Model 5 Problems with Races Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 5 EXERCISES
The following question based on time & distance topic of quantitative aptitude
(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 »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 »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 »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 »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.
GET time & distance PRACTICE TEST EXERCISES
Model 1 Basic Time & Distance using formula
Model 2 Vehicles in x/y of its usual speed
Model 3 Problems on average speed
Model 4 Time & Distance with Ratios
Model 5 Problems with Races
time & distance Shortcuts and Techniques with Examples
-
Model 1 Basic Time & Distance using formula
Defination & Shortcuts … -
Model 2 Vehicles in x/y of its usual speed
Defination & Shortcuts … -
Model 3 Problems on average speed
Defination & Shortcuts … -
Model 4 Time & Distance with Ratios
Defination & Shortcuts … -
Model 5 Problems with Races
Defination & Shortcuts …
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