Question
Tap 'A' can fill a tank 'X' in 16 minutes while it takes 24
minutes when there is a leakage tank 'Y'. If tap 'B' is 33(1/3)% less efficient than tap 'A', then find the time taken by tap 'B' to fill the leakage tank 'Y'. (Capacity of tanks 'X' and 'Y' are equal)Solution
Let the total capacity of the tank be 48 units [L.C.M. of 16 and 24]
So, the efficiency of tap 'A' = (48/16) = 3 units/minute
Efficiency of tap 'A' with leakage = [48 ÷ 24] = 2 units/minute
So, the efficiency of leakage = 3 - 2 = 1 unit/minute
Efficiency of tap 'B' = 3 × (2/3) = 2 units/minute
So, effective efficiency of tap 'B' along with leakage = 2 - 1 = 1 unit/minute
Therefore, required time = (48/1) = 48 minutes
114, 134, 174, 254, 414, ?Â
10, 20, 60, 300, ?, 23100
What will come in place of the question mark (?) in the following series?
17, 138, 307, 532, ?, 1182
6848 + 9215 – 7295 - ? = 5945 – 3244
12, 24, 46, 78, 120, ?
What will come in place of the question mark (?) in the following series?
121, 110, 97, ?, 61, 38
32, 42, 62, 92, ?, 182
418, 406, 430, 382, 478, ?, 670
11          20          ?             64          112        192
...250, 279, 311, 349, 396, ?
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