Model 5 Problems with Races Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 5 EXERCISES

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The following question based on time & distance topic of quantitative aptitude

Questions : In a race of 1000 m, A can beat B by 100m. In a race of 400 m, B beats C by 40m. In a race of 500m. A will beat C by

(a) 50 m

(b) 60 m

(c) 45 m

(d) 95 m

The correct answers to the above question in:

Answer: (d)

When A runs 1000m, B runs 900m.

When A runs 500m, B runs 450 m.

Again, when B runs 400m, C runs 360 m.

When B runs 450m, C runs

= $360/400 × 450$ = 405 metres

Required distance

= 500 - 405 = 95 metres

Practice time & distance (Model 5 Problems with Races) Online Quiz

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Read more problems on races Based Quantitative Aptitude Questions and Answers

Question : 1

A and B start running at the same time and from the same point around a circle. If A can complete one round in 40 seconds and B in 50 seconds, how many seconds will they take to reach the starting point simultaneously ?

a) 200

b) 2000

c) 90

d) 10

Answer: (a)

Required time

= LCM of 40 and 50 seconds

= 200 seconds

Question : 2

A is twice as fast as B, and B is thrice as fast as C is. The journey covered by C in 1$1/2$ hours will be covered by A in

a) 30 minutes

b) 10 minutes

c) 1 hour

d) 15 minutes

Answer: (d)

Time taken by C = t hours

Time taken by B = $t/3$ hours

and time taken by A = $t/6$ hours

Here, t = $3/2$ hours

∴ Required time taken by A

= $3/{2/6}$ hour = $1/4$ hour

= $(1/4 × 60)$ minutes = 15 minutes

Question : 3

A runs twice as fast as B and B runs thrice as fast as C. The distance covered by C in 72 minutes, will be covered by A in :

a) 24 minutes

b) 12 minutes

c) 16 minutes

d) 18 minutes

Answer: (b)

Ratio of the speed of A, B and C = 6 : 3 : 1

Ratio of the time taken

= $1/6 : 1/3$ : 1 = 1 : 2 : 6

Time taken by A

= $72/6$ =12 minutes

Question : 4

A is faster than B. A and B each walk 24 km. The sum of their speeds is 7 km/hr and the sum of times taken by them is 14 hours. Then A's speed is equal to:

a) 4 km/hr.

b) 7 km/hr.

c) 5 km/hr.

d) 3 km/hr.

Answer: (a)

Let, A's speed = x kmph.

B's speed = (7 - x) kmph

Time = $\text"Distance"/ \text"Speed"$

According to the question,

$24/x + 24/{7 - x} = 14$

$24({7 - x + x}/{x(7 - x)})$ = 14

${24 × 7}/{x(7 - x)}$ = 14

x (7 - x) = 12 = 4 × 3 or 3 × 4

x (7 - x) = 4 (7 - 4) or 3 (7 - 3)

x = 4 or 3

A's speed = 4 kmph.

Question : 5

Two persons ride towards each other from two places 55 km apart, one riding at 12km/hr and the other at 10 km/hr. In what time will they be 11 km apart?

a) 1 hour and 30 minutes

b) 2 hours and 45 minutes

c) 2 hours

d) 2 hours and 30 minutes

Answer: (c)

Relative speed

= 12 + 10 = 22 kmph

Distance covered

= 55 - 11 = 44 km

∴ Required time

= $(44/22)$ hours = 2 hours

Question : 6

Sarthak completed a marathon in 4 hours and 35 minutes. The marathon consisted of a 10 km run followed by 20 km cycle ride and the remaining distance again a run. He ran the first stage at 6 km/hr and then cycled at 16 km/ hr. How much distance did Sarthak cover in total, if his speed in the last run was just half that of his first run?

a) 35 km.

b) 45 km.

c) 40 km.

d) 5 km.

Answer: (a)

Let the total distance be x km.

Time = $\text"Distance"/ \text"Speed"$

According to the question,

$10/6 + 20/16 + {x - 30}/3$

= 4$35/60 = 4{7}/12$

$5/3 + 5/4 + x/3 - 10 = 55/12$

$x/3 + 5/3 + 5/4 - 10 = 55/12$

$x/3 + ({20 + 15 - 120}/12) = 55/12$

$x/3 - 85/12 = 55/12$

$x/3 = 85/12 + 55/12 = 140/12$

$x = 140/12 × 3$ = 35 km.

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