Aptitude Questions and Answers: 60 Solved Quantitative Problems from Easy to Hard
This guide is for students and freshers in India preparing for the aptitude round of campus placements and off-campus drives, and for anyone facing a numerical test in a bank, government or private-sector exam. It has 60 solved questions in three levels: the core formulas and quick methods, then the standard placement-test problems, and finally the multi-step questions that separate the top scorers. Each answer shows the method, not just the result, with shortcuts that save time under a timer. Work each one on paper before reading the answer, then take the mock test to practise at speed.
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What aptitude tests cover
- Arithmetic: percentages, profit and loss, simple and compound interest, ratios, averages, mixtures and partnership. These make up the largest share of most placement tests.
- Time-based problems: time and work, pipes and cisterns, speed, trains, and boats and streams.
- Number system: divisibility, remainders, unit digits, HCF and LCM, and simplification.
- Counting and chance: permutations, combinations and probability, usually at an easy to moderate level.
- Data interpretation and mensuration: reading tables and charts quickly, and areas and volumes of common shapes.
How to prepare in four weeks
- Week 1: revise the formulas and learn the fraction-percentage table (1/8 = 12.5%, 1/6 = 16.67% and so on) until you know it without thinking.
- Week 2: solve 20 to 30 questions a day topic by topic, starting with percentages, profit and loss and ratios, because many other topics build on them.
- Week 3: move to mixed, timed sets. Aim for about one minute per easy question and two for harder ones.
- Week 4: take full mock tests, then spend as long reviewing mistakes as you spent on the test. Keep an error notebook of the question types you get wrong.
Strategy on test day
- Read the last line of the question first, so you know exactly what is being asked before you start calculating.
- Skip a question that is taking more than two minutes and come back to it; one hard question should not cost you three easy ones.
- Use the options: eliminate the ones that are impossible (wrong units, too large, not a whole number when it must be) and test the rest.
- Check whether the test has negative marking before guessing. Most campus tests today do not, but some do.
Aptitude interview questions: Easy questions
20 questions1. What is the fastest way to find a percentage of a number in your head?
Break the percentage into easy parts. 10% is the number divided by 10, 5% is half of that, and 1% is the number divided by 100. For 15% of 640: 10% is 64 and 5% is 32, so 15% is 96. Also remember that x% of y equals y% of x: 8% of 50 is the same as 50% of 8, which is 4. This trick turns many awkward calculations into easy ones.
2. Which fraction-to-percentage values should you memorise, and why?
Learn 1/2 = 50%, 1/3 = 33.33%, 1/4 = 25%, 1/5 = 20%, 1/6 = 16.67%, 1/7 = about 14.29%, 1/8 = 12.5%, 1/9 = about 11.11%, 1/11 = about 9.09% and 1/12 = 8.33%. They let you switch between percentages and fractions instantly. For example, 37.5% of 480 is 3/8 of 480: 480/8 = 60, and 60 x 3 = 180. Using the fraction is much faster than multiplying by 0.375.
3. How do you calculate profit percent and loss percent?
Profit or loss is always measured on the cost price (CP) unless the question says otherwise. Profit % = (selling price - cost price)/cost price x 100. If a shopkeeper buys a fan for Rs 800 and sells it for Rs 920, the profit is Rs 120, and the profit percent is 120/800 x 100 = 15%. A common mistake is dividing by the selling price, which gives 13.04% here.
4. What is the difference between marked price, discount and selling price?
The marked price (MP) is the price printed on the item. A discount is a reduction on the marked price, and the selling price (SP) is what the customer pays. If a shirt is marked Rs 2,500 and sold at a 20% discount, the discount is Rs 500 and the SP is Rs 2,000. Remember: discount is calculated on the marked price, while profit is calculated on the cost price.
5. How do you calculate simple interest?
Simple interest (SI) = P x R x T/100, where P is the principal, R the yearly rate in percent and T the time in years. On Rs 12,000 at 8% for 3 years, SI = 12,000 x 8 x 3/100 = Rs 2,880, and the amount repaid is Rs 14,880. In simple interest the interest is the same every year, because it is always calculated on the original principal.
6. How do you divide an amount in a given ratio?
Add the parts of the ratio, find the value of one part, then multiply. To divide Rs 7,200 between two people in the ratio 4:5, the total is 9 parts, so one part is 7,200/9 = Rs 800. The shares are 4 x 800 = Rs 3,200 and 5 x 800 = Rs 4,000. Check that the shares add back to the total.
7. How can you find an average quickly using an assumed mean?
Pick a round number close to the values, add up how far each value is from it, and adjust. For 47, 52, 49, 55 and 57, assume 50. The differences are -3, +2, -1, +5 and +7, which add to +10. Divide by the count: 10/5 = 2. So the average is 50 + 2 = 52. This avoids adding large numbers and reduces errors.
8. A can finish a job in 10 days and B in 15 days. How long will they take together?
Use the LCM method. Take the total work as the LCM of 10 and 15, which is 30 units. A does 30/10 = 3 units a day and B does 30/15 = 2 units a day, so together they do 5 units a day. The job takes 30/5 = 6 days. The same answer comes from 1/10 + 1/15 = 1/6, but working in units is faster and avoids fractions.
9. How do you convert between km/h and m/s?
To convert km/h to m/s, multiply by 5/18; to convert m/s to km/h, multiply by 18/5. So 72 km/h = 72 x 5/18 = 20 m/s, and 15 m/s = 15 x 18/5 = 54 km/h. The factor comes from 1 km = 1,000 m and 1 hour = 3,600 seconds. Convert before you calculate whenever a question mixes metres and kilometres.
10. A train 150 m long passes a pole in 10 seconds. What is its speed?
To pass a pole, the train covers its own length, because the pole has no length. Speed = 150/10 = 15 m/s. Converting, 15 x 18/5 = 54 km/h. If the train passes a platform instead, it must cover its own length plus the platform length.
11. How do you find the unit digit of a large power, such as 7^45?
Unit digits of powers repeat in short cycles. For 7, the unit digits go 7, 9, 3, 1 and then repeat every 4 powers. Divide the power by 4: 45 = 4 x 11 + 1, so the remainder is 1 and the unit digit is the first in the cycle, 7. If the remainder is 0, take the last value in the cycle. Digits 0, 1, 5 and 6 always keep their own unit digit.
12. What are the divisibility rules for 3, 4, 8, 9 and 11?
Divisible by 3 or 9: the sum of the digits is divisible by 3 or 9. By 4: the last two digits form a number divisible by 4. By 8: the last three digits form a number divisible by 8. By 11: the difference between the sums of alternate digits is 0 or a multiple of 11. For 2,728: (2 + 2) - (7 + 8) = -11, so 2,728 is divisible by 11, and indeed 2,728 = 11 x 248.
13. How are the HCF and LCM of two numbers related?
For any two numbers, the product of the numbers equals the product of their HCF and LCM. For 12 and 18, the HCF is 6 and the LCM is 36, and 6 x 36 = 216 = 12 x 18. This lets you find a missing value quickly: if the HCF of two numbers is 6, their LCM is 36 and one number is 12, the other is 6 x 36/12 = 18. The rule works for two numbers, not three.
14. A father is 30 years older than his son. In 5 years he will be three times as old as the son. How old is the son now?
Let the son be S, so the father is S + 30. In 5 years: S + 35 = 3(S + 5). That gives S + 35 = 3S + 15, so 2S = 20 and S = 10. The son is 10 and the father is 40. Check: in 5 years they will be 15 and 45, and 45 is three times 15. The age difference never changes, which is often the key to age problems.
15. What is the difference between permutations and combinations?
Permutations count arrangements, where order matters; combinations count selections, where order does not. Choosing 3 people from 7 for a committee is a combination: 7C3 = 35. Choosing a president, secretary and treasurer from 7 is a permutation, because the roles differ: 7P3 = 7 x 6 x 5 = 210. Ask yourself whether swapping two chosen items gives a new result. If it does, use permutations.
16. Two dice are thrown. What is the probability that the sum is 7?
Two dice have 6 x 6 = 36 equally likely outcomes. The pairs that add to 7 are (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1), which is 6 outcomes. The probability is 6/36 = 1/6. A sum of 7 is the most likely total with two dice. Listing outcomes is the safest method for small dice and coin problems.
17. Which area and perimeter formulas come up most in aptitude tests?
Rectangle: area = length x breadth, perimeter = 2(l + b). Square: area = side^2, perimeter = 4 x side. Circle: area = pi r^2, circumference = 2 pi r. Triangle: area = half x base x height. With pi = 22/7, a circle of radius 7 cm has an area of 154 sq cm and a circumference of 44 cm. Questions often choose radii in multiples of 7 so the answers come out whole.
18. What is BODMAS, and how does it change the answer?
BODMAS sets the order of operations: Brackets, Orders (powers and roots), Division and Multiplication from left to right, then Addition and Subtraction from left to right. For 18 + 6 / 3 x 2 - 4: first 6/3 = 2, then 2 x 2 = 4, then 18 + 4 - 4 = 18. Working strictly left to right would give (24/3) x 2 - 4 = 12, which is wrong.
19. How do you calculate percentage change in data interpretation questions?
Percentage change = (new value - old value)/old value x 100. If sales rise from 240 to 300, the change is 60/240 x 100 = 25%. If they fall from 300 to 240, the change is 60/300 x 100 = 20%. The base is always the earlier value, which is why the rise and the fall give different percentages for the same gap.
20. A 20-litre mixture has milk and water in the ratio 3:1. How much of each is in it?
Total parts = 3 + 1 = 4, so one part is 20/4 = 5 litres. Milk = 3 x 5 = 15 litres and water = 1 x 5 = 5 litres. Mixture questions usually add or remove one ingredient next, so always find the actual quantities first. If 5 litres of water is added, the ratio becomes 15:10 = 3:2.
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Aptitude interview questions: Medium questions
20 questions21. What is the shortcut for two successive percentage changes?
For changes of a% and then b%, the net change is a + b + ab/100, using negative values for decreases. A 20% rise followed by a 10% fall gives 20 - 10 + (20 x -10)/100 = 8%, a net rise of 8%. Equal changes in opposite directions, such as +10% then -10%, always give a net loss of a^2/100 = 1%. Do not simply add the percentages.
22. If the price of a product rises by 25%, by how much must consumption fall so that spending stays the same?
Spending = price x quantity. If the price becomes 125% of the old price, quantity must become 100/125 of the old quantity, so it falls by 25/125 x 100 = 20%. The general rule: for a rise of r%, cut consumption by r/(100 + r) x 100 percent. For a fall of r%, consumption can rise by r/(100 - r) x 100 percent.
23. A trader marks his goods 40% above cost and gives a 25% discount. What is his profit percent?
Take the cost price as 100. The marked price is 140. After a 25% discount, the selling price is 140 x 0.75 = 105. The profit is 5 on 100, so 5%. Using multipliers is fastest: 1.40 x 0.75 = 1.05. Many students subtract 25 from 40 and wrongly get 15%.
24. A man sells two phones at Rs 9,000 each, gaining 10% on one and losing 10% on the other. What is the overall result?
When two items are sold at the same price with an equal percent gain and loss, there is always an overall loss of x^2/100 percent, here 100/100 = 1%. Checking with numbers: the cost prices are 9,000/1.1 = about Rs 8,181.82 and 9,000/0.9 = Rs 10,000. The total cost is about Rs 18,181.82 against a selling price of Rs 18,000, a loss of about Rs 181.82, which is 1% of the cost.
25. What is the difference between compound and simple interest for two years?
For 2 years, CI - SI = P x (R/100)^2. On Rs 10,000 at 10%, the difference is 10,000 x 0.01 = Rs 100. This happens because in the second year compound interest also earns interest on the first year's interest: 10% of Rs 1,000 = Rs 100. For 3 years the difference is P(R/100)^2 x (3 + R/100).
26. How do you calculate compound interest for 2 years?
Amount = P(1 + R/100)^T, and CI = amount - P. For Rs 15,000 at 10% for 2 years compounded annually: amount = 15,000 x 1.1 x 1.1 = Rs 18,150, so CI = Rs 3,150. A quick check: year 1 interest is Rs 1,500, and year 2 interest is Rs 1,650 (10% of 16,500). Adding them gives Rs 3,150.
27. If A:B = 2:3 and B:C = 4:5, what is A:B:C?
Make the common term B the same in both ratios. B is 3 in the first and 4 in the second, so multiply the first ratio by 4 and the second by 3: A:B = 8:12 and B:C = 12:15. So A:B:C = 8:12:15. Multiplying by the other ratio's B value always works, and you can simplify at the end if possible.
28. Class A has 30 students with an average of 60 marks and class B has 20 students with an average of 75. What is the combined average?
Use totals, not a simple average of averages. The total for A is 30 x 60 = 1,800 and for B is 20 x 75 = 1,500, so 3,300 marks for 50 students. The combined average is 3,300/50 = 66. The simple mean of 60 and 75 would be 67.5, which is wrong because the classes have different sizes. The combined average always lies closer to the larger group's average.
29. A and B together finish a job in 12 days. A alone takes 20 days. How long does B take alone?
B's daily rate = combined rate - A's rate = 1/12 - 1/20. With an LCM of 60: 5/60 - 3/60 = 2/60 = 1/30. So B alone takes 30 days. In units: total work 60, together 5 units a day, A 3 units, so B does 2 units a day and needs 60/2 = 30 days.
30. A pipe fills a tank in 6 hours, but a leak empties it in 12 hours. How long does the tank take to fill with the leak?
The net rate is the filling rate minus the emptying rate: 1/6 - 1/12 = 2/12 - 1/12 = 1/12 of the tank per hour. So the tank fills in 12 hours. In units: let the tank be 12 units; the pipe adds 2 units an hour and the leak removes 1, so the net is 1 unit an hour and filling takes 12 hours.
31. A car goes from A to B at 40 km/h and returns at 60 km/h. What is the average speed for the round trip?
For equal distances at speeds x and y, average speed = 2xy/(x + y) = 2 x 40 x 60/100 = 48 km/h. You can check with a distance of 120 km: going takes 3 hours, returning takes 2 hours, so 240 km in 5 hours = 48 km/h. The simple mean, 50 km/h, is wrong because the car spends more time at the slower speed.
32. Two trains 120 m and 180 m long run in opposite directions at 54 km/h and 36 km/h. How long do they take to cross each other?
In opposite directions, relative speed = sum of speeds = 90 km/h = 90 x 5/18 = 25 m/s. To cross completely, they cover the sum of their lengths: 120 + 180 = 300 m. Time = 300/25 = 12 seconds. If they ran in the same direction, the relative speed would be the difference, 18 km/h = 5 m/s, and crossing would take 60 seconds.
33. A boat goes downstream at 18 km/h and upstream at 12 km/h. Find the speed of the boat and of the stream.
Downstream speed = boat + stream and upstream speed = boat - stream. So boat speed = (18 + 12)/2 = 15 km/h and stream speed = (18 - 12)/2 = 3 km/h. These two formulas solve most boats and streams questions once the downstream and upstream speeds are known.
34. What is the remainder when 2^100 is divided by 7?
Find a power of 2 that leaves remainder 1: 2^3 = 8 = 7 + 1, so 2^3 leaves remainder 1. Then 2^99 = (2^3)^33 also leaves remainder 1, and 2^100 = 2^99 x 2 leaves remainder 1 x 2 = 2. Looking for a power that leaves remainder 1 (or -1) is the fastest way to handle large powers.
35. How do you find the number of factors of a number such as 360?
Write the number as a product of prime powers, add 1 to each power and multiply. 360 = 2^3 x 3^2 x 5^1, so the number of factors is (3 + 1)(2 + 1)(1 + 1) = 4 x 3 x 2 = 24. This works because each factor chooses a power of 2 from 0 to 3, a power of 3 from 0 to 2, and a power of 5 from 0 to 1.
36. How many zeros are at the end of 100! (100 factorial)?
Each trailing zero needs a factor of 10 = 2 x 5. There are always more 2s than 5s, so count the 5s: 100/5 = 20 numbers give one 5, and 100/25 = 4 numbers give an extra 5. Total = 20 + 4 = 24 zeros. For larger numbers, keep dividing by 125, 625 and so on, adding the whole-number parts.
37. In what ratio must rice at Rs 40/kg be mixed with rice at Rs 55/kg to get a mixture worth Rs 50/kg?
Use alligation: (dearer price - mean price):(mean price - cheaper price) gives the ratio of cheaper to dearer. Here (55 - 50):(50 - 40) = 5:10 = 1:2. So mix 1 kg of the Rs 40 rice with 2 kg of the Rs 55 rice. Check: (40 + 110)/3 = Rs 50 per kg.
38. A invests Rs 40,000 for 12 months and B invests Rs 60,000 for 6 months. How is a profit of Rs 35,000 shared?
Profit is shared in the ratio of capital x time. A: 40,000 x 12 = 4,80,000; B: 60,000 x 6 = 3,60,000. The ratio is 4:3, so A gets 4/7 x 35,000 = Rs 20,000 and B gets 3/7 x 35,000 = Rs 15,000. Forgetting the time factor and sharing 2:3 by capital alone is the most common error.
39. In how many ways can the letters of the word LEVEL be arranged?
LEVEL has 5 letters with L repeated twice and E repeated twice. The number of arrangements = 5!/(2! x 2!) = 120/4 = 30. Dividing by the factorial of each repeat removes arrangements that look the same, such as swapping the two Ls.
40. A bag has 5 red and 3 blue balls. Two balls are drawn together. What is the probability that both are red?
Count ways: choosing 2 red from 5 is 5C2 = 10, and choosing any 2 from 8 is 8C2 = 28. Probability = 10/28 = 5/14. Drawing one at a time without replacement gives the same answer: 5/8 x 4/7 = 20/56 = 5/14.
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Aptitude interview questions: Hard questions
20 questions41. A population rises 10% in the first year, 20% in the second and then falls 25% in the third. What is the overall change?
Multiply the factors: 1.10 x 1.20 x 0.75 = 0.99. So the population ends at 99% of the start, a 1% decrease, even though the changes add up to +5%. Successive changes always multiply. This is the most common trap in multi-step percentage questions.
42. A dishonest shopkeeper uses a 900 g weight instead of 1 kg and also sells at 10% above cost price. What is his real gain percent?
For every 900 g he gives, he charges the cost of 1,000 g plus 10%, that is 1,100 units for goods that cost him 900 units. Gain = (1,100 - 900)/900 x 100 = 22.22%. The two effects multiply: (1,000/900) x 1.1 = 1.2222. Adding 11.11% and 10% would give 21.11%, which is wrong.
43. A sum doubles in 5 years at compound interest. In how many years will it become 8 times?
Under compound interest, the sum keeps doubling every 5 years: 2 times after 5 years, 4 times after 10 and 8 times after 15. So the answer is 15 years. Under simple interest the answer would differ: doubling in 5 years means the interest is 20% a year, so 8 times (7 times the principal as interest) would take 35 years.
44. Find the compound interest on Rs 20,000 for 1 year at 10% per annum, compounded half-yearly.
Half-yearly compounding halves the rate and doubles the number of periods: 5% for 2 periods. Amount = 20,000 x 1.05^2 = 20,000 x 1.1025 = Rs 22,050, so CI = Rs 2,050. With annual compounding it would be Rs 2,000. The extra Rs 50 is the interest on the first half-year's interest.
45. A is twice as efficient as B. Together they finish a job in 18 days. How long would A take alone?
Let B do x units a day, so A does 2x and together they do 3x. If the job is W units, 3x x 18 = W, so W = 54x. A alone needs 54x/2x = 27 days, and B alone needs 54 days. Efficiency ratios turn into daily units directly, which avoids fractions.
46. 12 men working 8 hours a day build a wall in 10 days. How many days will 16 men take working 6 hours a day?
Use M1 x H1 x D1 = M2 x H2 x D2 for the same work. 12 x 8 x 10 = 960 man-hours. With 16 men for 6 hours, that is 96 man-hours a day, so 960/96 = 10 days. If the work also changes, divide each side by its amount of work: M1H1D1/W1 = M2H2D2/W2.
47. Pipes A and B fill a tank in 10 and 15 hours, and pipe C empties it in 30 hours. If all three are opened together, how long will the tank take to fill?
Take the tank as 30 units (the LCM). A adds 3 units an hour, B adds 2 and C removes 1, so the net is 4 units an hour. Time = 30/4 = 7.5 hours. Writing each pipe as units an hour, with a minus sign for outlets, handles any number of pipes.
48. In a 100 m race, A beats B by 10 m and B beats C by 10 m. By how much does A beat C?
When A finishes 100 m, B has run 90 m. When B runs 100 m, C runs 90 m, so C runs 90 m for every 100 m of B. When B has run 90 m, C has run 90 x 90/100 = 81 m. So A beats C by 100 - 81 = 19 m, not 20 m. Race questions are ratios of distance in the same time.
49. Two runners on a 400 m circular track run at 5 m/s and 3 m/s from the same point. How often do they meet if they run in the same direction, and in opposite directions?
In the same direction, the faster runner gains 2 m/s and must gain a full lap (400 m) to meet again: every 400/2 = 200 seconds. In opposite directions they close the gap at 8 m/s: every 400/8 = 50 seconds. The rule: time between meetings = track length/relative speed.
50. A train crosses a 300 m platform in 30 seconds and a man standing on the platform in 15 seconds. Find the length and speed of the train.
Crossing the man takes the train's own length L: L = 15v. Crossing the platform takes L + 300: L + 300 = 30v. Subtracting, 300 = 15v, so v = 20 m/s = 72 km/h, and L = 15 x 20 = 300 m. The 15 extra seconds are the time taken to cover the platform alone.
51. Find the smallest number that leaves remainder 3 when divided by 5 and remainder 4 when divided by 7.
List numbers leaving remainder 3 on division by 5: 3, 8, 13, 18, 23 and so on. Check each against division by 7: 18 = 2 x 7 + 4. So the answer is 18. Larger solutions repeat every LCM(5, 7) = 35: 18, 53, 88, and so on. Listing the larger divisor's sequence first is even faster.
52. What is the largest number that divides 1,305, 4,665 and 6,905 leaving the same remainder each time?
If each number leaves the same remainder r, the divisor divides their differences: 4,665 - 1,305 = 3,360, 6,905 - 4,665 = 2,240 and 6,905 - 1,305 = 5,600. The HCF of 3,360, 2,240 and 5,600 is 1,120 (3 x 1,120, 2 x 1,120 and 5 x 1,120). So the answer is 1,120, and the common remainder is 1,305 - 1,120 = 185.
53. The ages of A and B are in the ratio 4:5. Eight years ago the ratio was 3:4. Find their present ages.
Let the ages be 4k and 5k. Eight years ago: (4k - 8)/(5k - 8) = 3/4. Cross-multiplying, 16k - 32 = 15k - 24, so k = 8. A is 32 and B is 40. Check: eight years ago they were 24 and 32, a ratio of 3:4.
54. A container has 40 litres of milk. 4 litres is replaced with water, and this is done three times in all. How much milk is left?
Each step keeps (40 - 4)/40 = 0.9 of the milk. After three steps, milk = 40 x 0.9^3 = 40 x 0.729 = 29.16 litres. The formula is: final = initial x (1 - removed/total)^n. Subtracting 4 litres three times (getting 28 litres) is wrong, because later removals take out a mixture, not pure milk.
55. A committee of 3 is chosen from 5 men and 4 women. In how many ways can it include at least one woman?
Use the complement: all committees minus those with no women. All committees = 9C3 = 84. All-men committees = 5C3 = 10. So at least one woman = 84 - 10 = 74. Counting "pick one woman, then any two others" gives 4 x 8C2 = 112, which double-counts committees with two or three women.
56. Six people sit around a round table. In how many arrangements do A and B not sit together?
Circular arrangements of 6 people = (6 - 1)! = 120. Arrangements with A and B together: treat them as one unit, giving 5 units in (5 - 1)! = 24 ways, and A and B can swap, so 48. Not together = 120 - 48 = 72. Using "total minus the unwanted case" is usually easier than counting the wanted case directly.
57. A die is thrown three times. What is the probability of getting at least one six?
Use the complement. The probability of no six in one throw is 5/6, so no six in three throws is (5/6)^3 = 125/216. At least one six = 1 - 125/216 = 91/216, about 42%. Whenever a question says "at least one", calculating "none" and subtracting from 1 is the quickest route.
58. Two cards are drawn without replacement from a pack of 52. What is the probability that both are aces?
The first card is an ace with probability 4/52. Given that, the second is an ace with probability 3/51. Multiply: 4/52 x 3/51 = 12/2,652 = 1/221. Using combinations gives the same result: 4C2/52C2 = 6/1,326 = 1/221.
59. A cylinder of radius 6 cm and height 15 cm is melted and recast into cones of radius 3 cm and height 5 cm. How many cones are made?
Volume is conserved. Cylinder = pi x 6^2 x 15 = 540 pi. Each cone = (1/3) x pi x 3^2 x 5 = 15 pi. Number of cones = 540 pi/15 pi = 36. In melting and recasting problems, set the total volume before equal to the total volume after; pi always cancels.
60. A company's revenue grew from Rs 200 crore to Rs 242 crore in two years at a steady rate. What was the yearly growth rate?
Steady growth compounds: 200 x (1 + r)^2 = 242, so (1 + r)^2 = 1.21 and 1 + r = 1.1. The rate is 10% a year. The simple average, 21%/2 = 10.5%, overstates it. In data interpretation questions, check whether growth is meant to be simple or compound before you calculate.
Practice questions for November 2026
New every month3 questions from the Aptitude mock test, the way they appear in it. A new set comes on the 1st of every month. Pick your answer, then open it.
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Simplify: 25 + 7 x 3 - 35
- 61
- 11
- -199
- 18
Show the answer
Answer: 11. Multiply first (BODMAS): 7 x 3 = 21. Then 25 + 21 - 35 = 11.
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A man drives from home to office at 45 km/h and returns by the same road at 90 km/h. What is his average speed for the whole journey?
- 67.5 km/h
- 62 km/h
- 60 km/h
- 57 km/h
Show the answer
Answer: 60 km/h. For equal distances, average speed = 2xy/(x + y) = 2 x 45 x 90/135 = 60 km/h, not the simple mean.
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A bag has Re 1, 50 paise and 25 paise coins in the ratio 3:4:5. The total value is Rs 156.25. How many 50 paise coins are there?
- 200
- 125
- 75
- 100
Show the answer
Answer: 100. Value per set of 3, 4, 5 coins = 3 + 4/2 + 5/4 = 6.25 rupees. Sets = 156.25/6.25 = 25, so 50 paise coins = 4 x 25 = 100.
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