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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.
| 151. |
The interest for a principle of 4,500 which gives an amount of 5,000 at end of certain period is (i) Rs 500 (ii) Rs 200 (iii) 20% (iv) 15% |
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Answer» (i) Rs 500 Interest = Amount – Principle = Rs 5000 – Rs 4500 = Rs 500 |
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| 152. |
Which among the following is the simple interest for the principle of Rs 1,000 for one year at the rate of 10% interest per annum? (i) Rs 200 (ii) Rs 10 (iii) Rs 100 (iv) Rs 1,000 |
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Answer» (iii) Rs 100 Interest = \(\frac{pnr}{100}\) = \(\frac{1000\times1\times10}{100}\) = Rs 100 |
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| 153. |
A sum of Rs 48,000 was lent out at simple interest and at the end of 2 years and 3 months the total amount was Rs 55,560. Find the rate of interest per year. |
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Answer» Given Principal P = Rs 48,000 Time n = 2 years 3 months = 2 + \(\frac{3}{12}\) years = 2 + \(\frac{1}{4}\) years = (\(\frac{8}{4}\) + \(\frac{1}{4}\)) years = \(\frac{9}{4}\) years Amount A = Rs 55,660 A = p + 1 55660 = 48000 + I I = 55660 – 48000 = Rs 7660 ∴ Interest for \(\frac{9}{4}\) years = Rs 7660 Simple interest = \(\frac{pnr}{100}\) 7660 = 48000 x \(\frac{9}{4}\) x \(\frac{r}{100}\) r = \(\frac{7660\times4\times100}{9\times48000}\) = 7.09 % = 7 % Rate of interest = 7 % Per annum. |
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| 154. |
On what sum of money lent out at 9% per annum for 6 years does the simple interest amount to Rs 810? |
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Answer» Given simple interest I = Rs 810 Let the sum of money (Principal) be P Rate of interest r = 9 % Per annum. Time n = 6 years I = \(\frac{pnr}{100}\) 810 = \(\frac{P\times\,6\times\,9}{100}\) P = \(\frac{810\times100}{6\times9}\) P = Rs 1500 Sum of money required = Rs 1500 |
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| 155. |
In a camp, there were 500 soldiers. 60 more soldiers joined them. What percent of the earlier (original) number have joined the camp. |
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Answer» Number of soldiers = 500 More joined them = 60 Percentage to join the earlier = 60/500 x 100 = 12% |
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| 156. |
A tin contains 20 liters of petrol. Due to leakage, 3 liters of petrol is lost. What percent is still present in the tin ? |
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Answer» Total petrol in tin = 20 litres last due to leakage = 3 litres Balance petrol in tin = (20 – 3) = 17 litres Percentage of petrol in tin = 17/20 x 100 = 85% |
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| 157. |
An alloy of copper and zinc contains 45% copper and the rest is zinc. Find the weight of zinc in 20 kg of the alloy. |
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Answer» Total weight of alloy = 20 kg Weight copper = 20 x 45% = 20 x 45/100 = 9 kg Weight of zinc = (total weight of alloy – weight of copper) = 20 – 9 = 11 kg |
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| 158. |
An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy. |
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Answer» Mass of alloy = 1kg = 1000 gm Mass of copper in alloy \(=\frac{1000\times32}{100}\) = 320 gm Mass of nickel in alloy \(=\frac{1000\times40}{100}\) = 400 gm So, amount of zinc in alloy = 1000 - (320 + 400) = 1000 – 720 = 280 gm |
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| 159. |
A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teaachers in the school is female? |
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Answer» Number of students in school = 300 Number of boys = 142 ∴ Number of girls = 300 – 142 = 158 Number of teachers in school = 30 No. of male teachers = 12 ∴ No. of female teachers = 30 – 12 = 18 Total no. of students and teachers = 300+30 = 330 Total numbers of females = 158+18 = 176 Percentage of females in school \(=\frac{176}{330}\times100\) = \(\frac{160}{3}\)% |
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| 160. |
A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teachers in the school is female? |
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Answer» The given details are, In a school, number of students are = 300 Number of boys = 142 Number of girls = 300 – 142 = 158 In a school, number of teachers are = 30 Number of male teachers are = 12 Number of female teachers are = 30 – 12 = 18 Total number of students and teachers is = 300+30 = 330 Total numbers of female in the school is = 158+18 = 176 Percentage of female in the school = (176/330) × 100 = 160/3% ∴ Total of 160/3% are female in the school. |
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| 161. |
Gunpowder contains 75% nitre and 10% sulphur. Find the amount of the gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.3 kg sulphur? |
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Answer» Amount of nitre in gunpowder = 9 kg Percentage of nitre in gunpowder = 75% Let amount of gunpowder = x kg So, \(=\frac{X\times75}{100}\) = 9 \(= {X} =\frac{9\times100}{75}\) = 12 kg Amount of sulphur in gunpowder = 2.3 kg Percentage of sulphur in gunpowder = 10% Let amount of gunpoder = x kg So, \(=\frac{X\times10}{100}\) = 2.3 \(={X} =\frac{2.3\times100}{10}\) = 23 kg |
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| 162. |
Gunpowder contains 75% nitre and 10% sulphur. Find the amount of gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.5 kg sulphur? |
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Answer» Let the amount of gunpowder which carries 9 kg nitre = Z ∴ 75% of Z = 9 kg ⇒ Z × 75/100 = 9 ⇒ Z = 9 × 100/75 ⇒ Z = 9 × 4/3 ⇒ Z = 12 kg Now, Let the amount of gunpowder which carries 2.5 kg sulphur = K ∴ 10% of K = 2.5 kg ⇒ K × 10/100 = 2.5 ⇒ K = 2.5 × 100/10 ⇒ K = 2.5 × 10 ⇒ K = 25 kg |
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| 163. |
Gunpowder contains 75% of nitre and 10% of sulphur, and the rest of it is charcoal. Find the amount of charcoal in 8 kg of gunpowder. |
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Answer» The amount of gunpowder = 8 kg = 8000g … [WKT 1 kg = 1000g] Quantity of nitre in gunpowder = 75% of 8000g = (75/100) × (8000) = (75 ×80) = 6000g = 6 kg Quantity of sulphur in gunpowder = 10% of 8000g = (10/100) × 8000 = (10 × 80) = 800g = 0.8 kg The quantity of charcoal in gunpowder = [8000 – (6000 + 800)] = [8000 – 6800] = 1200g = 1.2 kg ∴the quantity of charcoal in 8 kg of gunpowder is 1.2 kg |
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| 164. |
A number decreased by 8% gives 69. The number is(a) 80 (b) 75 (c) 85 (d) none of these |
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Answer» (b) 75 Because, Let the required number be x Now, = x – 8% of x = 69 = [x – {(8/100) × x}] = 69 = [x – {8x/100}] = 69 = [x – {2x/25}] = 69 = [(25x – 2x)/25] = 69 = [23x/25] = 69 = x = (69 × 25)/23 = x = (75) |
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| 165. |
Find the number a, if(i) 8.4% of a is 42(ii) 0.5 of a is 3(iii) of a is 50(iv) 100% of a is 100 |
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Answer» (i) 8.4% of a is 42 \(=\frac{(a\times8.4)}{100}\)=42 = a =\(\frac{42\times100}{8.4}=\)\(=\frac{42\times100\times10}{84}\)=500 (ii) 0.5 of a is 3 \(=\frac{a\times0.5}{100}\)=3 \(=a=\frac{3\times100}{0.5}=\frac{3\times100\times10}{5}\)=600 (iii) \(\frac{1}{2}\)of a is 50 \(({a}\times\frac{1}{2})\)=50 = a = \({50\times2}\)= 100 (iv) 100% of a is 100 \(=\frac{a\times100}{100}\) =100 = a =\(\frac{100\times100}{100}\) =100 |
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| 166. |
The population of a village decreased by 12%. If the original population was 25,000, find the population after decrease ? |
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Answer» Original population = 25,000 Decrease in population = 12% Population after decrease = 25,000 – 12% of 25,000 = 25,000 – 12/100 x 25,000 = 25,000 – 3,000 = 22,000 |
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| 167. |
What percent of 1 day is 36 minutes?(a) 25% (b) 2.5% (c) 3.6% (d) 0.25% |
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Answer» (b) 2.5% Because, Assume x be the percent of 1 day is 36 minutes We know that, 1 day = 24 hours 1 hour = 60 minutes = 60 × 24 = 1440 minutes Now, = (x % of 1440) = 36 = (x/100) × 1440 = 36 = x = (36 × 100) /1440 = x = 2.5% |
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| 168. |
(3/4) as rate percent is(a) 7.5% (b) 75% (c) 0.75% (d) none of these |
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Answer» (b) 75% Because, = (3/4) × 100 = (300/4) = 75% |
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| 169. |
A lawyer willed his 3 sons Rs 250000 to be divided into portions 30%, 45% and 25%. How much did each of them inherit? |
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Answer» Given total amount with the lawyer = Rs. 250000 = (30/100) × 250000 = 7500000/100 = Rs. 75000 = (45/1000) × 250000 = 11250000/100 = Rs. 112500 Third son’s inheritance = 25% of 250000 = (25/100) × 250000 = 6250000/100 = Rs. 62500 |
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| 170. |
A bag contains 8 red balls, 11 blue balls and 6 green balls. Find the percentage of blue balls in the bag. |
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Answer» Total ball = 8 + 11 + 6 = 25 Blue balls = 11 Reqd. percentage = (11/25) x 100 = 44% |
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| 171. |
\(\frac{3}{5}=?\)A. 30%B. 40%C. 45%D. 60% |
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Answer» 3/5 = (3/5 × 100) % = (3 × 20) % = 60% |
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| 172. |
Aman obtained 410 marks out of 500 in CBSE XII examination while his brother Anish gets 536 marks out of 600 in IX class examination. Find whose performance is better? |
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Answer» Given Aman’s marks in CBSE XII = 410/500 Percentage of marks obtained by Aman = (410/500) × 100 = 82% Given that Anish’s marks in CBSE IX = 536/600 Percentage of marks obtained by Anish = (536/600) × 100 = 89.33% Clearly 89.33 > 82 Therefore, Anish’s performance is better than Aman’s |
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| 173. |
Express each of the following as a percentage:(i) \(\frac{9}{25}\)(ii) \(\frac{3}{125}\)(iii) \(\frac{12}{5}\) |
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Answer» (i) \(\frac{9}{25}=(\frac{9}{25}\times100)\%\) = (9 x 4) % = 36% (ii) \(\frac{3}{125}=(\frac{3}{125}\times100)\)% = 2.4% (iii) \(\frac{12}{5}=(\frac{12}{5}\times100)\) = (12 x 20) % = 240% |
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| 174. |
Express each of the following as a percentage:(i) 9/25(ii) 3/125(iii) 12/5 |
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Answer» We know the number has to be multiplied by 100 we get, (i) 9/25 = ((9/25) × 100) % = ( 9 × 4) % = 36% (ii) 3/125 = ((3/125) × 100) % = 2.4% (iii) 12/5 = ((12/5) × 100) %) = (12 × 20) % = 240 % |
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| 175. |
Express: 4 : 5 as a percentage; |
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Answer» 4 : 5 = 4 /5 = (4 /5 x 100) % [Because 100% = 1] = 80% |
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| 176. |
Express: 56% as a ratio. |
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Answer» 56% means, 56 divided by 100. So, 56% = 56/100 = 14/25 = 14:25 |
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| 177. |
Rohit deposits 12% of his income in a bank. He deposited Rs 1440 in the bank during 1997. What was his total income for the year 1997? |
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Answer» Percentage deposit by Rohit in bank = 12% of total income Money actually deposited by him = Rs. 1440 Let total income of Rohit = Rs. X So, \({X} =\frac{X\times12}{100}\) =1440 \({X} = \frac{1440\times100}{12}\) =Rs. 12000 ∴ Total income of Rohit = Rs.12000 |
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| 178. |
If 34% of a number is 85, find the number. |
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Answer» Let the number = Z ∴ 34% of Z = 85 ⇒ 34/100 x Z = 85 ⇒ Z = 85 x 100/34 ⇒ Z = 5 x 100/2 ⇒ Z = 250 |
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| 179. |
Radha earns 22% of her investment. If she earns Rs 187, Then how much did she invest? |
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Answer» Percentage earn of Radha = 22% of investment Let total investment = Rs. X So, \(=\frac{X\times22}{100}\) =187 \(=X =\frac{187\times100}{22}\) =350 ∴ Total investment made by Radha = Rs.350 |
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| 180. |
Mr. Sidhana saves 28% of his income. If he saves as 840 per month, find his monthly income. |
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Answer» Let Mr. Sidhana’s monthly income be x |
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| 181. |
A man saves 18% of his monthly income. If he saves ₹ 3780 per month, what is his monthly income? |
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Answer» Let us consider ₹ x as his monthly income ∴ saving = 18% of ₹ x 3780 = 18/100 × x 3780 = 9/50 × x x = 3780 × 50/9 x = 21000 ∴ The monthly income Is ₹ 21000 |
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| 182. |
Radha earns 22% of her investment. If she earns Rs 187, then how much did she invest? |
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Answer» Given, percentage Radha earns = 22% of investment So, let us consider total investment be Rs x By calculating, (x/100) × 22 = 187 By cross multiplying we get, (x/100) = 187/22 x = (187×100)/22 = 850 ∴ Radha’s total investment is Rs 850. |
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| 183. |
A man saves 18% of his monthly income. If he saves Rs. 3780 per month, what is his monthly income? |
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Answer» Let Rs. Z his monthly income. ∴ Saving = 18% of Rs. Z ⇒ 3780 = 18 /100 × Z ⇒ 3780 = 9 /50 × Z ⇒ Z = 3780 × 50/9 ⇒ Z = 420 × 50 [Because 420 × 9 = 3780] ⇒ Z = 21000 Therefore, his monthly income is Rs 21000/- |
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| 184. |
If 16% of number is 72, find the number. |
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Answer» Let the number = Z ∴ 16% of Z is 72. ⇒ 16 /100 × Z = 72 ⇒ 16 Z = 7200 ⇒ Z = 7200 /16 ⇒ Z = 450 |
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| 185. |
If 16% of a number is 72, find the number. |
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Answer» Let us consider a number x ∴ 16% of x = 72 16/100 × x =72 16x = 7200 x= 7200/16 x=450 ∴ The number is 450 |
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| 186. |
Find:7.5 % of 600 m |
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Answer» 7.5 % of 600 m We have, = (7.5/100) × 600 = (7.5/1) × 6 = (7.5 × 6) = 45 m |
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| 187. |
Find:20 % of ₹ 132 |
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Answer» 20 % of ₹ 132 We have, = (20/100) × 132 = (1/5) × 132 = (132/5) = ₹26.4 |
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| 188. |
Find 4 ½ % of ₹ 3600 |
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Answer» Here, 4 ½ % = (9/2) × 1/100 = 9/200 Now, 9/200 of 3600 = (9/200) × 3600 = 162 |
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| 189. |
Find:25 % of ₹ 76 |
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Answer» 25 % of ₹ 76 We have, = (25/100) × 76 = (1/4) × 76 = ₹19 |
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| 190. |
Find:0.6 % of 45 |
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Answer» 0.6 % of 45 The above question can be written as, = (0.6/ 100) × 45 = (6/1000) × 45 = (6/200) × 9 = (54/200) = 0.27 |
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| 191. |
Find:8.5 % of 5 kg |
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Answer» 8.5 % of 5 kg We have, = (8.5/100) × 5 = (85/1000) × 5 = (425/1000) = 0.425 kg = 425 g … [∵ 1 kg = 1000g] |
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| 192. |
Find:20 % of 12 liters |
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Answer» 20 % of 12 liters We have, = (20/100) × 12 = (1/5) × 12 = (12/5) = 2.4 liters |
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| 193. |
Ravi purchased a pair of shoes costing Rs. 950. Calculate the total amount to be paid by him, if the rate of sales tax is 7%. |
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Answer» Sales price = Rs. 950 Sales tax = 6/100 x 950 = 570/10 = 7 Total amount paid by Ravi = Rs. 950 + Rs. 57 = Rs.1007. |
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| 194. |
Find each of the following:(i) 7% of Rs 7150(ii) 40% of 400 kg(iii) 20% of 15.125 liters(iv) 3 (1/3) % of 90 km(v) 2.5% of 600 meters |
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Answer» (i) Given 7% of Rs 7150 = (7/100) × 7150 = Rs 500.50 (ii) Given 40% of 400 kg = (40/100) × 400 = 160 kg (iii) Given 20% of 15.125 liters = (20/100) × 15.125 = 3.025 liters (iv) Given 3 (1/3) % of 90 km We know that 3 (1/3) = (10/3) = (10/300) × 90 = 3 km |
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| 195. |
A motorist travelled 122 kilometers before his first stop. If he had 10%of his journey to complete at this point, how long was the total ride? |
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Answer» Given details are, Motorist total distance travelled before first stop = 122 km Journey completed at first stop = 10 % Let us consider total ride to be travelled be ‘x’ km So, by calculating (x/100) × 10 = 122 By cross multiplying we get, x/100 = 122/10 x = (122 × 100)/10 = 1220 km ∴ Motorist total ride is 1220 km. |
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| 196. |
A motorist travelled 122 kilometres before his first stop. If he had 10% of his journey to complete at this point, how long was the total ride? |
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Answer» Total distance travelled before first stop = 122 km Distance completed at first stop = 10 % Let total distance to be travelled = x km So, \(= {X} =\frac{122\times100}{10}\) 1220 km ∴ Total distance = 1220 km |
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| 197. |
Find 20% more than Rs.200. |
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Answer» Consider 20% of 200 = (20/100) × 200 = Rs 40 Therefore 20% more than Rs 200 = 200 + 40 = Rs 240 |
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| 198. |
Find 10% less than Rs.150. |
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Answer» Consider 10% of 150 = (10/100) × 150 = Rs 15 Therefore 10% less than Rs 150 = 150 – 15 = Rs 135 |
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| 199. |
Ashu had 24 pages to write. By the evening, he had completed 25% of his work. How many pages were left? |
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Answer» Given total number of pages Ashu had to write = 24 = (25/100) × 24 = 600/100 = 6 = 18 pages |
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| 200. |
In an examination, Nitin gets 98 marks. This amounts to 56% of the maximum marks. What are the maximum marks? A. 75 B. 150 C. 175 D. 225 |
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Answer» Let the maximum marks = Z ∴ 56% of Z = 98 ⇒ Z × 56/100 = 98 ⇒ Z = 98 × 100/56 ⇒ Z = 7 × 100/4 ⇒ Z = 175 |
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