M7MBA wrote: ↑

Sun Feb 21, 2021 12:11 am

Yesterday, Candice and Sabrina trained for a bicycle race by riding around an oval track. They both began riding at the same time from the track's starting point. However, Candice rode at a faster pace than Sabrina, completing each lap around the track in 42 seconds, while Sabrina completed each lap around the track in 46 seconds. How many laps around the track had Candice completed the next time that Candice and Sabrina were together at the starting point?

A.  21
B.  23
C.  42
D.  46
E. 483

Answer: B

Source: Official Guide

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

M7MBA wrote: ↑

Sun Feb 21, 2021 12:11 am

Yesterday, Candice and Sabrina trained for a bicycle race by riding around an oval track. They both began riding at the same time from the track's starting point. However, Candice rode at a faster pace than Sabrina, completing each lap around the track in 42 seconds, while Sabrina completed each lap around the track in 46 seconds. How many laps around the track had Candice completed the next time that Candice and Sabrina were together at the starting point?

A.  21
B.  23
C.  42
D.  46
E. 483

Answer: B

Source: Official Guide

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