Question

A and B work at digging a ditch alternately for a day each. If A can dig a ditch in \(a\) days and B can dig in \(b\) days, will work get done faster if A begins the work?
I. \(n\) is a positive integer such that \(n\left(\frac{1}{a} + \frac{1}{b}\right) = 1\)
II. \(b > a\)

• A. Both statements needed
• B. Statement II alone sufficient
• C. Statement I alone sufficient
• D. Cannot be answered even with both

Hint:

Look at the work done in one pair of days (one day of A plus one day of B). Statement I says a whole number of such pairs finishes the job exactly.

Answer 1

The Correct Option is C

Step 1: Work done in a pair of days:
In one day A does \(\frac{1}{a}\) of the ditch and in one day B does \(\frac{1}{b}\). So two consecutive days (one of each) complete \(\frac{1}{a}+\frac{1}{b}\) of the job, whoever starts.

Step 2: Use statement I:
\(n\left(\frac{1}{a}+\frac{1}{b}\right) = 1\) means that exactly \(n\) pairs of days finish the ditch, that is exactly \(2n\) days. Before the last day (after \(2n-1\) days) the work done is less than 1, because the last day's share is positive. So the job ends exactly at the end of day \(2n\), whether A or B begins. The time is the same, so the work is not done faster if A begins. The question gets a definite answer from statement I alone.

Step 3: Statement II alone:
If \(b > a\), A is the quicker worker, so starting with A is never slower. But it can be equal (as in \(a=3, b=6\), where both orders take 4 days) or strictly faster (as in \(a=2, b=3\): A first takes \(2\frac{1}{3}\) days, B first takes \(2\frac{1}{2}\) days). So statement II alone gives no definite answer.

Final Answer:
The correct option is (3): Statement I alone sufficient.