Question
Tap A can fill tank X in 20 minutes, but with a leakage
(tank Y), it takes 33(1/3) minutes to fill. Tap B operates at 20% less efficiency than Tap A. Determine the time required for Tap B to fill the tank Y with the leakage. (Capacity of tanks 'X' and 'Y' are equal)Solution
Let the total capacity of the tank be 100 units [L.C.M. of 20 and (100/3) ] So, the efficiency of tap 'A' = (100/20) = 5 units/minute Efficiency of tap 'A' with leakage = [100 ÷ (100/3) ] = 3 units/minute So, the efficiency of leakage = 5 - 3 = 2 units/minute Efficiency of tap 'B' = 5 X 0.8 = 4 units/minute So, effective efficiency of tap 'B' along with leakage = 4 - 2 = 2 units/minute Thererfore, required time = (100/2) = 50 minutes
(31.9)3 + (34.021)² - (16.11)3 - (42.98)² = ?
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
What approximate value should come in the place of question mark in the following questions?
(598.5 ÷ 29.9) + (401.2 ÷ 20.1) = ?
(16.16 ×  34.98) + 14.15% of 549.99 = ? + 124.34
What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)...
960.02 of 238.89 ÷ 144.01 of 79.97 = ?
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
320.98 + 49.99% of (261.09 + 138.98) = ?
? = 29.91% of 49.94% of ((56.25 × 4.09 + 23.01 × 7.89) × 99.98)
20.02% of (95.96 × 104.01 – 56.02 × 64.04) – ? = 12.02 × 39.96 + 103.03Â
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