A vehicle before overhauling requires $$\frac{5}{6}$$​ h service time every 90 days, while after overhauling it requires $$\frac{5}{6}$$​ h service time every 120 days. What fraction of the pre-overhauling services time is saved in the latter case?

Correct option is A

Given:
A vehicle requires:
Before overhauling:

$$\frac{5}{6}$$​ hours of service every 90 days.
After overhauling:

$$\frac{5}{6}$$​ hours of service every 120 days.

Concept Used:

Total service time required over a common period before and after overhauling.

A convenient common period is the Least Common Multiple (LCM) of 90 and 120 days, which is 360 days.

Solution:

Number of Service Events Over 360 Days:
Before overhauling (every 90 days):
Number of services =$$\frac{360 \text{ days}}{90 \text{ days/service}} = 4 \text{ services}$$​
After overhauling (every 120 days):
Number of services =$$\frac{360 \text{ days}}{120 \text{ days/service}} = 3 \text{ services}$$​
Total Service Time Over 360 Days:
Before overhauling:
Total service time =$$4 \text{ services} \times \frac{5}{6} \text{ hours/service} = \frac{20}{6} \text{ hours} = \frac{10}{3} \text{ hours}$$​
After overhauling:
Total service time =$$3 \text{ services} \times \frac{5}{6} \text{ hours/service} = \frac{15}{6} \text{ hours} = \frac{5}{2} \text{ hours}$$​

Time saved = Total time before - Total time after $$= \frac{10}{3} \text{ hours} - \frac{5}{2} \text{ hours}$$​

Time saved =$$\frac{20}{6} - \frac{15}{6} = \frac{5}{6} \text{ hours}$$​
Fraction saved =$$\frac{\text{Time saved}}{\text{Total time before}} = \frac{ \frac{5}{6} \text{ hours} }{ \frac{10}{3} \text{ hours} } = \frac{5}{6} \times \frac{3}{10} = \frac{15}{60} = \frac{1}{4}$$​​
The fraction of the pre-overhauling service time that is saved after overhauling is $$\frac 14$$.​