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Sum
Six taps can fill an empty cistern in 8 hours. How much more time will be taken, if two taps go out of order? Assume, all the taps supply water at the same rate.
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Solution
Total no. of taps = 6
Out of order taps = 2
Taps in working condition = 6 – 2 = 4
6 taps can fill an empty cistern in = 8 hours
1 tap can fill an empty cistern in = 6 x 8 hours
4 taps can fill an empty cistern in = `48/4` = 12 hours
More time taken when 2 taps are out of order = 12 – 8 = 4 hour
Concept: Concept for Unitary Method (With Only Direct Variation Implied)
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