Question
The speed of a boat in still water is 50% more than the
speed of the stream such that the boat takes 3 hours to cover 81 km downstream. If the speed of the boat in still water is reduced by 20%, then find the distance covered by the boat in 5 hours upstream.Solution
ATQ, Let the speed of the stream be 10x km/hr. So, speed of the boat in still water = 1.5×10x=15x km/hr ATQ, 
Speed of stream = 10x = 10.8 km/hr Now, Reduced speed of boat in still water = 15x × 0.8 = 12 km/hr Therefore, required distance = (12−10.8)×5 = 6 km
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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