Read the data carefully and answer the questions.
There are 900 employees in a company and there are two buildings P & Q in the company. There are three departments in each building i.e., HR, Production & Management. ⁷⁵⁄₄% of total employee in P are in Management and ²⁰⁰⁄₇% of total employee in Q are in Production. Sum of total employee in Management in P & total employee in Production in Q is 210. ⁴⁰⁰⁄₂₁% of total employee in Q are in Management and 50% of total employee in P are in HR.
Total HR employees in P is what percent more than that in Production in Q.
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Answer: (D) 100%
Explanation
Let the total employees in building P be P and building Q be Q.
P + Q = 900
Management in P = (75/4)% of P = (3/16) P
Production in Q = (200/7)% of Q = (2/7) Q
Given: (3/16) P + (2/7) Q = 210
Substituting Q = 900 - P gives P = 480 and Q = 420.
HR in P = 50% of 480 = 240
Production in Q = (2/7) × 420 = 120
Percentage excess = ((240 - 120) ÷ 120) × 100 = 100%
52Quantitative Aptitude
Read the data carefully and answer the questions.
There are 900 employees in a company and there are two buildings P & Q in the company. There are three departments in each building i.e., HR, Production & Management. ⁷⁵⁄₄% of total employee in P are in Management and ²⁰⁰⁄₇% of total employee in Q are in Production. Sum of total employee in Management in P & total employee in Production in Q is 210. ⁴⁰⁰⁄₂₁% of total employee in Q are in Management and 50% of total employee in P are in HR.
Find the ratio of total employee in Management in Q to total employee in Production in P.
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Answer: (A) 8:15
Explanation
Total employees in building P = 480 and in building Q = 420.
Management in Q = (400/21)% of 420 = (4/21) × 420 = 80
Production in P = 480 - 240 (HR) - 90 (Management) = 150
Ratio of Management in Q to Production in P = 80 : 150 = 8 : 15
53Quantitative Aptitude
Read the data carefully and answer the questions.
There are 900 employees in a company and there are two buildings P & Q in the company. There are three departments in each building i.e., HR, Production & Management. ⁷⁵⁄₄% of total employee in P are in Management and ²⁰⁰⁄₇% of total employee in Q are in Production. Sum of total employee in Management in P & total employee in Production in Q is 210. ⁴⁰⁰⁄₂₁% of total employee in Q are in Management and 50% of total employee in P are in HR.
If in other building R, total employee are 360 and total employee in Production in R are 25% more than total employee in Management in Q, then find total employee in HR & Management in R is how much less than total employee in HR & Management in P.
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Answer: (E) 70
Explanation
Total employees in building R = 360
Production in R = 1.25 × Management in Q = 1.25 × 80 = 100
Total HR and Management in R = 360 - 100 = 260
Total HR and Management in P = 240 + 90 = 330
Difference = 330 - 260 = 70
54Quantitative Aptitude
Read the data carefully and answer the questions.
There are 900 employees in a company and there are two buildings P & Q in the company. There are three departments in each building i.e., HR, Production & Management. ⁷⁵⁄₄% of total employee in P are in Management and ²⁰⁰⁄₇% of total employee in Q are in Production. Sum of total employee in Management in P & total employee in Production in Q is 210. ⁴⁰⁰⁄₂₁% of total employee in Q are in Management and 50% of total employee in P are in HR.
Find the average number of employees in Production in P & Q?
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Answer: (B) 135
Explanation
Production employees in P = 150
Production employees in Q = 120
Average number of Production employees = (150 + 120) ÷ 2 = 135
55Quantitative Aptitude
Read the data carefully and answer the questions.
There are 900 employees in a company and there are two buildings P & Q in the company. There are three departments in each building i.e., HR, Production & Management. ⁷⁵⁄₄% of total employee in P are in Management and ²⁰⁰⁄₇% of total employee in Q are in Production. Sum of total employee in Management in P & total employee in Production in Q is 210. ⁴⁰⁰⁄₂₁% of total employee in Q are in Management and 50% of total employee in P are in HR.
If out of total employee in HR in P & Q, ratio of male to female is 5:3 and 7:4 respectively, then find total male in HR from both buildings?
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Answer: (C) 290
Explanation
Total HR in P = 240
Male HR in P = (5/8) × 240 = 150
Total HR in Q = 220
Male HR in Q = (7/11) × 220 = 140
Total male HR employees = 150 + 140 = 290
56Quantitative Aptitude
Read the following table carefully and answer the questions given below. The following table shows the number of employees working in 5 different companies and ratio of number of employees working from home and working from office.
| Company | Number of employees | Working from Home: Working from Office |
|---|---|---|
| A | 1500 | 2:3 |
| B | 800 | 7:9 |
| C | 600 | 1:1 |
| D | 1200 | 13:11 |
| E | 450 | 1:2 |
What is the average of number of employees working from home in company A, C and E together?
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Answer: (A) 350
Explanation
The total number of employees working from home in each company is determined by multiplying the total employee count by the ratio fraction for work from home. For company A, employees working from home equal 1500 × (2 / 5) = 600. For company C, employees working from home equal 600 × (1 / 2) = 300. For company E, employees working from home equal 450 × (1 / 3) = 150. The sum of these employees across companies A, C, and E is 600 + 300 + 150 = 1050. Taking the average over 3 companies gives 1050 / 3 = 350.
57Quantitative Aptitude
Read the following table carefully and answer the questions given below. The following table shows the number of employees working in 5 different companies and ratio of number of employees working from home and working from office.
| Company | Number of employees | Working from Home: Working from Office |
|---|---|---|
| A | 1500 | 2:3 |
| B | 800 | 7:9 |
| C | 600 | 1:1 |
| D | 1200 | 13:11 |
| E | 450 | 1:2 |
Number of employees working from office in company B is what % of number of employees working from office in company C?
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Answer: (C) 150%
Explanation
The number of employees working from office is calculated using the office ratio part out of the total ratio sum for each company. For company B, office employees equal 800 × (9 / 16) = 450. For company C, office employees equal 600 × (1 / 2) = 300. To find the percentage of company B's office employees relative to company C's office employees, calculate (450 / 300) × 100 = 150%.
58Quantitative Aptitude
Read the following table carefully and answer the questions given below. The following table shows the number of employees working in 5 different companies and ratio of number of employees working from home and working from office.
| Company | Number of employees | Working from Home: Working from Office |
|---|---|---|
| A | 1500 | 2:3 |
| B | 800 | 7:9 |
| C | 600 | 1:1 |
| D | 1200 | 13:11 |
| E | 450 | 1:2 |
Find the ratio of number of employees working from home in company A and company C together to number of employees working from office in company E and company B together?
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Answer: (E) 6:5
Explanation
Employees working from home in company A equal 1500 × (2 / 5) = 600, and in company C equal 600 × (1 / 2) = 300, giving a combined total of 600 + 300 = 900. Employees working from office in company E equal 450 × (2 / 3) = 300, and in company B equal 800 × (9 / 16) = 450, giving a combined total of 300 + 450 = 750. The ratio of work-from-home employees in companies A and C together to work-from-office employees in companies E and B together is 900 : 750 = 6 : 5.
59Quantitative Aptitude
Read the following table carefully and answer the questions given below. The following table shows the number of employees working in 5 different companies and ratio of number of employees working from home and working from office.
| Company | Number of employees | Working from Home: Working from Office |
|---|---|---|
| A | 1500 | 2:3 |
| B | 800 | 7:9 |
| C | 600 | 1:1 |
| D | 1200 | 13:11 |
| E | 450 | 1:2 |
Ratio of male and female in company E is 5:4 and ratio of male and female working from home in that company only is 7:8, then find the number of females who working from office.
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Answer: (B) 120
Explanation
The total number of female employees in company E is determined using the male to female ratio 5:4, yielding 450 × (4 / 9) = 200. The number of work-from-home employees in company E is 450 × (1 / 3) = 150, of which female work-from-home employees constitute 150 × (8 / 15) = 80. Subtracting female work-from-home employees from the total female employees gives 200 - 80 = 120 female employees working from office.
60Quantitative Aptitude
Read the following table carefully and answer the questions given below. The following table shows the number of employees working in 5 different companies and ratio of number of employees working from home and working from office.
| Company | Number of employees | Working from Home: Working from Office |
|---|---|---|
| A | 1500 | 2:3 |
| B | 800 | 7:9 |
| C | 600 | 1:1 |
| D | 1200 | 13:11 |
| E | 450 | 1:2 |
Find the difference between number of employees in company D and C together to number of employees in company A and E together.
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Answer: (D) 150
Explanation
To find the required difference, sum the employee counts for companies D and C, then for companies A and E. The combined employee strength of company D and company C is 1200 + 600 = 1800. The combined employee strength of company A and company E is 1500 + 450 = 1950. The difference between these two combined sums is 1950 - 1800 = 150.
61Quantitative Aptitude
A train leaves station A at 9:00 AM. The train reaches station B at 10:30 PM. If the distance between station A and station B is 486 km, then find the speed of the train (in meter per sec)?
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Answer: (A) 10
Explanation
The total duration between 9:00 AM and 10:30 PM is 13.5 hours. Using the speed formula, Speed = Distance / Time, the speed of the train in kilometers per hour is 486 / 13.5 = 36 km/h. Converting speed from km/h to meters per second requires multiplying by 5 / 18, giving 36 × (5 / 18) = 10 m/s.
62Quantitative Aptitude
Amar and Sameer started a partnership with investment of Rs.21000 and Rs.25000 respectively. After 6 months, Sameer withdraws 20% and Tarun joined the business with Rs.30000. If at the end of 1.5 years, profit share of Sameer is Rs.1300, then find the total profit?
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Answer: (C) Rs.3760
Explanation
With a total duration of 1.5 years (18 months), Amar's investment-time product is 21000 × 18 = 378,000. Sameer invests 25000 for 6 months and 20000 for the remaining 12 months, giving (25000 × 6) + (20000 × 12) = 390,000. Tarun joins after 6 months with 30000 for 12 months, giving 30000 × 12 = 360,000. The profit ratio among Amar, Sameer, and Tarun is 378,000 : 390,000 : 360,000 = 63 : 65 : 60, totaling 188 parts. Since Sameer's share of 65 parts equals 1300, 1 part equals 20, making the total profit 188 × 20 = 3760.
63Quantitative Aptitude
A machine can produce 12 bolts in a second. In such speed of production, it takes 15 minutes to complete a job. If another machine, having speed of production 50% more than first machine, then find the time taken to complete 8 same types of job?
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Answer: (D) 1 hour 20 min
Explanation
Time taken by the first machine to complete 1 job is 15 minutes. Since the second machine's production speed is 50% greater than the first machine, its speed ratio relative to the first machine is 1.5 : 1. Because time is inversely proportional to production speed, time taken by the second machine to complete 1 job is 15 / 1.5 = 10 minutes. Time required by the second machine to complete 8 identical jobs is 8 × 10 = 80 minutes, which equals 1 hour 20 minutes.
64Quantitative Aptitude
A seller sold 120 articles. The marked price of each article is Rs.80 and he offers a discount of 10% on each article. If he made total profit of Rs.840, then find the cost price of each article?
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Answer: (B) Rs.65
Explanation
The selling price of each article after a 10% discount on the marked price of 80 is 80 × 0.90 = 72. The total profit made on 120 articles is 840, which yields a profit of 840 / 120 = 7 per article. Subtracting the profit per article from the selling price per article gives the cost price of each article: 72 - 7 = 65.
65Quantitative Aptitude
A boat covers 160 km from point A to point B and then return to point A in 15 hours. If the speed of stream increased by 8 kmph, then boat takes 4 hours to reach point B from A. Find the speed of boat in still water?
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Answer: (A) 24 kmph
Explanation
Let boat speed in still water be u and stream speed be v. Downstream speed with stream speed increased by 8 km/h is 160 / 4 = 40 km/h, giving u + v + 8 = 40, so u + v = 32 km/h. Original downstream travel time is 160 / 32 = 5 hours, leaving an original upstream travel time of 15 - 5 = 10 hours. Original upstream speed is u - v = 160 / 10 = 16 km/h. Adding u + v = 32 and u - v = 16 yields 2u = 48, so u = 24 km/h.
66Quantitative Aptitude
A can complete a piece of work in 8 days, A and B together can complete same work in 4.8 days and A, B and C together complete the work in ⁴⁰⁄₁₁ days. Find the time taken by A and C together to complete the work?
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Answer: (A) ¹²⁰⁄₂₃ days
Explanation
Work done by A in 1 day = 1/8.
Work done by A and B together in 1 day = 1/4.8 = 5/24.
Work done by A, B, and C together in 1 day = 11/40.
Work done by C in 1 day = 11/40 - 5/24 = (33 - 25)/120 = 8/120 = 1/15.
Work done by A and C together in 1 day = 1/8 + 1/15 = (15 + 8)/120 = 23/120.
Therefore, time taken by A and C together to complete the work = 120/23 days.
67Quantitative Aptitude
Monthly income of Kunal is 25% more than Shubham. Kunal spends 25% of his income on house rent, 25% of remaining on food and rest he saves. If difference between monthly savings of Kunal and monthly income of Shubham is Rs.1900, then find monthly income of Kunal?
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Answer: (D) Rs.8000
Explanation
Let Shubham's monthly income be 100x.
Then Kunal's monthly income = 125x.
Kunal's expenditure on rent = 25% of 125x = 31.25x.
Remaining income = 125x - 31.25x = 93.75x.
Kunal's savings = 75% of 93.75x = 70.3125x.
Difference between Shubham's income and Kunal's savings = 100x - 70.3125x = 29.6875x.
Given 29.6875x = 1900, solving gives x = 64.
Monthly income of Kunal = 125 × 64 = Rs.8000.
68Quantitative Aptitude
In a solution of sugar syrup and water, quantity of sugar syrup is 30% more than quantity of water. If 20% of quantity of solution of taken out and 11.5-liter water is added into the mixture, then the ratio of water and sugar syrup becomes 15: 8. Find the initial quantity of water?
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Answer: (B) 10 liters
Explanation
Let initial water quantity be 10x liters and sugar syrup be 13x liters.
After removing 20% of the solution, remaining water = 8x liters and remaining sugar syrup = 10.4x liters.
Adding 11.5 liters of water makes the water quantity (8x + 11.5) liters.
The new ratio of water to sugar syrup is (8x + 11.5) / 10.4x = 15 / 8.
Solving 8 × (8x + 11.5) = 15 × 10.4x gives 64x + 92 = 156x, so 92x = 92 and x = 1.
Initial quantity of water = 10 × 1 = 10 liters.
69Quantitative Aptitude
The length of rectangle is 16% more than its breadth. If perimeter of square, whose side is equal to length of rectangle is 8 cm more than perimeter of rectangle, then find the area of rectangle?
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Answer: (E) 725 cm²
Explanation
Let breadth of the rectangle be B cm.
Length L = 1.16B cm.
Perimeter of square with side L = 4L = 4.64B cm.
Perimeter of rectangle = 2(L + B) = 2(2.16B) = 4.32B cm.
Difference in perimeters = 4.64B - 4.32B = 0.32B cm.
Given 0.32B = 8, we get B = 25 cm.
Length L = 1.16 × 25 = 29 cm.
Area of rectangle = L × B = 29 × 25 = 725 cm².
70Quantitative Aptitude
A man invests Rs. (P + 2000) on simple interest at 20% per annum for 4 years. If he invests Rs. (2P- 4000) in same scheme for the same time, then the interest earned would be Rs.8000 more. Find the value of P?
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Answer: (A) Rs.16000
Explanation
Interest from first investment = (P + 2000) × 20 × 4 / 100 = 0.8(P + 2000).
Interest from second investment = (2P - 4000) × 20 × 4 / 100 = 0.8(2P - 4000).
Difference in interest = 0.8(2P - 4000) - 0.8(P + 2000) = 0.8(P - 6000).
Given 0.8(P - 6000) = 8000, solving gives P - 6000 = 10000, so P = Rs.16000.
71Quantitative Aptitude
What approximate value should come in the place of question (?) mark:
(360.91)^½ - 10.99 × 3.03 = (419.82)/(?) - 21.91 × 7.02
What approximate value should come in the place of question (?) mark:
? × (16.03% of 249.99 + 24.02% of 500.02) = (27.99)² + √(256.05)
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Answer: (E) 5
Explanation
Approximating the numbers gives 16% of 250 = 40 and 24% of 500 = 120, making the sum inside the bracket 160.
On the right side, 28² = 784 and √256 = 16, giving a sum of 800.
Solving ? × 160 = 800 yields ? = 5.
77Quantitative Aptitude
What approximate value should come in the place of question (?) mark:
(225 + ?)/(25) + 25% of 442.4 = 24.4 × 5
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Answer: (C) 60
Explanation
Approximating the terms gives 25% of 442.4 = 110.6 and 24.4 × 5 = 122.
Substituting these into the equation gives (225 + ?) / 25 + 110.6 = 122.
Subtracting 110.6 from 122 gives (225 + ?) / 25 = 11.4.
Multiplying by 25 gives 225 + ? = 285, so ? = 60.
78Quantitative Aptitude
The average weight of a class of 28 students is 22 kg. If the teacher's weight is also included the average weight increases by 2 kg. Furthermore, if the average weight of the teacher's family having a wife and a daughter is 52 kg and the daughter's weight is 40% less than her mother's weight. then find the weight of teacher's wife?
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Answer: (B) 47.5 kg
Explanation
Teacher's weight = Old Average + New Total Members × Increase = 22 + 29 × 2 = 80 kg.
Total weight of the teacher's 3-member family = 3 × 52 = 156 kg.
Sum of weights of wife and daughter = 156 - 80 = 76 kg.
If wife's weight is W, daughter's weight is 0.6W.
W + 0.6W = 76, which gives 1.6W = 76, so W = 47.5 kg.
79Quantitative Aptitude
A man leaves Rs.23200 to be divided among 2 sons, 3 daughters, & 2 nephews. If each daughter receives 3 times as much as each nephew, and each son receives 3 times as much as each daughter, how much does each son receive?
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Answer: (A) Rs.7200
Explanation
Let each nephew's share be x.
Each daughter receives 3x and each son receives 3 × 3x = 9x.
Total amount = 2(9x) + 3(3x) + 2(x) = 18x + 9x + 2x = 29x.
Setting 29x = 23200 gives x = 800.
Each son receives 9x = 9 × 800 = 7200 rupees.
80Quantitative Aptitude
The total cost of painting a rectangular wall at Rs.188.5 per square ft. is Rs.7540. the length of the wall is 3 ft. more than the breadth of the wall, then breadth of wall is what percent of length of wall?
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Answer: (D) 62.5%
Explanation
Area of wall = Total Cost / Rate = 7540 / 188.5 = 40 sq ft.
Let breadth be b ft, then length is b + 3 ft.
Solving b(b + 3) = 40 gives b² + 3b - 40 = 0, which yields b = 5 ft and length = 8 ft.
Breadth as percentage of length = (5 / 8) × 100 = 62.5%.
81Quantitative Aptitude
What should come in the place of question (?) mark in following number series.
21, ?, 129, 237, 381, 561
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Answer: (D) 57
Explanation
The differences between consecutive known terms are 237 - 129 = 108, 381 - 237 = 144, and 561 - 381 = 180.
The second difference is constant at 36.
Working backward, the difference preceding 108 is 72, and the first difference is 36.
Missing term = 21 + 36 = 57.
82Quantitative Aptitude
What should come in the place of question (?) mark in following number series.
12, 13.6, 16.8, 21.6, ?, 36
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Answer: (A) 28
Explanation
The differences between consecutive terms are:
13.6 - 12 = 1.6
16.8 - 13.6 = 3.2
21.6 - 16.8 = 4.8
The sequence of differences increases by 1.6 each time.
The next difference is 4.8 + 1.6 = 6.4.
Missing term = 21.6 + 6.4 = 28.
83Quantitative Aptitude
What should come in the place of question (?) mark in following number series.
116, 342, 512, 634, ?, 766
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Answer: (B) 716
Explanation
The differences between consecutive terms are:
342 - 116 = 226 = 15² + 1
512 - 342 = 170 = 13² + 1
634 - 512 = 122 = 11² + 1
The next difference follows 9² + 1 = 82.
Missing term = 634 + 82 = 716.
84Quantitative Aptitude
What should come in the place of question (?) mark in following number series.
16, ?, 512, 1024, 1024, 512
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Answer: (C) 128
Explanation
The multipliers between consecutive terms from right to left are:
1024 to 512 is × 0.5
1024 to 1024 is × 1
512 to 1024 is × 2
Doubling the multiplier for preceding terms gives × 4 and × 8.
Missing term = 16 × 8 = 128.
85Quantitative Aptitude
What should come in the place of question (?) mark in following number series.
4, 9, 28, 113, 566, ?
What approximate value will come in the place of question (?) mark in the following given questions.
(? × 2.12) + 37.5% of 959.96 = 5600.05 ÷ 6.85
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Answer: (D) 220
Explanation
Approximating the values in the equation gives:
5600.05 ÷ 6.85 ≈ 5600 ÷ 7 = 800
37.5% of 959.96 ≈ (3 ÷ 8) × 960 = 360
2.12 ≈ 2
Substituting these into the equation:
(? × 2) + 360 = 800
? × 2 = 800 - 360
? × 2 = 440
? = 220
88Quantitative Aptitude
What approximate value will come in the place of question (?) mark in the following given questions.
39.86% of 37.5% of 3999.86 + 19.82% of 30.15% of 499.89 = ?
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Answer: (B) 630
Explanation
Approximating the percentages and values:
39.86% ≈ 40% = 2 ÷ 5
37.5% = 3 ÷ 8
3999.86 ≈ 4000
First term = (2 ÷ 5) × (3 ÷ 8) × 4000 = (3 ÷ 20) × 4000 = 600
What approximate value will come in the place of question (?) mark in the following given questions.
62.5% of 319.95 - 20% of 784.8 + 43.75% of 480 = ?
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Answer: (A) 253
Explanation
Approximating the values and using fractional equivalents:
62.5% = 5 ÷ 8
319.95 ≈ 320
(5 ÷ 8) × 320 = 200
20% of 784.8 ≈ 0.20 × 785 = 157
43.75% = 7 ÷ 16
(7 ÷ 16) × 480 = 210
Substituting into the expression:
? = 200 - 157 + 210 = 253
90Quantitative Aptitude
What approximate value will come in the place of question (?) mark in the following given questions.
25.05³ - 44.95² = (16.32 × 3.96³) + 4.06 × ?
Read the data carefully and answer the questions.
There are total 'X' number of students in a class of three different (A, B & C) sections. 40% of total students are in section A and the ratio of total students in section A to C is 10: 9. The ratio of boys to girls in section B is 7: 5 and 60% of total students in section A are boys. There are 40 girls in section C and total 170 girls in the class.
Find difference between total boys and girls in the class?
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Answer: (A) 160
Explanation
Let X be the total students in the class.
Section A students = 0.40X
Section A : Section C ratio = 10 : 9, so Section C students = 0.36X
Section B students = X - 0.40X - 0.36X = 0.24X
Boys in Section A = 60% of 0.40X = 0.24X
Girls in Section A = 40% of 0.40X = 0.16X
Boys to Girls ratio in Section B = 7 : 5
Girls in Section B = (5 ÷ 12) × 0.24X = 0.10X
Boys in Section B = (7 ÷ 12) × 0.24X = 0.14X
Girls in Section C = 40
Total Girls in class = 170
0.16X + 0.10X + 40 = 170
0.26X = 130
X = 500
Calculating total numbers:
Total students X = 500
Section A: Boys = 120, Girls = 80
Section B: Boys = 70, Girls = 50
Section C: Total = 180, Girls = 40, Boys = 140
Total Boys = 120 + 70 + 140 = 330
Total Girls = 170
Difference between total boys and total girls = 330 - 170 = 160
94Quantitative Aptitude
Read the data carefully and answer the questions.
There are total 'X' number of students in a class of three different (A, B & C) sections. 40% of total students are in section A and the ratio of total students in section A to C is 10: 9. The ratio of boys to girls in section B is 7: 5 and 60% of total students in section A are boys. There are 40 girls in section C and total 170 girls in the class.
Total boys in section B are what percent less than total girls in section A?
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Answer: (B) 12.5%
Explanation
From the class distribution:
Total students X = 500
Section A total students = 200
Girls in Section A = 40% of 200 = 80
Section B total students = 0.24 × 500 = 120
Boys in Section B = (7 ÷ 12) × 120 = 70
Percentage by which boys in Section B are less than girls in Section A:
Percentage = ((80 - 70) ÷ 80) × 100 = (10 ÷ 80) × 100 = 12.5%
95Quantitative Aptitude
Read the data carefully and answer the questions.
There are total 'X' number of students in a class of three different (A, B & C) sections. 40% of total students are in section A and the ratio of total students in section A to C is 10: 9. The ratio of boys to girls in section B is 7: 5 and 60% of total students in section A are boys. There are 40 girls in section C and total 170 girls in the class.
Find average number of boys in all three sections?
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Answer: (C) 110
Explanation
From the class distribution:
Boys in Section A = 120
Boys in Section B = 70
Boys in Section C = 140
Total boys in all three sections = 120 + 70 + 140 = 330
Average number of boys = 330 ÷ 3 = 110
96Quantitative Aptitude
Read the data carefully and answer the questions.
There are total 'X' number of students in a class of three different (A, B & C) sections. 40% of total students are in section A and the ratio of total students in section A to C is 10: 9. The ratio of boys to girls in section B is 7: 5 and 60% of total students in section A are boys. There are 40 girls in section C and total 170 girls in the class.
Find ratio of total boys in section C to total girls in all three sections?
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Answer: (E) 14:17
Explanation
Let total students in class be X.
Section A = 0.40X.
Since the ratio of Section A to Section C is 10:9, Section C = 0.36X.
Section B = X - 0.40X - 0.36X = 0.24X.
Girls in Section A = 40% of 0.40X = 0.16X.
Girls in Section B = (5/12) × 0.24X = 0.10X.
Girls in Section C = 40.
Total girls = 0.16X + 0.10X + 40 = 170.
0.26X = 130, so X = 500.
Section C total students = 0.36 × 500 = 180.
Boys in Section C = 180 - 40 = 140.
Ratio of boys in Section C to total girls in the class = 140 : 170 = 14 : 17.
97Quantitative Aptitude
Ram and Sohan entered into business investing Rs. 36000 and 72000 respectively. After six months, Sohan left and Aman entered into the business investing Rs. 32000. If at the end of the year, profit share of Aman was Rs. 5680 then find the total profit earned by all of them together.
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Answer: (B) Rs. 31240
Explanation
Investment-time product for Ram = 36000 × 12 = 432000.
Investment-time product for Sohan = 72000 × 6 = 432000.
Investment-time product for Aman = 32000 × 6 = 192000.
Ratio of profit share Ram : Sohan : Aman = 432000 : 432000 : 192000 = 9 : 9 : 4.
Total profit units = 9 + 9 + 4 = 22.
Aman share = 4 units = Rs. 5680.
Total profit = 5680 × (22 / 4) = Rs. 31240.
98Quantitative Aptitude
Atul invested 62.5% of his monthly income in scheme A offering 20% C.I. (compounded annually) for 2 years and rest amount in scheme B offering 12.5% per annum simple interest for same period. If the difference between CI and SI is Rs. 2430 after 2 years, then find the monthly income(approx.) of Atul.
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Answer: (C) Rs 13407
Explanation
Let monthly income be P.
Investment in Scheme A = 0.625P.
Effective CI rate for 2 years at 20% = 20 + 20 + (20 × 20) / 100 = 44%.
Interest from Scheme A = 0.44 × 0.625P = 0.275P.
Investment in Scheme B = 0.375P.
Effective SI rate for 2 years at 12.5% = 25%.
Interest from Scheme B = 0.25 × 0.375P = 0.09375P.
Average of 25 numbers is 96. If average of first 11 numbers is 50 while average of last 12 numbers is 80 and 12ᵗʰ and 13ᵗʰ numbers are (2x + 26) and (3x + 14) respectively. Find the 13ᵗʰ number.
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Answer: (B) 524
Explanation
Sum of all 25 numbers = 25 × 96 = 2400.
Sum of first 11 numbers = 11 × 50 = 550.
Sum of last 12 numbers = 12 × 80 = 960.
Sum of 12th and 13th numbers = 2400 - 550 - 960 = 890.
(2x + 26) + (3x + 14) = 890
5x + 40 = 890
5x = 850
x = 170.
13th number = 3(170) + 14 = 524.
100Quantitative Aptitude
Aman's friend Mayank gave him Rs. 14000 and Naman gave to Aman 30% less than what Mayank gave to Aman. If Chetan gave 50% more than Naman to Aman, then the amount received by Aman from Chetan is what percent more than that of amount which is received by Aman from Mayank?
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Answer: (D) 5%
Explanation
Amount from Mayank = Rs. 14000.
Amount from Naman = 0.70 × 14000 = Rs. 9800.
Amount from Chetan = 1.50 × 9800 = Rs. 14700.
