Question
The efficiency of inlet pipe 'P' is 60% of the
efficiency of another inlet pipe 'Q'. When both pipes are opened together, they fill a tank in 4 hours. How long will it take for pipe 'Q', operating at 80% of its efficiency, to fill a tank that is 50% larger in capacity than the original tank?Solution
Let the efficiency of pipe 'Q' be '5a' litres/hour. So, efficiency of 'P' = 0.6 X 5a = '3a' litres/hour Total capacity of given tank = (3a + 5a) X 4 = '32a' litres Capacity of the tank to be filled by pipe 'Q' = 1.5 X 32a = '48a' litres New efficiency of pipe 'Q' = 0.8 X 5a = '4a' litres/hour Required time = (48a/4a) = 12 hours
If p = 24 - q - r and pq + r(q + p) = 132, then find the value of (p² + q² + r²).
((99.9 - 20.9)² + (99.9 + 20.9)² )/(99.9 x 99.9 + 20.9 x 20.9) = ?
...
Find the value of the given expression-
(4x+4 -5× 4x+2) / 15×4x – 22×4x
If 4x² + y² = 40 and x y = 6, then find the value
of 2x + y?
If p = 40 - q - r and pq + r(q + p) = 432, then find the value of (p² + q² + r²).
47.98 × 4.16 + √325 × 12.91 + ? = 79.93 × 5.91
If x + y = 4 and (1/x) + (1/y) = 24/7, then the value of (x3 + y3).
- If p = 20 - q - r and pq + r(p + q) = 154, then find the value of (p² + q² + r²).
If a = (√2 - 1)1/3, then the value of (a-1/a)3 +3(a-1/a) is:
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