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 Post subject: Help on Arithmetic Pg 74-75: Q 58 and Q59Posted: Thu Mar 10, 2011 4:49 pm

Joined: Fri Feb 25, 2011 10:46 am
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 Post subject: Re: Help on Arithmetic Pg 74-75: Q 58 and Q59Posted: Sun Mar 13, 2011 7:03 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
#58.

Original ratio is:
Blue / Red = 5/7

New ratio:
B + 3 / R - 9 = 3/2

This means we have two equations.
The first, through cross-multiplication, becomes:
7B = 5R
B = 5R/7
The second, through cross multiplication, becomes:
3(R-9) = 2(B+3)
3R-27 = 2B + 6
3R = 2B + 33
Using substitution, we can replace B in this one with 5R/7
3R = 2(5R/7) + 33
3R = 10R/7 + 33
I'd multiply both sides by 7 to get rid of the denominator
21R = 10R + 231
11R = 231
R = 21
Remember, we're looking for the number of Red pens AFTER the change, when we've taken away 9 of them, and the R represents the red pens before this, so we need to subtract 9 from the 21 to get to our answer, 12 (D).

59.
We need to find out which of the sets of two numbers represent a number that is potentially an original and a number that is between 8 and 9 times it.

I. 13. 8 times 13 is 104, and 9 times 13 is 117. 107 falls in this range, so this is a possible.
II. 48. 8 times 48 is 384. Since 379 is less than this, II is not an option.
III. 67. 67 times 8 is 536. 67 times 9 is 603. Since 591 falls in this range, this is a possible as well.
Thus, our answer is I and III, or D.

Veritas Help

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