CAT Exam Model Paper 7 with solutions for free online practice

Question : 17 of 60

Marks: +1, -0

Anshul and Nitish run between point A and point B which are 6 km apart. Anshul starts at 10 a.m. from A, reaches B, and returns to A. Nitish starts at 10:30 a.m. from B, reaches A, and comes back to B. Their speeds are constant with Nitish’s speed being twice that of Anshul's. While returning to their starting points they meet at a point which is exactly midway between A and B. When do they meet for the first time?

$10 : 33 {1}/{3}$ am
$10 : 37 {2}/{3}$ am
10 : 33 am
$10 : 33 {2}/{3}$ am

Validate

Solution:

Let the speed of Anshul be v km/hr.
So the speed of Nitish would be 2v km/hr.
Time taken by Anshul and Nitish to reach exactly midway between A and B, while returning to their 9 9 starting points =

9/\v

hrs and

9/{2\v}

hrs respectively,
Anshul started

1/2

hr early. Hence,

{9}/{\v}= {9}/{2 \v}+ {1}/{2} \v=9

km/hr
Distance covered by Anshul till 10:30 a.m.

=9 ×{1}{2}=4.5 {\km}

Time required by Anshul and Nitish to cover the remaining 6 - 4.5 = 1.5 km for their first meeting

= {1.5}/{\v+2 \v}= {1.5}/{9+18} {\hrs}= {1}/{18} {\hrs}

= {60}/{18} {\minutes}=3 {1}/{3} {\minutes}

Time when they meet first

=10 : (30+3 {1}/{3} )=10 : 33 {1}/{3} \ {\a.\m}

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