UPSC CDS 1 2022 Math Paper

Given:
If the manwalks@3 km/hr, he reaches 40 min early
If the manwalks@2 km/hr, he reaches 40 min late
Formula Used:
Total distacne = Speed × Time
Caluclation:
Let the distance between residence and office be D
The usual speed is S

km

hr

and usual time be T hr
Now, According to the question
D=3(T−40∕60)‌‌⋯−(ii)
And D=2(T+40∕60)‌‌⋯ (iii)
From (ii) and (iii), we get
3(T−40∕60)=2(T+40∕60)
⇒3T−2=2T+4∕3
⇒T=(2+4∕3)hr
⇒T=10∕3hr
From (ii)
D=3[(10∕3)−(40∕60)]km
⇒D=3[(10∕3)−(2∕3)]km
⇒D=3(8∕3)km
⇒D=8km
∴ The distance between residence and office is 8km.