Question

Cyclists set off from points A and B towards each other at the same speed. After a while they met. How many times should be taken away
speed to one of the cyclists so that the time in which they meet are reduced by half, provided that the speed of another cyclist
remained the same?​

Answer 1

Answer:

New speed of another cyclist = 3x

Speed of the another increased by 3 times.

Step-by-step explanation:

Distance between point A and B be y.

Initial speed of cyclist = x

Distance = Speed * time

Relative Speed in Opposite direction = sum of speeds = 2x

$time \: travel \: by \: each \:until \: \: they \: met \: \\ = \frac{y}{2 \times x}$

Now, time in which they met is reduced to half.

$time \: = \frac{y}{2x} \times \frac{1}{2} = \frac{y}{4x}$

one speed doesn't change =x

another one change and new speed = k

Relative speed = x + k

Distance = Speed * time

$y = (x + k) \times \frac{y}{4x}$

$k = 3x$

New speed of another cyclist = 3x

Speed of the another increased by 3 times.

Answer 2

Shortcut

Time reduced to 1/2

i.e Time ratio of first to second case = 2:1

Time is inversely proportional to Speed ratio

Speed ratio = 1 : 2

Relative speed of first case = 2x

Speed ratio = 1:2 = 2x : 4x

Relative speed of second case = 4x

x+k = 4x

k = 3x

For Formula and variable x, k check the answer by jeffarz01