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April 27th, 2018, 10:14 AM  #1 
Senior Member Joined: Aug 2014 From: India Posts: 343 Thanks: 1  How to solve this "time and distance" concept problem?
A and B start together from city P to city Q with respective speeds in the ratio 3:2 . If B takes $2\dfrac{1}{2}$ hours more than A to reach Q, then how much time does A take to reach Q? Ans: 5 hours I am stuck at this problem, I know distance =speed x time. Respective speed means 3x 2x = x 
April 27th, 2018, 10:44 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,262 Thanks: 1198 
respective speeds in the ratio 3:2 means that if A travels with speed $v$ then B travels with speed $\dfrac 2 3 v$ let $d$ be the distance between P and Q $\dfrac{d}{\frac 2 3 v}= \dfrac{d}{v}+\dfrac 5 2$ $\dfrac 3 2 \dfrac d v  \dfrac d v = \dfrac 5 2$ $\dfrac 1 2 \dfrac d v = \dfrac 5 2$ $T = \dfrac d v = 5$ and $T$ is the time it takes A to get from P to Q 
April 27th, 2018, 11:54 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,095 Thanks: 1905 
As the speeds of A and B are in the ratio 3:2, their times are in the ratio 2:3 respectively. Those times differ by 1, which you are told represents 2 1/2 hours, so A's time is 2 times 2 1/2 hours, which is 5 hours. This assumes that they take the same route from city P to city Q (or routes of the same length). 

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