Quantitative Aptitude — MCQ Practice Questions & Answers

Question 1

A train travels 240 km in 4 hours. If it increases its speed by 20%, how much distance will it cover in 5 hours at the new speed?

Accepted Answer

Correct answer: B. 360 km — Step 1: Calculate the original speed The train travels 240 km in 4 hours, so we divide distance by time. 	ext{Original Speed} = \frac{240 	ext{ km}}{4 	ext{ hours}} = 60 	ext{ km/h} Step 2: Calculate the new speed after 20% increase The speed increases by 20%, so we multiply the original speed by 1.20. 	ext{New Speed} = 60 	imes 1.20 = 72 	ext{ km/h} Step 3: Calculate distance covered in 5 hours at new speed Distance equals speed multiplied by time. 	ext{Distance} = 72 	ext{ km/h} 	imes 5 	ext{ hours} = 360 	ext{ km} **The train will cover 360 km in 5 hours at the new speed.**

Answer

A. 300 km

Answer

C. 330 km

Answer

D. 288 km

Question 2

A man walks from point A to point B at 4 km/h and returns from B to A at 6 km/h. If the total journey takes 5 hours, what is the distance between A and B?

Accepted Answer

Correct answer: B. 12 km — Step 1: Define Variables and Set Up Equations Let the distance between A and B be d km. Time to go from A to B at 4 km/h is \frac{d}{4} hours, and time to return from B to A at 6 km/h is \frac{d}{6} hours. 	ext{Total time} = \frac{d}{4} + \frac{d}{6} = 5 Step 2: Find Common Denominator and Simplify The common denominator of 4 and 6 is 12. Rewrite each fraction with denominator 12. \frac{3d}{12} + \frac{2d}{12} = 5 Step 3: Solve for Distance Combine the fractions on the left side and solve for d . \frac{5d}{12} = 5 5d = 5 	imes 12 5d = 60 d = 12 	ext{ km} **The distance between A and B is 12 km.**

Answer

A. 10 km

Answer

C. 11 km

Answer

D. 15 km

Question 3

Two cyclists start simultaneously from the same point and travel in opposite directions on a circular track of 600 m. If their speeds are 8 m/s and 10 m/s respectively, after how much time will they meet again at the starting point?

Accepted Answer

Correct answer: C. 200 seconds — Step 1: Find the time for each cyclist to complete one full lap [Cyclist 1 completes one lap] 	ext{Time}_1 = \frac{	ext{Track length}}{	ext{Speed}_1} = \frac{600}{8} = 75 	ext{ seconds} [Cyclist 2 completes one lap] 	ext{Time}_2 = \frac{	ext{Track length}}{	ext{Speed}_2} = \frac{600}{10} = 60 	ext{ seconds} Step 2: Find the LCM of their lap times [The cyclists will meet at the starting point when the time elapsed is a common multiple of both lap times] 	ext{LCM}(75, 60) = 	ext{LCM of lap times} [Finding prime factorization: 75 = 3 × 5², 60 = 2² × 3 × 5] 	ext{LCM}(75, 60) = 2^2 	imes 3 	imes 5^2 = 4 	imes 3 	imes 25 = 300 	ext{ seconds} Step 3: Verify the answer [In 300 seconds, Cyclist 1 completes] 	ext{Laps}_1 = \frac{300}{75} = 4 	ext{ laps} [In 300 seconds, Cyclist 2 completes] 	ext{Laps}_2 = \frac{300}{60} = 5 	ext{ laps} [Both return to the starting point after completing whole laps

Answer

A. 100 seconds

Answer

B. 150 seconds

Answer

D. 300 seconds

Question 4

A boat takes 6 hours to travel 120 km downstream and 8 hours to travel the same distance upstream. What is the speed of the boat in still water?

Accepted Answer

Correct answer: A. 17.5 km/h — Step 1: Find the downstream speed The boat travels 120 km downstream in 6 hours. 	ext{Downstream speed} = \frac{120}{6} = 20 	ext{ km/h} Step 2: Find the upstream speed The boat travels 120 km upstream in 8 hours. 	ext{Upstream speed} = \frac{120}{8} = 15 	ext{ km/h} Step 3: Calculate the speed of boat in still water The speed of boat in still water is the average of downstream and upstream speeds. 	ext{Speed of boat} = \frac{	ext{Downstream speed} + 	ext{Upstream speed}}{2} = \frac{20 + 15}{2} = \frac{35}{2} = 17.5 	ext{ km/h} **The speed of the boat in still water is 17.5 km/h.**

Answer

B. 15 km/h

Answer

C. 16 km/h

Answer

D. 18 km/h

Question 5

A car travels at 60 km/h for the first half of the distance and at 90 km/h for the second half of the distance. What is the average speed for the entire journey?

Accepted Answer

Correct answer: A. 72 km/h — Step 1: Define Variables and Total Distance Let the total distance be D km. The car travels \frac{D}{2} km at each speed. 	ext{Total Distance} = D 	ext{ km} Step 2: Calculate Time for Each Half For the first half at 60 km/h: t_1 = \frac{D/2}{60} = \frac{D}{120} 	ext{ hours} For the second half at 90 km/h: t_2 = \frac{D/2}{90} = \frac{D}{180} 	ext{ hours} Step 3: Calculate Total Time and Average Speed Total time for the journey: t_{	ext{total}} = \frac{D}{120} + \frac{D}{180} = \frac{3D}{360} + \frac{2D}{360} = \frac{5D}{360} = \frac{D}{72} 	ext{ hours} Average speed is total distance divided by total time: 	ext{Average Speed} = \frac{D}{t_{	ext{total}}} = \frac{D}{\frac{D}{72}} = 72 	ext{ km/h} **The average speed for the entire journey is 72 km/h.**

Answer

B. 75 km/h

Answer

C. 70 km/h

Answer

D. 74 km/h

Question 6

A shopkeeper buys notebooks at ₹40 each and sells them at ₹50 each. If he sells 120 notebooks in a day, what is his total profit?

Accepted Answer

Correct answer: A. ₹1,200 — Step 1: Calculate Total Cost Price The shopkeeper buys 120 notebooks at ₹40 each. 	ext{Total Cost Price} = 120 	imes 40 = ₹4,800 Step 2: Calculate Total Selling Price The shopkeeper sells 120 notebooks at ₹50 each. 	ext{Total Selling Price} = 120 	imes 50 = ₹6,000 Step 3: Calculate Total Profit Profit is the difference between selling price and cost price. 	ext{Total Profit} = 	ext{Selling Price} - 	ext{Cost Price} = 6,000 - 4,800 = ₹1,200 **The shopkeeper's total profit for selling 120 notebooks in a day is ₹1,200.**

Answer

B. ₹1,500

Answer

C. ₹1,800

Answer

D. ₹2,000

Question 7

A trader bought a watch for ₹800. He marked it at 60% above the cost price but offered a discount of 20% on the marked price. What is his profit percentage?

Accepted Answer

Correct answer: A. 28% — Step 1: Calculate the Marked Price The marked price is 60% above the cost price of ₹800. 	ext{Marked Price} = 800 + (60\% 	imes 800) = 800 + 0.60 	imes 800 = 800 + 480 = ₹1280 Step 2: Calculate the Selling Price A discount of 20% is offered on the marked price of ₹1280. 	ext{Selling Price} = 1280 - (20\% 	imes 1280) = 1280 - 0.20 	imes 1280 = 1280 - 256 = ₹1024 Step 3: Calculate the Profit Percentage The profit is the difference between selling price and cost price, divided by cost price and multiplied by 100. 	ext{Profit} = 1024 - 800 = ₹224 	ext{Profit Percentage} = \frac{	ext{Profit}}{	ext{Cost Price}} 	imes 100 = \frac{224}{800} 	imes 100 = 0.28 	imes 100 = 28\% **The trader made a profit of 28% on the watch.**

Answer

B. 32%

Answer

C. 36%

Answer

D. 40%

Question 8

Two articles were sold for ₹1,000 each. One was sold at a profit of 25% and the other at a loss of 25%. What is the overall profit or loss percentage?

Accepted Answer

Correct answer: A. 6.25% loss — Step 1: Find the Cost Price of the article sold at 25% profit [Let CP₁ be the cost price of first article. Selling price = ₹1,000 at 25% profit] SP_1 = CP_1 + 0.25 	imes CP_1 = 1.25 	imes CP_1 = 1000 CP_1 = \frac{1000}{1.25} = \frac{1000 	imes 100}{125} = \frac{100000}{125} = 800 Step 2: Find the Cost Price of the article sold at 25% loss [Let CP₂ be the cost price of second article. Selling price = ₹1,000 at 25% loss] SP_2 = CP_2 - 0.25 	imes CP_2 = 0.75 	imes CP_2 = 1000 CP_2 = \frac{1000}{0.75} = \frac{1000 	imes 100}{75} = \frac{100000}{75} = \frac{4000}{3} \approx 1333.33 Step 3: Calculate overall profit or loss percentage [Total Cost Price and Total Selling Price] 	ext{Total CP} = CP_1 + CP_2 = 800 + \frac{4000}{3} = \frac{2400 + 4000}{3} = \frac{6400}{3} 	ext{Total SP} = 1000 + 1000 = 2000 	ext{Loss} = \

Answer

B. 6.25% profit

Answer

C. No profit, no loss

Answer

D. 5% loss

Quantitative Aptitude - MCQ Practice Questions