Quantitative Aptitude: 5-Chapter Comprehensive Practice & 75 Verified MCQs

Question 1

What is the unit digit of the exponential number 7^105?

Accepted Answer

C. 7. The unit digit of powers of 7 follows a repeating cyclicity of 4: 7^1 = 7, 7^2 = 9, 7^3 = 3, 7^4 = 1. Dividing the power 105 by 4 gives 105 = (4 * 26) + 1 (remainder 1). Therefore, the unit digit corresponds to 7^1, which is 7.

Question 2

What is the divisibility test for checking if a large integer is divisible by 11?

Accepted Answer

B. The difference between the sum of digits at odd places and the sum of digits at even places must be either 0 or a multiple of 11.. A number is divisible by 11 if the absolute difference between the sum of digits at odd positions and the sum of digits at even positions (calculated from right to left) is either 0 or divisible by 11. For example, in 1331: (1 + 3) - (3 + 1) = 4 - 4 = 0, so 1331 is divisible by 11.

Question 3

The HCF of two numbers is 12, and their LCM is 240. If one of the numbers is 48, what is the other number?

Accepted Answer

A. 60. For any two positive integers A and B, the mathematical identity holds: Product of Numbers = HCF * LCM. Therefore, A * B = 12 * 240 = 2,880. Given A = 48, we have B = 2,880 / 48 = 60.

Question 4

If the price of a commodity increases by 25%, by what percentage must a household reduce its consumption so that the total expenditure remains unchanged?

Accepted Answer

B. 20%. Let initial price = 100, initial consumption = 100 (Expenditure = 10,000). The new price becomes 125. To keep expenditure at 10,000, new consumption = 10,000 / 125 = 80. The reduction in consumption is 100 - 80 = 20%. Alternatively, use formula: [R / (100 + R)] * 100 = [25 / 125] * 100 = 20%.

Question 5

A shopkeeper marks an item 40% above its cost price and offers a discount of 20% on the marked price. What is the net profit percentage?

Accepted Answer

B. 12%. Let Cost Price (CP) = 100. Marked Price (MP) = 140. A 20% discount on 140 is 0.20 * 140 = 28. Selling Price (SP) = 140 - 28 = 112. Net Profit = SP - CP = 112 - 100 = 12%. Using successive change formula: a + b + (ab/100) = +40 - 20 - (800/100) = 20 - 8 = +12%.

Question 6

What is the difference between compound interest (compounded annually) and simple interest on a principal of $10,000 for 2 years at an annual interest rate of 8%?

Accepted Answer

A. $64. The formula for the difference between CI and SI for 2 years is: Difference = P * (R / 100)^2. Here, P = 10,000 and R = 8. Difference = 10,000 * (8 / 100)^2 = 10,000 * (64 / 10,000) = $64.

Question 7

Person A can complete a work in 12 days, and Person B can complete the same work in 24 days. How many days will they take to complete the work working together?

Accepted Answer

B. 8 days. Total work = LCM(12, 24) = 24 units. A's efficiency = 24 / 12 = 2 units/day. B's efficiency = 24 / 24 = 1 unit/day. Combined efficiency = 2 + 1 = 3 units/day. Time required = Total work / Combined efficiency = 24 / 3 = 8 days.

Question 8

Two trains of lengths 150 m and 250 m are traveling towards each other on parallel tracks at speeds of 54 km/h and 90 km/h respectively. In how many seconds will they completely cross each other after meeting?

Accepted Answer

A. 10 seconds. Total distance to cross = Sum of lengths = 150 + 250 = 400 m. When moving towards each other, relative speed is the sum of speeds = 54 + 90 = 144 km/h. Convert to m/s: 144 * (5 / 18) = 40 m/s. Time = Distance / Relative Speed = 400 / 40 = 10 seconds.

Question 9

In what ratio must a merchant mix rice costing $40/kg with rice costing $60/kg to obtain a mixture worth $48/kg?

Accepted Answer

A. 3 : 2. By the Rule of Alligation: Ratio = (Higher Price - Mean Price) / (Mean Price - Lower Price) = (60 - 48) / (48 - 40) = 12 / 8 = 3 / 2 = 3 : 2.

Question 10

What is the remainder when 2^50 is divided by 7?

Accepted Answer

C. 4. By modular arithmetic: 2^1 = 2, 2^2 = 4, 2^3 = 8 ≡ 1 (mod 7). Because 2^3 ≡ 1 (mod 7), we express 50 in terms of multiples of 3: 50 = (3 * 16) + 2. Therefore, 2^50 = (2^3)^16 * 2^2 ≡ (1)^16 * 4 ≡ 4 (mod 7). Remainder = 4.

Question 11

A sum of money doubles itself in 5 years at simple interest. In how many years will it become 4 times of itself at the same rate?

Accepted Answer

B. 15 years. If a sum doubles (Amount = 2P), Simple Interest (SI) = P. This SI of P is earned in 5 years. For the sum to become 4 times (Amount = 4P), the total SI needed is 3P. Because SI is directly proportional to time: Time = 3 * 5 = 15 years.

Question 12

Pipe A can fill a tank in 10 hours, and Pipe B can empty the full tank in 15 hours. If both pipes are opened simultaneously, in how many hours will the empty tank be filled?

Accepted Answer

B. 30 hours. Let tank capacity = LCM(10, 15) = 30 units. Inlet Pipe A rate = +30 / 10 = +3 units/hour. Outlet Pipe B rate = -30 / 15 = -2 units/hour. Net flow rate when both open = +3 - 2 = +1 unit/hour. Time to fill empty tank = 30 / 1 = 30 hours.

Question 13

What is the average of the first 50 positive natural numbers (1, 2, 3, ..., 50)?

Accepted Answer

B. 25.5. The sum of the first N natural numbers is [N * (N + 1)] / 2. The average is Sum / N = (N + 1) / 2. For N = 50: Average = (50 + 1) / 2 = 51 / 2 = 25.5.

Question 14

If A : B = 2 : 3 and B : C = 4 : 5, what is the combined continuous ratio A : B : C?

Accepted Answer

A. 8 : 12 : 15. To combine A : B = 2 : 3 and B : C = 4 : 5, multiply the first ratio by 4 (the B-term of second) and the second ratio by 3 (the B-term of first): A : B = 8 : 12, and B : C = 12 : 15. Combining gives A : B : C = 8 : 12 : 15.

Question 15

How many trailing zeros does the factorial number 100! contain?

Accepted Answer

B. 24. Trailing zeros are formed by factors of 10 (2 * 5). In any factorial, factors of 2 far exceed factors of 5. By Legendre's Formula, the number of factors of 5 in N! is: ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ = 20 + 4 + 0 = 24 trailing zeros.

Question 16

If the price of sugar increases by 25%, by what percentage must a household reduce its consumption so that the total expenditure remains constant?

Accepted Answer

A. 20%. Formula: Reduction % = [r / (100 + r)] * 100. For r = 25: [25 / 125] * 100 = (1/5) * 100 = 20%. If original price was 100 for 100 units (expenditure = 10,000), new price is 125. New consumption = 10,000 / 125 = 80 units (20% reduction).

Question 17

A shopkeeper sells an article for $840, incurring a loss of 16%. What was the cost price (CP) of the article?

Accepted Answer

A. $1,000. Selling Price (SP) = CP * (100 - Loss%) / 100. Therefore, 840 = CP * (84/100) => CP = 840 * (100 / 84) = $1,000.

Question 18

A merchant marks up his goods by 40% above Cost Price and then offers a discount of 20% on the Marked Price. What is his net profit percentage?

Accepted Answer

B. 12%. Let CP = 100. Marked Price MP = 100 * 1.40 = 140. Selling Price SP = 140 * (1 - 0.20) = 140 * 0.80 = 112. Net Profit = 112 - 100 = 12%. Successive formula: a + b + (ab/100) = +40 - 20 - (40*20/100) = 20 - 8 = 12%.

Question 19

What single equivalent discount corresponds to two successive discounts of 20% and 10%?

Accepted Answer

B. 28%. Successive discount formula: d_eq = d1 + d2 - (d1 * d2 / 100). Here, 20 + 10 - (20 * 10 / 100) = 30 - 2 = 28%. Starting from 100: 100 * 0.80 * 0.90 = 72. Discount = 100 - 72 = 28%.

Question 20

A dishonest dealer professes to sell his goods at cost price, but uses a false weight of 950 grams instead of 1 kilogram (1000 grams). What is his actual profit percentage?

Accepted Answer

B. 5.26%. Formula: Gain % = [Error / (True Value - Error)] * 100. Error = 1000 - 950 = 50g. True value - Error = 950g. Gain % = (50 / 950) * 100 = 100 / 19 ≈ 5.263%.

Question 21

If the Cost Price of 15 articles is equal to the Selling Price of 12 articles, what is the profit percentage?

Accepted Answer

B. 25%. Given: 15 * CP = 12 * SP => SP / CP = 15 / 12 = 5 / 4. Profit % = [(SP - CP) / CP] * 100 = [(5 - 4) / 4] * 100 = (1/4) * 100 = 25%.

Question 22

A person sells two wristwatches for $1,200 each. On one he gains 20% and on the other he loses 20%. What is his overall transaction result?

Accepted Answer

B. 4% Loss. When two articles are sold at the SAME Selling Price, one at x% gain and other at x% loss, there is ALWAYS an overall loss given by: Loss % = (x / 10)^2 % = (20 / 10)^2 = 4% loss.

Question 23

In an election between two candidates, the winner received 58% of the total valid votes and won by a majority of 4,800 votes. What was the total number of valid votes polled?

Accepted Answer

B. 30,000. Winner = 58%, Loser = 100% - 58% = 42%. Difference (majority) = 58% - 42% = 16%. 16% of Total = 4,800 => Total = 4,800 / 0.16 = 30,000 votes.

Question 24

The population of a town increases by 10% annually. If the current population is 121,000, what was its population 2 years ago?

Accepted Answer

A. 100,000. Let population 2 years ago be P. Current Population = P * (1 + 10/100)^2 = P * (1.1)^2 = P * 1.21. 121,000 = 1.21 * P => P = 121,000 / 1.21 = 100,000.

Question 25

A student needs 35% marks to pass an examination. If he obtains 145 marks and fails by 30 marks, what are the maximum total marks of the exam?

Accepted Answer

A. 500. Passing marks = 145 + 30 = 175 marks. Given 35% of Total = 175. Total Marks = 175 / 0.35 = 175 * (100 / 35) = 5 * 100 = 500.

Question 26

A trader bought a cycle for $2,500 and spent $500 on its repair and repainting. He then sold it for $3,600. What is his profit percentage?

Accepted Answer

A. 20%. Total effective Cost Price = Purchase Price + Overhead Repair Expenses = 2500 + 500 = $3,000. Selling Price = $3,600. Profit = 3600 - 3000 = $600. Profit % = (600 / 3000) * 100 = 20%.

Question 27

If A's salary is 20% less than B's salary, by what percentage is B's salary more than A's salary?

Accepted Answer

B. 25%. Let B = 100. Then A = 80. B is greater than A by: (100 - 80) / 80 * 100 = (20 / 80) * 100 = 1/4 * 100 = 25%. Formula: [r / (100 - r)] * 100 = [20 / 80] * 100 = 25%.

Question 28

A vendor sells lemons at the rate of 5 for $14, gaining 40%. At what rate did he buy them (cost price per dozen)?

Accepted Answer

B. $24 per dozen. SP of 1 lemon = 14 / 5 = $2.80. Gain = 40%. CP of 1 lemon = SP / 1.40 = 2.80 / 1.40 = $2.00. Cost of 1 dozen (12 lemons) = 12 * $2.00 = $24.00.

Question 29

By selling an umbrella for $300, a shopkeeper gains 20%. During an off-season clearance, he sells it at a loss of 10%. What is the clearance sale price?

Accepted Answer

A. $225. First find CP: CP = 300 / 1.20 = $250. At 10% loss, clearance SP = 250 * (1 - 0.10) = 250 * 0.90 = $225.

Question 30

What is the successive percentage change if an item's price is increased by 30% and then decreased by 30%?

Accepted Answer

B. 9% decrease. Net % change = +30 - 30 - (30 * 30 / 100) = -900 / 100 = -9% (a decrease of 9%). If 100 -> 130 -> 130 * 0.70 = 91 (9% net drop).

Question 31

A principal sum of money invested at Simple Interest doubles itself in 8 years. What is the annual rate of interest?

Accepted Answer

B. 12.5%. When money doubles, Interest SI = Principal P. Formula: SI = (P * R * T) / 100 => P = (P * R * 8) / 100 => R = 100 / 8 = 12.5% per annum.

Question 32

What is the difference between Compound Interest and Simple Interest on a principal of $10,000 for 2 years at 8% per annum?

Accepted Answer

A. $64. Formula for difference between CI and SI for 2 years: Diff = P * (R / 100)^2. Here, Diff = 10,000 * (8 / 100)^2 = 10,000 * 0.0064 = $64.

Question 33

A sum of money placed at Compound Interest doubles itself in 5 years. In how many years will it become 8 times of itself at the same compound rate?

Accepted Answer

A. 15 years. Under Compound Interest, growth is exponential: P -> 2P in 5 years (2^1 in 5 yrs). 8 times = 2^3 times. Therefore, Time = 3 * 5 = 15 years.

Question 34

What is the effective annual rate of interest corresponding to a nominal interest rate of 10% per annum compounded half-yearly?

Accepted Answer

A. 10.25%. Half-yearly rate r = 10 / 2 = 5% per 6 months. For 1 year (2 periods): Effective Rate = (1 + 0.05)^2 - 1 = 1.1025 - 1 = 0.1025 = 10.25%.

Question 35

At what rate percent per annum simple interest will a sum of money become 4 times itself in 15 years?

Accepted Answer

A. 20%. Amount A = 4P => Simple Interest SI = A - P = 3P. SI = (P * R * T) / 100 => 3P = (P * R * 15) / 100 => 3 = 15R / 100 => R = 300 / 15 = 20%.

Question 36

Find the Compound Interest on $8,000 at 15% per annum for 2 years and 4 months, compounded annually.

Accepted Answer

A. $3,109. Time = 2 years + 4/12 years = 2 years + 1/3 year. Amount = 8000 * (1 + 15/100)^2 * (1 + (1/3 * 15)/100) = 8000 * (1.15)^2 * (1 + 0.05) = 8000 * 1.3225 * 1.05 = 8000 * 1.388625 = $11,109. CI = 11,109 - 8000 = $3,109.

Question 37

A machine was purchased for $50,000. Its value depreciates at the rate of 10% per annum. What will be its residual salvage value after 3 years?

Accepted Answer

A. $36,450. Depreciation formula: Value = P * (1 - r/100)^n = 50,000 * (1 - 0.10)^3 = 50,000 * (0.9)^3 = 50,000 * 0.729 = $36,450.

Question 38

What principal sum will yield $1,320 as simple interest in 3 years at 5.5% per annum?

Accepted Answer

A. $8,000. Formula: P = (SI * 100) / (R * T). P = (1320 * 100) / (5.5 * 3) = 132,000 / 16.5 = $8,000.

Question 39

The simple interest on a sum of money for 3 years at 8% per annum is $960. What would be the compound interest on the same sum at the same rate and for the same time?

Accepted Answer

A. $1,038.85. SI = (P * 8 * 3) / 100 = 0.24P = 960 => P = 960 / 0.24 = $4,000. CI Amount = 4000 * (1 + 0.08)^3 = 4000 * 1.259712 = $5,038.85. CI = 5038.85 - 4000 = $1,038.85.

Question 40

If a certain sum amounts to $756 in 2 years and to $873 in 3.5 years at simple interest, what is the principal and annual rate of interest?

Accepted Answer

A. Principal = $600, Rate = 13%. Interest in 1.5 years (3.5 - 2) = 873 - 756 = $117. Interest for 1 year = 117 / 1.5 = $78. Interest for 2 years = 78 * 2 = $156. Principal = Amount(2 yr) - SI(2 yr) = 756 - 156 = $600. Rate = (78 / 600) * 100 = 13%.

Question 41

What is the Rule of 72 used for in financial arithmetic?

Accepted Answer

B. Estimating the number of years required to double an investment at a fixed annual compound interest rate: Years ≈ 72 / (Interest Rate %). Rule of 72 provides a quick mental approximation for compound doubling time: t ≈ 72 / r. At 8% interest, money doubles in approx 72 / 8 = 9 years (exact value is ln(2)/ln(1.08) ≈ 9.006 yrs).

Question 42

A sum of $12,000 deposited at compound interest becomes double after 5 years. What will it be after 20 years?

Accepted Answer

A. $192,000. 20 years = 4 cycles of 5 years (20 / 5 = 4). Each cycle doubles the capital. Final Amount = 12,000 * (2^4) = 12,000 * 16 = $192,000.

Question 43

What is the formula for the difference between Compound Interest and Simple Interest for 3 years at rate R%?

Accepted Answer

B. P * (R/100)^2 * [ (300 + R) / 100 ]. CI_3 - SI_3 = P * (R/100)^2 * (3 + R/100) = P * (R/100)^2 * [(300 + R)/100]. This is the standard 3-year competitive examination shortcut formula.

Question 44

A father divides $2,602 between two sons aged 14 and 16 so that they receive equal amounts when each turns 18, under 4% compound interest. How much does the younger son get?

Accepted Answer

A. $1,250. Younger son has 4 years until 18; elder has 2 years. Let parts be A and B. A * (1.04)^4 = B * (1.04)^2 => A / B = 1 / (1.04)^2 = 1 / (26/25)^2 = 625 / 676. Total parts = 625 + 676 = 1,301. Younger son A = (625 / 1301) * 2602 = 625 * 2 = $1,250.

Question 45

At what annual compound interest rate will $5,000 amount to $5,832 in 2 years?

Accepted Answer

A. 8%. 5832 = 5000 * (1 + R/100)^2 => (1 + R/100)^2 = 5832 / 5000 = 11664 / 10000 = 1.1664. Taking square root: 1 + R/100 = 1.08 => R/100 = 0.08 => R = 8%.

Question 46

If A : B = 2 : 3 and B : C = 4 : 5, what is the combined ratio A : B : C?

Accepted Answer

A. 8 : 12 : 15. Equalize the common term B: LCM(3, 4) = 12. Multiply first ratio by 4: A:B = 8:12. Multiply second ratio by 3: B:C = 12:15. Combined: A : B : C = 8 : 12 : 15.

Question 47

What is the third proportional to 9 and 12?

Accepted Answer

A. 16. Third proportional x to a and b satisfies: a : b = b : x => 9 : 12 = 12 : x => 9x = 144 => x = 144 / 9 = 16.

Question 48

What is the Mean Proportional between 4 and 64?

Accepted Answer

A. 16. Mean proportional between a and b is the geometric mean: x = sqrt(a * b) = sqrt(4 * 64) = sqrt(256) = 16.

Question 49

In what ratio must tea worth $60 per kg be mixed with tea worth $65 per kg so that the mixture is worth $62 per kg?

Accepted Answer

A. 3 : 2. Rule of Alligation: (Cheaper Quantity) / (Dearer Quantity) = (Dearer Price - Mean Price) / (Mean Price - Cheaper Price) = (65 - 62) / (62 - 60) = 3 / 2. Ratio is 3 : 2.

Question 50

A container contains 40 liters of pure milk. From this, 4 liters are removed and replaced with water. This process is repeated 2 more times (3 times total). How much pure milk remains?

Accepted Answer

A. 29.16 liters. Formula: Remaining Liquid = Total * (1 - x / Total)^n = 40 * (1 - 4/40)^3 = 40 * (0.9)^3 = 40 * 0.729 = 29.16 liters.

Question 51

A and B enter into a partnership. A contributes $5,000 for 8 months and B contributes $6,000 for 5 months. How should a profit of $3,500 be divided?

Accepted Answer

A. A: $2,000, B: $1,500. Profit ratio = (Capital_A * Time_A) : (Capital_B * Time_B) = (5000 * 8) : (6000 * 5) = 40,000 : 30,000 = 4 : 3. Total parts = 7. A's share = (4/7) * 3500 = $2,000; B's share = (3/7) * 3500 = $1,500.

Question 52

Two vessels A and B contain milk and water in the ratio 4 : 3 and 2 : 3 respectively. In what ratio should the liquids be mixed to obtain a new mixture containing half milk and half water (1 : 1)?

Accepted Answer

A. 7 : 5. Focus on milk fraction: Vessel A = 4/7. Vessel B = 2/5. Target mixture = 1/2. By Alligation: (1/2 - 2/5) / (4/7 - 1/2) = (1/10) / (1/14) = 14 / 10 = 7 / 5. The ratio is 7 : 5.

Question 53

The ratio of ages of two persons is 5 : 7. Eighteen years ago, their ages were in the ratio 8 : 13. What are their present ages?

Accepted Answer

A. 50 years and 70 years. Let present ages be 5x and 7x. 18 years ago: (5x - 18) / (7x - 18) = 8 / 13 => 13(5x - 18) = 8(7x - 18) => 65x - 234 = 56x - 144 => 9x = 90 => x = 10. Present ages = 5(10) = 50 and 7(10) = 70 years.

Question 54

Seats for Mathematics, Physics, and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50%, and 75% respectively. What will be the new ratio of seats?

Accepted Answer

A. 2 : 3 : 4. Let initial seats be 50, 70, 80. Math = 50 * 1.40 = 70. Physics = 70 * 1.50 = 105. Biology = 80 * 1.75 = 140. New ratio = 70 : 105 : 140. Dividing by 35: 2 : 3 : 4.

Question 55

A bag contains coins of $1, 50 cents, and 25 cents in the ratio 3 : 4 : 8. If the total money in the bag is $140, find the number of 50-cent coins.

Accepted Answer

A. 80. Let coins be 3x ($1), 4x (50¢ = $0.50), and 8x (25¢ = $0.25). Total value = 3x($1.00) + 4x($0.50) + 8x($0.25) = 3x + 2x + 2x = 7x dollars. 7x = 140 => x = 20. Number of 50-cent coins = 4x = 4 * 20 = 80 coins.

Question 56

If x : y = 3 : 4, what is the value of (5x - 2y) / (7x + 2y)?

Accepted Answer

A. 7 / 29. Substitute x = 3 and y = 4 directly into the homogeneous expression: Numerator = 5(3) - 2(4) = 15 - 8 = 7. Denominator = 7(3) + 2(4) = 21 + 8 = 29. Result = 7 / 29.

Question 57

A merchant blends two varieties of coffee costing $250/kg and $300/kg in the ratio 3 : 2 and sells the blend at $312/kg. What is his profit percentage?

Accepted Answer

A. 15.56%. Weighted Cost Price of blend = (3 * 250 + 2 * 300) / (3 + 2) = (750 + 600) / 5 = 1350 / 5 = $270/kg. SP = $312/kg. Profit = 312 - 270 = $42. Profit % = (42 / 270) * 100 = 140 / 9 ≈ 15.56%.

Question 58

What number must be added to each of the numbers 6, 14, 18, and 38 so that the resulting numbers form a proportion?

Accepted Answer

A. 2. Let number be x: (6 + x) / (14 + x) = (18 + x) / (38 + x). Test x = 2: (6+2)/(14+2) = 8/16 = 1/2. (18+2)/(38+2) = 20/40 = 1/2. Both equal 1/2! Thus x = 2.

Question 59

In a mixture of 60 liters, the ratio of milk and water is 2 : 1. If this ratio is to be made 1 : 2, what quantity of water should be added?

Accepted Answer

A. 60 liters. Initial: Milk = (2/3)*60 = 40 L; Water = (1/3)*60 = 20 L. Water is added, so milk remains constant at 40 L. In the new ratio 1 : 2, milk is 1 part = 40 L. Therefore, new water (2 parts) = 2 * 40 = 80 L. Water to add = 80 - 20 = 60 liters.

Question 60

Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should these metals be mixed so that the resulting alloy is 15 times as heavy as water?

Accepted Answer

A. 3 : 2. By Alligation: Gold density = 19, Copper density = 9, Alloy density = 15. Ratio of Gold to Copper = (15 - 9) / (19 - 15) = 6 / 4 = 3 / 2.

Question 61

A can complete a piece of work in 12 days and B can do it in 18 days. If they work together, in how many days will they finish the work?

Accepted Answer

A. 7.2 days (7 1/5 days). Let total work = LCM(12, 18) = 36 units. Efficiency of A = 36/12 = 3 units/day. Efficiency of B = 36/18 = 2 units/day. Combined efficiency = 3 + 2 = 5 units/day. Time required = 36 / 5 = 7.2 days.

Question 62

Pipe A can fill a tank in 10 hours and Pipe B can empty the full tank in 15 hours. If both pipes are opened simultaneously, in how many hours will the empty tank be filled?

Accepted Answer

A. 30 hours. Tank capacity = LCM(10, 15) = 30 units. Inlet A rate = +3 units/hr. Outlet B rate = -2 units/hr. Net rate = +3 - 2 = +1 unit/hr. Time to fill = 30 / 1 = 30 hours.

Question 63

A train 150 meters long is traveling at a speed of 54 km/h. How many seconds will it take to completely cross a stationary electric pole?

Accepted Answer

A. 10 seconds. Convert speed from km/h to m/s: Speed = 54 * (5/18) = 15 m/s. Crossing a pole requires the train to cover its own length: Distance = 150 m. Time = Distance / Speed = 150 / 15 = 10 seconds.

Question 64

Two trains of lengths 120m and 180m are traveling on parallel tracks in opposite directions with speeds of 40 km/h and 32 km/h. In how many seconds will they pass each other completely?

Accepted Answer

A. 15 seconds. Total distance to clear = Length1 + Length2 = 120 + 180 = 300 meters. Relative speed in opposite directions = S1 + S2 = 40 + 32 = 72 km/h. In m/s: 72 * (5/18) = 20 m/s. Time = 300 / 20 = 15 seconds.

Quantitative Aptitude: 5-Chapter Comprehensive Practice & 75 Verified MCQs

Quantitative Aptitude: 5-Chapter Comprehensive Practice & 75 Verified MCQs

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