Yesterday, Candice and Sabrina trained for a bicycle race by riding ar

Before solving the question, let's make sure we fully understand what the question tells us AND the implications.

Given: Candice can complete one lap in 42 seconds.
So, after 42 seconds, Candice will be at the starting point (and she will have completed 1 lap).
After 84 seconds, Candice will be at the starting point (and she will have completed 2 laps).
After 126 seconds, Candice will be at the starting point (and she will have completed 3 laps).
After 168 seconds, Candice will be at the starting point (and she will have completed 4 laps).
etc...
If we let C = the number of laps Candice has completed, then 42C = Candice's total running time


Given: Sabrina can complete one lap in 46 seconds.
So, after 46 seconds, Sabrina will be at the starting point (and she will have completed 1 lap).
After 92 seconds, Sabrina will be at the starting point (and she will have completed 2 laps).
After 138 seconds, Sabrina will be at the starting point (and she will have completed 3 laps).
After 184 seconds, Sabrina will be at the starting point (and she will have completed 4 laps).
etc...
If we let S = the number of laps Sabrina has completed, then 46S = Sabrina's total running time

How many laps around the track had Candice completed the next time that Candice and Sabrina were together at the starting point?
If Candice and Sabrina were together at the starting point, then we know that : 42C = 46S
Aside: Keep in mind that, in order for both people to be at the starting point, C and S must both be positive integers

So, we're looking for the smallest possible integer value of C such that: 42C = 46S
To make things a bit easier on ourselves, let's divide both sides of the equation by 2 to get: 21C = 23S
At this point, we can see that the smallest possible (positive) values of C and S are C = 23 and S = 21.

In other words, after Candice completes 23 laps, and Sabrina completes 21 laps, both runners will be together at the starting point.

Answer: B

Cheers,
Brent
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Candice —42 seconds
Sabrina —46 seconds
—> LCM ( 42, 46) = 2*21*23 = 966 seconds

They will meet together at the starting point after 966 seconds.
—> 966/42 = 23 laps ( for Candice)

Answer (B)

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Candice completes each round in 42 seconds, so she will be back at the starting point after every 42 seconds, so after 42*1, then 42*2 and so on.
Similarly Sabrina will be back at the starting point after every 46 seconds..46*1 , then 46*2 and so on.

So we look for TIME that would be multiple of 42 and 46, or in other words we are looking for the LCM.
LCM(42,46) = LCM({2*3*7},{2*23})=2*3*7*23=42*23=21*46

Now, Candice would have done \(\frac{42*23}{42}\) or 23 laps at the rate of 42 secs per lap.
Similarly, Sabrina would have done \(\frac{46*21}{46}\) or 21 laps at the rate of 46 secs per lap.


B. 23
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chetan2u VeritasKarishma could you explain this question, please.

If Candice had taken 1 second and Sabrina 2 seconds to complete a lap then the ratio of their speeds would have been 2:1 (inverse of their respective timings) and they would have met at the starting point after Candice had completed 2 laps and Sabrina 1 lap. By the same token:
Candice's speed: Sabrina's speed = 46:42 = 23:21 and they will meet at the starting point after Candice and Sabrina complete 23 and 21 laps respectively.
ANS: B

They both start together from S.

At what time points will candice reach S again?

Candice reaches S again after: 42 sec (1 lap done), 2*42 secs (2 laps done), 3*42 secs (3 laps done), 4*42 secs ... (all multiples of 42)
Sabrina reaches S again after: 46 sec (1 lap done), 2*46 sec (2 laps done), 3*46 secs (3 laps done), 4*46 secs ... (all multiples of 46)

When will they both reach S together again? At a time which is the first common multiple of 42 and 46.
LCM of 42 and 46 is 42 * 23. At this time, Candice would have completed 23 laps.

Answer (B)
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This question can be easily solved if we have the basics right.

Lets assume, Speed of Candice is S1 = D/46....................(1)
Speed of Sabrina is S2= D/48........................(2)
Where D is the total distance of the oval track.

Now, think what was constant throughout when they were running.
Their speed was variable and so is the distance they covered...only think constant was time(They ran for the same interval of time).

So Now, T = D1/S1 and T = D2/S2 .................................................................(3)
D1 and D2 are the distances traveled by Candice and Sabrina respectively.

Now, we can equate - equation (3)

so D1/S1 = D2/S2.................(4)

Substitute value of (1) and (2) in (4)

D1/(D/46) = D2/(D/48)
On solving D1/D2 = 23/21
Thus the answer will be a multiple of 23, Hence B

Solution:

Let’s let C = the integer number of laps that Candice completes; since her rate is 42 seconds per lap, then the expression 42C is the time when Candice passes the starting point. For example, when C = 1, then 42C = 42, and she has passed the starting point at the 42-second mark. When C = 2, then 42C = 84, and she has again passed the starting point at the 84-second mark.

Similarly, if we let S = the integer number of laps that Sabrina completes, then 46S is the time when Sabrina passes the starting point. For example, when S = 2, then 46S = 92, and she has passed the starting point at the 92-second mark.

We want to ascertain the number of laps that Candice has completed when the two runners again pass the starting point simultaneously. Thus, we equate the two expressions:

42C = 46S

21C = 23S

We see that we have equality if C = 23 and S = 21. Thus, Candice will have completed 23 laps when the two women pass the starting point simultaneously.

Answer: B
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Given :
Candice completes one round in 42s.
Sabrina completes one round in 46s.

To Find:
Number of rounds Candice has completed when they both meet for the second time.(First time is at start time).

Process:
If we find the time at which they both meet, we can easily find number of rounds Candice has completed.
A lot of people have explained the proper lengthy method, but let's look at shortcut.

So suppose when they meet, Candice has completed x rounds and Sabrina y.
Time taken for both is same.
42*x = 46*y
46 = 2*23
42 = 2*21
2*21*x = 2*23*y
As 23 is prime number, we can say that x(number of rounds Candice completes) will be a multiple of 23.
So option A, C are out.

Put x=23 and we get y=21. Both are whole numbers.
Hence second time they meet will be after 23 rounds.
B is the correct answer.

Concept: They will both meet at the Starting Point at the Exact Time that is the LCM of the Time it takes them to complete 1 lap Each

Find the LCM (42 seconds and 46 seconds) = 2 * 3 * 7 * 23 = 966 seconds will pass when they meet at the Starting Point.


In Those 966 Seconds, at a Speed of 42 Seconds per lap, Candice will finish 23 laps (966 / 42 = 23)

-B-

Each 46 seconds Sabrina is 4 seconds ahead.
\(\frac{46}{4}\) =11,5
So after 11,5 laps they are together, but since we want them at the start, we need a round number
11,5 * 2 = 23
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This answer was very helpful! Except, the speeds should have been 1/42 and 1/46 (as in given in the problem).

Hi Brent,

May i ask you why you chose 23 over 46? Nowhere in the question is the word least/lowest number of tracks that Candice had biked?

Thanks in Advance!

Note the words “ the next time that ” at the end of the question stem.

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IMO B

Time taken by Candice for a lap = 42
Time taken by Sabrina for a lap = 46

Time taken for both of them to meet at the start = LCM (42,46) = 42 * 23

Since, Candice completes 1 lap every 42 seconds
No. of laps completed in 42*23 seconds = [42*23]/42 = 23 laps when they meet at the start

Intially we need to figure out the LCM of the time taken = 23*21*2

Then for figuring out the no. of circles taken by Candice = 23*21*2/ 21*2

Therefore IMO B

I did it like that. Candice needs 4 seconds less than Sabrina. Therefore, I divided n* 46/5 until I get an integer --> 46/4 isn't an integer. Then again 2*46/4 now I got the integer 23.

I do not know if this works always, but for me it made sense to me in this case and was really fast.

Candice is coming at starting point in every 42 secs and Sabrina in every 46 secs
When do they come together-- when the time is a common multiple of 42 and 46 sec
This leads to the idea of calculating the lcm
lcm (42,46) by divisibility method = 2*21*23 (take the hcf out and multiply hcf with coprime parts)
in this time laps by candice = 2*21*23/42 = 23 laps