Question

A boy covers a distance of 25 km in 4 hours partly on foot at the rate of 3.5 kmph and partly by cycle at 9 kmph. Find the distance covered on foot.
(a). 7 km
(b). 6.75 km
(c). 5.42 km
(d). None of these

Answer 1

Hint: We are given the total distance and the total time and the speeds of each of going on foot and by cycle. Assign a variable to the distance travelled by foot. Hence, equate the expressions of individual times to total time and find the distance covered on foot from it.

Complete step-by-step answer:
A boy covers a certain distance partly by walk and partly by cycle in a certain period. We need to find the distance, he covered on foot.
Let x be the distance covered by the boy on foot and y be the distance he covered by cycle.
The total distance he covered is 25 km. Hence, we have:
\[x + y = 25\]
\[y = 25 - x..........(1)\]
We know that the formula of speed is the distance covered divided by the time taken.
\[{\text{Speed = }}\dfrac{{{\text{Distance}}}}{{{\text{Time}}}}\]
Then, the time is distance divided by speed.
\[{\text{Time = }}\dfrac{{{\text{Distance}}}}{{{\text{Speed}}}}...........(2)\]
The time he walked on foot at the rate of 3.5 kmph is then given as follows:
\[{t_1} = \dfrac{x}{{3.5}}\]
The time he cycled at the rate of 9 kmph is given using equation (2) as follows:
\[{t_2} = \dfrac{y}{9}\]
The total time taken is 4 hours, hence, we have:
\[{t_1} + {t_2} = 4\]
\[\dfrac{x}{{3.5}} + \dfrac{y}{9} = 4\]
Substituting equation (1) in the above equation, we have:
\[\dfrac{x}{{3.5}} + \dfrac{{25 - x}}{9} = 4\]
Simplifying, we have:
\[\dfrac{{9x + 87.5 - 3.5x}}{{31.5}} = 4\]
Cross-multiplying, we have:
\[9x + 87.5 - 3.5x = 4 \times 31.5\]
Simplifying, we have:
\[5.5x + 87.5 = 126\]
Solving for x, we have:
\[5.5x = 126 - 87.5\]
\[5.5x = 38.5\]
\[x = \dfrac{{38.5}}{{5.5}}\]
\[x = 7km\]
Hence, the distance he covered on foot is 7 km.
Hence, option (a) is the correct answer.

Note: We have two equations with two unknowns. You can also solve for the distance covered by cycle first and then substitute to find the distance covered on foot.