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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.
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