I got these right, but is there an easier/faster way to get to the answers?
For Q13, I did a quick run through of multiples of 2 & 3 that will give me 288.
For Q11, I just plugged in the answer, whats a quicker way?
Thanks!
Quicker way to tackle these?
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- beeparoo
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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
Cheers
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
Cheers
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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.
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.
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
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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