A takes 2 hours more than B to cover a distance of 40 km. If A doubles his speed, he takes 1$$\frac12$$​ hours morethan B to cover 80 km. To cover a distance of 100 km, how much time (in hours) will B take, travelling at his same speed?

Correct option is B

Given :

Distance₁ = 40 km

A takes 2 hours more than B for 40 km

When A doubles his speed, for 80 km he takes 1$$\dfrac{1}{2}$$​ = 1.5 hours more than B

Find time taken by B to cover 100 km at same speed

Formula Used :
​$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$​​
Solution :

Let speed of A = a km/h
Let speed of B = b km/h

Condition 1 (40 km):

​$$\frac{40}{a} = \frac{40}{b} + 2 \quad \text{……(1)}$$​​

Condition 2 (80 km, A doubles speed):

New speed of A = 2a

​$$\frac{80}{2a} = \frac{80}{b} + 1.5$$​​

​$$\frac{40}{a} = \frac{80}{b} + 1.5 \quad \text{……(2)}$$​​

Subtract (2) from (1):

​$$\left(\frac{40}{b} + 2\right) - \left(\frac{80}{b} + 1.5\right) = 0$$​​

​$$\frac{40}{b} =$$​0.5

b = 80 km/h
Required Time (B for 100 km):

​$$\text{Time} = \frac{100}{80} =$$​ 1$$\frac 14$$ hours