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Quicker way to tackle these?

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by beeparoo » Sat Nov 24, 2007 2:12 pm
For Q13 (first one):

Use a simple factor-tree approach:
288 splits into 2 x 144 (a prime and perfect square of 12)

Upon realizing the above, I know the prime factors are now:
2 * (2 * 2 * 3) * (2 * 2 * 3) = (2^5)*(3^2)

Therefore x = 5, y = 2

Plugging these values into (2^(x-1))*(3^(y-2)) is
2^4 * 3^0 = 16

++

For Q11 (second question):

F/J = 3x/2x

Adding $40 for the month of FEB now changes the ratio in the following way:
F'/J = (3x+40)/2x, which the question claims is equivalent to 5/3

Equate, simplify and re-express in terms of x.
(x = 120)

Noting from the question that you are looking for January's costs, you know that JAN = 2x = 2*120 = 420

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by cris » Sat Nov 24, 2007 8:01 pm
I think there is an easiest way.


q11.the difference between 5/3 and 3/2 is 1/6

then multiply 40 x 6= 240
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by xcise_science » Sun Nov 25, 2007 1:27 pm
Thanks. Why did you multiply by 6?
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by me4mba » Sun Nov 25, 2007 4:40 pm
I think the key to answering Q13 is to know that:

x^(a-b) = x^a / x^b

So 2^(x-1) = 2^x / 2 and 3^(y-2) = 3^y / 9

Put the two components together, and you'll see the question is asking for:
(2^x)(3^y) / 18 it's already given that (2^x)(3^y) = 288

so really, the question is asking 288/18. The answer is 16.
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by me4mba » Sun Nov 25, 2007 4:49 pm
For Q11, you'll have to set up 2 equations based on the given ratios.

I) F/J = 3/2

II) (F+40) / J = 5/3

We have 2 equations and 2 variables, so we can solve for the variables. Since we're looking for J, I would substitute F = 3J/2 (derived from equation I) to equation II. So you'll have

3J/2 + 40 = 5J/3

Sove for J, J = 240
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by xcise_science » Sun Nov 25, 2007 6:27 pm
great, Thanks!!
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