IMAT 2011 Q10 [Security Guards]

Two security guards, Dave and Geoff, are patrolling an airbase. Dave passes the front gate every 8 minutes. Geoff passes the front gate every 15 minutes. They have just set off on their individual routes at the start of their shift.
How long will it be before they meet up at the front gate again?

A. 1h 00mins

B. 1h 30mins

C. 2h 00mins

D. 2h 30mins

E. 3h 00mins

Dave passes every 8 minutes, and Geoff passes every 15 minutes.
What do we need to find: How long will it take them to meet again?
What should we do: We need to find the least common multiple (LCM) of both of their times, so we can find the earliest point where they met each other again.

But what is the LCM?

The Least Common Multiple (LCM) is also referred to as the Lowest Common Multiple (LCM) and Least Common Divisor (LCD). For two integers a and b, denoted LCM(a,b), the LCM is the smallest positive integer that is evenly divisible by both a and b. For example, LCM(2,3) = 6 and LCM(6,10) = 30.

The LCM of two or more numbers is the smallest number evenly divisible by all numbers in the set.

As we are not allowed to use a calculator during the exam, we can solve the question by listing multiples or by prime factorization, if you aren’t familiar with these methods, please navigate to our Appendix.

To solve this question, I will use both methods as an example.

The easier one would be to list the multiples, it will allow you to find and recognize the answer using visual elements:

Dave: 8,16,24,32,40,48,56,64,72,80,88,96,104,112,(120)
Geoff: 15,30,45,60,75,90,105,(120)
\frac{120}{60} = 2\ hours

Using the quicker method, we can use prime factorization and find the LCM:

LCM(8,15):
8=2x2x2
15=3x5
LCM: 2x2x2x3x5 = \frac{120}{60} = 2\ hours