Question
A boat takes 20 hours to travel 260 km downstream. The
speed of the boat in still water is 125% greater than the speed of the stream. If the boat completes a round trip from point 'P' to point 'Q' and back to 'P' in 18 hours, determine the distance between points 'P' and 'Q'.Solution
Downstream speed of the boat = (260/20) = 13 km/hr Let the speed of the stream be '4x' km/hr So, speed of boat in still water = 4x X (2.25) = '9x' km/hr So, (4x + 9x) = 13 So, 13x =13 So, 'x' = 1 So, speed of the boat in still water = 9 km/h The upstream and downstream speed of the boat are 5 km/h and 13 km/h respectively. Let the distance between points 'P' and 'Q' be 'A' km. ATQ, (A/13) + (A/5) = 18 Or, (18A/65) = 18 So, 'A' = 65
`sqrt(1297)` + 189.99 =?
90.004% of 9500 + 362 = ?
(74.76 ÷ 12.11 X ?)% of 239.89 = 600.19
- 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.)
1784.04 - 483.98 + 464.98 ÷ 15.06 = ?3
If tan θ + cot θ = 16, then find the value of tan2θ + cot2θ.
480 ÷ 10 + 18 % of 160 + ? * 9 = 60 * √36
(95.89% of 625.15 + 36.36% of 499.89) ÷ 6.02 = ? – 269.72
11.89 × 2.10 × 4.98 × 4.03 ÷ 7.98 of 15.03 = ?
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