Walking at \(\frac{5}{6}\) th of his normal speed from his home, Shahid is 25 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and his office is :
- AOption (a)
118 minutes
- BOption (b)
119 minutes
- COption (c)
125 minutes
- DOption (d)
128 minutes
View answer & explanationHide answer & explanation
Correct answer
125 minutes
Why this answer is correct
Shortcut Trick
Speed ratio = 5 : 6 → Time ratio = 6 : 5 (since distance is constant).
Difference in time ratio parts = 6 − 5 = 1 unit.
Given 1 unit = 25 minutes.
Usual time = 5 units = 5 × 25 = 125 minutes.
∴ The correct answer is 125 min.
Alternate Method
Given: New speed = 5/6 of normal speed, Time delay = 25 minutes.
Formula Used: Distance = Speed × Time. If distance is constant, Speed ∝ 1/Time.
⇒ Let normal speed be S and usual time be T.
⇒ Normal Distance = S × T
⇒ New Distance = (5/6)S × (T + 25)
⇒ S × T = (5/6)S × (T + 25)
⇒ T = 5/6(T + 25)
⇒ 6T = 5T + 125
⇒ T = 125 minutes.
∴ The correct answer is 125 min.
Additional Information
Inverse Proportion of Speed and Time
When the distance is constant, the ratio of speeds is inversely proportional to the ratio of time taken: S1/S2 = T2/T1.
Late/Early Concept Shortcut
Usual Time = (Late Time × Numerator) ÷ (Difference between Numerator and Denominator) = (25 × 5) ÷ (6 − 5) = 125 min.
Average Speed
If a person covers two equal distances at speeds x and y, the average speed is 2xy/(x+y).