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Problem Solving — algebra and arithmetic
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by GMATST » Mon Mar 21, 2011 10:29 am
I just can't figure out how to set up the equation properly to solve for this:

Three types of pencils J, K, and L cost $0.05, $0.10, and $0.25. If a box of 32 of these pencils costs a total of $3.40 and if there are twice as many K pencils as L pencils in the box, how many J pencils are there?

A) 6
B) 12
C) 14
D) 18
E 20
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by VivianKerr » Mon Mar 21, 2011 12:55 pm
.05j + .10k + .25l = $3.40

j + k + l = 32

2l = k

Plug the third equation into the 1st and 2nd equations and simplify:

.05j + .2l + .25l = 3.40
.05j + .45l = 3.40
5j + 45l = 340
j + 9l = 68

j + 2l + l = 32
j + 3l = 32

Now we have two equations with two variables. We can solve for j.

j + 9l = 68
- (j + 3l = 32)

6l = 36
l = 6

Plug l back in to solve for j.

j + 9(6) = 68
j + 54 = 68
j = 14
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by force5 » Mon Mar 21, 2011 1:00 pm
Nicely done Vivian

IMO= 14
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by luiscarlos59 » Fri Mar 25, 2011 8:25 am
Hi
Im not 100% sure

But i increased the decimals on the equations to look something like this
5j+10k+25l=340

And got J=20. L=8 and k=4

Which satisfies the above equation..
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by luiscarlos59 » Fri Mar 25, 2011 7:02 pm
My answer is different. Whats the OA?

Mariana do you know it?
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by GMATST » Mon Mar 28, 2011 9:20 am
The OA is 14
Thanks for your effort!

Vivian, I was just wondering why 'if there are twice as many K pencils as L pencils' the equation is 2L = K as opposed to 2K = L. Your equation works and yields the right answer but can you please explain to me the concept behind this? I seem to struggle with the wording of questions sometimes.
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