Question
Tap 'A' can fill a tank 'X' in 12 minutes while it takes 18
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 36 units [L.C.M. of 12 and 18]
So, the efficiency of tap 'A' = (36/12) = 3 units/minute
Efficiency of tap 'A' with leakage = [36 ÷ 18] = 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 = (36/1) = 36 minutes
√1764 + 35 × 8 + 39 = ?2
18% of 200 - 16% of 150 = ?
25% of 30% of 3/5 of 14500 =?
2(1/3) + 2(5/6) – 1(1/2) = ? – 6(1/6)
7/3 of 4/5 of 15/56 of ? = 83
What will come in place of the question mark (?) in the following expression?
40% of 150 – ?% of 80 = 25% of 400
555.05 + 55.50 + 5.55 + 5 +0.55 = ?
64.5% of 800 + 36.4% of 1500 = (?)² + 38
What will come in the place of question mark (?) in the given expression?
25% of 1280 + (41 × 4) = ?2
Simplify the following expression:
((32)4 - 1)/33×31× (210+1)
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