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Algebra

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Algebra [#permalink] New post 27 Jun 2010, 00:38
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Hi, can someone please help to explain the step-by-step instructions (in detail) in the most simplest way and clear to understand.


1-
The ratio of two quantities is 3 to 4. If each of the quantities is increased by 5, what is that ratio of these two new quantities?

2-
Drum x is 1/2 full of oil and Drum Y, which has twice the capacity of Drum X, is 2/3 full of oil. If all of the oil in Drum X is poured into Drum Y, then Drum Y will be filled to what fraction of its capacity?
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Re: Algebra [#permalink] New post 27 Jun 2010, 02:11
1. The new ratio cant be determined. You can plug in numbers.

Old ratio \frac{3}{4}

Add 5 to both numerator & denominator

\frac{(3+5)}{(4+5)}=\frac{8}{9}

Now consider \frac{9}{12} (remember its the same ratio as \frac{3}{4} )

Add 5 to both

\frac{(9+5)}{(12+5)}=\frac{14}{17}
This gives a different result, hence we cant determine the new ratio with the limited given info.


2. Lets assume that capacity of drum X is x & that of drum Y is y.
we need to know that
\frac{x}{2} + \frac{4x}{3} is what proportion of 2x.

On evaluating, it will give \frac{11}{12}, which shold be the answer.


Consider Kudos, if my input was helpful.
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Re: Algebra [#permalink] New post 28 Jun 2010, 22:48
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The ratio of two quantities is 3 to 4. If each of the quantities is increased by 5, what is that ratio of these two new quantities?


We can't determine this. We could have 3 and 4 items or 3000 and 4000. Adding 5 in the first case drastically changes the ratio while adding 5 in the second case will have a negligble effect on the ratio.

However, if we were told that the second quantity doubled but the first remained the same, then the new ratio would be 3 to 8. Or, if we were told that the first quantity doubled but the second remained constant, then the new ratio would be 6 to 4 or 3 to 2.

Quote:
Drum x is 1/2 full of oil and Drum Y, which has twice the capacity of Drum X, is 2/3 full of oil. If all of the oil in Drum X is poured into Drum Y, then Drum Y will be filled to what fraction of its capacity?


We can solve algebraically or by picking numbers. Because Drum Y is 2/3 full of oil, we should pick for its capacity a number that is divisible by 3--let's pick 300. Y is 2/3 full, so there is 200 units of oil in it. Drum Y has twice the capacity of that of X. Thus, Drum X's capacity is 1/2 of 300 or 150. Drum X is 1/2 full so it has 75 units of oil in it. So, if we pour those 75 units of oil into Drum Y, then Drum Y will have 200 + 75 = 275 units of oil. So, Drum Y will be filled to 275/300 or 11/12 of its capacity. Although that looks long-winded, if you are organized in your scratchwork, you can answer really quickly by picking numbers:

---------------------X--------Y
capacity----------150------300
amount of oil-----75------200
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Re: Algebra [#permalink] New post 02 Nov 2011, 11:26
2) Picking smart numbers like 6 or 100 will give the answer.

Lets say Drum x capacity is 100. Drum Y capacity will be 200.

Oil in drum x is 50
Oil in drum y is 100* 2/3 = 400/3

Fraction of oil in y after x is emptied into y is = (50 + 400/3) / 200
= 550/600
= 11/12
Answer is c
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Re: Algebra [#permalink] New post 05 Nov 2011, 04:17
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multiply the denominator of all fractions to get pick smarter number. we have \frac{1}{2} & \frac{1}{3}, so pick 6 for X (smaller drum).

let X has capacity 6, and Y has 12.
X is half filled = 3
Y is 2/3 filled = 8

Y's capacity after pouring = \frac{3+8}{12} = \frac{11}{12}
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Re: Algebra   [#permalink] 05 Nov 2011, 04:17
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