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 : In a 100m race, Kamal defeats Bimal by 5 seconds. If the speed of Kamal is 18 Kmph, then the speed of Bimal is

(a) 14.5 kmph

(b) 14 kmph

(c) 14.4 kmph

(d) 15.4 kmph

The correct answers to the above question in:

Answer: (c)

Time taken by Kamal

= $100/{18 × 5/18}$ = 20 seconds

Time taken by Bimal

= 20 + 5 = 25 seconds

Bimal's speed

= $100/25$ =4 m/sec = ${4 × 18}/5$ kmph

= 14.4 kmph.

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Read more problems on races Based Quantitative Aptitude Questions and Answers

Question : 1

Rubi goes to a multiplex at the speed of 3 km/hr to see a movie and reaches 5 minutes late. If she travels at the speed of 4 km/hr she reaches 5 minutes early. Then the distance of the multiplex from her starting point is

a) 5 km.

b) 5 metre

c) 2 metre

d) 2 km.

Answer: (d)

Distance between starting point and multiplex = x metre

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

According to the question,

$x/3 - x/4 = {5 + 5}/60$

${4x - 3x}/12 = 1/6$

$x/12 = 1/6$

$x = 12/6$ = 2 km.

Question : 2

In a race of one kilometre, A gives B a start of 100 metres and still wins by 20 seconds. But if A gives B a start of 25 seconds, B wins by 50 metres. The time taken by A to run one kilometre is

a) $500/29$ seconds

b) $700/29$ seconds

c) $1200/29$ seconds

d) 17 seconds

Answer: (a)

Let A take x seconds in covering 1000m and b takes y seconds

According to the question,

x + 20 = $900/1000$ y

x + 20 = ${9y}/10$ ...(i)

and, $950/1000$ x + 25 = y ...(ii)

From equation (i),

${10x}/9 + 200/9 = y$

${10x}/9 + 200/9 = {950x}/1000 + 25$

${10x}/9 + 200/9 = {19x}/20 + 25$

${10x}/9 - {19x}/20 = 25 - 200/9$

${200x - 171x}/180 = {225 - 200}/9$

${29x}/180 = 25/9$

$x = 25/9 × 180/29 = 500/29$ seconds.

Question : 3

In a one-kilometre race A, B and C are the three participants. A can give B a start of 50 m. and C a start of 69 m. The start, which B can allow C is

a) 20 m.

b) 18 m.

c) 19 m.

d) 17 m.

Answer: (a)

Let the time taken to complete the race by A,B, and C be x minutes.

Speed of A = $1000/x$,

B = ${1000 - 50}/x = 950/x$

C = ${1000 - 69}/x = 931/x$

Now, time taken to complete the race by

B = $1000/{950/x} = {1000 × x}/950$

and distance travelled by C in

${1000x}/950$ min = ${1000x}/950 × 931/x$ = 980 km.

B can allow C

= 1000 - 980 = 20 m

Question : 4

In a race of 800 metres, A can beat B by 40 metres. In a race of 500 metres, B can beat C by 5 metres. In a race of 200 metres, A will beat C by

a) 1.19 metre

b) 1.27 metre

c) 12.7 metre

d) 11.9 metre

Answer: (d)

According to the question,

When A runs 800 metres, B runs 760 metres

When A runs 200 metres, B runs

= $760/800 × 200$ = 190 metres

Again, when B runs 500 metres, C runs 495 metres.

When B runs 190 metres, C runs

= $495/500 × 190$ = 188.1 metres

∴ Hence, A will beat C by

200 - 188.1 = 11.9 metres in a race of 200 metres.

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