Question

Mr. Kumar drives to work at an average speed of 48 km per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. How far is his office?

• A. $30km$
• B. $40km$
• C. $45km$
• D. $48km$

Answer 1

Answer: (B)

$40km$

Let the total distance be $xkm.$

Time taken to cover remaining $40\%$ of $x$ distance $T_1=\frac{40\times x}{100\times 48}$

But given time taken to cover first $60\%$ of $x$ distance is $T_2=T_1+\frac{10}{60}hrs$

$\Rightarrow$ $T_2=\frac{60\times x}{100\times 48}$

$\therefore$ $\frac{60\times x}{100\times 48}=\frac{40\times x}{100\times 48}+\frac{10}{60}$

$\Rightarrow$ $\frac{60\times x}{100\times 48}-\frac{40\times x}{100\times 48}=\frac{10}{60}$

$\Rightarrow$ $\frac{20x}{4800}=\frac{10}{60}$

$\Rightarrow$ $x=\frac{10}{60}\times \frac{4800}{20}$

$\Rightarrow$ $x=40km$