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24 oz of juice P and 25 oz of juice V are mixed to make smot

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24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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New post 26 Apr 2011, 06:17
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24 oz of juice P and 25 oz of juice V are mixed to make smothies X and Y . The ratio of p to V in smothie X is 4 is to 1 and that in Y is 1is to 5.How many ounces of juice P are contained in the smothie X?

A. 5
B. 10
C. 15
D. 20
E. 25
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Re: mixture problem  [#permalink]

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New post 26 Apr 2011, 17:52
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shalu wrote:
24 oz of juice P and 25 oz of juice V are mixed to make smothies X and Y . The ratio of p to V in smothie X is 4 is to 1 and that in Y is 1is to 5.How many ounces of juice P are contained in the smothie X?

5
10
15
20
25


There isn't much of mixture concept you need here. It is actually based on ratios.
Smoothie X -> Juice P: 4n oz, Juice V: n oz
Juice P left for Smoothie Y: 24 - 4n
Juice V left for Smoothie Y: 25 - n
Smoothie Y ratio of P/V = \(\frac{1}{5} = \frac{(24-4n)}{(25-n)}\)
n = 5
Juice P in smoothie X = 4*5 = 20 oz
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Re: mixture problem  [#permalink]

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New post 26 Apr 2011, 06:58
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shalu wrote:
24 oz of juice P and 25 oz of juice V are mixed to make smothies X and Y . The ratio of p to V in smothie X is 4 is to 1 and that in Y is 1is to 5.How many ounces of juice P are contained in the smothie X?

5
10
15
20
25


Equation 1: Xsmoothie + Ysmoothie = 24 + 25 = 49 oz
Using the ratio of 4 is to 1 for Xsmoothie then, we know 4/(1+4) is P juice in Xsmoothie
Using the ratio of 1 is to 5 for Ysmoothie, we know 1/(1+5) is P jiuce in Ysmoothie


Equation 2: (4/5)X + (1/6)Y = 24

Let us now solve for x:
(4/5)X + (1/6)(49-X) = 24
24x + 5(49-x) = (24)(30)
24x + 245-5x = (24)(30)
19x = 720 -245
19x = 475
x=25

There is (4/5)(25) in Xsmoothie.
Answer is 20!
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Re: mixture problem  [#permalink]

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New post 26 Apr 2011, 06:58
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shalu wrote:
24 oz of juice P and 25 oz of juice V are mixed to make smothies X and Y . The ratio of p to V in smothie X is 4 is to 1 and that in Y is 1is to 5.How many ounces of juice P are contained in the smothie X?


Let there be x units of smothies X and y units of smothies Y.
For X:
P = 4x/5
V= x/5

For Y
P= y/6
V= 5y/6

Also, 4x/5 + y/6 = 24 ( 24 oz of P )--------(1)
x/5 + 5y/6 = 25 ( 25 oz of V)--------(2)
Solving above two equations: x= 25, y= 24

Hence, ounces of juice P contained in smothie X = 4x/5 => 4*25/5 = 20.

OA. 20.
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Re: mixture problem  [#permalink]

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New post 26 Apr 2011, 07:04
Easy way to solve this question is start from the answer and then conform the information provided in the question.

we can start from option D i.e 20 ... as a quantity of juice P in X because it is the only one option that gets divided by 4 is 20 ... since in the X the juice P to V ratio is 4:1

this gives us that quantity of juice P in X = 20 therefore quantity of Juice V will be 5 ... hence ratio = 4:1

This will lead to quantity of juice P in X = 4 and quantity of Juice V = 20 ... hence ratio 1:5

if we calculate total Juice P = 24 and total of juice V = 25
it fits because totals are same as what mentioned in the question ...

thus ans is D
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Re: mixture problem  [#permalink]

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New post 26 Apr 2011, 08:03
2
p1 + p2 = 24

v1 + v2 = 25


p1 = 4v1

p2 = v2/5

4v1 + v2/5 = 24

v1 + v2 = 25

4v2 - v2/5 = 76

19v2/5 = 76 => v2 = 20

=> v1 = 5

=> p1 = 20

Answer - D
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Re: mixture problem  [#permalink]

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New post 26 Apr 2011, 19:54
Karishma
Pls see this. Lets evaluate the ratio of X : Y.
If I mix X and Y smoothie then they should yield the ratio of p:q = 24:25.

The net ratio is 24:25 or 0.96 (mean)
The ratio of X is 4:1 or 4.0
The ratio of Y is 1:5 or 0.2

Since the ratio of Y is near the weighted mean, when I combine X and Y then Y > X i.e. the quantity of Y is more than that of X -----(1)

However, I am getting X = 25 oz and Y = 24 oz by using algebra - exactly opposite of inference (1). I thought that quantity that is closer to mean will pull the mean - the quantity of Y should be more. Did I miss something?

On the other hand equations yield-

In X : p and q are in ratio 4a:a
In Y : p and q are in ratio b:5b

Now, 4a + b = 24 (total quantity of juice P)
a + 5b = 25 (total quantity of juice Q)

Solving we get a = 5, b = 4
Hence X = 5a = 5 * 5 = 25 oz
Y = 6b = 6*4 = 24 oz
or X : Y = 25 : 24

VeritasPrepKarishma wrote:
There isn't much of mixture concept you need here. It is actually based on ratios.
Smoothie X -> Juice P: 4n oz, Juice V: n oz
Juice P left for Smoothie Y: 24 - 4n
Juice V left for Smoothie Y: 25 - n
Smoothie Y ratio of P/V = \(\frac{1}{5} = \frac{(24-4n)}{(25-n)}\)
n = 5
Juice P in smoothie X = 4*5 = 20 oz
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Re: mixture problem  [#permalink]

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New post 27 Apr 2011, 18:42
VeritasPrepKarishma wrote:

24/25 doesn't have any physical meaning.


Thanks ! So for using the "balance" I can use any one of the ingredients in the mix.
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Re: 24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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New post 22 Nov 2013, 07:45
6
3
shalu wrote:
24 oz of juice P and 25 oz of juice V are mixed to make smothies X and Y . The ratio of p to V in smothie X is 4 is to 1 and that in Y is 1is to 5.How many ounces of juice P are contained in the smothie X?

A. 5
B. 10
C. 15
D. 20
E. 25


Pretty straightforward.

For X: P/V = 4X/X
For Y: P/V = Y/5Y

Then 4x + y = 24
Also x+5y=25
So x=5
4x = 20

Answer is D

Hope it helps
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Re: 24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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New post 18 Apr 2014, 20:43
24-P
25-V

Solutions X and Y

X - P:V = 4:1

Y - P:V = 1:5

Now since total P and V are consumed to make X and Y, so

4X + Y = 24

X + 5Y = 25

X=5

We need to know what is 4X

Hence 20
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Re: 24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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New post 23 Apr 2014, 12:31
TGC wrote:
24-P
25-V

Solutions X and Y

X - P:V = 4:1

Y - P:V = 1:5

Now since total P and V are consumed to make X and Y, so

4X + Y = 24

X + 5Y = 25

X=5

We need to know what is 4X

Hence 20


Same way I did it.

Bunuel could you please validate?

Thanks
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Re: 24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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New post 23 Apr 2014, 21:47
3
1
Mixture I

P solution = p

V solution = v

Mixture II

P solution = 24-p

V solution = 25-v

Total = 49 ounces

Ratio in Mixture I

\(\frac{p}{v} = \frac{4}{1}\)

Ratio in Mixture II

\(\frac{24-p}{25-v} = \frac{1}{5}\)

Solving both equations above,

p = 20

Answer = D
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Re: 24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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New post 02 Aug 2015, 17:19
shalu wrote:
24 oz of juice P and 25 oz of juice V are mixed to make smothies X and Y . The ratio of p to V in smothie X is 4 is to 1 and that in Y is 1is to 5.How many ounces of juice P are contained in the smothie X?

A. 5
B. 10
C. 15
D. 20
E. 25


This is how I did it and just want to make sure that this is a correct approach and not just a co-incidence.

x = p/v = 4/1
y = p/v = 1/5

so asusming that total solution is a multiple of = 5 ( taking 1 from first fractions denominator and 5 from second denominator so 5x1 = 5)

So x = 4x5/1x4 = 20/4
y = 1x5/5x5 = 5/25

Hence P in X = 20
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Re: 24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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New post 02 Aug 2015, 22:34
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neeraj609 wrote:
shalu wrote:
24 oz of juice P and 25 oz of juice V are mixed to make smothies X and Y . The ratio of p to V in smothie X is 4 is to 1 and that in Y is 1is to 5.How many ounces of juice P are contained in the smothie X?

A. 5
B. 10
C. 15
D. 20
E. 25


This is how I did it and just want to make sure that this is a correct approach and not just a co-incidence.

x = p/v = 4/1
y = p/v = 1/5

so asusming that total solution is a multiple of = 5 ( taking 1 from first fractions denominator and 5 from second denominator so 5x1 = 5)

So x = 4x5/1x4 = 20/4
y = 1x5/5x5 = 5/25

Hence P in X = 20


Not sure what you did here. Cannot say if it is correct.
"total solution is a multiple of = 5 ( taking 1 from first fractions denominator and 5 from second denominator so 5x1 = 5)" doesn't sound correct.
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Re: 24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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refer the attachment for solution
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24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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New post 12 Nov 2016, 22:38
shalu wrote:
24 oz of juice P and 25 oz of juice V are mixed to make smothies X and Y . The ratio of p to V in smothie X is 4 is to 1 and that in Y is 1is to 5.How many ounces of juice P are contained in the smothie X?

A. 5
B. 10
C. 15
D. 20
E. 25


total ounces in X are divisible by 5
total ounces in Y are divisible by 6
if X+Y=49, then
X must have a ones digit of 5
Y must have a ones digit of 4
only possible option for X=25 ounces
25*4/5=20 ounces of P in X
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Re: 24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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New post 04 Sep 2017, 06:49
I somehow did (4/5)*24=19.2 and since it was close to 20, assumed it to be the answer. Was my approach on the right track?
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24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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New post Updated on: 08 Sep 2017, 06:13
I understand the answer is D.

My approach was to create as many equations as possible:
X+Y=49 (Eq 1)
Px+Py=24 (Eq 2)
Vx+Vy=25 (Eq 3)
X=Px/Vy=4/1=Px=4Vy (Eq 4)
Y=Px/Vy=1/5=Px=Vy/5 (Eq 5)

Then I saw, I can substitute Px=4Py in Eq 2.
I did that to get the following:
4Py+Py=24
5Py=24
Py=4.8
Px+4.8=24
Px=24-4.8=19.2.

Please help me understand where is my understanding wrong/what is the concept here/where am I going wrong?

Originally posted by aashishagarwal2 on 08 Sep 2017, 02:23.
Last edited by aashishagarwal2 on 08 Sep 2017, 06:13, edited 1 time in total.
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Re: 24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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New post 08 Sep 2017, 04:22
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Total juice = 24+25=49 (X+Y=49)
P juice in X is 4/5X.
V juice in Y is 1/6Y
Then
4/5X+1/6Y=24
CALCULATE X=25.
IN X, JUICE P equals to 4/5*25=20.



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Re: 24 oz of juice P and 25 oz of juice V are mixed to make smot  [#permalink]

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