On a track for remote controlled racing cars

Question

On a track for remote controlled racing cars, racing car A completes the track in 55 seconds, while racing car B completes it in 35 seconds. If they both start at the same time after how many seconds will they be side by side again.

A. 275
B. 325
C. 385
D. 425
E. None of these

A. 275
B. 325
C. 385
D. 425
E. None of these

Abhishek009 · Accepted Answer

Cumulative lap timings of Car A and Car B is as follows -

Car A = { 55 , 110 , 165 , 220 , 275 , 330 ,   385   , 440..........}

Car B = { 35, 70, 105 , 140 , 175 , 210 , 245 , 280 , 315 , 350 ,   385  , 420........}

Since, both are independent events, they will start again at 385 seconds ( from starting ) , hence, LCM is preferred here..

Abhishek009 · Answer

Time required in seconds will be LCM ( 35, 55 ) = 385
 
Hence, answer will be (C) 385

voccubd · Answer

Sorry I can't understand. Can you please explain in details or suggest me any resource material on this ? I found another similar problem in the internet where L.C.M wasn't the answer.

gracie · Answer

55=11*5
35=7*5
lcm=11*7*5=385

rohit8865 · Answer

Bunuel/Experts

What will be the approach if question mentions that Car b started 10 minuter after Car A??

THanks
Rohit

Blackbox · Answer

rohit8865 - I am no expert but here is my dig on it. Speed of Car A = 1/35 Speed of Car B = 1/55 Convert 10 minutes to seconds. That will be 600 secs Distance traveled by Car A in 600 secs = 600/55 = 120/11 laps. Relative speed = \frac{1}{35}-\frac{1}{55} So, Relative speed = \frac{4}{385} Relative distance = 120/11 laps. Time when they meet = \frac{\frac{120}{11}}{\frac{4}{385}} = 1050 secs or 17.5 mins. Hope I am correct

cbh · Answer

55x=35y (point in time when they run side by side)
x=35y/55 (integer, seconds)

x= 7*5*y/5*11, then y = 11, so 385
you can read it as in 385 secs car A(x) will 7 complete tracks and car B 11

KarishmaB · Answer

They will be side-by-side again when the faster car covers exactly one circle more than the slower car. Assuming the track length to be 35*11 meters, speed of car A is 7 m/sec and speed of car B is 11 meters/sec. Car B will cover the extra 35*11 meters (one full circle) in 35*11/4 secs = 96.25 secs

The problem is different when you have to find the time when they meet up again at the starting point.

Car A is at the starting point after: 55 sec, 110 sec, 165 sec, 220 sec ... etc
Car B is at the starting point after: 35 sec, 70 sec, 105 sec, 140 sec ... etc
So we need to take LCM here. The time when both would be at the starting point: 35*11 = 385 secs

Since the question has only 385 as an option, it must specify side by side at the starting point.

fogarasm · Answer

Rehashing an old question here—I agree with Karishma above—how are we to know that they have to be 'side by side' at the starting point? I must be missing something here, since the vast majority of people are getting this question right... but I would assume that in the processing of passing each other (which the quicker car does before the slower car has completed a second lap), they must've been side by side (and thus, the answer would be none of these, since it happened much earlier). Curious to hear how others knew how to interpret this.

timothy1212 · Answer

Let's do it as ratios, where 55x=35y x,y are number of laps that the cars will do, so we get x/y=35/55 and than just multiply one of them with the seconds it takes it to complete for example 7*55=385

Paras96 · Answer

LCM (35,55)=385

Hence C

itowhidul876 · Answer

If the question intended to know how many seconds after, for the first time they meet then the answer will be 96.25.
385 is one of the answer then.at 385 sec it will be their 4th meet up.

Posted from my mobile device

btsaami · Answer

If LCM method doesn't strike, we can used elimination.

Sa/Sb=35/55 i.e 7/11

Time at which they meet will be a multiple of 7 and 11. Check from options. B and D are not multiple of 11 (don't need to calculate for 7)

From A and C, we see that 275 is not a multiple of 7. Hence, 385 is the answer.

bumpbot · Answer

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On a track for remote controlled racing cars

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