Question
A man rows 45 km downstream and 27 km upstream in a
total of 6 hours. If the speed of the stream is 3 km/h, find the speed of the boat in still water.Solution
Let boat speed in still water = b km/h Downstream speed = b + 3, upstream speed = b − 3 Equation: [45/(b+3)] + [27/(b−3)] = 6 45(b−3) + 27(b+3) = 6(b²− 9) 45b − 135 + 27b + 81 = 6b²− 54 72b − 54 = 6b²− 54 6b²− 72b = 0 ⇒ 6b(b−12) = 0 ⇒ b = 12.
25.31% of 5199.90 + (19.9 × 17.11) + 46.021 =? + 168.98
1560.182 ÷ √168 + √143 * √224 – 4649.87 ÷ 30.883= ?    Â
1242.12 ÷ √530 + 1139.89 ÷ 14.91 = ? + 45.39
[(2/3 of 899.79) + 25% of 500.21] × (√195.77 + 30.03% of 399.79) = ?
416.021 ÷ 3.782 + 13.012 × 24.987 =?
...√31684.11 × √728.9 – (19.02)2 = ? × 4.99Â
12.03 x 4.21 +19.95% of 300.05 =Â ?
58.03% of 1499.99 - ? % of 699.95 = 394.04
Solve the given equation for ?. Find the approximate value.
(3/5 of 399.78 + 7/9 of 360.15) ÷ (√63.94 + 25% of 383.94) = ?
...1299.999 ÷ 325.018 × 24.996 = ?