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Two circles, one with radius 10 inches and the other with ra [#permalink]

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03 Nov 2013, 21:28

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A

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E

Difficulty:

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Question Stats:

67% (02:45) correct
33% (02:47) wrong based on 221 sessions

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Two circles, one with radius 10 inches and the other with radius 4 inches, are tangent at point Q. Two insects start crawling at the same time from point Q: one along the larger circle at 3π inches per minute, the other along the smaller circle at 2.5π inches per minute. How much time has elapsed when the two insects meet again at point Q?

Answer Since the insects are crawling along the boundaries of the circles, calculate the circumferences. The circumference of the larger circle is 20π inches, and that of the smaller circle is 8π inches.

Next, use the speeds to figure out how long each insect takes to walk around its respective circle exactly once. In each case, RT = D (the circumference).

For the insect on the larger circle, 3πT = 20π. This insect takes 20 / 3 minutes to crawl around the larger circle. For the insect on the smaller circle, 2.5πT = 8π, or, doubling both sides, 5πT = 16π. This insect takes 16 / 5 minutes to crawl around the larger circle. (Don’t simplify the fractions unless / until you have to.)

From this point, you can either work backwards from the answers or use arithmetic.

Working backwards: The correct answer must be a multiple of both 20 / 3 and 16 / 5. How to test this? Turn the answers into fractions that have the same denominator.

(A) 15 minutes. This answer turns into 45 / 3. 45 is not a multiple of 20, so this can’t be the right answer. (B) 30 minutes. This answer turns into 90 / 3. 90 is not a multiple of 20, so this can’t be the right answer. (C) 40 minutes. This answer turns into 120 / 3. 120 is a multiple of 20, so test 16 / 5. 40 minutes turns into 200 / 5. 200 is not a multiple of 16, so this can’t be the right answer. (D) 60 minutes. This answer turns into 180 / 3. 180 is a multiple of 20, so test 16 / 5. 60 minutes turns into 300 / 5. 300 is not a multiple of 16, so this can’t be the right answer. (E) 80 minutes. This answer turns into 240 / 3. 240 is a multiple of 20, so test 16 / 5. 80 minutes turns into 400 / 5. 400 is a multiple of 16, so this is the right answer.

The correct answer is (E).

A pure arithmetic solution is possible but the calculations are pretty laborious.

Turn the times of the insects into fractions with common denominators: the two insects take 100 / 15 and 48 / 15 minutes, respectively, to crawl around the circles. Find the least common multiple (LCM) of 100 / 15 and 48 / 15. Use a shortcut: the LCM will be 1/15 of the LCM of the numerators, 100 and 48.

Since 100 = (2)(2)(5)(5) and 48 = (2)(2)(2)(2)(3), the LCM of 100 and 48 must contain four 2’s, one 3, and two 5’s: (2)(2)(2)(2)(3)(5)(5) = 1200. The LCM of 100 / 15 and 48 / 15 is therefore 1200 / 15 = 400 / 5 = 80 minutes. It will take the insects 80 minutes, or 1 hour and 20 minutes, to meet up again.

Re: Two circles, one with radius 10 inches and the other with ra [#permalink]

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03 Nov 2013, 21:31

AccipiterQ wrote:

Two circles, one with radius 10 inches and the other with radius 4 inches, are tangent at point Q. Two insects start crawling at the same time from point Q: one along the larger circle at 3π inches per minute, the other along the smaller circle at 2.5π inches per minute. How much time has elapsed when the two insects meet again at point Q?

Answer Since the insects are crawling along the boundaries of the circles, calculate the circumferences. The circumference of the larger circle is 20π inches, and that of the smaller circle is 8π inches.

Next, use the speeds to figure out how long each insect takes to walk around its respective circle exactly once. In each case, RT = D (the circumference).

For the insect on the larger circle, 3πT = 20π. This insect takes 20 / 3 minutes to crawl around the larger circle. For the insect on the smaller circle, 2.5πT = 8π, or, doubling both sides, 5πT = 16π. This insect takes 16 / 5 minutes to crawl around the larger circle. (Don’t simplify the fractions unless / until you have to.)

From this point, you can either work backwards from the answers or use arithmetic.

Working backwards: The correct answer must be a multiple of both 20 / 3 and 16 / 5. How to test this? Turn the answers into fractions that have the same denominator.

(A) 15 minutes. This answer turns into 45 / 3. 45 is not a multiple of 20, so this can’t be the right answer. (B) 30 minutes. This answer turns into 90 / 3. 90 is not a multiple of 20, so this can’t be the right answer. (C) 40 minutes. This answer turns into 120 / 3. 120 is a multiple of 20, so test 16 / 5. 40 minutes turns into 200 / 5. 200 is not a multiple of 16, so this can’t be the right answer. (D) 60 minutes. This answer turns into 180 / 3. 180 is a multiple of 20, so test 16 / 5. 60 minutes turns into 300 / 5. 300 is not a multiple of 16, so this can’t be the right answer. (E) 80 minutes. This answer turns into 240 / 3. 240 is a multiple of 20, so test 16 / 5. 80 minutes turns into 400 / 5. 400 is a multiple of 16, so this is the right answer.

The correct answer is (E).

A pure arithmetic solution is possible but the calculations are pretty laborious.

Turn the times of the insects into fractions with common denominators: the two insects take 100 / 15 and 48 / 15 minutes, respectively, to crawl around the circles. Find the least common multiple (LCM) of 100 / 15 and 48 / 15. Use a shortcut: the LCM will be 1/15 of the LCM of the numerators, 100 and 48.

Since 100 = (2)(2)(5)(5) and 48 = (2)(2)(2)(2)(3), the LCM of 100 and 48 must contain four 2’s, one 3, and two 5’s: (2)(2)(2)(2)(3)(5)(5) = 1200. The LCM of 100 / 15 and 48 / 15 is therefore 1200 / 15 = 400 / 5 = 80 minutes. It will take the insects 80 minutes, or 1 hour and 20 minutes, to meet up again.

The correct answer is (E).

here's how I solved it:

the larger circle has a radius of 20pi, the smaller has a radius of 8pi, in order to meet at the tangent point both insects would have to have completed a rotation at the same time. Based on the speeds I eliminated A through D, since none of them resulted in a complete rotation for the insect trundling along the smaller circle, that left E. To check my work I plugged in the time (80 minutes) and multiplied that by the larger-circle bugs speed (3pi), to come up with 240", or in other words back at the tangent point to meet with its companion.

Re: Two circles, one with radius 10 inches and the other with ra [#permalink]

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07 Nov 2013, 09:31

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Two circles, one with radius 10 inches and the other with radius 4 inches, are tangent at point Q. Two insects start crawling at the same time from point Q: one along the larger circle at 3π inches per minute, the other along the smaller circle at 2.5π inches per minute. How much time has elapsed when the two insects meet again at point Q?

First we figure out the circumference of the circles so we can determine how long it takes to complete one revolution from point Q to point Q.

C (large) = 20pi C (small) = 8pi

R (large) = 3pi R (small) = 5/2pi

It takes 20/3 minutes to make one revolution around the larger circle It takes 8/(5/2) = 16/5 minutes to make one revolution around the smaller circle

The time it takes them to intersect can be determined by finding the LCM of both times. Get the denominators of both numbers equal one another.

Re: Two circles, one with radius 10 inches and the other with ra [#permalink]

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10 Nov 2014, 05:30

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Two circles, one with radius 10 inches and the other with ra [#permalink]

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17 Nov 2014, 08:31

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There are two methods to solving this problem: a long way and a short way.

The long way:

We know the insect on the smaller circle makes \(\frac{5}{16}\) revolutions per minute, and the insect on the larger circle makes \(\frac{3}{20}\) revolutions per minute.

In order for \(\frac{5}{16}*t\) to be an integer, \(t\) must be a multiple of 16. Likewise, \(t\) must also be a multiple of 20. We now have to find the LCM of 16 & 20, and to do that, we can use the nifty GMAT trick.

Find the prime factorization of 16: 2*2*2*2 Find the prime factorization of 20: 2*2*5

Notice the factors in common are 2*2. Remove these from each prime factorization, and multiply what is left with the ones in common. (2*2)(5)(2*2) = 20*4 = 80.

Therefore, 80 is the LCM and must be the answer. 80 minutes = 1 hour and 20 minutes.

Answer: E

---

The Short Way:

This one can also be solved much quicker by just checking the possibilities.

\(\frac{5}{16}\) *15 minutes = not an integer; \(\frac{5}{16}\) * 30 minutes = not an integer; \(\frac{5}{16}\) * 40 minutes = not an integer; \(\frac{5}{16}\) * 60 minutes = not an intetger; \(\frac{5}{16}\) * 80 minutes = 25[/m] <-- this is our only integer. Therefore it must be the correct answer.

Re: Two circles, one with radius 10 inches and the other with ra [#permalink]

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17 Nov 2014, 11:17

1. circumference of a circle: 2Rpi circumference of circle A = 2(10)pi = 20pi circumference of circle B = 2(4)pi = 8 pi

2. the insect on circle A will take 20/3 minutes to get back to point Q (6 minutes 40 seconds) the insect on circle B will take 8/2.5 minutes to get back to point Q (3 minutes 12 seconds)

3. look at answer choices A. in 15 minutes, insect A will not be on point Q because 15 minutes does not divide evenly by the 6 minutes 40 sec that it takes him to make 1 lap B. in 30 minutes, insect A will not be on point Q for the same reason as above C. in 40 minutes, insect A will not be on point Q for the same reason as above D. in 60 minutes, insect A will be on point Q (6min40sec X 9 = 60 minutes). insect B will not be on point Q though because 60 minutes does not divide evenly by 3 min 12 sec. E. in 80 minutes, insect A will be on point Q (6min40sec x 12 = 80 minutes). insect B will be on point Q as well (3min12sec x 25 = 80 minutes)

Re: Two circles, one with radius 10 inches and the other with ra [#permalink]

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26 Dec 2015, 00:30

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: Two circles, one with radius 10 inches and the other with ra [#permalink]

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29 Dec 2015, 06:08

distance in this case should be same as that of circumference of the circle circumference of the bigger circle = 2pir =20pi circumference of the smaller circle = 2pir1=8pi time taken by the first insect to complete the circle = 20pi/3pi=20/3 time taken by the second insect to complete the circle = 80pi/25pi=16/5 thus time taken to meet again at the starting point lcm = 1200/15 =80 minutes = 1 hour 20 minutes

Two circles, one with radius 10 inches and the other with ra [#permalink]

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21 Sep 2017, 10:34

OFFICIAL SOLUTION FROM MANHATTAN

Since the insects are crawling along the boundaries of the circles, calculate the circumferences. The circumference of the larger circle is 20pi inches, and that of the smaller circle is 8pi inches.

Next, use the speeds to figure out how long each insect takes to walk around its respective circle exactly once. In each case, RT = D (the circumference).

For the insect on the larger circle, 3piT = 20pi. This insect takes 20 / 3 minutes to crawl around the larger circle. For the insect on the smaller circle, 2.5piT = 8pi, or, doubling both sides, 5piT = 16pi. This insect takes 16 / 5 minutes to crawl around the larger circle. (Don’t simplify the fractions unless / until you have to.)

From this point, you can either work backwards from the answers or use arithmetic.

Working backwards: The correct answer must be a multiple of both 20 / 3 and 16 / 5. How to test this? Turn the answers into fractions that have the same denominator.

(A) 15 minutes. This answer turns into 45 / 3. 45 is not a multiple of 20, so this can’t be the right answer. (B) 30 minutes. This answer turns into 90 / 3. 90 is not a multiple of 20, so this can’t be the right answer. (C) 40 minutes. This answer turns into 120 / 3. 120 is a multiple of 20, so test 16 / 5. 40 minutes turns into 200 / 5. 200 is not a multiple of 16, so this can’t be the right answer. (D) 60 minutes. This answer turns into 180 / 3. 180 is a multiple of 20, so test 16 / 5. 60 minutes turns into 300 / 5. 300 is not a multiple of 16, so this can’t be the right answer. (E) 80 minutes. This answer turns into 240 / 3. 240 is a multiple of 20, so test 16 / 5. 80 minutes turns into 400 / 5. 400 is a multiple of 16, so this is the right answer.

The correct answer is (E).

A pure arithmetic solution is possible but the calculations are pretty laborious.

Turn the times of the insects into fractions with common denominators: the two insects take 100 / 15 and 48 / 15 minutes, respectively, to crawl around the circles. Find the least common multiple (LCM) of 100 / 15 and 48 / 15. Use a shortcut: the LCM will be 1/15 of the LCM of the numerators, 100 and 48.

Since 100 = (2)(2)(5)(5) and 48 = (2)(2)(2)(2)(3), the LCM of 100 and 48 must contain four 2’s, one 3, and two 5’s: (2)(2)(2)(2)(3)(5)(5) = 1200. The LCM of 100 / 15 and 48 / 15 is therefore 1200 / 15 = 400 / 5 = 80 minutes. It will take the insects 80 minutes, or 1 hour and 20 minutes, to meet up again.

The correct answer is (E).
_________________

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