Practice Problems on races - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-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)
(b)
(c)
(d)
Required time
= LCM of 40 and 50 seconds
= 200 seconds
Q-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)
(b)
(c)
(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
Q-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)
(b)
(c)
(d)
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
Q-4) 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)
(b)
(c)
(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.
Q-5) 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)
(b)
(c)
(d)
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.
Q-6) 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)
(b)
(c)
(d)
Relative speed
= 95 - 75 = 15 kmph
Required time
= $\text"Distance"/ \text"Relative speed"$
= $5/15$ hours = $5/15 × 60$ minutes
= 20 minutes
Q-7) Walking at the rate of 4 kmph a man covers certain distance in 2 hrs 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in how many minutes ?
(a)
(b)
(c)
(d)
2 hours 45 minutes
= $(2 + 45/60)$ hours
= $(2 + 3/4)$ hours = $11/4$ hours
Distance = Speed × Time
= 4 × $11/4$ = 11 km.
Time taken in covering 11 km at 16.5 kmph
= $11/{16.5}$ hour
= $({11 × 10 × 60}/165)$ minutes
= 40 minutes
Q-8) 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)
(b)
(c)
(d)
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.
Q-9) In a race of 200 metres, B can give a start of 10 metres to A, and C can give a start of 20 metres to B. The start that C can give to A, in the same race, is
(a)
(b)
(c)
(d)
According to the question,
Since, When B runs 200 m metres, A runs 190 metres
When B runs 180 metres, A runs
= $190/200 × 180$ = 171 metres
When C runs 200m, B runs 180 metres.
Hence, C will give a start to A by
= 200 - 171 = 29 metres
Q-10) 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)
(b)
(c)
(d)
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.