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Comment on Curt & Omar Paint a Fence
How is it that the Lowest
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,
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
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.