Model 4 Time & Distance with Ratios Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 5 EXERCISES

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The following question based on time & distance topic of quantitative aptitude

Questions : 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) 3 : 4

(b) 4 : 3

(c) 4 : 5

(d) 3 : 5

The correct answers to the above question in:

Answer: (a)

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

Practice time & distance (Model 4 Time & Distance with Ratios) Online Quiz

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Read more problems with ratios Based Quantitative Aptitude Questions and Answers

Question : 1

Two trains started at the same time, one from A to B and the other from B to A. If they arrived at B and A respectively 4 hours and 9 hours after they passed each other, the ratio of the speed of the two trains was

a) 3 : 2

b) 4 : 3

c) 5 : 4

d) 2 : 1

Answer: (a)

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 = $s_1t_1 + S_2t_2$

Required ratio of the speed of two trains

= $√{9}/√{4} = 3/2$ or 3 : 2

Question : 2

A car travels 80 km. in 2 hours and a train travels 180 km. in 3 hours. The ratio of the speed of the car to that of the train is :

a) 3 : 2

b) 3 : 4

c) 4 : 3

d) 2 : 3

Answer: (d)

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

∴ Speed of car : Speed of train

= $80/2 : 180/3$ = 40 : 60 = 2 : 3

Question : 3

It takes 8 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) 3 : 2

b) 3 : 4

c) 4 : 3

d) 2 : 3

Answer: (b)

Using Rule 1,

Speed of train = x kmph

Speed of car = y kmph

Case I,

$120/x + {600 - 120}/y$ = 8

$120/x + 480/y$ = 8

$15/x + 60/y$ = 1 ...(i)

Case II,

$200/x + 400/y$ = 8 hours 20 minutes

$200/x + 400/y = 8{1}/3$ hours = $25/3$

$8/x + 16/y = 1/3$

$24/x + 48/y$ = 1 ...(ii)

$15/x + 60/y = 24/x + 48/y$

$24/x - 15/x = 60/y - 48/y$

$9/x = 12/y$

$x/y = 9/12 = 3/4$ = 3 : 4

Question : 4

The ratio of length of two trains is 5 : 3 and the ratio of their speed is 6 : 5. The ratio of time taken by them to cross a pole is

a) 11 : 8

b) 25 : 18

c) 27 : 16

d) 5 : 6

Answer: (b)

Required ratio

= $5/6 : 3/5$

${30 × 5}/6 : {30 × 3}/5$ = 25 : 18

Question : 5

The speed of A and B are in the ratio 3 : 4. A takes 20 minutes more than B to reach a destination. In what time does A reach the destination ?

a) 2 hours

b) 2$2/3$ hours

c) 1$2/3$ hours

d) 1$1/3$ hours

Answer: (d)

Ratio of speed = 3 : 4

Ratio of time taken = 4 : 3

Let the time taken by A and B be 4x hours and 3 x hours respectively.

Then, $4x - 3x = 20/60 ⇒ x = 1/3$

Time taken by A = 4x hours

= $(4 × 1/3)$ hours = 1$1/3$ hours

Using Rule 9,

Here, $S_1 = 3x, S_2 = 4x$

$t_2 = y, t_1 = y + 20/60 = y + 1/3$

$S_1t_1 = S_2t_2$

$3x(y + 1/3) = 4xy$

3y + 1 = 4y, y = 1

∴ Time taken by A

= 1 + $1/3 = 1{1}/3$ hours

Question : 6

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) 2

b) 3

c) 5

d) 1

Answer: (b)

The winner will pass the other, one time in covering 1600m.

Hence, the winner will pass the other 3 times in completing 5km race.

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