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Time & Speed
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CAT 2025 Lesson : Time & Speed - Product Constancy

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4. Product Constancy

We know that
Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}

∴ When the product of two variables is constant, i.e. Distance in this case, then

- if one of time or speed increases by
ab\dfrac{a}{b} of itself, the other decreases by ab+a\dfrac{a}{b + a} of itself

- if one of time or speed decreases by
ab\dfrac{a}{b} of itself, the other increases by aba\dfrac{a}{b - a} of itself

Example 9

If a train decreases its speed from 8080 km/hr to 6060 km/hr, then it takes 55 more hours to travel from Chennai to Mumbai. What is the distance between Chennai and Mumbai?

Solution

Method 1

The distance between Chennai and Mumbai is constant. When the train decreases its speed from 8080 km/hr to 6060 km/hr, time taken increases by 55 hours.

When the speed decreases by
2020 km/hr, i.e.14\dfrac{1}{4} times its usual speed, there will be an increase of141\dfrac{1}{4 - 1} =13\dfrac{1}{3} times the initial time taken, which is given to be 55 hours.

Where the initial time taken is
tt, 13×t=5\dfrac{1}{3} \times t = 5 t=15 t = 15 hours

As the train takes
1515 hours while travelling at 8080 km/hr,

Distance between cities
=80×15=1200= 80 \times 15 = 1200 km

Alternatively

Ratio of speeds
=80:60=4:3= 80 : 60 = 4 : 3

Ratio of time taken is the reciprocal of ratio of speeds.

∴ Ratio of time taken
=3:4= 3 : 4
Let the initial time and new time taken be
3x\bm{3x} and 4x\bm{4x} respectively.
4x3x4x - 3x = x=5x = 5 hours

∴ Distance
=80×3x=80×3×5= 80 \times 3x = 80 \times 3 \times 5 = 1,200\bm{1,200} km

Answer:
12001200 km

Example 10

When Musafar travels at three-fifth his usual speed, he reaches office 1616 minutes late. How many minutes does he usually take to cover this distance?

Solution

Let the usual time taken to cover the distance be t.

This decrease of 25\dfrac{2}{5} of usual speed will result in increase of 252=23\dfrac{2}{5 - 2} = \dfrac{2}{3} of the usual time.

23×t=16\dfrac{2}{3} \times t = 16

t=24 t = \bm{24} minutes

Alternatively

Ratio of speeds
=1:35=5:3= 1 : \dfrac{3}{5} = 5 : 3

∴ Ratio of time taken
=3:5= 3 : 5

Let the time taken be
3x3x and 5x5x respectively.
5x3x=165x - 3x = 16
x=8 x = 8

He usually takes
3x=3×8=243x = 3 \times 8 = \bm{24} minutes to travel to office.

Answer:
2424 minutes

Example 11

Ram always leaves home for his office at 88 am. If he travels at 4040 km/hr, he is 1818 minutes late. If he travels at 6060 km/hr, then he is 1212 minutes early. What is the time at which Ram is required to report for work?

Solution

Increase in speed from 4040 km/hr to 6060 km/hr has reduced time by 12+18=3012 + 18 = 30 minutes.

As Speed increased by
12\dfrac{1}{2} of itself, time would decrease by 12+1=13\dfrac{1}{2 + 1} = \dfrac{1}{3} of itself.

∴ Time taken at 40 km/hr
=3×30=90= 3 \times 30 = 90 minutes

As Ram is late by
1818 minutes at 4040 km/hr, to be on time, he should reach in 9018=7290 - 18 = 72 minutes

∴ Ram's reporting time
=9:12= 9:12 am

Alternatively

Ratio of speeds
=40:60=2:3= 40 : 60 = 2 : 3
∴ Ratio of time taken
=3:2= 3 : 2

Let the time taken in the two scenarios be
3x3x and 2x2x respectively.
3x2x=12+183x - 2x = 12 + 18
x=30 x = 30

When he travels at
4040 km/hr, he takes 3x=903x = 90 minutes.
As he is
1818 minutes late in this case, he is supposed to reach office in 9018=7290 - 18 = 72 minutes.

∴ Ram's reporting time
=9:12= 9:12 am

Answer:
9:129:12 am

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