Question

A worker reaches his workplace 15 minutes late when he walks at a speed of 4km/hr from his house . The next day he increases his speed by 2 km/hr and reaches his work place on time . Find the distance from his house to his workplace ?

Answer 1

Given :-

→ The man reaches 15 mins late at speed of 4 km/hr

→ Next day increases his speed by 2 km/hr and reaches on time

To Find :-

→ Distance b/w his house and the workplace

Solution :-

→ Let that distance be x

★ Day 1 ★

→ Speed = 4 km/hr

→ Distance covered = x km

→ Time taken will be

$\sf \implies Time = \dfrac{Distance}{Speed}$

$\sf \implies \dfrac{x}{4} \; hrs$

★ Day 2 ★

→ Speed = 4 km/hr + 2 km/hr = 6 km/hr

→ Distance Covered = x km

→ Time taken will be

$\sf \implies \dfrac{x}{6} \; hrs$

→ As he was 15 mins late , let's covert it into hours

$\sf \implies 1 \; min = \dfrac{1}{60} \; hrs$

$\sf \implies 15\;mins = \dfrac{15}{60} = \dfrac{1}{4} \; hrs$

→ If we will add 1/4 hrs to time taken on day 2 we will get the time taken on Day 1 . So, the equation will be

$\sf \implies \dfrac{x}{6} + \dfrac{1}{4} = \dfrac{x}{4}$

→ Let's solve this equation !

$\sf \implies \dfrac{x}{4} - \dfrac{x}{6} = \dfrac{1}{4}$

$\sf \implies \dfrac{3x-2x}{12} = \dfrac{x}{y}$

$\sf \implies \dfrac{x}{12} = \dfrac{1}{4}$

$\sf \implies 4x = 12$

$\sf \implies x = \dfrac{12}{4}$

${\huge{\boxed{\underline{\mathrm{\blue{x=3}}}}}}}$

∴ The distance b/w his house and the workplace is 3 km

Answer 2

Answer:

let S be the distance, V be his second days speed, and U be his first days speed⇒ V= 6Km/hr, U=4Km/hr, T(time)=15min=0.25hr$S=[\frac{V+U}{2} ] t\\S=[ \frac{6+4}{2} ] 0.25\\S=\left[\frac{10}{2} ] 0.25\\\\\\S=\left[5] 0.25\\= 1.25Km$

Therefore the distance from his house to workplace is 1.25km