Practice GIKI Engineering Computing questions with answers and explanations.
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Daily wage equals total earnings divided by days worked.
Thus, 3,600 ÷ 12 = Rs. 300.
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A's share is 3 out of 8 total parts.
Thus, A receives 3/8 × 4,800 = Rs. 1,800.
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Total work is proportional to workers × hours × days.
So days = 8 × 6 × 15 ÷ (12 × 5) = 12.
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Twenty-four workers complete 1/4 in 10 days, so the remaining 3/4 needs 30 equivalent days at the same workforce.
Only 20 days remain, requiring 36 workers, so 12 extra are needed.
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In four days, each works twice.
Completed work is 2/15 + 2/20 = 4/30 + 3/30 = 7/30.
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Let B's rate be 1 unit and A's 3 units, total 4 units.
A alone takes 8 × 4/3 = 32/3 days.
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The machine produces 60/5 = 12 units per hour.
In 8 hours it produces 12 × 8 = 96 units.
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Their combined rate is 1/12 + 1/18 + 1/36 = 1/6.
Therefore, they finish in 6 days.
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If 40% takes 12 days, 10% takes 3 days.
The remaining 60% takes 18 days.
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Let B's rate be 2 units and A's rate be 3 units, total 5 units.
If 5 units finish in 6 days, B alone takes 6 × 5/2 = 15 days.
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The leak rate is 1/9 - 1/12 = 1/36 tank per hour.
Thus, the leak alone empties the tank in 36 hours.
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Their daily rate is 1/16 + 1/24 = 5/48.
In 4 days they do 5/12, so 7/12 remains.
