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# There are two kinds of coffee - coffee P and coffee R.

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Director
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There are two kinds of coffee - coffee P and coffee R.  [#permalink]

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15 Dec 2005, 03:44
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There are two kinds of coffee - coffee P and coffee R. Laxie has 24kg of coffee P and 25kg of cofee R. She made two mixtures of coffee using coffee P and coffee R. The ratio of coffee P to coffee R in mixture X is 4:1, and the ratio of coffee P to coffee R in mixture Y is 1:5. How many kg's of mixture X did she make?
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Director
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Re: PS - Laxie's coffee blend [#permalink]

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15 Dec 2005, 05:37
gamjatang wrote:
There are two kinds of coffee - coffee P and coffee R. Laxie has 24kg of coffee P and 25kg of cofee R. She made two mixtures of coffee using coffee P and coffee R. The ratio of coffee P to coffee R in mixture X is 4:1, and the ratio of coffee P to coffee R in mixture Y is 1:5. How many kg's of mixture X did she make?

By brute force
P and R lets take some coffe in proportion 4:1 so that the remainder was in ratio 1:5
24 25
8 2 remainder 16 and 23 nope
12 3 remainder 12 and 22 nope
16 4 ............... 8 and21 nope
20 5 .............. 4 and 20 bingo
answer is 25 kg of mixture X
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15 Dec 2005, 05:45
On the second thought there must be a shorter way to solve
let me try we have 49kg total and two mixtures
49=5X+6Y---->49-5X=6Y
taking numbers let x be 2 -->49-10=39 39/6 not integer
...................................3--->49-15=34 34/6not integer
....................................4--->49-20=29 ........
....................................5---->49-25=24 bingo
this may be easier way for someone
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15 Dec 2005, 07:32
25

let:
weight of mixure X: 5x
weight of mixture Y: 6y

4x + y = 24
x + 5y = 25

solve for x
x = 5

5x = 25
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15 Dec 2005, 08:39
duttsit wrote:
25

let:
weight of mixure X: 5x
weight of mixture Y: 6y

4x + y = 24
x + 5y = 25

solve for x
x = 5

5x = 25
very nice approach!
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15 Dec 2005, 11:44
duttsit wrote:
25

let:
weight of mixure X: 5x
weight of mixture Y: 6y

4x + y = 24
x + 5y = 25

solve for x
x = 5

5x = 25

I'm a little confused about this one, not sure how to come up with the two equations. Why is ok to set mixture X and Y equal to 24 and 25, respectively (24 is the weight of blend P, and 25 is the weight of blend R).
I'm sure you are right, I just need help understanding it. Thanks
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15 Dec 2005, 21:14
yb wrote:
duttsit wrote:
25

let:
weight of mixure X: 5x
weight of mixture Y: 6y

4x + y = 24
x + 5y = 25

solve for x
x = 5

5x = 25

I'm a little confused about this one, not sure how to come up with the two equations. Why is ok to set mixture X and Y equal to 24 and 25, respectively (24 is the weight of blend P, and 25 is the weight of blend R).
I'm sure you are right, I just need help understanding it. Thanks

24 kg is weight of P we have 4 parts from this 24 kg in mixture X and 1 part in mixture Y.Hence we can right the equation 24=5X(or if you want G any letter you like)
25 kg is weight of R we have 1 part from this 25 kg in mixture X and 5 parts in mixture Y.Hence we can right the equation 25=6Y(or if you want Z any letter you like)
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15 Dec 2005, 21:26
Yurik79 wrote:
duttsit wrote:
25

let:
weight of mixure X: 5x
weight of mixture Y: 6y

4x + y = 24
x + 5y = 25

solve for x
x = 5

5x = 25
very nice approach!

By the way duttsit how do you know that 4x+y should be 25?
4x + y = 24 --->here you can't know that it is equal to 24
x + 5y = 25 same is here))
All we know from question stem is that 4X+Y=mixture X and X+5Y=mixture Y
4x + y = 24
x + 5y = 25 if we knew that why to solve further this is the answer))
VP
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15 Dec 2005, 23:10
Yurik79 wrote:
Yurik79 wrote:
By the way duttsit how do you know that 4x+y should be 24?
4x + y = 24 --->here you can't know that it is equal to 24
x + 5y = 25 same is here))
All we know from question stem is that 4X+Y=mixture X and X+5Y=mixture Y
4x + y = 24
x + 5y = 25 if we knew that why to solve further this is the answer))

4x + y is total P we have and we know it 24

in mixture X: 4/5 th part is P
in mixture Y: 1/6 th part is P

total P = 24

now if X = 5x, P is 4x
in Y = 6y, P is y
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15 Dec 2005, 23:21
duttsit wrote:
Yurik79 wrote:
Yurik79 wrote:
By the way duttsit how do you know that 4x+y should be 24?
4x + y = 24 --->here you can't know that it is equal to 24
x + 5y = 25 same is here))
All we know from question stem is that 4X+Y=mixture X and X+5Y=mixture Y
4x + y = 24
x + 5y = 25 if we knew that why to solve further this is the answer))

4x + y is total P we have and we know it 24

in mixture X: 4/5 th part is P
in mixture Y: 1/6 th part is P

total P = 24

now if X = 5x, P is 4x
in Y = 6y, P is y

4x + y is total P not right because 4x+y=total mixture X
Total P=5X and total R=5Y
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15 Dec 2005, 23:28
Let Laxie confirm the answer since it's her coffee.
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16 Dec 2005, 08:46
duttsit wrote:
25

let:
weight of mixure X: 5x
weight of mixture Y: 6y

4x + y = 24
x + 5y = 25

solve for x
x = 5

5x = 25

I agree with duttsit's approach. Just need to clarify a little bit:
Let x be the times of the ratio 4:1 in mixture X
let y be the times of the ratio 1:5 in mixture Y
----> the amount of coffee P = 4x+ y = 24
coffee R= x+5y = 25

Exactly what duttsit did. Well, I can't drink that much coffee so I give them all to duttsit
16 Dec 2005, 08:46
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