# CAT Practice : Speed Time, Races

Lightning McQueen starts later but still catches up with King and Chick. And you thought CAT prep was boring.

## Speed Time - Cars

Q.3: Three cars leave A for B in equal time intervals. They reach B simultaneously and then leave for Point C which is 240 km away from B. The first car arrives at C an hour after the second car. The third car, having reached C, immediately turns back and heads towards B. The first and the third car meet a point that is 80 km away from C. What is the difference between the speed of the first and the third car?
1. 60 kmph
2. 80 kmph
3. 20 kmph
4. 40 kmph

Choice A. 60 kmph

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## Detailed Solution

${{AB \over V_1} {-} {AB \over V_2} {=} {AB \over V_2} {-} {AB \over V_3}}$
${{240 \over V_1} {-} {240 \over V_2}}$ = 1

v3 = 2v1

Let v1, v2 and v3 be the speeds of the cars.
Condition I states that the cars leave in equal intervals of time and arrive at the same time. Or, the difference in the time taken between cars 1 and 2 should be equal to the time taken between cars 2 and 3.

We get ${{AB \over V_1} {-} {AB \over V_2} {=} {AB \over V_2} {-} {AB \over V_3}}$

As the second car arrived at C an hour earlier than the first, we get a second equation ${{240 \over V_1} {-} {240 \over V_2}}$ = 1

The third car covered 240 + 80 kms when the first one covered 240 – 80 kms. Therefore, ${{320 \over V_3} {=} {160 \over V_1}}$

This gives us v3 = 2v1

From condition 1, we have ${{AB \over V_1} {-} {AB \over V_2} {=} {AB \over V_2} {-} {AB \over V_3}}$
Substituting v3 = 2v1, this gives us ${{AB \over V_1} {-} {AB \over V_2} {=} {AB \over V_2} {-} {AB \over 2V_1}}$
or ${{3 x AB \over 2 x V_1} {=} {2 x AB \over V_2}}$
or v2 = ${4 x v_1 \over 3}$

Solving ${{240 \over v_1} {-} {240 \over v_2}}$ = 1, we get ${60 \over v_1}$ =1 or, v1 = 60 kmph
=> v2 = 80 kmph and v3 = 120 kmph

Correct Answer: 60 kmph

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## More questions from Speed Time, Races

"Life is a race ... if you don't run fast ... you will be like a broken undaa" - Veerusahasra Buddhi