(solved)Question 6 SSC-CGL 2018 June 4 Shift 2

A takes 30 minutes longer than B to cover a distance of 15 km at a certain speed. However, if A doubles his speed, he covers the same distance in an hour less time than B. What is the speed of B (in km / h)?
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Question 6
SSC-CGL 2018 June 4 Shift 2

A takes 30 minutes longer than B to cover a distance of 15 km at a certain speed. However, if A doubles his speed, he covers the same distance in an hour less time than B. What is the speed of B (in km / h)?

  1. 5
  2. 6
  3. 6-1/2
  4. 5-1/2

Method 1

Journey of A has been given in two ways. In one case "A" travels at normal speed, but in the second case he travels at double speed.

If time taken by B is $\displaystyle T$ hours, then it is given that "A" takes T + 1/2 hour during the journey at normal speed, whereas he takes T - 1 hours during his journey at double the speed. Distance remains same. So, time taken during the first case should be twice the time taken during the second case. We can write

$\displaystyle \underset{\text{(at normal speed)}}{T + 1/2} = \underset{\text{(at doubled speed)}}{2(T - 1)}$

Solving, we get $\displaystyle T = \frac 52$

Hence, speed of B = $\displaystyle \frac{\text{distance}}{\text{time}}$ $\displaystyle = \frac{15}{5/2} = 6 $ kph answer!

Method 2

If speed of B = x kph, then time taken by B is $\displaystyle \bigg(\frac{15}{x}\bigg)$ hours.

If speed of A is $\displaystyle v_A$, then, by distance = speed x time, we have

$\displaystyle 15 = v_A \times \bigg(\frac{15}{x} + \frac{30}{60}\bigg)\text{ . . . (1)}$

From the second condition, we have

$\displaystyle 15 = 2v_A \times \bigg(\frac{15}{x} - 1\bigg)\text{ . . . (2)}$

Equating (1) and (2), and solving for x, we get x = 6 kph answer.

Method 3

Let speed of A be $\displaystyle a$ kph, and speed of B be $\displaystyle b$ kph. They travel 15 km in two ways. distance and time are proportional, so we can consider just the first km of their entire journey.

After first kilometer of their first journeys, "A" would have been 2 minutes behind.

$\displaystyle \frac{1}{a} - \frac{1}{b} = \frac{2}{60} \text{. . . (1)}$

For the second journey, A would have been 4 minutes ahead after first km of their journeys, so

$\displaystyle \frac{1}{b} - \frac{1}{2a} = \frac{4}{60} \text{. . . (2)}$

We need $\displaystyle b$, so multiply eq(2) by 2 and add to (1), to finally get $\displaystyle b = 6$ kph answer!

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