Question

A and B together can complete a task in 15 days. If A works half as efficiently as he usually does and B works thrice as efficiently as he usually does, then the task gets completed in 8 days. How many days would A take to do the task if he works alone?

• A. \(33\frac{1}{3}\)
  (B) \(26\frac{2}{3}\)
• B. \(26\frac{2}{3}\)
• C. 40
• D. 30

Hint:

Using a convenient number for the total work (like the LCM of the days given) helps avoid complex fractions and makes solving the rate equations easier.

Answer 1

The Correct Option is A

Step 1: Set up equations based on work rates. Let A's normal rate of work be 'a' units/day and B's be 'b' units/day. Let the total work be W. From the first condition: They complete the task in 15 days. So, their combined rate is \( a+b = \frac{W}{15} \). Let's assume W=120 units (LCM of 15 and 8 is 120). Eq (1): \( a+b = \frac{120}{15} = 8 \).

Step 2: Set up the second equation with changed efficiencies. A's new rate = \( a/2 \). B's new rate = \( 3b \). They complete the task in 8 days. Eq (2): \( \frac{a}{2} + 3b = \frac{120}{8} = 15 \).

Step 3: Solve the system of linear equations for 'a' and 'b'. From Eq (1), \( b = 8 - a \). Substitute this into Eq (2): \[ \frac{a}{2} + 3(8 - a) = 15 \] \[ \frac{a}{2} + 24 - 3a = 15 \] \[ 24 - 15 = 3a - \frac{a}{2} \] \[ 9 = \frac{6a - a}{2} = \frac{5a}{2} \] \[ 5a = 18 \Rightarrow a = \frac{18}{5} = 3.6 \] A's rate is 3.6 units/day.

Step 4: Calculate the time A would take to complete the task alone. Total Work = 120 units. A's rate = 3.6 units/day. Time for A alone = \( \frac{\text{Total Work}}{\text{A's Rate}} = \frac{120}{3.6} = \frac{1200}{36} = \frac{100}{3} \). Time = \( 33\frac{1}{3} \) days.