Quantitative Aptitude Questions & Answers

Q.1101 Easy Profit and Loss

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?

A ₹1,200

B ₹1,500

C ₹1,800

D ₹2,000

Correct Answer: A. ₹1,200

EXPLANATION

Step 1: Calculate Total Cost Price

The shopkeeper buys 120 notebooks at ₹40 each.

\text{Total Cost Price} = 120 \times 40 = ₹4,800

Step 2: Calculate Total Selling Price

The shopkeeper sells 120 notebooks at ₹50 each.

\text{Total Selling Price} = 120 \times 50 = ₹6,000

Step 3: Calculate Total Profit

Profit is the difference between selling price and cost price.

\text{Total Profit} = \text{Selling Price} - \text{Cost Price} = 6,000 - 4,800 = ₹1,200

The shopkeeper's total profit for selling 120 notebooks in a day is ₹1,200.

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Q.1102 Hard Speed Distance Time

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?

A 72 km/h

B 75 km/h

C 70 km/h

D 74 km/h

Correct Answer: A. 72 km/h

EXPLANATION

Step 1: Define Variables and Total Distance

Let the total distance be \(D\) km. The car travels \(\frac{D}{2}\) km at each speed.

\text{Total Distance} = D \text{ 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} \text{ hours}

For the second half at 90 km/h:

t_2 = \frac{D/2}{90} = \frac{D}{180} \text{ hours}

Step 3: Calculate Total Time and Average Speed

Total time for the journey:

t_{\text{total}} = \frac{D}{120} + \frac{D}{180} = \frac{3D}{360} + \frac{2D}{360} = \frac{5D}{360} = \frac{D}{72} \text{ hours}

Average speed is total distance divided by total time:

\text{Average Speed} = \frac{D}{t_{\text{total}}} = \frac{D}{\frac{D}{72}} = 72 \text{ km/h}

The average speed for the entire journey is 72 km/h.

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Q.1103 Medium Speed Distance Time

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?

A 17.5 km/h

B 15 km/h

C 16 km/h

D 18 km/h

Correct Answer: A. 17.5 km/h

EXPLANATION

Step 1: Find the downstream speed

The boat travels 120 km downstream in 6 hours.

\text{Downstream speed} = \frac{120}{6} = 20 \text{ km/h}

Step 2: Find the upstream speed

The boat travels 120 km upstream in 8 hours.

\text{Upstream speed} = \frac{120}{8} = 15 \text{ 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.

\text{Speed of boat} = \frac{\text{Downstream speed} + \text{Upstream speed}}{2} = \frac{20 + 15}{2} = \frac{35}{2} = 17.5 \text{ km/h}

The speed of the boat in still water is 17.5 km/h.

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Q.1104 Hard Speed Distance Time

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?

A 100 seconds

B 150 seconds

C 200 seconds

D 300 seconds

Correct Answer: C. 200 seconds

EXPLANATION

Step 1: Find the time for each cyclist to complete one full lap

[Cyclist 1 completes one lap]

\text{Time}_1 = \frac{\text{Track length}}{\text{Speed}_1} = \frac{600}{8} = 75 \text{ seconds}

[Cyclist 2 completes one lap]

\text{Time}_2 = \frac{\text{Track length}}{\text{Speed}_2} = \frac{600}{10} = 60 \text{ 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]

\text{LCM}(75, 60) = \text{LCM of lap times}

[Finding prime factorization: 75 = 3 × 5², 60 = 2² × 3 × 5]

\text{LCM}(75, 60) = 2^2 \times 3 \times 5^2 = 4 \times 3 \times 25 = 300 \text{ seconds}

Step 3: Verify the answer

[In 300 seconds, Cyclist 1 completes]

\text{Laps}_1 = \frac{300}{75} = 4 \text{ laps}

[In 300 seconds, Cyclist 2 completes]

\text{Laps}_2 = \frac{300}{60} = 5 \text{ laps}

[Both return to the starting point after completing whole laps

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Q.1105 Medium Speed Distance Time

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?

A 10 km

B 12 km

C 11 km

D 15 km

Correct Answer: B. 12 km

EXPLANATION

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.

\text{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 \times 12

\[5d = 60\]

d = 12 \text{ km}

The distance between A and B is 12 km.

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Q.1106 Easy Speed Distance Time

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?

A 300 km

B 360 km

C 330 km

D 288 km

Correct Answer: B. 360 km

EXPLANATION

Step 1: Calculate the original speed

The train travels 240 km in 4 hours, so we divide distance by time.

\text{Original Speed} = \frac{240 \text{ km}}{4 \text{ hours}} = 60 \text{ 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.

\text{New Speed} = 60 \times 1.20 = 72 \text{ km/h}

Step 3: Calculate distance covered in 5 hours at new speed

Distance equals speed multiplied by time.

\text{Distance} = 72 \text{ km/h} \times 5 \text{ hours} = 360 \text{ km}

The train will cover 360 km in 5 hours at the new speed.

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