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The number of students who attend a school could be divided

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by VJesus12 » Sat Jan 05, 2019 6:47 am

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The number of students who attend a school could be divided among 10, 12, or 16 buses, such that each bus transports an equal number of students. What is the minimum number of students that could attend the school?

(A) 120
(B) 160
(C) 240
(D) 320
(E) 480

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT
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Source: — Problem Solving |

by Brent@GMATPrepNow » Sat Jan 05, 2019 8:13 am
VJesus12 wrote:The number of students who attend a school could be divided among 10, 12, or 16 buses, such that each bus transports an equal number of students. What is the minimum number of students that could attend the school?

(A) 120
(B) 160
(C) 240
(D) 320
(E) 480
The number of students who attend a school could be divided among 10, 12, or 16 buses, such that each bus transports an equal number of students.
This tells us that the TOTAL number of students is a multiple of 10, 12 and 16

What is the minimum number of students that could attend the school?
This whole question is a clever way to ask "What is the LEAST common multiple of 10, 12, and 16?"

Since the answer choices are written is ASCENDING order, we can just start with answer choice A and keep checking answers until we find a value that is a multiple of 10, 12, and 16

(A) 120. This is NOT divisible by 16. ELIMINATE
(B) 160. This is NOT divisible by 12. ELIMINATE
(C) 240. This is divisible by 10, 12, and 16

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Fri Feb 01, 2019 5:38 pm
VJesus12 wrote:The number of students who attend a school could be divided among 10, 12, or 16 buses, such that each bus transports an equal number of students. What is the minimum number of students that could attend the school?

(A) 120
(B) 160
(C) 240
(D) 320
(E) 480

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT
We need to determine the LCM of 10, 12, and 16

10 = 2 x 5

12 = 2^2 x 3

16 = 2^4

So the LCM is 2^4 x 3 x 5 = 16 x 3 x 5 = 240.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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