Model 6 Change in speed Vs Change with time travel Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 6 EXERCISES

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The following question based on trains topic of quantitative aptitude

Questions : 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) 35

(b) 36$2/3$

(c) 40

(d) 20

The correct answers to the above question in:

Answer: (c)

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.

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Practice trains (Model 6 Change in speed Vs Change with time travel) Online Quiz

Shortcut : 2

Posted by : BHUSHAN
Posted Date: 2021-10-19 10:35:55

d1 = d2 s*(s+3)*40/3

= s*(s-2)* 40/2

s=12

put value of s in equation of d: d=s*(s+3)*40/3*60...divide by equation by 60 to convert minute into hour

d = 40-KM

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Read more changing speed with time travel Based Quantitative Aptitude Questions and Answers

Question : 1

A train covers a distance of 10 km in 12 minutes. If its speed is decreased by 5 km/hr, the time taken by it to cover the same distance will be :

a) 13 minutes 20 sec

b) 13 minutes

c) 11 minutes 20 sec

d) 10 minutes

Answer: (a)

Using Rule 1,

Speed of train = $\text"Distance"/\text"Time"$

= $10/{12/60}$ kmph

= ${10 × 60}/12 = 50$ kmph

New speed = 45 kmph

∴ Required time = $10/45$ hour

= $2/9 × 60$ minutes

= $40/3$ minutes or 13 minutes 20 seconds

Question : 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) 4

b) 3

c) 1

d) 5

Answer: (a)

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

Question : 3

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) 8 km

b) 3 km

c) 2 km

d) 5 km

Answer: (b)

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

Question : 4

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) 4 km

b) 6 km

c) 4.5 km

d) 5 km

Answer: (a)

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

Question : 5

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) 7 km

b) 12 km

c) 16 km

d) 6 km

Answer: (b)

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

Question : 6

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) 15

b) 19

c) 21

d) 13

Answer: (b)

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

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