Aptitude Q&A (Set 1with Explanations)
Q1. A sum of ₹12,000 amounts to ₹15,600 in 3 years at simple interest. Find the rate of interest per annum.
Solution:
• Amount (A) = ₹15,600, Principal (P) = ₹12,000
• SI = A – P = 15,600 – 12,000 = ₹3,600
• Formula: SI = (P × R × T)/100
• 3,600 = (12,000 × R × 3)/100
• R = (3,600 × 100)/(36,000) = 10%
Answer: Rate = 10% per annum
Q2. A man walks at 5 km/h instead of 4 km/h and reaches his office 15 minutes early. Find the distance.
Solution:
• Time saved = 15 min = 15/60 = 0.25 hr
• Let distance = d.
• Time at 4 km/h = d/4, at 5 km/h = d/5
• Difference = d/4 – d/5 = 0.25
• (5d – 4d)/20 = 0.25 → d/20 = 0.25 → d = 5 km
Answer: Distance = 5 km
Q3. The cost price of 15 pens is equal to the selling price of 12 pens. Find the gain %.
Solution:
• Let CP = ₹1 per pen → CP(15 pens) = ₹15
• Given SP(12 pens) = ₹15 → SP per pen = 15/12 = ₹1.25
• Profit per pen = 1.25 – 1 = ₹0.25
• Profit % = (Profit/CP) × 100 = (0.25/1) × 100 = 25%
Answer: Gain = 25%
Q4. A shopkeeper allows 10% discount on goods and still gains 20%. If SP = ₹216, find CP.
Solution:
• Let CP = 100 → SP = 120 (since 20% gain).
• If SP is after 10% discount, then MP = SP/0.9 = 120/0.9 = 133.33.
• Now given SP = 216.
• Scale CP = (100/120) × 216 = 180.
Answer: Cost Price = ₹180
Q5. A tank can be filled by a pipe in 6 hours and emptied by another in 8 hours. If both are opened, how long will it take to fill?
Solution:
• Filling pipe’s 1-hr work = 1/6
• Emptying pipe’s 1-hr work = 1/8
• Net 1-hr work = 1/6 – 1/8 = (4 – 3)/24 = 1/24
• Total time = 24 hrs
Answer: Tank fills in 24 hours
Q6. A sum triples itself in 12 years at simple interest. In how many years will it double?
Solution:
• Tripling → A = 3P → SI = 2P in 12 years.
• Rate = (SI × 100)/(P × T) = (2P × 100)/(P × 12) = 200/12 = 16.67%
• For doubling → SI = P.
• Time = (SI × 100)/(P × Rate) = (P × 100)/(P × 16.67) = 6 yrs
Answer: It will double in 6 years
Q7. The average age of 8 men is increased by 2 years when one of them aged 30 is replaced by a new man. Find the new man’s age.
Solution:
• Increase in average = 2 yrs for 8 men = total increase of 16 yrs.
• New man’s age = Old man’s age + increase = 30 + 16 = 46 yrs
Answer: New man = 46 years
Q8. The ratio of incomes of A and B is 3:5 and their expenditures are 4:7. If A saves ₹50 and B saves ₹100, find their incomes.
Solution:
• Let incomes = 3x, 5x; expenditures = 4y, 7y
• Savings → 3x – 4y = 50 …(1)
• 5x – 7y = 100 …(2)
• Multiply (1) × 5 → 15x – 20y = 250
• Multiply (2) × 3 → 15x – 21y = 300
• Subtract → y = 50
• From (1): 3x – 200 = 50 → 3x = 250 → x = 83.33
• Income A = 3x = 250, Income B = 5x = 416.67
Answer: A = ₹250, B = ₹416.67
Q9. A sum of money is invested at 20% per annum compounded annually. It amounts to ₹864 after 3 years. Find principal.
Solution:
• Formula: A = P(1 + R/100)^T
• 864 = P(1.2)^3 = P × 1.728
• P = 864/1.728 = 500
Answer: Principal = ₹500
Q10. A train crosses a man in 12 seconds and a bridge 180 m long in 42 seconds. Find the length of train.
Solution:
• Let train length = L
• Speed = L/12 m/s
• Time for bridge = (L + 180)/(L/12) = 42
• (L + 180) × 12 = 42L
• 12L + 2160 = 42L → 30L = 2160 → L = 72 m
Answer: Length = 72 meters
Q11. Two numbers are in the ratio 4 : 7. If their sum is 220, find the numbers.
Solution:
• Let numbers = 4x and 7x.
• Sum: 4x + 7x = 11x = 220 → x = 20.
• Numbers: 4x = 80 and 7x = 140.
Answer: 80 and 140.
Q12. A shopkeeper mixes two varieties of rice costing ₹40/kg and ₹60/kg in the ratio 3:2. What is the cost price per kg of the mixture?
Solution:
• Weighted average = (3×40 + 2×60) / (3+2) = (120 + 120)/5 = 240/5 = 48.
Answer: ₹48 per kg.
Q13. A car covers a distance of 150 km partly at 50 km/h and partly at 75 km/h. If total time taken is 3 hours, find the distance covered at 75 km/h.
Solution:
• Let distance at 75 km/h = x km, at 50 km/h = 150 − x.
• Time = (150 − x)/50 + x/75 = 3.
• Multiply by 150 (LCM): 3(150 − x) + 2x = 450 → 450 − 3x + 2x = 450 → −x = 0 → x = 0.
• That says all distance at 50 km/h — but check arithmetic: better multiply properly.
Redo: Multiply equation by 150: 3(150 − x) + 2x = 450 → 450 − 3x + 2x = 450 → (−x) = 0 → x = 0.
Interpretation: Given times, implies travel entirely at 50 km/h fits 150/50 = 3 hours. So distance at 75 km/h is 0 km. (Problem is consistent.)
Answer: 0 km (i.e., whole trip at 50 km/h).
Q14. If 8 men can complete a piece of work in 15 days, how many men are required to finish it in 6 days (assuming same efficiency)?
Solution:
• Work ∝ men × days. Let total work = W.
• 8 × 15 = 120 man-days = W. To finish in 6 days, required men = W / 6 = 120 / 6 = 20.
Answer: 20 men.
Q15. From a deck of 52 cards, one card is drawn at random. What is the probability that it is a king or a heart?
Solution:
• Number of kings = 4. Number of hearts = 13. Intersection (king of hearts) = 1.
• P(king ∪ heart) = (4 + 13 − 1) / 52 = 16/52 = 4/13.
Answer: 4/13.
Q16. Find the least number which when divided by 9, 12 and 15 leaves remainder 7 in each case.
Solution:
We seek N such that N ≡ 7 (mod 9), (mod 12), (mod 15). Let N − 7 be divisible by 9,12,15 → LCM(9,12,15) =
• 9 = 3², 12 = 2²·3, 15 = 3·5 → LCM = 2²·3²·5 = 180.
So N − 7 = 180k. Least positive k = 1 → N = 180 + 7 = 187.
Answer: 187.
Q17. A number increased by 20% and then decreased by 20%. What is the net percentage change?
Solution:
• Let initial value = 100. After +20% → 120. After −20% → 120 × 0.8 = 96.
• Net change = 96 − 100 = −4 → a 4% decrease.
Answer: 4% decrease.
Q18. How many distinct 4-letter words (real or nonsense) can be formed from the letters of the word “APPLE” if letters cannot be repeated?
Solution:
• “APPLE” letters: A, P, P, L, E → counts: P twice, others once. We need 4-letter arrangements without repeating a letter more times than available. Cases: include one P or include both Ps.
Case 1: Use both Ps + 2 of {A,L,E} → choose 2 from 3 = 3 choices. For each choice, letters multiset has 4 distinct positions with two identical Ps → permutations = 4! / 2! = 12. So total = 3 × 12 = 36.
Case 2: Use exactly one P + choose 3 from {A,L,E} (i.e., all three) → letters are P,A,L,E all distinct → permutations = 4! = 24.
Total = 36 + 24 = 60.
Answer: 60.
Q19. A bag contains 5 red and 7 blue balls. Two balls are drawn without replacement. What is the probability both are red?
Solution:
• P(first red) = 5/12. Then P(second red) = 4/11. So probability = (5/12)×(4/11) = 20/132 = 5/33.
Answer: 5/33.
Q20. A sum of ₹10,000 is lent at 5% per annum simple interest. How much interest will be earned in 2 years 6 months?
Solution:
• Time = 2.5 years. SI = P×R×T/100 = 10000×5×2.5/100 = 10000×12.5/100 = 1250.
Answer: ₹1,250.
Q21. The ages of two siblings are in the ratio 5 : 7. If the sum of their ages is 96 years, find both ages.
Solution:
• Let ages = 5x5x5x and 7x7x7x.
• Sum: 5x+7x=12x=96⇒x=85x + 7x = 12x = 96 \Rightarrow x = 85x+7x=12x=96⇒x=8.
• Ages: 5x=405x = 405x=40 years and 7x=567x = 567x=56 years.
Answer: 40 years and 56 years.
Q22. 12 men can complete a job in 10 days. How many men are required to finish it in 8 days (same efficiency)?
Solution:
• Total work = 12×10=12012 \times 10 = 12012×10=120 man-days.
• Required men = 120/8=15120/8 = 15120/8=15.
Answer: 15 men.
Q23. Two trains of lengths 100 m and 150 m are running in opposite directions at speeds 54 km/h and 36 km/h respectively. How long will they take to cross each other?
Solution:
• Relative speed = 54+36=9054 + 36 = 9054+36=90 km/h. Convert to m/s: 90×1000/3600=2590 \times 1000/3600 = 2590×1000/3600=25 m/s.
• Distance to cover = 100+150=250100 + 150 = 250100+150=250 m.
• Time =250/25=10= 250/25 = 10=250/25=10 seconds.
Answer: 10 seconds.
Q24. What is the amount and compound interest on ₹8,000 for 2 years at 10% per annum compounded annually?
Solution:
• Amount =P(1+r)t=8000×(1.1)2=8000×1.21=9680= P(1 + r)^t = 8000 \times (1.1)^2 = 8000 \times 1.21 = 9680=P(1+r)t=8000×(1.1)2=8000×1.21=9680.
• Compound interest =9680−8000=1680= 9680 - 8000 = 1680=9680−8000=1680.
Answer: Amount = ₹9,680, CI = ₹1,680.
Q25. A square has side 10 cm. Find the length of its diagonal.
Solution:
• Diagonal of square =side×2=102= \text{side} \times \sqrt{2} = 10\sqrt{2}=side×2=102 cm.
• If decimal needed: 102≈10×1.414=14.1410\sqrt{2} \approx 10 \times 1.414 = 14.14102≈10×1.414=14.14 cm.
Answer: 10210\sqrt{2}102 cm (≈ 14.14 cm).
Q26. Three fair coins are tossed. What is the probability of getting at least one head?
Solution:
• Total outcomes = 23=82^3 = 823=8.
• Probability of no head (all tails) = 1/81/81/8.
• So probability of at least one head =1−1/8=7/8= 1 - 1/8 = 7/8=1−1/8=7/8.
Answer: 7/8.
Q27. How many distinct 5-letter arrangements can be formed from the letters of the word “LEVEL”?
Solution:
• Letters: L, E, V, E, L → counts: L ×2, E ×2, V ×1.
• Number of distinct permutations =5!2!×2!=1204=30= \dfrac{5!}{2! \times 2!} = \dfrac{120}{4} = 30=2!×2!5!=4120=30.
Answer: 30.
Q28. Pipe A fills a tank in 9 hours, pipe B fills it in 12 hours, and pipe C empties it in 20 hours. If all three are opened together, how long to fill the tank?
Solution:
• A’s 1-hr work = 1/91/91/9. B’s = 1/121/121/12. C (emptying) = −1/20-1/20−1/20.
• Net 1-hr work =1/9+1/12−1/20= 1/9 + 1/12 - 1/20=1/9+1/12−1/20. Find LCM 180:
=(20+15−9)/180=26/180=13/90= (20 + 15 - 9)/180 = 26/180 = 13/90=(20+15−9)/180=26/180=13/90.
• Time to fill =1÷(13/90)=90/13≈6.923= 1 \div (13/90) = 90/13 \approx 6.923=1÷(13/90)=90/13≈6.923 hours ≈ 6 hours 55 minutes.
Answer: 90/1390/1390/13 hours ≈ 6 h 55 min.
Q29. A mixture contains milk and water in the ratio 7:3. If total mixture is 20 litres, how many litres of milk are there?
Solution:
• Milk fraction =7/(7+3)=7/10= 7/(7+3) = 7/10=7/(7+3)=7/10.
• Milk =7/10×20=14= 7/10 \times 20 = 14=7/10×20=14 litres.
Answer: 14 litres of milk.
Q30. The average of 9 numbers is 20. One of the numbers is 12 and is replaced by xxx. After replacement the average becomes 21. Find xxx.
Solution:
• Original total = 9×20=1809 \times 20 = 1809×20=180.
• New total = 9×21=1899 \times 21 = 1899×21=189.
• New total = original total − 12 + xxx: 180−12+x=189⇒x=21180 - 12 + x = 189 \Rightarrow x = 21180−12+x=189⇒x=21.
Answer: 21.
Q31. A shop gives a discount of 15% on the marked price. If the marked price is ₹400, what is the selling price and the discount amount?
Solution:
• Discount = 15% of 400 = 0.15×400=₹600.15 \times 400 = ₹600.15×400=₹60.
• Selling Price (SP) = Marked Price − Discount = 400−60=₹340400 - 60 = ₹340400−60=₹340.
Answer: Discount = ₹60, SP = ₹340.
Q32. If the population of a town increases by 10% per year, what will be the population after 2 years if current population is 50,000?
Solution:
• Growth factor per year = 1+10%=1.11 + 10\% = 1.11+10%=1.1.
• After 2 years: 50,000×1.12=50,000×1.21=60,50050{,}000 \times 1.1^2 = 50{,}000 \times 1.21 = 60{,}50050,000×1.12=50,000×1.21=60,500.
Answer: 60,500 people.
Q33. Find the LCM of 14, 20 and 28.
Solution:
• Prime factors: 14=2×7, 20=22×5, 28=22×7.14=2\times7,\;20=2^2\times5,\;28=2^2\times7.14=2×7,20=22×5,28=22×7.
• Take highest powers: 22,51,712^2, 5^1, 7^122,51,71.
• LCM = 4×5×7=1404 \times 5 \times 7 = 1404×5×7=140.
Answer: 140.
Q34. A retailer marks up price by 25% on cost price and then offers 10% discount on marked price. If cost price is ₹800, find his profit percent.
Solution:
• Marked Price (MP) = 800×1.25=₹1000800 \times 1.25 = ₹1000800×1.25=₹1000.
• After 10% discount, SP = 1000×0.9=₹9001000 \times 0.9 = ₹9001000×0.9=₹900.
• Profit = 900−800=₹100900 - 800 = ₹100900−800=₹100. Profit% = (100/800)×100=12.5%(100/800)\times100 = 12.5\%(100/800)×100=12.5%.
Answer: 12.5% profit.
Q35. The sum of three consecutive even numbers is 78. Find the numbers.
Solution:
• Let numbers = x,x+2,x+4x, x+2, x+4x,x+2,x+4. Sum: 3x+6=78⇒3x=72⇒x=243x + 6 = 78 \Rightarrow 3x = 72 \Rightarrow x = 243x+6=78⇒3x=72⇒x=24.
• Numbers: 24, 26, 28.
Answer: 24, 26, 28.
Q36. Two pipes fill a cistern in 15 minutes and 20 minutes respectively. A third pipe can empty it in 30 minutes. If all three are opened together, how long will it take to fill the cistern?
Solution:
• A’s 1-min work = 1/151/151/15. B’s = 1/201/201/20. C (emptying) = −1/30-1/30−1/30.
• Net 1-min work = 1/15+1/20−1/301/15 + 1/20 - 1/301/15+1/20−1/30. LCM 60: (4+3−2)/60=5/60=1/12(4 + 3 - 2)/60 = 5/60 = 1/12(4+3−2)/60=5/60=1/12.
• Time to fill = 121212 minutes.
Answer: 12 minutes.
Q37. A die is rolled twice. What is the probability that the sum of the two outcomes is 7?
Solution:
• Total outcomes = 6×6=366 \times 6 = 366×6=36.
• Favorable pairs summing to 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 outcomes.
• Probability = 6/36=1/66/36 = 1/66/36=1/6.
Answer: 1/6.
Q38. If x+1x=5x + \dfrac{1}{x} = 5x+x1=5, find x2+1x2x^2 + \dfrac{1}{x^2}x2+x21.
Solution:
• Square both sides: (x+1x)2=x2+1x2+2\left(x + \dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} + 2(x+x1)2=x2+x21+2.
• So 25=x2+1x2+2⇒x2+1x2=2325 = x^2 + \dfrac{1}{x^2} + 2 \Rightarrow x^2 + \dfrac{1}{x^2} = 2325=x2+x21+2⇒x2+x21=23.
Answer: 23.
Q39. A wire of length 44 cm is bent to form a rectangle such that the length is twice the breadth. Find the dimensions.
Solution:
• Let breadth = bbb, length = 2b2b2b. Perimeter =2(length+breadth)=2(2b+b)=6b= 2(\text{length} + \text{breadth}) = 2(2b + b) = 6b=2(length+breadth)=2(2b+b)=6b.
• Given perimeter =44⇒6b=44⇒b=44/6=7.3‾= 44 \Rightarrow 6b = 44 \Rightarrow b = 44/6 = 7.\overline{3}=44⇒6b=44⇒b=44/6=7.3 cm = 7137\frac{1}{3}731 cm.
• Length = 2b=14232b = 14\frac{2}{3}2b=1432 cm. Or in fraction: breadth =22/3= 22/3=22/3 cm, length =44/3= 44/3=44/3 cm.
Answer: Breadth = 22/322/322/3 cm (≈7.33 cm), Length = 44/344/344/3 cm (≈14.67 cm).
Q40. A man buys an article for ₹600 and sells it at a loss of 12%. What is the selling price?
Solution:
• Loss of 12% → he receives 88%88\%88% of cost. SP = 600×0.88=₹528600 \times 0.88 = ₹528600×0.88=₹528.
Answer: ₹528.
Q41. A sum of money becomes four times in 6 years at simple interest. Find the rate of interest per annum.
Solution:
• If principal = PPP, amount after 6 years = 4P4P4P. So simple interest =4P−P=3P= 4P - P = 3P=4P−P=3P over 6 years.
• SI formula: SI=P×R×T100\text{SI} = \dfrac{P \times R \times T}{100}SI=100P×R×T.
• 3P=P×R×6100⇒3=6R100⇒R=3006=50%3P = \dfrac{P \times R \times 6}{100} \Rightarrow 3 = \dfrac{6R}{100} \Rightarrow R = \dfrac{300}{6} = 50\%3P=100P×R×6⇒3=1006R⇒R=6300=50%.
Answer: 50% per annum.
Q42. A train 120 m long crosses a platform in 30 seconds at a speed of 72 km/h. Find the length of the platform.
Solution:
• Speed = 727272 km/h = 72×10003600=2072 \times \tfrac{1000}{3600} = 2072×36001000=20 m/s.
• In 30 s distance covered = 20×30=60020 \times 30 = 60020×30=600 m. That equals (train length + platform length).
• Platform length = 600−120=480600 - 120 = 480600−120=480 m.
Answer: 480 meters.
Q43. A and B can do a piece of work in 12 and 20 days respectively. They work together for 4 days, then A leaves. How many more days will B take to finish the remaining work?
Solution:
• A’s 1-day work = 1/121/121/12. B’s = 1/201/201/20. Together = 1/12+1/20=(5+3)/60=8/60=2/151/12 + 1/20 = (5+3)/60 = 8/60 = 2/151/12+1/20=(5+3)/60=8/60=2/15 per day.
• Work done in 4 days = 4×2/15=8/154 \times 2/15 = 8/154×2/15=8/15. Remaining = 1−8/15=7/151 - 8/15 = 7/151−8/15=7/15.
• B’s rate = 1/201/201/20 → time = (7/15)÷(1/20)=(7/15)×20=28/3=913(7/15) \div (1/20) = (7/15) \times 20 = 28/3 = 9\frac{1}{3}(7/15)÷(1/20)=(7/15)×20=28/3=931 days = 9 days 8 hours.
Answer: 9139\frac{1}{3}931 days (9 days 8 hours).
Q44. A box contains 6 red, 4 blue and 5 green balls. One ball is drawn at random. What is the probability it is blue or green?
Solution:
• Total balls = 6+4+5=156 + 4 + 5 = 156+4+5=15. Blue or green favourable = 4+5=94 + 5 = 94+5=9.
• Probability = 9/15=3/59/15 = 3/59/15=3/5.
Answer: 3/5.
Q45. The ratio of the investments of A, B and C is 4 : 5 : 6. Profit of ₹3,000 is distributed accordingly. What is C’s share?
Solution:
• Total parts = 4+5+6=154 + 5 + 6 = 154+5+6=15. C’s share = 6/15×3000=(2/5)×3000=12006/15 \times 3000 = (2/5)\times3000 = 12006/15×3000=(2/5)×3000=1200.
Answer: ₹1,200.
Q46. A rectangular park has length 50 m and breadth 30 m. A path of uniform width is made outside the park increasing the area by 520 m². Find the width of the path.
Solution:
• Let width = xxx. New length = 50+2x50 + 2x50+2x, new breadth = 30+2x30 + 2x30+2x.
• Increase in area = (50+2x)(30+2x)−50×30=520(50+2x)(30+2x) - 50\times30 = 520(50+2x)(30+2x)−50×30=520.
• Expand: 1500+100x+60x+4x2−1500=5201500 + 100x + 60x + 4x^2 - 1500 = 5201500+100x+60x+4x2−1500=520 → 160x+4x2=520160x + 4x^2 = 520160x+4x2=520.
• Simplify: divide by 4 → 40x+x2=13040x + x^2 = 13040x+x2=130 → x2+40x−130=0x^2 + 40x - 130 = 0x2+40x−130=0.
• Solve quadratic: x=−40±1600+5202=−40±21202x = \dfrac{-40 \pm \sqrt{1600 + 520}}{2} = \dfrac{-40 \pm \sqrt{2120}}{2}x=2−40±1600+520=2−40±2120.
• 2120≈46.05\sqrt{2120} \approx 46.052120≈46.05. Positive root: (−40+46.05)/2≈3.025(-40 + 46.05)/2 \approx 3.025(−40+46.05)/2≈3.025 m (approx).
• Width ≈ 3 m (practical rounding).
Answer: Approximately 3 m.
Q47. In an election, candidate A got 54% of valid votes. If invalid votes were 2% of total votes and total votes were 50,000, how many votes did A get?
Solution:
• Invalid votes = 2% of 50,000 = 100010001000. Valid votes = 50,000−1,000=49,00050,000 - 1,000 = 49,00050,000−1,000=49,000.
• A’s votes = 54%54\% 54% of valid = 0.54×49,000=26,4600.54 \times 49,000 = 26,4600.54×49,000=26,460.
Answer: 26,460 votes.
Q48. Solve for xxx: 3x+1=813^{x+1} = 813x+1=81.
Solution:
• 81=3481 = 3^481=34. So 3x+1=34⇒x+1=4⇒x=33^{x+1} = 3^4 \Rightarrow x + 1 = 4 \Rightarrow x = 33x+1=34⇒x+1=4⇒x=3.
Answer: x=3x = 3x=3.
Q49. A company increases the price of a product by 20% and then offers a discount of 10%. What is the net % increase from original price?
Solution:
• Let original price = 100. After +20% → 120. After −10% on 120 → 120×0.9=108120 \times 0.9 = 108120×0.9=108.
• Net change = 108−100=8108 - 100 = 8108−100=8 → 8% increase.
Answer: 8% increase.
Q50. How many ways can 4 students be seated in a row of 6 chairs?
Solution:
• Choose 4 chairs out of 6: 6C4=15 {}^6C_4 = 156C4=15. For each selection, arrange 4 students: 4!=244! = 244!=24.
• Total = 15×24=36015 \times 24 = 36015×24=360. (Alternatively: permutations P(6,4)=6×5×4×3=360P(6,4)=6\times5\times4\times3=360P(6,4)=6×5×4×3=360).
Answer: 360 ways.
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