Practice FAST-NU CS Engineering questions with answers and explanations.
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Three taps together have three times the rate of one tap.
One tap therefore takes 3 × 6 = 18 hours.
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The remaining part is 3/4 of the tank.
At 8 hours per full tank, 3/4 takes 6 hours.
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Daily wage equals total earnings divided by days worked.
Thus, 3,600 ÷ 12 = Rs. 300.
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Their combined daily rate is 1/10 + 1/15 = 1/6.
Therefore, they complete the job in 6 days.
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B's rate is 1/8 - 1/12 = 1/24.
Therefore, B alone needs 24 days.
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Worker-days remain constant for the same job.
Thus, 15 × 20 ÷ 12 = 25 workers.
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Twice the efficiency means twice the daily work rate.
Therefore, A takes half of 18 days, which is 9 days.
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In 5 days, A completes 5/20 = 1/4 of the job.
Therefore, 3/4 remains.
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If half the work takes 6 days, the full work takes twice as long.
Thus, the total time is 12 days.
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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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A composite number has more than two positive factors.
The number 4 has factors 1, 2, and 4.
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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.
