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EIV 4th Ed. Ch. 10 problem set Question 8 -- please help

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EIV 4th Ed. Ch. 10 problem set Question 8 -- please help

by futjim » Tue Jul 06, 2010 4:22 pm
Q8)
If c + d = 11 and c and d are positive integers, which of the following is a possible value for 5c + 8d?

- 55
- 61
- 69
- 83
- 88

The explanation at the back offered two approaches. I understand the first one but don't understand the second:

...you can notice that consecutive values of 5c + 8d differ by 3. In other words, every possible value of 5c + 8d equals a multiple of 3 plus some constant...

Question one: what is the rule here? What happens when a Multiple of 5, MO5, is added to a Multiple of 8, MO8? (I tried looking in the number properties book and found nothing)

I've also tried working this relationship out with numbers and found that if c & d in the above equation is different then the resulting number isn't a MO3. However, if c=d then the result is MO3 + some constant.

Can someone show me the light?

Thank you in advance.
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by jeremy8 » Tue Jul 06, 2010 6:04 pm
Not sure how to do it conceptually, but plugging numbers works very quickly.

Just try all the combinations of c+d=11 knowing they're both positive integers and plug them into 5c+8d

1+10, you know d can't be 10 because that's already greater than any of the answer choices, so you try 50+8=58, nowhere in the answer choices.

2+9, again d can't be 9, so you try 45+16=61. That's answer B.

This might actually be even faster than the algebra, although I'd really like to see that from someone.
But they always advise not to get stuck on trying to conceptually sovle a problem on the actual test, even though it's great to do so in practice.

You only have 2 minutes, so if you don't understand after a couple of seconds, try to see if there's a way of plugging numbers. In this case, you have your answer in a couple of seconds.
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by indiantiger » Tue Jul 06, 2010 6:18 pm
Given c+d = 11

need to solve for 5c+8d = ?

we can write 5c+8d as 5c+5d+3d = 5(c+d) + 3d ----(1)

now try plugging in c+d = 11 in (1)

this becomes 55+3d

A)rules out A
B) 55+6 = 61 (d could be 2, lets keep this one)
C) 55+14 = 69 (14 is not divisible by 3, ruled out)
D) 55+ 28 = 83(28 is not divisible by 3, ruled out)
E) 55+33 = 88 (for this d = 11 and c =0, as we know on GMAT 0 is neither +ve or -ve, ruled out)

Please do correct me if I have done any step wrong.

Answer is B
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by futjim » Tue Jul 06, 2010 7:05 pm
> Jeremy8, Thanks for the advice. I'm not very good at picking numbers coz I always seem to choose numbers that don't work and I get rather frustrated by it. Admittedly, when I do choose good numbers things work out very smoothly. This is something I need to train myself on.

> Indiantiger, Thank you so much. This is a very unique way of looking at 5c + 8d.
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