Question
Taps Q and R can fill a tank in 24 minutes and 32
minutes, respectively. The efficiency of tap S is 75% greater than that of tap Q. If tap R's efficiency is increased by 66â…”%, calculate the time it would take for taps S and the modified R to fill the tank together.Solution
Let the total capacity of the tank be 96 units (LCM of 24 and 32) . So, the efficiency of tap 'Q' = (96/24) = 4 units/minute So, the efficiency of tap 'R' = (96/32) = 3 units/minute So, the efficiency of tap 'S' = 4 X 1.75 = 7 units/minute Increased efficiency of tap 'R' = 3 X (5/3) = 5 units/minute So, combined efficiency of taps 'R' and 'S' = 7 + 5 = 12 units/minute Therefore, required time = (96/12) = 8 minutes
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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