# Question: Curt & Omar Paint a Fence

## Comment on Curt & Omar Paint a Fence

### How is it that the Lowest

How is it that the Lowest common denominator of C, 2C AND 8 is 8C? How do we figure out such a value?

Aman ### Another name for lowest

Another name for lowest common denominator is least common multiple (LCM).

The multiples of C are: C, 2C, 3C, 4C, 5C, 6C, 7C, 8C, 9C,...etc
The multiples of 2C are: 2C, 4C, 6C, 8C, 10C, 12C,...etc

Some multiples of 8 are: 8, 16, 24, 32,...
Hold on. None of the multiples of 8 are the same as the multiples of C and 2C.
So, if we're to find a COMMON multiple, we'll have to examine multiples of 8 that also have a C term.

Multiples of 8 (with a C term) are: 8C, 16C, 24C, 32C, 40C,...

At this point, we can see that the LCM of C, 2C and 8 is 8C (since 8C is the smallest value that all three sets of multiples have IN COMMON.

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,

i got the answer correct but I think my working out was a bit unorthodox.

I determined from the question 2C = O and S = 8

Therefore, created C + O + S = 4

Taking S out added 8 hours onto 4 hours meaning:
C + O = 12

C + 2C(substitution from above) = 12

Therefore, 3C = 12

As there was no answer choice of 4 and considering it is Curt by himself I concluded the answer would be 12.

Would this be a problem in other similar questions doing things like this? ### I'm not entirely sure I

I'm not entirely sure I understand your approach.

What do C, O and S refer to?

I'm assuming these variables represent the time (in hours) it takes each person to paint the fence. However, if this is the case, your equation, C + O + S = 4, doesn't make any sense, since you already stated that S = 8

That is, if we replace S with 8, we get: C + O + 8 = 4, which means C + O = -4

I suggest that you don't use that approach.

Cheers,
Brent

### So my only weakness on this

So my only weakness on this problem is that it took way too long for me to figure out how to write "Curt can paint a fence in half the time that Omar can".

It took me a long time to figure out how to write that out algebraically. I first just thought okay, Curt's rate is faster, but then I realized the rate is 1/time, so I had to kind of reverse things since a shorter time would result in a larger value for the rate. So to make it easier I compared just the Time values for each person rather than the rate.

So I new Curt's time would be smaller than Omars, so C<O.

Then setting them equal to each other C = O I took a long time to figure out if C = 1/2 O or C = 2O. I finally figured it out that it needed to be C=1/2O but the wording really trips me up for some reason. ### The wording here is quite

The wording here is quite tricky, since it doesn't follow the same pattern as other questions.

For example, consider the following: "Curt's age is HALF Omar's age"
In this example, we can easily conclude that Curt's age = (0.5)(Omar's age)

Compare that to the given sentence: "Curt can paint a fence in HALF the time that Omar can".
We can write: Curt's work TIME = (0.5)(Omar's work TIME)
However, if we want to compare RATES, we must write: Curt's work RATE = (2)(Omar's work RATE)

One way to confirm whether your algebraic expression is correct is to test an easy set of values.