Practice Time and distance - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) You arrive at your school 5 minutes late if you walk with a speed of 4 km/h, but you arrive 10 minutes before the scheduled time if you walk with a speed of 5 km/h. The distance of your school from your house (in km) is
(a)
(b)
(c)
(d)
If the required distance be = x km, then
$x/4 - x/5 = {10 + 5}/60$
${5x - 4x}/20 = 1/4$
$x/20 = 1/4 ⇒ x= 1/4 × 20$ = 5 km.
Using Rule 10,If a man travels at the speed of $s_1$, he reaches his destination $t_1$ late while he reaches $t_2$ before when he travels at $s_2$ speed, then the distance between the two places is D = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
Here, $S_1 = 4, t_1 = 5, S_2 = 5, t_2$= 10
Distance =${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
=${(4 × 5)(5 + 10)}/{5 - 4}$
=20 × $15/60$ = 5 kms
Q-2) Walking at $3/4$ of his usual speed, a man is 1$1/2$ hours late. His usual time to cover the same distance, (in hours) is
(a)
(b)
(c)
(d)
Time and speed are inversely proportional.
$4/3$ of usual time –usual time = $3/2$
$1/3$ usual time = $3/2$
Usual time = ${3 × 3}/2$
$9/2 = 4{1}/2$ hours
Using Rule 8,
Here, A = 3, B = 4, t= $3/2$
Usual time = $\text"A"/\text"Diff of A and B"$ × time
= $3/{(4 - 3)} × 3/2 = 4{1}/2$ hrs.
Q-3) A man goes from A to B at a uniform speed of 12 kmph and returns with a uniform speed of 4 kmph His average speed (in kmph) for the whole journey is :
(a)
(b)
(c)
(d)
Using Rule 5,
If two equal distances are covered at two unequal speed of x kmph and y kmph,
then average speed = $({2xy}/{x + y})$
= ${2 × 12 × 4}/{12 + 4} = 96/16$ = 6 kmph
Q-4) A man goes to a place on bicycle at speed of 16 km/hr and comes back at lower speed. If the average speed is 6.4 km/hr in total journey, then the return speed (in km/hr) is :
(a)
(b)
(c)
(d)
Let the speed of cyclist while returning be x kmph.
Average speed = ${2 × 16 × x}/{16 + x}$
6.4 = ${32x}/{16 + x}$
6.4 × 16 + 6.4x = 32x
32x - 6.4x = 6.4 × 16
25.6x = 6.4 × 16
$x = {6.4 × 16}/{25.6}$ = 4 kmph.
Q-5) A and B run a 5 km race on a round course of 400 m. If their speed are in the ratio 5 : 4, the number of times, the winner passes the other, is
(a)
(b)
(c)
(d)
The winner will pass the other, one time in covering 1600m.
Hence, the winner will pass the other 3 times in completing 5km race.
Q-6) A certain distance is covered by a cyclist at a certain speed. If a jogger covers half the distance in double the time, the ratio of the speed of the jogger to that of the cyclist is
(a)
(b)
(c)
(d)
Using Rule 1,
Let speed of cyclist = x kmph
& Time = t hours
Distance = ${xt}/2$ while time = 2t
∴ Required ratio = ${xt}/{2 × 2t}$ : x = 1 : 4
Q-7) A train starts from A at 7 a.m. towards B with speed 50 km/h. Another train starts from B at 8 a.m. with speed 60 km/h towards A. Both of them meet at 10 a.m. at C. The ratio of the distance AC to BC is
(a)
(b)
(c)
(d)
AC = Distance covered by train starting from A in 3 hours
= 50 × 3 = 150 km
BC = Distance covered by train starting from B in 2 hours
= 60 ×2 = 120 km
∴ AC : BC = 150 : 120 = 5 : 4
Q-8) A truck covers a distance of 550 metre in one minute where as a bus covers a distance of 33 km in $3/4$ hour. Then the ratio of their speeds is :
(a)
(b)
(c)
(d)
Speed of truck
= ${550\text"metre"}/{60\text"second"} = (55/6)$ m./sec.
Speed of bus
= ${33 × 1000 \text"metre"}/{3/4 × 60 × 60 \text"second"} = 440/36$ m./sec.
∴ Required ratio = $55/6 : 440/36$
= 55 × 6 : 440 = 3 : 4
Q-9) A cyclist, after cycling a distance of 70 km on the second day, finds that the ratio of distance covered by him on the first two days is 4 : 5. If he travels a distance of 42 km. on the third day, then the ratio of distance travelled on the third day and the first day is :
(a)
(b)
(c)
(d)
Using Rule 1,
Distance covered on the first day
= $4/5 × 70$ = 56 km
∴ Required ratio = 42 : 56 = 3 : 4
Q-10) It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the car is :
(a)
(b)
(c)
(d)
Let the speed of train be x kmph. and the speed of car be y kmph.
Time = $\text"Distance"/ \text"Speed"$
According to the question,
$120/x + 480/y$ = 8
$15/x + 60/y$ = 1 ...(i)
and, $200/x + 400/y = 25/3$
$8/x + 16/y = 1/3$
$24/x + 48/y$ = 1 ...(ii)
From equations (i) and (ii),
$24/x + 48/y = 15/x + 60/y$
$24/x - 15/x = 60/y - 48/y$
$9/x = 12/y$
$x/y = 9/12 = 3/4$ = 3 : 4
Q-11) A car travels from A to B at the rate of 40 km/h and returns from B to A at the rate of 60 km/ h. Its average speed during the whole journey is
(a)
(b)
(c)
(d)
Here, distance is same.
Average speed = ${2xy}/{x + y}$
= $({2 × 40 × 60}/{40 + 60})$ kmph.
= $({2 × 40 × 60}/100)$ kmph.
= 48 kmph.
Q-12) When Alisha goes by car at 50 kmph, she reaches her office 5 minutes late. But when she takes her motorbike, she reaches 3 minutes early. If her office is 25 kms away, what is the approximate average speed at which she rides her motorbike ?
(a)
(b)
(c)
(d)
Difference of time
= 5 + 3 = 8 minutes
= $8/60$ hour = $2/15$ hour
If the speed of motorbike be x kmph, then
$25/50 - 25/x = 2/15$
$25/x = 1/2 - 2/15$
$25/x = {15 - 4}/30 = 11/30$
11x = 25 × 30
$x = {25 × 30}/11 = 750/11$
= 68.18 kmph ≈ 68 kmph
Q-13) A train moves with a speed of 30 kmph for 12 minutes and for next 8 minutes at a speed of 45 kmph. Find the average speed of the train:
(a)
(b)
(c)
(d)
Using Rule 3,
Average speed
= $\text"Total distance"/\text"time taken"$
= ${30 × 12/60 + 45 × 8/60}/{12/60 + 8/60}$
= 12 × 3 = 36 kmph
Q-14) Gautam travels 160 kms at 32 kmph and returns at 40 kmph. Then his average speed is
(a)
(b)
(c)
(d)
Here, distances are equal.
Average speed = $({2xy}/{x + y})$ kmph.
= $({2 × 32 × 40}/{32 + 40})$ kmph.
= $({2 × 32 × 40}/72)$ kmph.
= $(320/9)$ kmph. = 35.55 kmph.
Q-15) A man covers half of his journey at 6km/hr and the remaining half at 3km/hr. His average speed is
(a)
(b)
(c)
(d)
Using Rule 5,If a bus travels from A to B with the speed x km/h and returns from B to A with the speed y km/h,then the average speed will be $({2xy}/{x + y})$
If the same distance are covered at different speed of x kmph and y kmph,
the average speed of the whole journey is given by = $({2xy}/{x + y})$ kmph.
Required average speed
= ${2 × 6 × 3}/{6 + 3} = 36/9$ = 4 kmph
Q-16) A person started his journey in the morning. At 11 a.m. he covered $3/8$ of the journey and on the same day at 4.30 p.m. he covered $5/6$ of the journey. He started his journey at
(a)
(b)
(c)
(d)
Difference of time
= 4.30 p.m - 11.a.m.
= $5{1}/2$ hours $11/2$ hours
Distance covered in $11/2$ hrs
= $5/6 - 3/8 = {20 - 9}/24 = 11/24$ part
Since, $11/24$ part of the journey is covered in $11/2$ hours
$3/8$ part of the journey is covered in
= $11/2 × 24/11 × 3/8$
= $9/2$ hours = 4$1/2$ hours.
Clearly the person started at 6.30 a.m.
Q-17) The speed of a car is 54 km/hr. What is its speed in m/sec?
(a)
(b)
(c)
(d)
1 kmph = $5/18$ m/sec
54 kmph = $5/18 × 54$
= 15 m/sec.
Q-18) Gautam goes to office at a speed of 12 kmph and returns home at 10 kmph. His average speed is :
(a)
(b)
(c)
(d)
Here distances are same.
∴ Average speed = $({2xy}/{x + y})$ kmph
= $({2 × 12 × 10}/{12 + 10})$ kmph
= $(240/22)$ kmph = 10.9 kmph
Q-19) A and B are 20 km apart. A can walk at an average speed of 4 km/ hour and B at 6 km/hr. If they start walking towards each other at 7 a.m., when they will meet ?
(a)
(b)
(c)
(d)
Using Rule 11,Time taken by 1st man to reach B after meeting 2nd man at C is '$t_1$' and time taken by 2nd man to reach A after meeting 1st man at C is '$t_2$' then:${\text"Speed of 1st man"(s_1)}/{\text"Speed of 2nd man"(s_2)} = √{t_2/t_1}$Distance from A to B = $s_1t_1 + S_2t_2$
If A and B meet after t hours, then
4 t + 6 t = 20
10 t = 20 ⇒ t = $20/10$ = 2 hours.
Hence, both will meet at 9 a.m.
Q-20) A bus travels 150 km in 3 hours and then travels next 2 hours at 60 km/hr. Then the average speed of the bus will be
(a)
(b)
(c)
(d)
Total distance covered by the bus
= 150 km. + 2 × 60 km.
= (150 + 120) km.
= 270 km.
∴ Average speed = $\text"Total distance"/ \text"Time taken"$
= $270/5$ = 54 kmph.