Practice Changing speed with time travel - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) If a man reduces his speed to 2/ 3, he takes 1 hour more in walking a certain distance. The time (in hours) to cover the distance with his normal speed is :
(a)
(b)
(c)
(d)
Since man walks at $2/3$ of usual speed, time taken wil be $3/2$ of usual time.
$3/2$ of usual time
= usual time + 1 hour.
$(3/2 –1)$ of usual time = 1
usual time = 2 hours.
Q-2) A student goes to school at the rate of 2$1/2$ km/h and reaches 6 minutes late. If he travels at the speed of 3 km/h. he is 10 minutes early. The distance (in km) between the school and his house is
(a)
(b)
(c)
(d)
Let the required distance be x km.
$x/{5/2} - x/3 = 16/60$
${2x}/5 - x/3 = 4/15$
${6x - 5x}/15 = 4/15 ⇒ x = 4$ km.
Using Rule 10,
Here, $S_1 = 2{1}/2 , t_1 = 6, S_2 = 3, t_2$ = 10
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${5/2 × 3(6 + 10)}/{3 - 5/2}$
= $15 × 16/60$ km = 4 km
Q-3) If a train runs at 40 km/hour, it reaches its destination late by 11 minutes. But if it runs at 50 km/ hour, it is late by 5 minutes only. The correct time (in minutes) for the train to complete the journey is
(a)
(b)
(c)
(d)
If the distance be x km,
then $x/40 - x/50 = 6/60$
$x/4 - x/5 = 1$
x = 20 km.
Required time
= $(20/40)$ hour - 11 minutes
= $(1/2 × 60 - 11)$ minutes
= 19 minutes
Q-4) A man covered a certain distance at some speed. Had he moved 3 km per hour faster, he would have taken 40 minutes less. If he had moved 2 km per hour slower, he would have taken 40 minutes more. The distance (in km) is :
(a)
(b)
(c)
(d)
Let the distance be x km and initial speed be y kmph.
According to question,
$x/y - x/{y + 3} = 40/60$ ...(i)
and, $x/{y - 2} - x/y = 40/60$ ...(ii)
From equations (i) and (ii),
$x/y - x/{y + 3} = x/{y - 2} - x/y$
$1/y - 1/{y + 3} = 1/{y - 2} - 1/y$
${y + 3 -y}/{y(y + 3)} = {y - y + 2}/{y(y - 2)}$
3 (y - 2) = 2 (y + 3)
3y - 6 = 2y + 6
y = 12
From equation (i),
$x/12 - x/15 =40/60$
= ${5x - 4x}/60 = 2/3$
$x = 2/3 × 60$ = 40
Distance = 40 km.
Q-5) A boy is late by 9 minutes if he walks to school at a speed of 4 km/hour. If he walks at the rate of 5 km/hour, he arrives 9 minutes early. The distance to his school is
(a)
(b)
(c)
(d)
Let the required distance be x km.
According to the question,
$x/4 - x/5 = 18/60$
${5x - 4x}/20 = 3/10$
$x = 3/10 × 20$ = 6 km
Using Rule 10,
Here, $S_1 = 4, t_1 = 9, S_2 = 5, t_2$ = 9
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${(4 × 5)(9 + 9)}/{5 - 4}$
= $20 × 18/60 = $6 km
Q-6) When a person cycled at 10 km per hour he arrived at his office 6 minutes late. He arrived 6 minutes early, when he increased his speed by 2 km per hour. The distance of his office from the starting place is
(a)
(b)
(c)
(d)
Let the distance be x km.
$x/10 - x/12 = 12/60$
${6x - 5x}/60 = 1/5$
$x = 1/5 × 60$ = 12 km.
Using Rule 10,
Here, $S_1 = 10, t_1 = 6, S_2 = 12, t_2$ = 6
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${(10 × 12)(6 + 6)}/{12 - 10}$
= ${120 × 12}/2$
= 60 × $12/60$ km = 12km
Q-7) If a boy walks from his house to school at the rate of 4 km per hour, he reaches the school 10 minutes earlier than the scheduled time. However, if he walks at the rate of 3 km per hour, he reaches 10 minutes late. Find the distance of his school from his house.
(a)
(b)
(c)
(d)
Let the distance of school be x km, then
$x/3 - x/4 = 20/60$
$x/12 = 1/3 ⇒ x =12/3 = 4$ km
Using Rule 10,
Here, $S_1 = 3, t_1 = 10, S_2 = 4, t_2$ = 10
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${(3 × 4)(10 + 10)}/{4 - 3}$
= $12 × 20/60$ = 4 km
Q-8) Walking at 5 km/hr a student reaches his school from his house 15 minutes early and walking at 3 km/hr he is late by 9 minutes. What is the distance between his school and his house ?
(a)
(b)
(c)
(d)
Let the required distance be x km.
Then, $x/3 - x/5 = 24/60$
${5x - 3x}/15 = 2/5$
${2x}/3 = 2$ ⇒ 2x = 2 × 3
x = 3 km
Using Rule 10,
Here, $S_1 = 3, t_1 = 9, S_2 = 5, t_2$ = 15
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${(3 × 5)(9 + 15)}/{5 - 3}$
= $15/2 × 24/60$ = 3 km
Q-9) A student rides on bicycle at 8 km/hour and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/ hour and reaches school 5 minutes early. How far is the school from his house ?
(a)
(b)
(c)
(d)
Let x km. be the required distance.
Difference in time
= 2.5 + 5 = 7.5 minutes
= $7.5/60$ hrs. = $1/8$ hrs.
Now, $x/8 - x/10 = 1/8$
= ${5x - 4x}/40 = 1/8$
$x = 40/8$ = 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, $S_1 = 8, t_1 = 2.5, S_2 = 10, t_2 = 5$
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${(8 × 10)(2.5 + 5)}/{10 - 8}$
= $40 × {7.5}/60$ = 5 km
Q-10) Walking at a speed of 5 km/hr, a man reaches his office 6 minutes late. Walking at 6 km/hr, he reaches there 2 minutes early. The distance of his office is
(a)
(b)
(c)
(d)
Let the distance of the office be x km, then
$x/5 - x/6 = 8/60$
${6x - 5x}/30 = 2/15$
x = 2 × 2 = 4 km
Using Rule 10,
Here, $S_1 = 5, t_1 = 6, S_2 = 6, t_2$ = 2
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${(6 × 5)(6 + 2)}/{6 - 5}$
= $30 × 8/60$ = 4 km