
Que: Two cars A and B start simultaneously from two points P and Q with certain speeds towards each other. After reaching a point R, the speed of A decreases by 1/3. It then meets B at a pointS. where SQ = 2PR. If the speed of A had become 1/3 less at the mid point of RS, the cars would have met at T where ST = PR/4. Find RS:PR.
A. 5:1
B. 6:1
C. 8:1
D. 10:1
Official Approach as given in the solution: It finds the distance travelled by A had it retained its original speed till the time it meets B in both the cases. It then equates the ratio of distance travelled by A to that of B in both the cases.
I understand the above approach taken by the official solution,
but I fail to understand why the approach that I had originally taken is wrong. I would be really thankful if someone can point out the fault in my reasoning below:
Let time it takes for A to travel from P to R be t1 and that from R to S be t2.
Let the speed of A be A.
So the time it would take to travel from R to M would be t2/2.
As A had travelled RM with a speed 2/3 of A and now it travels the same distance at A, the new time taken for RM would be t2/3.
The time saved in RM would thus be t2/2  t2/3 = t2/6
A would travel the extra distance ST at speed 2/3 of A during this time.
So ST = t2/6 * (2/3)*A
and PR = A*t1
according to question ST = PR/4.
this gives t2/t1 = 9/4
As RS = (2/3 of A ) * t2
and PR = A * t1
RS/PR = 2/3 * 9/4 = 3/2
which is not the correct answer choice. 