A, b and c start simultaneously from x to y. A reaches y, turns back and meet b at a distance of 11 km from y. B reached y, turns back and meet c at a distance of 9 km from y. If the ratio of the speeds of a and c is 3:2, what is the distance between x and y?
Answers
Answer:
99 km
Step-by-step explanation:
Let the distance between x and y be '' d ''
Let the speeds of a, b and c be "Va", "Vb" and "Vc".
Given that a reaches y and turns back and meets b at distance of 11 km from y .
⇒ a has covered a distance of (d + 11) km in the same time where b has covered (d - 11) km.
⇒ d + 11 / Va = d - 11/Vb
⇒ Va/Vb = d + 11/d - 11------(1)
Given that b reaches y and turns back and meets c at distance of 9 km from y .
⇒ b has covered a distance of (d + 9) km in the same time where c has
covered (d - 9) km.
⇒ d + 9 / Vb = d - 9/Vc
⇒ Vb/Vc = d + 9/d - 9------(2)
From (1) and (2), we get
Va/Vc = (d + 11)(d + 9)/(d - 11)(d - 9)
But given that Va/Vc = 3/2
⇒(d + 11)(d + 9)/(d - 11)(d - 9) = 3/2
⇒ 2[d² + 20d + 99] = 3[d² - 20d + 99]
⇒ d² - 100d + 99 = 0
⇒(d - 1)(d - 99) = 0
⇒ d = 1 or d = 99
d cannot be 1 since there are midway points at 9 km and 11 km from y
where they meet with each other
therefore : d = 99 km