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

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

Questions : 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.

The correct answers to the above question in:

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.

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

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 : 2

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 : 3

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

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

Question : 4

A jeep is chasing a car which is 5km ahead. Their respective speed are 90 km/hr and 75 km/ hr. After how many minutes will the jeep catch the car ?

a) 20 min.

b) 25 min.

c) 24 min.

d) 18 min.

Answer: (a)

Relative speed

= 95 - 75 = 15 kmph

Required time

= $\text"Distance"/ \text"Relative speed"$

= $5/15$ hours = $5/15 × 60$ minutes

= 20 minutes

Question : 5

A can give 40 metres start to B and 70 metres to C in a race of one kilometre How many metres start can B give to C in a race of one kilometre ?

a) 31$1/4$ metre

b) 32 metre

c) 31$3/4$ metre

d) 30 metre

Answer: (a)

According to the question,

When A covers 1000m, B covers

= 1000 - 40 = 960 m

and C covers =1000 - 70 = 930 m

When B covers 960m, C covers 930 m.

When B covers 1000m, C covers

= $930/960 × 1000$ = 968.75 metre

Hence, B gives C a start of

= 1000 - 968.75 = 31.25 metre

Question : 6

Walking at $3/4$ of his usual speed, a man reaches his office 20 minutes late. Then his usual time for walking to his office is :

a) 30 minutes

b) 40 minutes

c) 45 minutes

d) 1 hour

Answer: (d)

Usual time = x minutes

New time = ${4x}/3$ minutes

$(Since, \text"Speed ∞ "\text"1"/\text"Time")$

According to the question,

${4x}/3 - x = 20$

$x/3 = 20$

x = 60 minutes i.e. 1 hour.

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