Given table shows the data of population in 5 different parks. Study the data carefully and answer the questions.
| Parks | Total Population | Female population |
|---|---|---|
| A | 400 | 150 |
| B | 500 | 200 |
| C | 700 | 350 |
| D | 800 | 450 |
| E | 900 | 500 |
(Total population = Male population + Female population)
If 20% of total population did not visit on a particular day in park A of which male population was 60% then what percent of total population in park B is male population who visited in park A?
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Answer: (B) 40.4%
Explanation
Park A total population = 400, female population = 150, male population = 250.
20% of park A population did not visit = 0.20 × 400 = 80.
Males who did not visit park A = 60% of 80 = 48.
Males who visited park A = 250 - 48 = 202.
Total population in park B = 500.
Required percentage = (202 / 500) × 100 = 40.4%.
2Quantitative Aptitude
Given table shows the data of population in 5 different parks. Study the data carefully and answer the questions.
| Parks | Total Population | Female population |
|---|---|---|
| A | 400 | 150 |
| B | 500 | 200 |
| C | 700 | 350 |
| D | 800 | 450 |
| E | 900 | 500 |
(Total population = Male population + Female population)
What is average of male population in park B, C and D?
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Answer: (D) 333.33
Explanation
Male population = Total population - Female population.
Park B male population = 500 - 200 = 300.
Park C male population = 700 - 350 = 350.
Park D male population = 800 - 450 = 350.
Average male population = (300 + 350 + 350) / 3 = 1000 / 3 = 333.33.
3Quantitative Aptitude
Given table shows the data of population in 5 different parks. Study the data carefully and answer the questions.
| Parks | Total Population | Female population |
|---|---|---|
| A | 400 | 150 |
| B | 500 | 200 |
| C | 700 | 350 |
| D | 800 | 450 |
| E | 900 | 500 |
(Total population = Male population + Female population)
By what percent female population in park D is more or less than the male population in park E?
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Answer: (E) 12.5%
Explanation
Female population in park D = 450.
Male population in park E = 900 - 500 = 400.
Percentage difference = ((450 - 400) / 400) × 100 = (50 / 400) × 100 = 12.5%.
4Quantitative Aptitude
Given table shows the data of population in 5 different parks. Study the data carefully and answer the questions.
| Parks | Total Population | Female population |
|---|---|---|
| A | 400 | 150 |
| B | 500 | 200 |
| C | 700 | 350 |
| D | 800 | 450 |
| E | 900 | 500 |
(Total population = Male population + Female population)
What is ratio of male population in park A & D together to female population in park B & E together?
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Answer: (A) 6:7
Explanation
Male population in park A = 400 - 150 = 250.
Male population in park D = 800 - 450 = 350.
Total males in A and D = 250 + 350 = 600.
Female population in park B = 200.
Female population in park E = 500.
Total females in B and E = 200 + 500 = 700.
Required ratio = 600 : 700 = 6 : 7.
5Quantitative Aptitude
Given table shows the data of population in 5 different parks. Study the data carefully and answer the questions.
| Parks | Total Population | Female population |
|---|---|---|
| A | 400 | 150 |
| B | 500 | 200 |
| C | 700 | 350 |
| D | 800 | 450 |
| E | 900 | 500 |
(Total population = Male population + Female population)
If 30 females from each park are above 80 years age then find the average of no. of females who are below or equal to the age of 80 years from all the parks.
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Answer: (C) 300
Explanation
Total female population across all parks = 150 + 200 + 350 + 450 + 500 = 1650.
Average female population per park = 1650 / 5 = 330.
Since 30 females in each park are above 80 years old:
Average of females ≤ 80 years = 330 - 30 = 300.
6Quantitative Aptitude
The ratio of ages of A and B 4 years ago was 5 : 3. The sum of present ages of A, B and C is 80 years. If present age of C is equal to sum of present ages of A and B. find the present age of A.
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Answer: (B) 24 years
Explanation
Given C = A + B and A + B + C = 80, substituting C yields 2(A + B) = 80, so A + B = 40. Four years ago, the sum of their ages was (A - 4) + (B - 4) = 40 - 8 = 32 years. Since the ratio of their ages 4 years ago was 5 : 3, A's age 4 years ago was (5 / 8) × 32 = 20 years. Adding 4 years gives A's present age as 20 + 4 = 24 years.
7Quantitative Aptitude
The ratio of speed of boat in still water to speed of stream is 8:1 . It takes 4 hours by boat to cover 54 km in downstream & 42 km in upstream. Find the downstream speed of boat.
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Answer: (D) 27 kmph
Explanation
Let the speed of the boat in still water be 8x and the speed of the stream be x. The downstream speed is 9x and the upstream speed is 7x. The total time taken is (54 / 9x) + (42 / 7x) = 4 hours. Simplifying gives (6 / x) + (6 / x) = 4, so 12 / x = 4, which yields x = 3. Therefore, the downstream speed of the boat is 9 × 3 = 27 kmph.
8Quantitative Aptitude
Manoj gave 60% of his salary to his wife and invested rest amount in mutual funds. His wife spends 30% amount on grocery and 20% on rent. From remaining amount, she purchased gold worth Rs. 18000. Find salary of Manoj.
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Answer: (A) Rs 60000
Explanation
Let Manoj's total salary be S. His wife receives 60% of S, which is 0.60S. She spends 30% on grocery and 20% on rent, totaling 50% of her received amount, leaving 50% of 0.60S = 0.30S. Given 0.30S = ₹18000, Manoj's salary S = 18000 / 0.30 = ₹60000.
9Quantitative Aptitude
the length & breadth of a rectangle is in ratio 4:7 . If perimeter is 88 cm . find area of rectangle.
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Answer: (C) 448 cm²
Explanation
Let the length be 4x and breadth be 7x. The perimeter is 2(4x + 7x) = 22x = 88 cm, which gives x = 4 cm. Thus, length = 16 cm and breadth = 28 cm. The area of the rectangle is length × breadth = 16 × 28 = 448 cm².
10Quantitative Aptitude
the radius of a circle is 14 cm . what is area of another circle having radius 1.5 times the actual circle
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Answer: (B) 1386 cm²
Explanation
The radius of the original circle is 14 cm, so the new radius is 1.5 × 14 = 21 cm. The area of the new circle is πr² = (22 / 7) × 21² = (22 / 7) × 441 = 1386 cm².
11Quantitative Aptitude
In the following two equations questions numbered (I) and (II) are given. You have to solve both equations and Give answer
I. x² - 7x + 12 = 0
II. y² - 8y + 12 = 0
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Answer: (E) If x = y or no relation can be established
Explanation
Solving equation I, x² - 7x + 12 = 0 gives roots x = 3 and x = 4. Solving equation II, y² - 8y + 12 = 0 gives roots y = 2 and y = 6. Comparing the roots, 3 > 2 but 3 < 6, so no relationship can be established between x and y.
12Quantitative Aptitude
In the following two equations questions numbered (I) and (II) are given. You have to solve both equations and Give answer
1. 2x² + x - 28 = 0
II. 2y² - 23y + 56 = 0
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Answer: (D) If y ≥ x
Explanation
Solving equation I, 2x² + x - 28 = 0 gives roots x = 3.5 and x = -4. Solving equation II, 2y² - 23y + 56 = 0 gives roots y = 3.5 and y = 8. Comparing every value of x with y gives -4 < 3.5, -4 < 8, 3.5 = 3.5, and 3.5 < 8, which yields y ≥ x.
13Quantitative Aptitude
In the following two equations questions numbered (I) and (II) are given. You have to solve both equations and Give answer
1. 2x² - 7x - 60 = 0
II. 3y² + 13y + 4 = 0
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Answer: (E) If x = y or no relation can be established
Explanation
Solving equation I, 2x² - 7x - 60 = 0 gives roots x = 7.5 and x = -4. Solving equation II, 3y² + 13y + 4 = 0 gives roots y = -4 and y = -1/3. Comparing x = 7.5 with y = -1/3 gives x > y, but comparing x = -4 with y = -1/3 gives x < y, so no relationship can be established between x and y.
14Quantitative Aptitude
In the following two equations questions numbered (I) and (II) are given. You have to solve both equations and Give answer
I.x² - 17x - 84 = 0
II.y² + 4y - 117 = 0
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Answer: (E) If x = y or no relation can be established
Explanation
Solving equation I, x² - 17x - 84 = 0 gives roots x = 21 and x = -4. Solving equation II, y² + 4y - 117 = 0 gives roots y = 9 and y = -13. Comparing x = 21 with both values of y gives x > y, but comparing x = -4 with y = 9 gives x < y, so no relationship can be established between x and y.
15Quantitative Aptitude
In the following two equations questions numbered (I) and (II) are given. You have to solve both equations and Give answer
I.x² = 81
II.(y - 9)² = 0
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Answer: (D) If y≥ x
Explanation
Solving equation I, x² = 81 gives roots x = 9 and x = -9. Solving equation II, (y - 9)² = 0 gives root y = 9. Comparing the roots, when x = 9, x = y, and when x = -9, x < y, which yields y ≥ x.
16Quantitative Aptitude
X liters of milk is taken out and replaced with water from a container having 240 liters milk. Now, 20% of the mixture is taken out and replaced with water. In final mixture, the difference in quantity of milk & water is 128 liters. Find X.
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Answer: (B) 10
Explanation
After replacing X liters of milk with water, the quantity of milk remaining is 240 - X liters. When 20% of the mixture is removed and replaced with water, 80% of the milk remains. Thus, final milk quantity = 0.8 × (240 - X) = 192 - 0.8X liters. Final water quantity = 240 - (192 - 0.8X) = 48 + 0.8X liters. The difference between milk and water is given as 128 liters: (192 - 0.8X) - (48 + 0.8X) = 128, which simplifies to 144 - 1.6X = 128, giving 1.6X = 16, hence X = 10.
17Quantitative Aptitude
A can do a work in 36 days while B can do the same in 48 days. If A work for 'x' days while B work for 'x+2' days then one-third of the work is complete. Find the value of x.
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Answer: (C) 6
Explanation
Using the work formula where Work = Rate × Time, A's rate is 1/36 per day and B's rate is 1/48 per day. Setting up the total work equation: x/36 + (x + 2)/48 = 1/3. Multiplying the entire equation by 144 gives 4x + 3(x + 2) = 48, which simplifies to 7x + 6 = 48, so 7x = 42, giving x = 6.
18Quantitative Aptitude
A shopkeeper marked the price of an article by 40% above cost price and gave discount of Rs. 224. On the final amount, he charged 10% tax. In the whole transaction, he earned Rs. 158.6. Find cost price of the article.
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Answer: (A) Rs 750
Explanation
Let C be the cost price. Marked price = 1.40C. Price after a discount of Rs. 224 = 1.40C - 224. Adding 10% tax makes the total amount collected 1.10 × (1.40C - 224) = 1.54C - 246.4. The total profit earned over cost price is 158.6, so (1.54C - 246.4) - C = 158.6. This gives 0.54C = 405, yielding C = 750.
19Quantitative Aptitude
Two friends Pinki and Rinki entered into a partnership by investing an amount of Rs 6000 and Rs 9000 respectively and ratio of their period of investment is 2: 3. Find the profit share of Pinki if profit share of Rinki is Rs 45,000.
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Answer: (B) Rs 20,000
Explanation
Profit share ratio equals the product of investment and time period. Ratio of profit = (6000 × 2) : (9000 × 3) = 12000 : 27000 = 4 : 9. Since Rinki's profit share of 9 units equals Rs. 45,000, 1 unit equals Rs. 5,000. Pinki's profit share is 4 units, which equals 4 × 5000 = Rs. 20,000.
20Quantitative Aptitude
Time taken to travel a certain distance of 'x' km at speed of 40 km/h is 2 hours more than the time taken to travel (x + 20) km at speed of 60 km/h . Find the time taken to travel (x + 40) km at the speed of 40 km/h ?
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Answer: (C) 8 hours
Explanation
Using Time = Distance / Speed, the given time difference equation is x/40 - (x + 20)/60 = 2. Multiplying by 120 gives 3x - 2(x + 20) = 240, which simplifies to x - 40 = 240, so x = 280 km. The required time to travel (x + 40) km at 40 km/h is (280 + 40) / 40 = 320 / 40 = 8 hours.
21Quantitative Aptitude
What approximate value should come in place of question mark (?) in the following question? Note: (you are not expected to calculate the exact value.)
111.01 + 41.23 + (4.96)² + (2.09)² = ?
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Answer: (C) 181
Explanation
Approximating each term to its nearest integer gives 111 + 41 + 5² + 2² = 111 + 41 + 25 + 4 = 181.
22Quantitative Aptitude
What approximate value should come in place of question mark (?) in the following question? Note: (you are not expected to calculate the exact value.)
109.07√? - (61)/(21.02) × ? = 47.96√?
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Answer: (A) 441
Explanation
Approximating the coefficients: 109 √? - (61 / 21) × ? = 48 √?. Rearranging terms gives (109 - 48) √? = (61 / 21) × ?, which simplifies to 61 √? = (61 / 21) × ?. Dividing both sides by 61 √? gives 1 = ? / (21 √?), which yields √? = 21, so ? = 21² = 441.
23Quantitative Aptitude
What approximate value should come in place of question mark (?) in the following question? Note:- (you are not expected to calculate the exact value.)
1332.89 + 171.928 + 17.01 + ?² = 1690.87
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Answer: (D) 13
Explanation
Rounding off to the nearest integers gives 1333 + 172 + 17 + ?² = 1691. Summing the constant values gives 1522 + ?² = 1691. Subtracting 1522 from 1691 gives ?² = 169, which yields ? = 13.
24Quantitative Aptitude
What approximate value should come in place of question mark (?) in the following question? Note:- (you are not expected to calculate the exact value.)
150.09% of 20 + (322.9)/(17.02) +√7 = (8.96)²
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Answer: (B) 1024
Explanation
Approximating the terms in the given equation gives 1.50 × 20 + 323 / 17 + √? = 9². Calculating each part yields 30 + 19 + √? = 81, which simplifies to 49 + √? = 81. Thus, √? = 32, so ? = 32² = 1024.
25Quantitative Aptitude
What approximate value should come in place of question mark (?) in the following question? Note:- (you are not expected to calculate the exact value.)
56.08% of 149.92 + √(28.02 × 6.98) - 11⅑% of 998.9 = ?
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Answer: (B) -13
Explanation
Approximating the expression using percentage and square root approximations: 56% of 150 + √(28 × 7) - 11.11% of 999 = ?. This gives 0.56 × 150 + √196 - (1/9) × 999 = 84 + 14 - 111 = -13.
26Quantitative Aptitude
Read the given information carefully and answer the following questions.
There are four hostels i.e. A, B, C and D.
In hostel A: Number of boys are 120 and number of girls are 30% more than that of number of girls in hostel B.
In hostel B: Number of boys are double than that of number of girls.
In hostel C: Number of boys are 100 more than that of boys in hostel A and total boys and girls in C is 1000.
In hostel D: Number of boys is 182 more than the number of boys in hostel A. Average of number of girls of hostel A and that of hostel D is 223. Number of boys in hostel D is 98 less that of boys in hostel B.
Difference between number of boys and girls in hostel D is what percent of difference between boys and girls in hostel B?
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Answer: (B) 58%
Explanation
In hostel B, boys = 400 and girls = 200, giving a difference of 400 - 200 = 200. In hostel D, boys = 302 and girls = 186, giving a difference of 302 - 186 = 116. The required percentage is (116 / 200) × 100 = 58%.
27Quantitative Aptitude
Read the given information carefully and answer the following questions.
There are four hostels i.e. A, B, C and D.
In hostel A: Number of boys are 120 and number of girls are 30% more than that of number of girls in hostel B.
In hostel B: Number of boys are double than that of number of girls.
In hostel C: Number of boys are 100 more than that of boys in hostel A and total boys and girls in C is 1000.
In hostel D: Number of boys is 182 more than the number of boys in hostel A. Average of number of girls of hostel A and that of hostel D is 223. Number of boys in hostel D is 98 less that of boys in hostel B.
Find difference between total number of boys and girls together in hostel A and that in hostel D?
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Answer: (A) 108
Explanation
The total number of students in hostel A is 120 boys + 260 girls = 380. The total number of students in hostel D is 302 boys + 186 girls = 488. The difference between total students in hostel D and hostel A is 488 - 380 = 108.
28Quantitative Aptitude
Read the given information carefully and answer the following questions.
There are four hostels i.e. A, B, C and D.
In hostel A: Number of boys are 120 and number of girls are 30% more than that of number of girls in hostel B.
In hostel B: Number of boys are double than that of number of girls.
In hostel C: Number of boys are 100 more than that of boys in hostel A and total boys and girls in C is 1000.
In hostel D: Number of boys is 182 more than the number of boys in hostel A. Average of number of girls of hostel A and that of hostel D is 223. Number of boys in hostel D is 98 less that of boys in hostel B.
Find ratio of total number of boys and girls together in hostel B to that of in hostel C?
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Answer: (A) 3:5
Explanation
Hostel B has 400 boys and 200 girls, giving a total of 600 students. Hostel C has a total of 1000 students. The ratio of total students in hostel B to hostel C is 600 : 1000 = 3 : 5.
29Quantitative Aptitude
Read the given information carefully and answer the following questions.
There are four hostels i.e. A, B, C and D.
In hostel A: Number of boys are 120 and number of girls are 30% more than that of number of girls in hostel B.
In hostel B: Number of boys are double than that of number of girls.
In hostel C: Number of boys are 100 more than that of boys in hostel A and total boys and girls in C is 1000.
In hostel D: Number of boys is 182 more than the number of boys in hostel A. Average of number of girls of hostel A and that of hostel D is 223. Number of boys in hostel D is 98 less that of boys in hostel B.
If 20 boys from each hostel left their hostel then find the average of remaining number of boys in all the hostels?
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Answer: (D) 240.5
Explanation
The initial number of boys in hostels A, B, C, and D are 120, 400, 220, and 302 respectively, totaling 1042. When 20 boys leave from each of the 4 hostels, a total of 80 boys leave. The remaining average number of boys is (1042 - 80) / 4 = 962 / 4 = 240.5.
30Quantitative Aptitude
Read the given information carefully and answer the following questions.
There are four hostels i.e. A, B, C and D.
In hostel A: Number of boys are 120 and number of girls are 30% more than that of number of girls in hostel B.
In hostel B: Number of boys are double than that of number of girls.
In hostel C: Number of boys are 100 more than that of boys in hostel A and total boys and girls in C is 1000.
In hostel D: Number of boys is 182 more than the number of boys in hostel A. Average of number of girls of hostel A and that of hostel D is 223. Number of boys in hostel D is 98 less that of boys in hostel B.
Total number of boys in hostel A and that of girls in hostel C is what percent more than number of boys in hostel B?
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Answer: (B) 125%
Explanation
The number of boys in hostel A is 120 and girls in hostel C is 780, giving a combined sum of 120 + 780 = 900. Comparing this to the 400 boys in hostel B, the percentage excess is [(900 - 400) / 400] × 100 = (500 / 400) × 100 = 125%.
31Quantitative Aptitude
Study the table given below carefully and answer the following questions.
Table shows the total (male + female) students in 3 different schools (A, B & C) and total (male + female) students in 10th class in these schools and total female students in 10th class in these schools.
School Total (male + female) students Total (male + female) students in 10th Total female students in 10th
A 450 60 35
B 360 80 30
C 240 64 16
Total male students in 10th in schools - A & B together are what percent of total (male + female) students in school - C?
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Answer: (B) 31.25%
Explanation
Male 10th students in school A = 60 - 35 = 25, and in school B = 80 - 30 = 50, totaling 75. The required percentage relative to the total 240 students in school C is (75 / 240) × 100 = 31.25%.
32Quantitative Aptitude
Study the table given below carefully and answer the following questions.
Table shows the total (male + female) students in 3 different schools (A, B & C) and total (male + female) students in 10th class in these schools and total female students in 10th class in these schools.
School Total (male + female) students Total (male + female) students in 10th Total female students in 10th
A 450 60 35
B 360 80 30
C 240 64 16
Find average student in 10th in schools - A, B & C.
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Answer: (D) 68
Explanation
Total students in 10th class in schools A, B, and C = 60 + 80 + 64 = 204. The average number of 10th class students per school is 204 / 3 = 68.
33Quantitative Aptitude
Study the table given below carefully and answer the following questions. Table shows the total (male + female) students in 3 different schools (A, B & C) and total (male + female) students in 10th class in these schools and total female students in 10th class in these schools.
School Total (male + female) students Total (male + female) students in 10th Total female students in 10th
A 450 60 35
B 360 80 30
C 240 64 16
If ratio of total male to total female students in school - A & B is 5 : 4 and 5 : 7 respectively, then find ratio of total male students in school - A & B together to total female students in 10th in schools - A & B together.
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Answer: (A) 80:13
Explanation
Total male students in school A = 450 × (5 / 9) = 250, and total male students in school B = 360 × (5 / 12) = 150, giving 400 total male students. Total female students in 10th in schools A and B together = 35 + 30 = 65. The ratio is 400 : 65 = 80 : 13.
34Quantitative Aptitude
Study the table given below carefully and answer the following questions.
Table shows the total (male + female) students in 3 different schools (A, B & C) and total (male + female) students in 10th class in these schools and total female students in 10th class in these schools.
School Total (male + female) students Total (male + female) students in 10th Total female students in 10th
A 450 60 35
B 360 80 30
C 240 64 16
Total male students in 10th in schools - B & C together are what percent more or less than total (male + female) students in 10th in school - A & B together?
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Answer: (A) 30%
Explanation
Male 10th students in schools B and C together = (80 - 30) + (64 - 16) = 50 + 48 = 98. Total 10th class students in schools A and B together = 60 + 80 = 140. The percentage difference is ((140 - 98) / 140) × 100 = (42 / 140) × 100 = 30%.
35Quantitative Aptitude
Study the table given below carefully and answer the following questions.
Table shows the total (male + female) students in 3 different schools (A, B & C) and total (male + female) students in 10th class in these schools and total female students in 10th class in these schools.
School Total (male + female) students Total (male + female) students in 10th Total female students in 10th
A 450 60 35
B 360 80 30
C 240 64 16
Male students in 10th in schools - A, B & C together are how much more or less than female students in 10th in schools - A, B & C together?
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Answer: (E) 42
Explanation
Total male students in 10th class across schools A, B, and C = (60 - 35) + (80 - 30) + (64 - 16) = 25 + 50 + 48 = 123. Total female 10th students = 35 + 30 + 16 = 81. The difference between them is 123 - 81 = 42.
36Quantitative Aptitude
What will come in place of (?) in the following questions?
100, 97, 92, 83, 68, ?
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Answer: (B) 45
Explanation
The differences between consecutive terms are -3, -5, -9, and -15. The second differences are 2, 4, and 6. Following this arithmetic pattern, the next second difference is 8, making the next term difference -15 - 8 = -23. Thus, the missing number is 68 - 23 = 45.
37Quantitative Aptitude
What will come in place of (?) in the following questions?
12, 129, 233, 324, ?, 467
The differences decrease by 13 at each step. The next difference must be 13 × 6 = 78.
Adding this difference to 324 gives 324 + 78 = 402.
Checking the next step: 402 + (13 × 5) = 402 + 65 = 467.
38Quantitative Aptitude
What will come in place of (?) in the following questions?
0.75, ?, 3, 6, 12, 24
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Answer: (A) 1.5
Explanation
The pattern in the series is a geometric progression with a common ratio of 2:
24 ÷ 12 = 2
12 ÷ 6 = 2
6 ÷ 3 = 2
Multiplying the first term by 2 gives 0.75 × 2 = 1.5.
Checking the next term: 1.5 × 2 = 3.
39Quantitative Aptitude
What will come in place of (?) in the following questions?
8, 7, 4, 6, 12, 30
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Answer: (D) 4
Explanation
The given series follows a pattern where each term is obtained by multiplying the preceding term by numbers increasing in steps of 0.5, starting from 0.5:
Applying this pattern, multiplying the first term 8 by 0.5 yields 4 as the missing number.
40Quantitative Aptitude
What will come in place of (?) in the following questions?
1, 2, 5, 16, 65, 7
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Answer: (C) 326
Explanation
The pattern for the sequence is:
1 × 1 + 1 = 2
2 × 2 + 1 = 5
5 × 3 + 1 = 16
16 × 4 + 1 = 65
To find the missing term, multiply the last term by 5 and add 1:
65 × 5 + 1 = 325 + 1 = 326.
41Quantitative Aptitude
A and B invested an amount of Rs. 5000 & Rs. 4000 respectively in two schemes offering simple interest at the rate of 10% p.a. & 12% p.a. respectively. If A invested for 2 years while B for 3 years, then find the difference between interest received by A and B.
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Answer: (D) Rs 440
Explanation
Using the simple interest formula SI = (P × R × T) / 100:
For A:
P = ₹5000, R = 10%, T = 2 years
SI = (5000 × 10 × 2) / 100 = ₹1000
For B:
P = ₹4000, R = 12%, T = 3 years
SI = (4000 × 12 × 3) / 100 = ₹1440
Difference in interest = 1440 - 1000 = ₹440.
42Quantitative Aptitude
A boat covers a distance of 50 kms in 2 hours in downstream. If speed of stream is 5 kmph, then find time taken by boat to cover 45 km in upstream and 50 km in downstream.
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Answer: (A) 5 hours
Explanation
Downstream speed = 50 / 2 = 25 km/h.
Speed of stream = 5 km/h.
Speed of boat in still water = 25 - 5 = 20 km/h.
Upstream speed = 20 - 5 = 15 km/h.
Time taken for 45 km upstream = 45 / 15 = 3 hours.
Time taken for 50 km downstream = 50 / 25 = 2 hours.
Total time = 3 + 2 = 5 hours.
43Quantitative Aptitude
In a firm, 60% of total employees are males and 40% of total female employees earn more than Rs. 35000. If 60% of total employees earn more than Rs. 35000, then what percent of male employees earn more than Rs 35000?
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Answer: (E) 73.33%
Explanation
Let the total number of employees be 100.
Male employees = 60
Female employees = 40
Total employees earning > ₹35000 = 60% of 100 = 60.
Female employees earning > ₹35000 = 40% of 40 = 16.
Male employees earning > ₹35000 = 60 - 16 = 44.
Percentage of male employees earning > ₹35000 = (44 / 60) × 100 = 73.33%.
44Quantitative Aptitude
The diagonal of a square is equal to height of cylinder of volume 500π cm³ . If radius of cylinder is 10cm , then find perimeter of square.
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Answer: (C) 10√2 cm
Explanation
Cylinder volume formula V = π × r² × h.
500π = π × 10² × h
h = 5 cm.
Diagonal of square = height = 5 cm.
Side of square a = 5 / √2 cm.
Perimeter of square = 4 × a = 4 × (5 / √2) = 20 / √2 = 10√2 cm.
45Quantitative Aptitude
A and B are inlet pipes which take 5 hours and 9 hours respectively alone to fill the tank. C is outlet pipe which alone can empty the tank in 15 hours. If all three pipes are opened simultaneously, then in what time the tank will be filled completely?
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Answer: (B) ⁴⁄₁₁ ¹⁄₁₁ hours
Explanation
Rate of pipe A = 1/5 per hour
Rate of pipe B = 1/9 per hour
Rate of pipe C = -1/15 per hour
Net filling rate per hour = 1/5 + 1/9 - 1/15
LCM of 5, 9, 15 is 45.
Net rate = (9 + 5 - 3) / 45 = 11/45 per hour.
Time required to fill the tank = 45/11 hours = 4 1/11 hours.
46Quantitative Aptitude
Vessel A contains milk & water in ratio 3 : 2 while vessel B contains milk & water in ratio 5 : 4. If these two mixtures are mixed in ratio 3 : 4. Find ratio of milk to water in final mixture.
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Answer: (A) 181 : 134
Explanation
In vessel A, milk fraction = 3/5, water fraction = 2/5.
In vessel B, milk fraction = 5/9, water fraction = 4/9.
For a 3 : 4 mixture ratio:
Total milk = 3 × (3/5) + 4 × (5/9) = 9/5 + 20/9 = 181/45.
Total water = 3 × (2/5) + 4 × (4/9) = 6/5 + 16/9 = 134/45.
Ratio of milk to water in final mixture = (181/45) : (134/45) = 181 : 134.
47Quantitative Aptitude
A work can be completed by A alone and B alone in 10 days and 12 days respectively. With the help of C, they complete the work in 5 days. In how many days can C alone complete the work?
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Answer: (C) 60 days
Explanation
Work rate formula: Rate of C = Rate of (A + B + C) - Rate of A - Rate of B.
Substituting the given time values into the work rate equation:
1/C = 1/5 - (1/10 + 1/12)
1/C = 1/5 - (6/60 + 5/60)
1/C = 1/5 - 11/60
1/C = 12/60 - 11/60 = 1/60
Taking the reciprocal yields C = 60 days.
48Quantitative Aptitude
A shopkeeper gives a discount of 25% on marked price of article A and CP of article B is 20% more than SP of article A. if shopkeeper sold article B at 15% profit and SP of article B is Rs 1140 more than SP of article A, then find MP of article A.
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Answer: (A) Rs 4000
Explanation
Let marked price of article A be MP.
Selling price of article A = MP × (1 - 0.25) = 0.75 × MP.
Cost price of article B = 1.20 × 0.75 × MP = 0.90 × MP.
Selling price of article B = 1.15 × 0.90 × MP = 1.035 × MP.
A bag contains 5 red, 4 green and 6 yellow balls. If 3 balls are drawn at random then find the probability of obtaining no yellow ball.
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Answer: (B) 12/65
Explanation
Total balls in the bag = 5 + 4 + 6 = 15 balls.
Number of non-yellow balls = 5 + 4 = 9 balls.
The probability of obtaining no yellow ball when drawing 3 balls is the number of ways to select 3 non-yellow balls divided by total ways to select 3 balls from 15.
Manoj invested for 2 more months than Saroj and Saroj invested Rs 8000 for 9 months. Find amount invested by Manoj, if profit sharing ratio of Manoj to Saroj is 11 : 8.
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Answer: (E) Rs 9000
Explanation
Profit sharing ratio equals the ratio of the products of investment amount and investment time period.
Saroj's investment product = 8000 × 9 = 72000.
Manoj invested for 2 more months than Saroj, so Manoj's investment period = 9 + 2 = 11 months.
Let Manoj's investment amount be M.
(M × 11) / 72000 = 11 / 8
11M / 72000 = 11 / 8
M / 72000 = 1 / 8
M = 72000 / 8 = 9000.
