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Solutions x, y and z are mixed at a ratio of 5:7:9

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Solutions x, y and z are mixed at a ratio of 5:7:9 [#permalink] New post 29 May 2006, 05:10
Solutions x, y and z are mixed at a ratio of 5:7:9 respectively in order to formulate one batch of 273 gallons of Brand X cleaning fluid. If the supplier of the solutions mixed up the order and supplied 20 gallons less of solution x, 28 gallons more of solution y and 54 gallons less of solution z, how many gallons of Brand X could be made?

a) 231
b) 189
c) 158
d) 156
e) 147

Can someone solve it for me? Thanks.
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 [#permalink] New post 29 May 2006, 05:35
Hallo raaja,
We need 273 gallons of brand X. Then brand X will contain 273/21=13 gallons of each part multiplied b y the respective ratio. Or x-65, y-91,z-117. When we take into account the confusion of the supplier, we have x 45,y-119, z-63. If we take z=9*7 then we need of each component 7 gallons or x-35,y-49, z-63 and the mixture would be 147 gallons in total
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 [#permalink] New post 30 May 2006, 04:06
Its E 147 Gallons of solution

Given 5x+7x+9x = 273 or x=13
Therefore for 273 Gallons we have 65, 91 and 117
But the supplier supplied us with 45, 119 and 63

Now this has to be in the ratio of 5:7:9
hence the final quantities required is 35, 49, 63 = 147
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 [#permalink] New post 30 May 2006, 05:25
I agree with E.

273/21 = 13 batches.

Based on the info there six less batches because 54/9.

13 - 6 = 7 batches.

7 x 21 = 147
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 [#permalink] New post 30 May 2006, 08:41
jaynayak wrote:
Its E 147 Gallons of solution

Given 5x+7x+9x = 273 or x=13
Therefore for 273 Gallons we have 65, 91 and 117
But the supplier supplied us with 45, 119 and 63

Now this has to be in the ratio of 5:7:9
hence the final quantities required is 35, 49, 63 = 147

Used same approach but it took me more than 3 minut to solve))getting rusty I guess
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 [#permalink] New post 30 May 2006, 10:11
BG wrote:
Hallo raaja,
We need 273 gallons of brand X. Then brand X will contain 273/21=13 gallons of each part multiplied b y the respective ratio. Or x-65, y-91,z-117. When we take into account the confusion of the supplier, we have x 45,y-119, z-63. If we take z=9*7 then we need of each component 7 gallons or x-35,y-49, z-63 and the mixture would be 147 gallons in total


I got E too, using the same approach. However, I made a couple silly errors along the way, glossing over the line that 28 gallons MORE of solution y was supplied.

Shows I gotta be more careful when reading to avoid silly mistakes.
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 [#permalink] New post 31 May 2006, 06:45
Thanks for all the responses. The OA is E.
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Re: problem on Mixture [#permalink] New post 31 May 2006, 09:26
raaja wrote:
Solutions x, y and z are mixed at a ratio of 5:7:9 respectively in order to formulate one batch of 273 gallons of Brand X cleaning fluid. If the supplier of the solutions mixed up the order and supplied 20 gallons less of solution x, 28 gallons more of solution y and 54 gallons less of solution z, how many gallons of Brand X could be made?

a) 231
b) 189
c) 158
d) 156
e) 147


= [{(273/21)(9) - 54}/9]21= 147

consider z cuz it is supplied by large reduction. therefore, the qty of z determines the qty of X.

if y were greately reduced, then it would be taken for the calculation.
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 [#permalink] New post 31 May 2006, 10:39
Good questiion.

I came up with E - 147.
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 [#permalink] New post 01 Jun 2006, 11:49
Here's the approach I followed :-

5a + 7a + 13a = 273 => a=13.

Thus, original quantity: x=65, y=91, z=117

New quantity: x=45, y=119, z=63.

Now, this is the point where you need to figure out, which one should be the limiting quantity. Thankfully, it's pretty easy in this case as z is the one that limits the other two.

Thus, if z=63, one part = 63/9 = 7 litres.
x= 5*7 = 35 litres, and y= 7*7 = 49 litres.

Therefore, total liquid = 63+35+49 = 147 litres.
  [#permalink] 01 Jun 2006, 11:49
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