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) 1 hour and 30 minutes
(b) 2 hours and 45 minutes
(c) 2 hours
(d) 2 hours and 30 minutes
The correct answers to the above question in:
Answer: (c)
Relative speed
= 12 + 10 = 22 kmph
Distance covered
= 55 - 11 = 44 km
∴ Required time
= $(44/22)$ hours = 2 hours
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Read more problems on races Based Quantitative Aptitude Questions and Answers
Question : 1
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 »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 : 2
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 »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 : 3
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 »Answer: (a)
Required time
= LCM of 40 and 50 seconds
= 200 seconds
Question : 4
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 »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.
Question : 5
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 »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 : 6
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 »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
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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