In each question two equations number (I) and (II) are given. You should solve both the equations and mark appropriate answer.
I. x² - 10x + 24 = 0 II. 5y² - 16y + 12 = 0
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Answer: (B) If x > y
Explanation
Solving equation I, x² - 10x + 24 = 0 gives (x - 6)(x - 4) = 0, so x = 6 or x = 4.
Solving equation II, 5y² - 16y + 12 = 0 gives (5y - 6)(y - 2) = 0, so y = 1.2 or y = 2.
Comparing all values, 6 > 2, 6 > 1.2, 4 > 2, and 4 > 1.2. Hence, x > y.
2Quantitative Aptitude
In each question two equations number (I) and (II) are given. You should solve both the equations and mark appropriate answer.
I. 2x² - 9x + 10 = 0 II. y² - 12y + 35 = 0
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Answer: (C) If x< y
Explanation
Solving equation I, 2x² - 9x + 10 = 0 gives (2x - 5)(x - 2) = 0, so x = 2.5 or x = 2.
Solving equation II, y² - 12y + 35 = 0 gives (y - 5)(y - 7) = 0, so y = 5 or y = 7.
Comparing all values, 2 < 5, 2 < 7, 2.5 < 5, and 2.5 < 7. Hence, x < y.
3Quantitative Aptitude
In each question two equations number (I) and (II) are given. You should solve both the equations and mark appropriate answer.
I. 6x² - 7x + 2 = 0 II. 10y² - 11y + 3 = 0
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Answer: (A) If x = y or no relation can be established
Explanation
Solving equation I, 6x² - 7x + 2 = 0 gives (3x - 2)(2x - 1) = 0, so x = 2/3 (≈ 0.67) or x = 1/2 (0.5).
Solving equation II, 10y² - 11y + 3 = 0 gives (5y - 3)(2y - 1) = 0, so y = 3/5 (0.6) or y = 1/2 (0.5).
When x = 2/3 and y = 3/5, x > y. However, when x = 1/2 and y = 3/5, x < y. Because both x > y and x < y occur depending on the roots chosen, no relation can be established.
4Quantitative Aptitude
In each question two equations number (I) and (II) are given. You should solve both the equations and mark appropriate answer.
I. 2x² +19x + 42 = 0 II. 2y² +7y + 5 = 0
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Answer: (C) If x< y
Explanation
Solving equation I, 2x² + 19x + 42 = 0 gives (2x + 7)(x + 6) = 0, so x = -3.5 or x = -6.
Solving equation II, 2y² + 7y + 5 = 0 gives (2y + 5)(y + 1) = 0, so y = -2.5 or y = -1.
Comparing all values, -3.5 < -2.5, -3.5 < -1, -6 < -2.5, and -6 < -1. Hence, x < y.
5Quantitative Aptitude
What approximate value come in the place of 'x' in the following questions.
(779.23)/(13) +4.98 + 23 = x% of 60.03
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Answer: (D) 147
Explanation
Approximating the terms in the given expression:
779.23 / 13 ≈ 780 / 13 = 60
4.98 ≈ 5
23 = 23
Sum of left side = 60 + 5 + 23 = 88.
Right side ≈ x% of 60 = 0.6x.
Solving 0.6x = 88 gives x = 88 / 0.6 = 880 / 6 = 146.67 ≈ 147.
6Quantitative Aptitude
What approximate value come in the place of 'x' in the following questions.
39.98% of 249.99 + √(196.04) = x² +55.93
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Answer: (D) 8
Explanation
Approximating the numbers:
39.98% of 249.99 ≈ 40% of 250 = 100
√(196.04) ≈ √196 = 14
55.93 ≈ 56
Substituting these into the equation:
100 + 14 = x² + 56
114 = x² + 56
x² = 58
x = √58 ≈ 7.61, which approximates to 8.
7Quantitative Aptitude
What approximate value come in the place of 'x' in the following questions.
20.99² - 24.97% of 220.02 - ? = 131.13
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Answer: (C) 255
Explanation
Approximating the given values:
20.99² ≈ 21² = 441
24.97% of 220.02 ≈ 25% of 220 = 55
131.13 ≈ 131
Setting up the equation:
441 - 55 - x = 131
386 - x = 131
x = 386 - 131 = 255
8Quantitative Aptitude
What approximate value come in the place of 'x' in the following questions.
24.21²+12.16=?x(49.99% of55.98)
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Answer: (C) 21
Explanation
Approximating the terms:
24.21² ≈ 24² = 576
12.16 ≈ 12
49.99% of 55.98 ≈ 50% of 56 = 28
Substituting into the equation:
576 + 12 = x × 28
588 = 28x
x = 588 / 28 = 21
9Quantitative Aptitude
The table given below shows the total number of red and blue balls in five different bags (A, B, C, D & E). It also shows the difference between red and blue balls and percentage of red balls with respect to total balls.
| Total bags | Difference between red and blue balls | Percentage of red balls with respect to total balls |
| ---------- | ------------------------------------: | --------------------------------------------------: |
| A | 60 | 62.5% |
| B | 198 | 66.5% |
| C | 188 | 70% |
| D | 76 | 60% |
| E | 330 | 72% |
The number of Red ball in A and E together is what percentage of number of blue balls in A and E together.
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Answer: (A) 230
Explanation
For bag A, percentage difference between red and blue balls is 2 × 62.5% - 100% = 25%.
Total balls in A = 60 / 0.25 = 240.
Red balls in A = 62.5% of 240 = 150, Blue balls in A = 90.
For bag E, percentage difference is 2 × 72% - 100% = 44%.
Total balls in E = 330 / 0.44 = 750.
Red balls in E = 72% of 750 = 540, Blue balls in E = 210.
Total red balls in A and E = 150 + 540 = 690.
Total blue balls in A and E = 90 + 210 = 300.
Percentage = (690 / 300) × 100 = 230%.
10Quantitative Aptitude
The table given below shows the total number of red and blue balls in five different bags (A, B, C, D & E). It also shows the difference between red and blue balls and percentage of red balls with respect to total balls.
| Total bags | Difference between red and blue balls | Percentage of red balls with respect to total balls |
| ---------- | ------------------------------------: | --------------------------------------------------: |
| A | 60 | 62.5% |
| B | 198 | 66.5% |
| C | 188 | 70% |
| D | 76 | 60% |
| E | 330 | 72% |
The sum of number of blue balls in B and number of red balls in C are how many less than number of red balls in E.
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Answer: (C) 10
Explanation
Blue balls in B = 201, red balls in C = 329, sum = 530. Red balls in E = 540. 540 - 530 = 10.
11Quantitative Aptitude
The table given below shows the total number of red and blue balls in five different bags (A, B, C, D & E). It also shows the difference between red and blue balls and percentage of red balls with respect to total balls.
| Total bags | Difference between red and blue balls | Percentage of red balls with respect to total balls |
| ---------- | ------------------------------------: | --------------------------------------------------: |
| A | 60 | 62.5% |
| B | 198 | 66.5% |
| C | 188 | 70% |
| D | 76 | 60% |
| E | 330 | 72% |
Find the ratio of number of red balls in A and E together to number of blue balls in A.
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Answer: (C) 23:3
Explanation
Total balls in A = 60 / (2 × 0.625 - 1) = 240.
Red balls in A = 150, Blue balls in A = 90.
Total balls in E = 330 / (2 × 0.72 - 1) = 750.
Red balls in E = 540.
Red balls in A and E together = 150 + 540 = 690.
Blue balls in A = 90.
Ratio = 690 : 90 = 23 : 3.
12Quantitative Aptitude
The table given below shows the total number of red and blue balls in five different bags (A, B, C, D & E). It also shows the difference between red and blue balls and percentage of red balls with respect to total balls.
| Total bags | Difference between red and blue balls | Percentage of red balls with respect to total balls |
| ---------- | ------------------------------------: | --------------------------------------------------: |
| A | 60 | 62.5% |
| B | 198 | 66.5% |
| C | 188 | 70% |
| D | 76 | 60% |
| E | 330 | 72% |
Find the average number of red balls in B, D and E.
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Answer: (D) 389
Explanation
Red balls in B = (198 / 0.33) × 0.665 = 600 × 0.665 = 399.
Red balls in D = (76 / 0.20) × 0.60 = 380 × 0.60 = 228.
Red balls in E = (330 / 0.44) × 0.72 = 750 × 0.72 = 540.
Average = (399 + 228 + 540) / 3 = 1167 / 3 = 389.
13Quantitative Aptitude
The table given below shows the total number of red and blue balls in five different bags (A, B, C, D & E). It also shows the difference between red and blue balls and percentage of red balls with respect to total balls.
| Total bags | Difference between red and blue balls | Percentage of red balls with respect to total balls |
| ---------- | ------------------------------------: | --------------------------------------------------: |
| A | 60 | 62.5% |
| B | 198 | 66.5% |
| C | 188 | 70% |
| D | 76 | 60% |
| E | 330 | 72% |
The number of balls in bag F is 25% more than red balls in E and the ratio of red balls in A and blue balls in F is 2:1. Find the red balls in F.
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Answer: (E) 600
Explanation
Red balls in E = 540.
Total balls in F = 1.25 × 540 = 675.
Red balls in A = 150.
Given ratio of red balls in A to blue balls in F is 2 : 1, blue balls in F = 150 / 2 = 75.
Red balls in F = Total balls in F - Blue balls in F = 675 - 75 = 600.
14Quantitative Aptitude
Read the information and answer the following question given below.
The information is about the staffs (lab assistant and professor) in three colleges.
For A - total lab assistant and total professor is 82.
For B - total lab assistant is 4 more than total lab assistant in C.
For C - total lab assistant is 6 less than total professor
Note - total professor in A is 44 less than total lab assistant in B.
Total staffs in C is 24 more than that in A and total professor in al the college is 102.
Find the total staff in B.
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Answer: (A) 90
Explanation
Total staff in C = Total staff in A + 24 = 82 + 24 = 106.
In C, lab assistants L_C = P_C - 6, so L_C + (L_C + 6) = 106.
2 L_C = 100, which gives L_C = 50 and P_C = 56.
Lab assistants in B = L_C + 4 = 50 + 4 = 54.
Professors in A = L_B - 44 = 54 - 44 = 10.
Professors in B = 102 - P_A - P_C = 102 - 10 - 56 = 36.
Total staff in B = L_B + P_B = 54 + 36 = 90.
15Quantitative Aptitude
Read the information and answer the following question given below.
The information is about the staffs (lab assistant and professor) in three colleges.
For A - total lab assistant and total professor is 82.
For B - total lab assistant is 4 more than total lab assistant in C.
For C - total lab assistant is 6 less than total professor
Note - total professor in A is 44 less than total lab assistant in B.
Total staffs in C is 24 more than that in A and total professor in al the college is 102.
Find the ratio of total professors in B to total lab assistant in A.
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Answer: (D) 1:2
Explanation
Let the total staff in A be L_A + P_A = 82. From the given conditions, total staff in C = 82 + 24 = 106 and L_C = P_C - 6, giving L_C = 50 and P_C = 56. Total lab assistants in B is L_B = L_C + 4 = 54. Total professors in A is P_A = L_B - 44 = 10, which gives L_A = 82 - 10 = 72. Total professors in B is P_B = 102 - 10 - 56 = 36. The required ratio of total professors in B to total lab assistants in A is 36 : 72 = 1 : 2.
16Quantitative Aptitude
Read the information and answer the following question given below.
The information is about the staffs (lab assistant and professor) in three colleges.
For A - total lab assistant and total professor is 82.
For B - total lab assistant is 4 more than total lab assistant in C.
For C - total lab assistant is 6 less than total professor
Note - total professor in A is 44 less than total lab assistant in B.
Total staffs in C is 24 more than that in A and total professor in al the college is 102.
The ratio of total staff is in college D to lab assistant in A is 3:4. Total lab assistant in D is 20 less than that in C. Find the professor in D.
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Answer: (A) 24
Explanation
From the college staff distribution, L_A = 72 and L_C = 50. The total staff in college D is (3 / 4) × 72 = 54. Total lab assistants in D is L_D = 50 - 20 = 30. The number of professors in college D is 54 - 30 = 24.
17Quantitative Aptitude
Read the information and answer the following question given below.
The information is about the staffs (lab assistant and professor) in three colleges.
For A - total lab assistant and total professor is 82.
For B - total lab assistant is 4 more than total lab assistant in C.
For C - total lab assistant is 6 less than total professor
Note - total professor in A is 44 less than total lab assistant in B.
Total staffs in C is 24 more than that in A and total professor in al the college is 102.
Total professor in A is what percentage of total staff in B.
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Answer: (B) 11.11%
Explanation
From the college staff distribution, professors in A is P_A = 10. Total staff in college B is L_B + P_B = 54 + 36 = 90. The required percentage is (10 / 90) × 100% = 11.11%.
18Quantitative Aptitude
Read the information and answer the following question given below.
The information is about the staffs (lab assistant and professor) in three colleges.
For A - total lab assistant and total professor is 82.
For B - total lab assistant is 4 more than total lab assistant in C.
For C - total lab assistant is 6 less than total professor
Note - total professor in A is 44 less than total lab assistant in B.
Total staffs in C is 24 more than that in A and total professor in al the college is 102.
Find the average of total lab assistant in A and C.
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Answer: (A) 61
Explanation
From the college staff distribution, lab assistants in college A is L_A = 72 and lab assistants in college C is L_C = 50. The average is (72 + 50) / 2 = 61.
19Quantitative Aptitude
The boat goes downstream X kilometer in two hours. The same distance covered by boat in upstream in three hours. If the speed of stream is of 3km/h, then find 2X?
A invest Rs. 6400 and B invest Rs. 7200 in a business. After 4 months, A withdrawal 30% of the initial investment and C join with 40% of initial investment of B. After one year, the total profit is 90000. Find the profit of A and C(in Rs.) approximate?
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Answer: (C) 44494
Explanation
Equivalent investment of A = (6400 × 4) + (0.70 × 6400 × 8) = 25600 + 35840 = 61440.
Equivalent investment of B = 7200 × 12 = 86400.
Equivalent investment of C = (0.40 × 7200 × 8) = 23040.
Total equivalent investment = 61440 + 86400 + 23040 = 170880.
Combined equivalent investment of A and C = 61440 + 23040 = 84480.
Combined profit share = (84480 / 170880) × 90000 ≈ 44494.
21Quantitative Aptitude
A jar contains milk and water. The ratio of milk and water is 8: x . When 5 liters of milk and 30 liters of water is added into the jar, then the ratio of milk to water becomes 1: 1 . Find x .
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Answer: (E) Can't Determined
Explanation
Let initial milk be 8k and initial water be xk. Adding 5 liters of milk and 30 liters of water gives (8k + 5) / (xk + 30) = 1 / 1, which simplifies to k(8 - x) = 25. Without additional information to fix the value of k, the value of x cannot be uniquely determined.
22Quantitative Aptitude
A rectangular park having twice length of the breadth. A half pathway width of 4 cm having area of is 352 m² . Find the length of the rectangular park (in meters)?
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Answer: (D) 176
Explanation
Let breadth = B and length L = 2B. For B = 88 meters, length L = 2 × 88 = 176 meters.
23Quantitative Aptitude
The speed of a train is 72 km/h and it crosses a platform in 25 seconds. If the train crosses an athlete whose speed is 18 km/h running in same direction of train in 10 seconds, then find the length of platform (in meters)?
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Answer: (D) 350
Explanation
Train speed = 72 × (5/18) = 20 m/s. Athlete speed = 18 × (5/18) = 5 m/s.
Relative speed with respect to the athlete running in the same direction = 20 - 5 = 15 m/s.
Length of train = 15 × 10 = 150 meters.
Total distance covered while crossing the platform in 25 seconds = 20 × 25 = 500 meters.
Length of platform = 500 - 150 = 350 meters.
24Quantitative Aptitude
The selling price of an article is Rs. 624 and a shopkeeper got a loss of 22% . If he marks up same article at 30% above the cost price and get a profit of Rs. 84, then find the discount percentage allowed by the shopkeeper?
Gopi invested Rs. X in scheme A at rate of 15% p.a. for two years. The amount received by Gopi from scheme A is reinvest in scheme B at the rate of 10% p.a. for two years on compound interest. If the difference between interest received from A and B is 32.4, then find X?
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Answer: (B) 1200 Rs.
Explanation
Simple interest received from scheme A for 2 years at 15% per annum is 0.30X, so the total amount reinvested in scheme B is 1.30X. The effective compound interest rate for 2 years at 10% per annum is 10 + 10 + (10 × 10)/100 = 21%. Thus, interest received from scheme B is 21% of 1.30X = 0.273X. The difference between the two interests is 0.30X - 0.273X = 0.027X. Given 0.027X = 32.4, solving for X yields X = 32.4 ÷ 0.027 = 1200.
26Quantitative Aptitude
7 men, 15 women can complete a piece of work in 15 days and the same work is completed by 18 men and 20 women in eight days. If X men and 20 women can complete the work in 10 days. Then find X.
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Answer: (D) 12
Explanation
Equating the work completed in both cases, 15 × (7M + 15W) = 8 × (18M + 20W), which simplifies to 105M + 225W = 144M + 160W. Rearranging terms gives 39M = 65W, or M = 5 and W = 3 in efficiency units. The total work done is 15 × (7 × 5 + 15 × 3) = 1200 units. For X men and 20 women to complete 1200 units of work in 10 days, 10 × (5X + 20 × 3) = 1200, which gives 5X + 60 = 120, leading to 5X = 60 and X = 12.
27Quantitative Aptitude
A set of four consecutive multiple of four, and its average is 30. If another series which is five consecutive even number series which lowest number is equal to second lowest multiple of these four consecutive multiples of four. Find the highest number of five consecutive even number series?
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Answer: (A) 36
Explanation
Let four consecutive multiples of four be a, a + 4, a + 8, and a + 12. Their average is (a + a + 12)/2 = a + 6 = 30, which gives a = 24. The four multiples are 24, 28, 32, and 36. The second lowest multiple is 28. The five consecutive even numbers starting from 28 are 28, 30, 32, 34, and 36. The highest number in this series is 36.
28Quantitative Aptitude
The sum of the present ages of P & Q and R & Q is 64 years and 56 years, respectively. The ratio of the age of P after 12 years to the age of Q six years ago is 4:X. If the age of R after 12 years is 32 years, then find X.
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Answer: (C) 3
Explanation
From the given condition, the age of R after 12 years is 32 years, so R = 32 - 12 = 20. Since R + Q = 56, the present age of Q is 56 - 20 = 36. Since P + Q = 64, the present age of P is 64 - 36 = 28. The age of P after 12 years is 28 + 12 = 40, and the age of Q six years ago is 36 - 6 = 30. The ratio of P after 12 years to Q six years ago is 40 : 30 = 4 : 3. Comparing with 4 : X yields X = 3.
29Quantitative Aptitude
The table given below shows the total number of wooden and mechanical pencils in five shops and it also shows the price of total mechanical and wooden pencils.
# Shop Pencil Data
| Shops | Total number of wooden and mechanical pencils | Price of total mechanical and wooden pencils |
| ----- | --------------------------------------------: | -------------------------------------------: |
| A | 38 | 490 |
| B | 47 | 415 |
| C | 112 | 1310 |
| D | 75 | 900 |
| E | 40 | 425 |
Note - Cost of wooden pencil and mechanical pencil is Rs 5 and Rs 20 respectively.
Find the ratio of cost of total mechanical pencil in A and B together to cost of total wooden pencil in D.
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Answer: (A) 16:5
Explanation
Let W and M be the number of wooden and mechanical pencils respectively.
Using the price equation 5W + 20M = Total Price and total count W + M = Total Pencils, M = (Total Price - 5 × Total Pencils) / 15.
For shop A:
M = (490 - 5 × 38) / 15 = 300 / 15 = 20
W = 38 - 20 = 18
Cost of mechanical pencils in A = 20 × ₹20 = ₹400
For shop B:
M = (415 - 5 × 47) / 15 = 180 / 15 = 12
W = 47 - 12 = 35
Cost of mechanical pencils in B = 12 × ₹20 = ₹240
Total cost of mechanical pencils in A and B = ₹400 + ₹240 = ₹640
For shop D:
M = (900 - 5 × 75) / 15 = 525 / 15 = 35
W = 75 - 35 = 40
Cost of wooden pencils in D = 40 × ₹5 = ₹200
Ratio = 640 : 200 = 16 : 5
30Quantitative Aptitude
The table given below shows the total number of wooden and mechanical pencils in five shops and it also shows the price of total mechanical and wooden pencils.
| Shops | Total number of wooden and mechanical pencils | Price of total mechanical and wooden pencils |
| ----- | --------------------------------------------: | -------------------------------------------: |
| A | 38 | 490 |
| B | 47 | 415 |
| C | 112 | 1310 |
| D | 75 | 900 |
| E | 40 | 425 |
Note - Cost of wooden pencil and mechanical pencil is Rs 5 and Rs 20 respectively.
Find the total mechanical pencil in C is what percentage of total wooden pencil in E.
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Answer: (B) 200
Explanation
Using the relation M = (Total Price - 5 × Total Pencils) / 15 and W = Total Pencils - M:
For shop C:
M = (1310 - 5 × 112) / 15 = 750 / 15 = 50
For shop E:
M = (425 - 5 × 40) / 15 = 225 / 15 = 15
W = 40 - 15 = 25
Percentage = (50 / 25) × 100 = 200%
31Quantitative Aptitude
The table given below shows the total number of wooden and mechanical pencils in five shops and it also shows the price of total mechanical and wooden pencils.
# Shop Pencil Data
| Shops | Total number of wooden and mechanical pencils | Price of total mechanical and wooden pencils |
| ----- | --------------------------------------------: | -------------------------------------------: |
| A | 38 | 490 |
| B | 47 | 415 |
| C | 112 | 1310 |
| D | 75 | 900 |
| E | 40 | 425 |
Note - Cost of wooden pencil and mechanical pencil is Rs 5 and Rs 20 respectively.
Find the average number of wooden pencil in A, mechanical pencil in B and E.
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Answer: (D) 15
Explanation
Using M = (Total Price - 5 × Total Pencils) / 15 and W = Total Pencils - M:
Wooden pencils in shop A = 38 - (490 - 190) / 15 = 38 - 20 = 18
Mechanical pencils in shop B = (415 - 235) / 15 = 12
Mechanical pencils in shop E = (425 - 200) / 15 = 15
Average = (18 + 12 + 15) / 3 = 45 / 3 = 15
32Quantitative Aptitude
The table given below shows the total number of wooden and mechanical pencils in five shops and it also shows the price of total mechanical and wooden pencils.
# Shop Pencil Data
| Shops | Total number of wooden and mechanical pencils | Price of total mechanical and wooden pencils |
| ----- | --------------------------------------------: | -------------------------------------------: |
| A | 38 | 490 |
| B | 47 | 415 |
| C | 112 | 1310 |
| D | 75 | 900 |
| E | 40 | 425 |
Note - Cost of wooden pencil and mechanical pencil is Rs 5 and Rs 20 respectively.
Shop B sold 20% of wooden pencil at the profit of 20% and rest at the profit of 50% . Find the total selling price of wooden pencil.
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Answer: (E) 252
Explanation
Using M = (415 - 5 × 47) / 15 = 12, the number of wooden pencils in shop B is 47 - 12 = 35.
The cost price per wooden pencil is ₹5.
20% of 35 = 7 pencils sold at 20% profit:
Selling price = 7 × ₹5 × 1.20 = ₹42
Remaining 80% = 28 pencils sold at 50% profit:
Selling price = 28 × ₹5 × 1.50 = ₹210
Total selling price = ₹42 + ₹210 = ₹252
33Quantitative Aptitude
The table given below shows the total number of wooden and mechanical pencils in five shops and it also shows the price of total mechanical and wooden pencils.
# Shop Pencil Data
| Shops | Total number of wooden and mechanical pencils | Price of total mechanical and wooden pencils |
| ----- | --------------------------------------------: | -------------------------------------------: |
| A | 38 | 490 |
| B | 47 | 415 |
| C | 112 | 1310 |
| D | 75 | 900 |
| E | 40 | 425 |
Note - Cost of wooden pencil and mechanical pencil is Rs 5 and Rs 20 respectively.
The ratio of total pencil in F to total pencil in C is 5: 7 and shop F has 30% more than wooden pencil than D. Find the mechanical pencils in F is what percentage more/less than wooden pencil sold by E.
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Answer: (D) 112%
Explanation
Total pencils in shop C = 112.
Total pencils in shop F = (5 / 7) × 112 = 80.
Wooden pencils in shop D = 75 - (900 - 5 × 75) / 15 = 75 - 35 = 40.
Wooden pencils in shop F = 40 × (1 + 0.30) = 52.
Mechanical pencils in shop F = 80 - 52 = 28.
Substituting back into the equation:
31 = ? - 12
? = 31 + 12 = 43
43Quantitative Aptitude
Directions (82-86): What approximate value should come in place of (?) in the following questions.
(2511.99 - ?)/(√(1368)) = {(26.99)² - 138.99 - (360.08 ÷ 3.99)} ÷ 9.99
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Answer: (A) 662
Explanation
Approximate the terms in the equation to their nearest convenient values:
2511.99 ≈ 2512
√1368 ≈ √1369 = 37
26.99 ≈ 27
138.99 ≈ 139
360.08 ÷ 3.99 ≈ 360 ÷ 4 = 90
9.99 ≈ 10
Now equate the left-hand side to RHS:
(2512 - ?) ÷ 37 = 50
2512 - ? = 50 × 37
2512 - ? = 1850
? = 2512 - 1850 = 662
44Quantitative Aptitude
What approximate value should come in place of (?) in the following questions.
(? ÷ 17.99) × 5.99 = 319.87% of 60
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Answer: (D) 576
Explanation
Approximating the values in the given expression gives (? ÷ 18) × 6 = 320% of 60. Simplifying the right side yields 3.2 × 60 = 192. Thus, ? ÷ 3 = 192, which gives ? = 576.
45Quantitative Aptitude
A spend 25% of his monthly income on rent and 20% of remaining in food, 40% remaining in traveling and rest he saved. If total monthly saving of A is Rs.10800, then find the monthly income of A. (in Rs.)?
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Answer: (B) 30000
Explanation
Let the total monthly income be 100%. After spending 25% on rent, 75% remains. Spending 20% of the remainder on food leaves 80% of 75%, which is 60%. Spending 40% of this remaining amount on traveling leaves 60% of 60%, which is 36%. Given that the remaining savings equal Rs. 10800, the equation 36% of Income = 10800 gives Income = 10800 × 100 / 36 = Rs. 30000.
46Quantitative Aptitude
Two trains, A and B, are moving at speeds of 60 km / h and 80 km / h respectively. Train A crosses a platform that is 150 meters long in 15 seconds, while Train B crosses a pole in 18 seconds. What the sum of length of both trains A and B (in meters)?
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Answer: (E) 500
Explanation
Speed of Train A is 60 km/h, which equals 50/3 m/s. Train A crosses a 150 m platform in 15 seconds, so (Length of A + 150) = (50/3) × 15 = 250 m, making Length of A equal to 100 m. Speed of Train B is 80 km/h, which equals 200/9 m/s. Train B crosses a pole in 18 seconds, so Length of B = (200/9) × 18 = 400 m. Sum of the lengths of both trains is 100 + 400 = 500 meters.
47Quantitative Aptitude
In a mixture, the ratio of milk to water is 5:1. When 20 liters of water is added into the mixture, then the ratio of milk to water becomes equal. Find the quantity of milk in the final mixture (in liter).
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Answer: (A) 25
Explanation
Let the initial quantities of milk and water be 5x and x liters respectively. When 20 liters of water is added, the new quantity of water becomes x + 20, while milk remains 5x. Since the final ratio of milk to water is 1:1, 5x = x + 20, which gives x = 5. Therefore, the quantity of milk in the final mixture is 5 × 5 = 25 liters.
48Quantitative Aptitude
A can complete 40% of a work in 8 days, while B takes four more days than that of A to complete the same work. Find the time taken by A and B together to complete the same work (in days)?
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Answer: (E) 10 ¹⁰⁄₁₁
Explanation
Since A completes 40% of the work in 8 days, A completes the total work in 8 ÷ 0.4 = 20 days. B takes 4 more days than A, so B completes the work in 24 days. Combined work done per day by A and B is 1/20 + 1/24 = 11/120. Therefore, time taken by A and B together to complete the work is 120/11 = 10 10/11 days.
49Quantitative Aptitude
The length of the rectangle is 20% more than its breadth. If the side of a square is equal to the breadth of rectangle, then find the respective ratio of area of square to that of rectangle?
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Answer: (A) 5:6
Explanation
Let the breadth of the rectangle be B. The length of the rectangle is 1.2 B. The side of the square is equal to the breadth B. Area of the square is B² and area of the rectangle is 1.2 B × B = 1.2 B². The ratio of area of square to area of rectangle is B² : 1.2 B², which simplifies to 1 : 1.2 or 5:6.
50Quantitative Aptitude
A boat covers 120 km in six hours in downstream. The respective ratio of speed of boat in still water and speed of stream is 3: 1. If speed of stream is increased by 80% , then find in how much time the boat will cover 84 km in upstream?
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Answer: (B) 14 hours
Explanation
Downstream speed is 120 / 6 = 20 km/h. Given the ratio of boat speed in still water to stream speed is 3:1, boat speed is 15 km/h and stream speed is 5 km/h. Increasing the stream speed by 80% makes the new stream speed 5 × 1.8 = 9 km/h. The new upstream speed becomes 15 - 9 = 6 km/h. Time taken to cover 84 km upstream is 84 / 6 = 14 hours.
