Solution

The speed of a car is 40% less than the speed of a bike.

Let’s assume the speed of a bike is ‘15y’.

speed of a car = 15y of (100-40)%

= 15y of 60%

= 9y

The sum of the speed of the bus and car is equal to the speed of the bike.

speed of the bus = 15y-9y = 6y

The time taken by a bus to cover (d-100) km distance in 9 hours.

(d-100)/6y = 9

(d-100) = 54y

d = (54y+100)    Eq.(i)

If the time taken by car to cover (d+80) km distance is 2 hours more than the time taken by bike to cover (d+260) km distance.

(d+80)/9y = 2+ (d+260)/15y

Put the value of ‘d’ from Eq.(i) in the above equation.

(54y+100+80)/9y = 2+ (54y+100+260)/15y

(54y+180)/9y = 2+ (54y+360)/15y

(54y+180)/9y - (54y+360)/15y = 2

After solving the above equation, 0.4y+20 = 24

0.4y = 24-20

0.4y = 4

y = 10

Put the value of ‘y’ in Eq.(i).

d = (54x10+100) = 540+100 = 640

(i) The value of ‘d’ is the multiple of 16.

The above given statement is true.

(ii) The speed of the bike is 180 km/h.

Speed of bike = 15y = 15x10 = 150 km/h

The above given statement is not true.

(iii) The time taken by the car to cover (d+80) km is 8 hours.

time taken by the car to cover (d+80) km = (d+80)/9y

= (640+80)/(9x10)

= 720/90

= 8

The above given statement is true.