECTS. So, the PERCENTAGE of candidates passed in only one subject = (84 + 76)% - 70% = 90% ∴ The percentage of candidates FAILED in both subjects = (100 – 90) % = 10%. |
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| 21. |
A man spends 60% of his income and saves the rest. If his income increases by 20% and spending decreases by 20%, then what is the percentage change in his savings?1). 30% increase2). 20% increase3). 40% decrease4). 80% increase |
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Answer» Let the initial INCOME of the man be Rs. X. We know that, Income (I) = Expenditure (E) + Savings (S) Initial expenditure = 60% of x = Rs. 0.6x Initial savings = x – 0.6x = Rs. 0.4x Income after increment of 20% = x + (20% of x) = Rs. 1.2x Expenditure after increment = 0.6x – (20% of 0.6x) = Rs. 0.48x Savings after increment = 1.2x – 0.48x = 0.72x ∴ % CHANGE in savings = [(0.72x – 0.4x)/0.4x] × 100 = 80% increase |
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| 22. |
Manoj has a certain amount of money with him. He can buy either 60 apples or 80 mangoes. He wants to spend only 40% of his money. So, he buys 28 mangoes and some apples. Find the number of apples purchased by Manoj.1). 52). 303). 34). 40 |
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Answer» Let’s assume Manoj has RS. X in total. AMOUNT required to buy either 60 APPLES = Amount required to by 80 MANGOES = X Amount spent in buying 28 mangoes $(= X \TIMES \frac{{28}}{{80}} = 0.35X)$ Amount Manoj wants to spend = 40% of X = 0.4X Remaining amount = 0.4X – 0.35X = 0.05X Number of apples that can be bought in remaining amount $(= 60 \times \frac{{0.05X}}{X} = 3)$ |
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| 23. |
A total of 2,60,000 students appeared for class 10th exam in a state. If a total of 1,97,600 students pass the exam, what will be the percentage of students who failed in the exam?1). 76%2). 24%3). 28%4). 26% |
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Answer» Total number of students appeared for CLASS 10th exam in a STATE = 2,60,000 Total number of student PASSED in the exam = 1,97,600 Total number of student failed in the exam = 2,60,000 - 1,97,600 = 62,400 Percentage of students who failed in the exam = (62400/260000) × 100 = 24% |
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| 24. |
If 9% of sale price of an article is equivalent to 11% of its cost price and 13% of its sale price exceeds 15% of the cost price by Rs. 4. Then what will be the cost price of the article?1). Rs. 4402). Rs. 4503). Rs. 4704). Rs. 600 |
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| 25. |
1). 88 and 672). 89 and 683). 28 and 74). 98 and 77 |
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Answer» LET marks secured by ONE student be X, then the score of other student will be x + 21 According to question x + 21 = 80% of (x + x + 21) ⇒ x + 21 = 1.6x + 16.8 ⇒ x = 7 x + 21 = 28 ∴ Marks obtained by them are 28 and 7 |
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| 26. |
1). 180002). 200003). 240004). 32000 |
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Answer» Let the total INCOME of Amit be RS. a ⇒ Amount donated to SCHOOL = a × (20/100) = 0.2a ⇒ Remaining amount LEFT with Amit = a – 0.2 = Rs. 0.8a ⇒ Amount deposited in the BANK = 0.8a × (20/100) = 0.16a ⇒ Total remaining amount left with Amit = 0.8a – 0.16a = 0.64a Now, 0.64a = 12800 ⇒ a = 20000 So, the total income of Amit is Rs. 20000 ∴ the correct option is 2) |
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| 27. |
Mohan earned Rs. 4000 per month. From the last month his income increased by 8%. Due to rise in price, his expenditure increased by 12% and his saving decreased by 4%. Find his initial expenditure and initial savings.1). Rs. 3000, Rs. 10002). Rs. 1000, Rs. 30003). Rs. 2000, Rs. 40004). Rs. 4000, Rs. 2000 |
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Answer» INITIAL expenditure of Mohan = E Initial savings of Mohan = S We KNOW, income = expenditure + savings Mohan earned Rs. 4000 per month ∴ E + S = 4000----(1) From the last month his income increased by 8%. ∴ New income: 4000 + ( 8% of 4000) = 4000 + 320 = Rs. 4320 His expenditure increased by 12% ∴ New expenditure: E + (12% of E) = E + 0.12E = 1.12E His saving decreased by 4% New savings: S – (4% of S) = S – 0.04S = 0.96S Again, income = expenditure + savings ⇒ 4320 = 1.12E + 0.96S----(2) Solving equation (1) & equation (2) we get, [To solve (1) & (2) multiply (1) with 0.96 or 1.12, then subtract (1) from (2)] E = Rs. 3000 & S = Rs. 1000 |
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| 28. |
Length and breadth of a rectangle are increased by 40% and 70% respectively. What will be the percentage increase in the area of rectangle?1). 1182). 1103). 1384). 128 |
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Answer» Let LENGTH and breadth of RECTANGLE be a and b respectively Initial AREA = length × Breadth = ab New length = a + 40% of a = 1.4a New breadth = b + 70% of b = 1.7b New area = (1.4a) × (1.7b) = 2.38ab ∴ Required % = {(2.38ab – ab/ab} × 100 = 138% |
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| 29. |
UPA government has decided to distribute laptops, mobile phones and tablet under its new campaign for the empowerment and elementary education for girls in schools and colleges. They distributed laptops for college girls, mobile for safety purposes and tablets to girls studying in classes 10 up-to classes 12. The state has total 35 Lakh girls, out of which 20% are not getting any education and 5% of the remaining are in college and 10% of the remaining are studying in class 10 to class 12, and the girls who are in PG got 4,20,000 total tablets, which contributes the 15% of the distribution, then find the number of laptops which were distributed. (The distribution is according to the number of girls). 266,0001). 110,0002). 280,0003). 120,0004). 140,000 |
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Answer» Total girls who are GETTING education = 80% of 35 Lakh = 28 Lakh Total girls in college = 5% of 28 lakh = 140,000 Total girls STUDYING in class 10 to class 12 = 10% of remaining Total girls studying in class 10 to class 12 = 10% of (2800000 – 140000) = 266,000 The girls who are in PG got 4,20,000 total tablets, which contributes the 15% of the distribution Let the total distribution be ‘X’ ⇒ 15% of x = 4,20,000 ⇒ x = 28 Lakh ⇒ number of laptops = number of girls studying in college ∴ Number of laptops distributed is 140,000. |
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| 30. |
All students can speak at least one of the two languages.Find the minimum number of students can speak both Tamil and Telugu.1). 122). 153). 184). 22 |
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| 31. |
The value of machine depreciates 15% every year. If its present value is Rs. 65,000. Find the value of machine after 7 years?Given that, (0.85)7 = 0.3201). Rs. 25092.52). Rs. 278343). Rs. 208004). Rs. 21390.5 |
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Answer» <P>The FORMULA for above CONDITION can be given as, Value of machine $(= P{\left({1 - \frac{r}{{100}}} \right)^n})$ Given that P = 65,000, n = 7 YEARS, r = 15% Value of machine $(= 65,000{\left({1 - \frac{{15}}{{100}}}\right)^7})$ ⇒ Value of machine = 65,000 × 0.320 ∴ Value of machine = Rs. 20800 |
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| 32. |
Chikara has 40% more pencils than Priya. Two days later the no of pencils with Chikara increased by 20% and that of Priya decreased by 10%. Again two days later, no of pencils with Chikara decreased to \(66\frac{2}{3}\% \) and that of Priya increased by \(33\frac{1}{3}\%\). Then-1). Chikara has more pencils than Priya2). Priya has more pencils than Chikara3). They both have same number of pencils4). Data insufficient |
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Answer» Let Priya has 100 pencils initially, therefore no of pencils with Chikara = 140 (40% more). Now,
i.e. Priya has more pencils than Chikara. |
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| 33. |
P is 10% less than Q. Q is 10% more than R. R is 20% less than S. If S is 1000, then what is the value of P?1). 6482). 7923). 6764). 736 |
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Answer» S = 1000 R = 1000 - (0.2 × 1000) = 1000 - 200 = Rs. 800 Q = 800 + (0.1 × 800) = 800 + 80 = Rs. 880 P = 880 - (0.1 × 880) = 880 - 88 = Rs. 792 ∴ P = Rs. 792 |
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| 34. |
1). 33.12). 303). 40.254). 35.76 |
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Answer» LET, SIDE of cube at first = x ∴ Volume of the cube at first = x3 ∴ side of the cube now = x + x × 10/100 = 1.1x ∴ Volume of the cube now = (1.1x)3 = 1.1331x ∴ percentage increase in the volume of the cube, $( \Rightarrow \FRAC{{1.331x - x}}{x} \times 100\% )$ ⇒ 0.331 × 100% ⇒ 33.1% |
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| 35. |
1). 2002). 3003). 4004). 600 |
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Answer» LET the MAXIMUM marks of the examination be x. Passing marks of the EXAM = 105 - 6 = 99 ∴ 33% of x = 99 $(\begin{ARRAY}{l} \Rightarrow \frac{{33}}{{100}} \times x = 99\\ \Rightarrow x = \frac{{99{\rm{\;}} \times 100}}{{33}} = 300\end{array})$ Hence, maximum marks of the exam is 300. |
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| 36. |
In an election between two candidates, the winning candidate got 70% of the valid votes and won by a majority of 9000 votes. If out of total votes polled 90% votes are valid, then what is the total number of votes polled?1). 300002). 250003). 225004). 27000 |
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Answer» VALID votes of winning candidate = 70% Valid votes of LOSING candidate = 30% Difference of valid votes of winning and losing candidates = 70% - 30% = 40% ⇒ 40% = 9000 ⇒ 10% = 2250 ⇒ 100% = 22500 = TOTAL valid votes ∴ Total no. of votes polled = 22500 × 100/90 = 25000 |
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| 37. |
In an examination, 22.5% of the student failed in science & 25% failed in French. If 30% failed in both the subjects, then the percentage of students who passed in both the subjects, was – 1). 81%2). 82.5%3). 85%4). 89% |
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Answer» Students Failed in science = n(A) = 22.5% Students failed in French = n(B) = 25% Students who failed in both the subjects = n (A ∩ B) = 30% As we know, n(A ∪ B) = n(A) + n(B) – n(A ∩ B) ⇒ n(A ∪ B) = 22.5 + 25 – 30 = 17.5% Students failed EITHER one or both SUBJECT = 17.5% ∴ PERCENTAGE of students who PASSED in both subjects = 100 – 17.5 = 82.5% Hence, the percentage of students who passed in both the subjects is 82.5% |
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| 38. |
1). 64,75,0002). 97,50,0003). 47,76,0004). 74,25,000 |
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Answer» POPULATION of the city = 75,00,000 Population of the city after FIRST YEAR = 7500000 × (120/100) Population of the city after SECOND year = 7500000 × (120/100) × (75/100) Population of the city after third year = 7500000 × (120/100) × (75/100) × (110/100) = 74,25,000 ∴ The population of the city after third year is 74,25,000. |
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| 39. |
If the price of a commodity is increased by 40% by what fraction must its consumption reduced so as to keep the same expenditure on its consumption?1). \(\frac{1}{4}\)2). \(\frac{2}{7}\)3). \(\frac{1}{5}\)4). \(\frac{2}{5}\) |
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Answer» Let the price of the commodity be Rs. 100/kg. The price of a commodity is increased by 40%. So, the new piece the commodity = Rs. 140/kg Let we could get x kg of the commodity initially for a certain amount of money and after increment of the price, we will get y kg of the commodity for the same amount of money. So, we can write now, x × 100 = y × 140 ⇒ y/x = 5/7 So, we can see that for the same amount of money we will get 2 kg less after increment in price where PREVIOUSLY we used to get 7 kg. ∴ The required CONSUMPTION reduced so as to KEEP the same expenditure on its consumption = 2/7. |
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| 40. |
If the price of an article is decreased by 20%, then to restore its former value the new price must be increased by what percent?1). 302). 503). 204). 25 |
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Answer» Let, the former value = Rs. x After DECREASE of 20%, NEW value = Rs.(x - x × 20/100) = Rs. 0.8x ∴ Required increase percentage, $(\RIGHTARROW \frac{{x - 0.8x}}{{0.8x}} \TIMES 100\%)$ ⇒ 2/8 × 100% ⇒ 25% |
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| 41. |
If 75% of (x - y) = 45% of (x + y), then what percent of x is y?1). 25%2). 100/3%3). 40%4). 400% |
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Answer» GIVEN, 75% of (X – y) = 45% of (x + y) ⇒ 0.75 × (x – y) = 0.45 × (x + y) ⇒ 0.75x – 0.75y = 0.45x + 0.45y ⇒ 0.75x – 0.45x = 0.45y + 0.75y ⇒ 0.3x = 1.2y ⇒ y = 0.25x ∴ y = 0.25x or 25% of x. |
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| 42. |
In a class of 120 students, 40% of the students individually got the number of sweets that are 30% of the total students and each of the remaining 60% of the students got the number of sweets that are 40% of the total number of students. How many sweets were distributed among 120 students?1). 52342). 51843). 59654). Can’t be determined |
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Answer» In a class of 120 STUDENTS, 40% of the students individually got number of sweets that are 30% of the total students and each of the remaining 60% of the students got number of sweets that are 40% of the total number of students. According to the question, 40% of the total 120 students = (40/100) × 120 = 48 So, No. of sweets received by each of these 48 students is 30% of the total students ⇒ number of sweets $(= \;48 \times \FRAC{{30}}{{100}} \times 120 = 1728)$ 60% of the total 120 students = (60/100) × 120 = 72 So, No. of sweets received by 72 each of these 72 students is 40% of the total students ⇒ number of sweets $(= 72 \times \frac{{40}}{{100}} \times 120 = 3456)$ ∴ Total No. of sweets received by 120 students = 1728 + 3456 = 5184 ∴ Numbers of Sweets distributed among 120 students were 5184. |
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| 43. |
Duranto Express has a capacity of 400 seats of which 12% is in the Executive class, rest being Chair car. During one journey the train was booked to 80% of its capacity. If Executive class was booked to 75% of its capacity, how many Chair car seats were empty during that journey?1). 752). 683). 734). 62 |
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Answer» The number of SEATS in the Executive class = 400 × (12/100) = 48 Then, the number of seats in Chair car = 400 – 48 = 352 During one journey the train was booked to 80% of its capacity. So, the no. of booked seats = 400 × (80/100) = 320 Executive class was booked to 75% of its capacity. Then, the number of seats booked in Executive class = 48 × (75/100) = 36 So, the number of booked seats in Chair car = 320 – 36 = 284 ∴ The number of seats in Chair car seats were EMPTY during that journey = 352 – 284 = 68. |
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| 44. |
1). Rs. 1200002). Rs. 1500003). Rs. 1800004). Rs. 200000 |
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Answer» LET his initial MONTHLY salary be Rs. ‘X’ New monthly salary = (100 + 20)% of x = 1.2x Initial monthly donation = 2% of x = 0.02x New monthly donation = 2% of 1.2x = 0.024x Now, INCREASE in monthly donation = Rs. 500 ⇒ 0.024x – 0.02x = 500 ⇒ x = 500/0.004 = 125000 ∴ His new monthly salary = 1.2x = 1.2 × 125000 = Rs. 150000 |
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| 45. |
The ratio of the number of boys and girls in Arya Public School is 4 : 15. If 20% of boys and 80% of girls are not getting scholarships for their studies, then find their percentage.1). 65.202). 66.443). 67.364). 68.89 |
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Answer» Here LET NUMBER of boys be 4x and number of GIRLS be 15x Total STUDENTS are: 4x + 15x = 19X Number of students with no scholarships are 20% boys + 80% girls i.e. 1/5(4x) + 4/5(15x) = 64x/5 So, percentage of students who are not getting scholarship out of total is {(64x/5)/19x} × 100 On solving we get 67.36% |
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| 46. |
1). 1002). 603). 804). 92 |
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Answer» Let, the number of STUDENTS APPLIED for the groups is ‘x’ Given, 40% of the students were allotted group A. ∴ Number of students allotted in group A = 0.40x ∴ Number of remaining students = x – 0.40x = 0.60x 75% of the remaining were given group B ∴ Number of students allotted in group B = (0.75) × (0.60x) = 0.45x And, number of students allotted in group C = 12 ∴ 0.40x + 0.45x + 12 = x ⇒ x – 0.85x = 12 ⇒ 0.15x = 12 ⇒ x = 12/0.15 = 80 |
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| 47. |
1). 80542). 92863). 100054). 11245 |
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Answer» Total no. of VOTERS = 108054 Voting PERCENTAGE = 83.33% ⇒ No. of VOTES = 83.33% of 108054 = 5/6 × 108054 = 90045 Now, Percentage of votes received by winner = 33.33% Percentage of votes received by runner-up = 22.22% ⇒ Percentage of votes by which the winner won = 33.33 – 22.22 = 11.11% ∴ No. of votes by which the winner won the election = 11.11% of 90045 = 1/9 × 90045 = 10005 |
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| 48. |
1). 11.762). 13.53). 17.84). 20.4 |
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Answer» Let the third number be ‘X’. FIRST number = x – (25% of x) = 0.75x Second number = x – (15% of x) = 0.85x ∴ REQUIRED % = [(0.85x – 0.75x)/0.85x] × 100 = 11.76% |
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| 49. |
Sum of two numbers is 190% of the larger number. If smaller number is 18, then what is the value of the larger number?1). 202). 253). 144). 26 |
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Answer» According to PROBLEM, ⇒ x × 190/100 = x + 18 ⇒ 1.9x = x + 18 ⇒ 0.9x = 18 ⇒ x = 20 ∴ The larger number = 20 |
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| 50. |
1). 20%2). 40%3). 50%4). 75% |
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Answer» Number of BOYS = 45 So, total number of GIRLS = 8/9 × 45 = 40 If number of girls who passed is y, Number of boys who passed will be 1.25y 45 - 1.25y = 20----- (1) Number of girls who passed = y = (45 - 20)/1.25 = 20 ∴ Pass PERCENTAGE of girls = 100 × 20/40 = 50% |
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