Question
A motorboat covers 48 km downstream and 30 km upstream
in 6 hours. Under the same conditions, it covers 64 km downstream and 20 km upstream also in 6 hours. Find the speed of the boat in still water and the speed of the stream.Solution
ATQ, Let downstream speed = D km/h, upstream speed = U km/h. From the data: 48/D + 30/U = 6 …(1) 64/D + 20/U = 6 …(2) Let 1/D = x and 1/U = y. Then: 48x + 30y = 6 …(1') 64x + 20y = 6 …(2') Subtract (1') from (2'): (64 − 48)x + (20 − 30)y = 0 16x − 10y = 0 ⇒ 16x = 10y ⇒ y = (16/10)x = (8/5)x. Substitute y into (1'): 48x + 30 × (8/5)x = 6 48x + 48x = 6 96x = 6 x = 6/96 = 1/16. So, D = 1/x = 16 km/h. Now y = (8/5)x = (8/5) × (1/16) = 8/80 = 1/10 ⇒ U = 10 km/h. Speed of boat in still water, B = (D + U)/2 = (16 + 10)/2 = 13 km/h. Speed of stream, S = (D − U)/2 = (16 − 10)/2 = 3 km/h. Answer: Boat speed = 13 km/h; stream speed = 3 km/h.
? × 5.5 = √1225 + 40% of 30% of 37.5% of 5000 – 63
26 X √25 + 15 - 80% of 120 = ?2
121/? = ?/144
- What will come in place of (?), in the given expression.
(5³ + 3²) × 2 = ? (125 × 12 × √8100) ÷ 150 = ?
40500 ÷ 30 × 42 – 45% of 2400 =? + 42100
(75 + 0.25 × 10) × 4 = ?2 - 14
Simplify:
What will come in the place of question mark (?) in the given expression?
(252 + 198) ÷ ? + 126 ÷ 3 = 144 ÷ 2
