To do a certain work, A and B work on alternate days with B beginning the work on the first day. A alone can complete the same work in 24 days. If the work gets completed in \(11 \frac{1}{3}\) days, then B alone can complete \(\rm \frac{7}{9}^{th}\) part of the original work in:

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SSC CGL 2020 Tier-I Official Paper 1 (Held On : 13 Aug 2021 Shift 1)
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  1. 4 days
  2. 6 days
  3. \(5 \frac{1}{2}\) days
  4. \(4 \frac{1}{2}\) days

Answer (Detailed Solution Below)

Option 2 : 6 days
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Detailed Solution

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Given:

A can complete the work in 24 days

A and B work on alternate days with B beginning the work, can complete the work in \(11 \frac{1}{3}\) days

Formula Used:

Total Work = Efficiency × Time

Calculation:

Let the total work be 24 units

⇒ Efficiency of A = 24/24 = 1 unit

According to the question,

A works on 2nd, 4th, 6th, 8th, 10th and (1/3) of 12th day

⇒ A works for 5(1/3) = 16/3 days

⇒ B works for = \(11 \frac{1}{3}\) - \(\frac{16}{3}\) = \(\frac{34-16}{3}\) = 6 days

⇒ Work done by A in 16/3 days = 16/3 units

Remaining work = 24 - (16/3) = 56/3 units

⇒ 56/3 units are completed by B in 6 days

(7/9)th part of 24 = (24 × 7)/9 = 56/3 units

∴ B alone will complete (7/9)th of the original work in 6 days.

Alternate Method

Given:

Time taken by A to finish a task alone = 24 days 

Calculation:

Let the total work be = 1 

A alone can finish the task in 24 days 

⇒ A's one-day work = 1/24 

A and B complete the whole task in = \(11 \frac{1}{3}\) days 

A and B work on alternate days, with B beginning so, we can say B will work only 6 days 

⇒ A will work only \(11 \frac{1}{3}\) - 6 = \(5 \frac{1}{3}\) days 

If A's one day work = 1/24 of work A completes in 1 day

⇒ A's \(5 \frac{1}{3}\) days work = 1/24 × \(5 \frac{1}{3}\) = 1/24 × 16/3 

⇒ 2/9 

Remaining work = 1 - 2/9 = 7/9 

∴ B does the 7/9th part of the work in 6 days. 

Note-

B, A, B, A, B, A, B, A, B, A, B, A/3

B completely 6-day work

That's why we have taken the 6 days work by B alone.

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