Adrian, Brian and Colin walk from P to Q (PQ=55 meters) at their resp

Question

Adrian, Brian and Colin walk from P to Q (PQ=55 meters) at their respective uniform speeds. Adrian sets off first followed by Brian and Colin at 5 second intervals. Brian and Colin catch up with Adrian at the same time and then the three proceed to Q. On reaching Q, Colin turns back and starts walking towards P, meeting Brian and Adrian 9 and 15 meters respectively from Q. What is Adrian's speed in meters per second?

A. 1
B. 1/2
C. 2
D. 2/3
E. None of the above

A. 1/4
B. 1/2
C. 2/3
D. 1
E. 2

chetan2u · Answer

Adrian sets off first followed by Brian and Colin at 5 second intervals  
 Let the speed of Adrian, Brian and Colin be A, B and C, and they meet each other after Colin travels for x seconds, that is Brian has traveled for x+5 and Adrian has traveled for x+10 
So  A(x+10)=B(x+5)=Cx...........A=\frac{Cx}{x+10}.....B=\frac{Cx}{x+5}

On reaching Q, Colin turns back and starts walking towards P, meeting Brian and Adrian 9 and 15 meters respectively from Q.

So A travels 55-15 or 40 and C travels 55+15 or 70 when they meet....
 \frac{40}{A}=\frac{70}{C}+10  as C starts 10 secs later.
Substitute  A=\frac{Cx}{x+10} ..
 \frac{40(x+10)}{Cx}=\frac{70}{C}+10.......40x+400=70x+10Cx...........10Cx=400-30x ...(i)

Also B travels 55-9 or 46 and C travels 55+9 or 64 when they meet....
 \frac{46}{B}=\frac{64}{C}+5  as C starts 5 secs later.
Substitute  B=\frac{Cx}{x+5} ..
 \frac{46(x+5)}{Cx}=\frac{64}{C}+5.......46x+230=64x+5Cx...........5Cx=230-18x....10Cx=460-36x ...(ii)

subtract ii from i....
 400-30x-(460-36x)=10Cx-10Cx=0......6x=60....x=10 secs

Now we can substitute x=10 in (i) to get  10Cx=400-30x...10*10C=400-30*10....C=1

Now when we know x=10 and C=1, substitute in  A=\frac{Cx}{x+10}=\frac{1*10}{10+10}=\frac{1}{2}

B

ShankSouljaBoi · Answer

How to do all this in 2 minutes, will definitely skip this one on test day.

chetan2u · Answer

This is more of a CAT question. Not likely to be tasted on actual GMAT

Posted from my mobile device

effatara · Answer

Let's say (a) Adrian's Brian's and Colin's speeds are S1, S2 and S3 respectively; (b) after they set off from P at 5-second intervals, all three meet at a point that is 'd' meters from P; and (c) the time taken by Adrian to cover 'd' meters is 't' seconds.

THE RATIO OF THE SPEEDS OF TWO MOVING OBJECTS ARE EQUAL TO (A) THE INVERSE OF THE RATIO OF THE TIMES TAKEN BY EACH TO COVER THE SAME DISTANCE; AND ALSO (B) THE RATIO OF THE DISTANCES EACH COVER IN THE SAME PERIOD OF TIME. Therefore:

(1) S1/S3 = (t-10)/t (As per Statement A)
 S1/S3 =  /  (As per Statement B)
 (t-10)/t = (40-d)/(70-d)....> t(40-d) = (t-10)(70-d)....> 3t+d = 70......(i)

(2)S2/S3 = (t-10)/(t-5) (as per Statement A)
 S2/S3 =  /  (As per Statement B)
 (t-10)/(t-5) = (46-d)/(64-d)....> (t-10)(64-d) = (t-5)(46-d)...> 18t+5d = 410.....(ii)

Solving (i) and (ii), we get d=10 meters and t=20 seconds. V1= d/t = 10/20 = 1/2 m/s. ANS: B

Kinshook · Answer

effatara

Given: Adrian, Brian and Colin walk from P to Q (PQ=55 meters) at their respective uniform speeds. Adrian sets off first followed by Brian and Colin at 5 second intervals. Brian and Colin catch up with Adrian at the same time and then the three proceed to Q. On reaching Q, Colin turns back and starts walking towards P, meeting Brian and Adrian 9 and 15 meters respectively from Q.

Asked: What is Adrian's speed in meters per second?

Let the Adrian's speed, Brian's speed & Colin's speed be x, y & z meter per second respectively.

When they meet after departing from P: 
Time taken by Adrian = t seconds
Distance travelled by Adrian = xt meters
Time taken by Brian = t-5 seconds
Distance travelled by Adrian = y(t-5) meters
Time taken by Colin = t-10 seconds
Distance travelled by Adrian = z(t-10) meters
xt = y(t-5) = z(t-10) = D meters
x = D/t; y = D/(t-5); z = D/(t-10)

When Colin turns back and meet Brian
Distance travelled by Colin = 55 + 9 = 64 meters
Distance travelled by Brian = 55 - 9 = 46 meters
Time taken by Colin = Time taken by Brian - 5
64/z = 46/y - 5
64(t-10)/D = 46(t-5)/D - 5
46(t-5) - 64(t-10) = 5D
410 - 18t = 5D

When Colin turns back and meet Adrian
Distance travelled by Colin = 55 + 15 = 70 meters
Distance travelled by Brian = 55 - 15 = 40 meters
Time taken by Colin = Time taken by Brian - 10
70/z = 40/x - 10
70(t-10)/D = 40t/D - 10 
700 - 30t = 10D

2(410-18t) = 820-36t = 700-30t
6t = 120; t = 20 seconds
5D = 410-18t = 410-360 = 50 
D = 10 meters

x = D/t = 10/20 = 1/2 meter per second

IMO B

Adrian, Brian and Colin walk from P to Q (PQ=55 meters) at their resp

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Adrian, Brian and Colin walk from P to Q (PQ=55 meters) at their resp

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards