Question

A car and a bus starts from point ‘A’. . After ‘T'hours bus is 48 km ahead of car, while after 8 hours distance
between bus and car is same as distance covered by bus
one hour. If relative speed of car and bus if they
move towards each other is 240, then how much distance can bus cover in
in (T-(1/2)] hour.
A. 320 km
E.256 km
B. 280 km
C.384 km
D.336 km

Answer 1

320 km a bus can cover in 1/2 hour

Answer 2

Answer:

The bus covers 384 km in [T-(1/2)] hours.

Solution is C. 384 km

Step-by-step explanation:

Given: Car and a bus starts from point ‘A’

Let the speed of car be x, and speed of bus be y.

Distance of car after T hours = speed X time = Tx

Distance of bus after T hours = speed X time = Ty

Given, after ‘T' hours bus is 48 km ahead of car,

i.e, Ty= 48+Tx

∴ Ty - Tx = - 48

i.e, T(y-x) = - 48

T = $\frac{-48}{y-x}$

Relative speed of car and bus if they  move towards each other is 240,

i.e,y-x = 240 ⇒ equation 1

∴ T = $\frac{-48}{240}$ = $\frac{-4}{20}$ = $\frac{-1}{5}$

Also given, after 8 hours distance  between bus and car is same as distance covered by bus  one hour,

distance covered by car and bus after 8 hours = 8x and 8y respectively,

and distance covered by bus in one hour = 1 X y = y

∴ 8y- 8x = y

∴ 8y - y - 8x = 0

∴ 7y - 8x  = 0  ⇒ equation 2

From equation 1, y-x = 240

∴ x = y - 240

Putting this value of x in equation 2

7y - 8(y - 240)  = 0

∴ 7y -8y + 1920 = 0

-y = -1920

Hence, y = 1920

Now, we have to find the distance covered by bus in (T-(1/2)] hour.

distance = speed X time

= y X (T-(1/2))

= 1920 X [(-1/5)-(1/2)]

= 1920 X ($\frac{-1}{5}$ - $\frac{1}{2}$)

= 1920 X $\frac{(-2)(-1)+(5)(1)}{-10}$

= 1920 X $\frac{-3+5}{-10}$

=-1920 X $\frac{-2}{-10}$

=-1920 X $\frac{2}{10}$

= 384

Therefore, the bus covers 384 km in [T-(1/2)] hours.