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# Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol

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Re: Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
KarishmaB wrote:
neeraj609 wrote:
Hi VeritasPrepKarishma

I learnt the method to solve mixture problems from your blog and i applied the same method here. Now not sure if that is correct way or not, but i got the answer. Can you please confirm if what I did was correct?

__________|_________|__________
1% 1.5% 2% 3%
x Y z

==> x(1.5-1) = y(2..0 - 1.5) + z(3.0-1.5)
==> x/2 = y/2 + 3z/2
==> x = y + 3z

Yes, great work! In essence, you are using 'the deviation method for mean' which is of course applicable to weighted mean too.

KarishmaB

A quick question about 'the deviation method for mean'

1%----1.5%----2%----%3
x-------avg-------y-------z

the distance between x and avg: 1 unit
the distance between y and avg: 1 unit
the distance between z and avg: 3 unit

3x
3y
z

What am I missing here?
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
custodio wrote:
KarishmaB wrote:
neeraj609 wrote:
Hi VeritasPrepKarishma

I learnt the method to solve mixture problems from your blog and i applied the same method here. Now not sure if that is correct way or not, but i got the answer. Can you please confirm if what I did was correct?

__________|_________|__________
1% 1.5% 2% 3%
x Y z

==> x(1.5-1) = y(2..0 - 1.5) + z(3.0-1.5)
==> x/2 = y/2 + 3z/2
==> x = y + 3z

Yes, great work! In essence, you are using 'the deviation method for mean' which is of course applicable to weighted mean too.

KarishmaB

A quick question about 'the deviation method for mean'

1%----1.5%----2%----%3
x-------avg-------y-------z

the distance between x and avg: 1 unit
the distance between y and avg: 1 unit
the distance between z and avg: 3 unit

3x
3y
z

What am I missing here?

You are trying to apply the scale method here which works only for average of two quantities, not weighted average of 3 quantities.
The distance is between 1% and 1.5%, which is 0.5%. x is the weight of 1% i.e. number of gallons at 1%.

So to use the deviations method, you need to do this:
x * 0.5 = shortfall
y*0.5 + z*1.5 = excess

x * 0.5 = y*0.5 + z*1.5
x = y + 3z
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
Given: ]Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume.

Asked: If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?

x + 2y + 3z / (x + y + z) = 1.5
x + 2y + 3z = 1.5 (x + y + z)
.5x - .5y - 1.5z = 0
x - y - 3z = 0
x = y + 3z

IMO A
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Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
KarishmaB wrote:
MSoS wrote:
Hi, would someone please so kind and explain the question:

Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. If x gallons of the q percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x + y + z gallons of a 1.5 percent grade, what is x in terms of y and z?

(a) y + 3z
(b) (y+z)/4
(c) 2y +3z
(d) 3y + z
(e) 3y + 4.5z

Thanks a lot...

A quick approach:

The question asks you for x in terms of y and z. Whatever values x, y and z can take, this relation should hold.
Since we mix 1%, 2% and 3% milk and get 1.5% milk, one way of mixing them could be that 1% and 2% are mixed in equal quantities (to give 1.5% milk) and 3% milk is not added at all. Which means x = 1, y = 1 and z = 0 should satisfy the relation between x, y and z.
The only relation that satisfies these values is (A).

Note: If multiple options satisfied these values, you could take another set of values e.g. x = 3, y = 0 and z = 1 and check out of the shortlisted options.

KarishmaB
Sorry to revive and old post, but I was hoping you could clarify the way you pick the numbers; I was picking a random set say x =5 ,y=3, and z=2.
However, I realized it came no where close to the posts here when plugging into any of the AC here. 5 =/ 9 in choice A for instance. Then I thought, I guess the numbers have to add up to 1.5% given in the stem, meaning the xyz that I picked have to combine to give 1.5% using the provided concentration for each?
x*1%+y*2% etc / (total gallons)?
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
1
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M838TE wrote:
KarishmaB wrote:
MSoS wrote:
Hi, would someone please so kind and explain the question:

Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. If x gallons of the q percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x + y + z gallons of a 1.5 percent grade, what is x in terms of y and z?

(a) y + 3z
(b) (y+z)/4
(c) 2y +3z
(d) 3y + z
(e) 3y + 4.5z

Thanks a lot...

A quick approach:

The question asks you for x in terms of y and z. Whatever values x, y and z can take, this relation should hold.
Since we mix 1%, 2% and 3% milk and get 1.5% milk, one way of mixing them could be that 1% and 2% are mixed in equal quantities (to give 1.5% milk) and 3% milk is not added at all. Which means x = 1, y = 1 and z = 0 should satisfy the relation between x, y and z.
The only relation that satisfies these values is (A).

Note: If multiple options satisfied these values, you could take another set of values e.g. x = 3, y = 0 and z = 1 and check out of the shortlisted options.

KarishmaB
Sorry to revive and old post, but I was hoping you could clarify the way you pick the numbers; I was picking a random set say x =5 ,y=3, and z=2.
However, I realized it came no where close to the posts here when plugging into any of the AC here. 5 =/ 9 in choice A for instance. Then I thought, I guess the numbers have to add up to 1.5% given in the stem, meaning the xyz that I picked have to combine to give 1.5% using the provided concentration for each?
x*1%+y*2% etc / (total gallons)?

We cannot pick numbers randomly here. They need to satisfy the given information.
We mix 1%, 2% and 3% milk and get 1.5% milk.
Can we mix 10 gallons of 3% milk with 1 gallon each of 1% and 2% and get 1.5% milk? No. In this case the concentration will be much closer to 3%.

There are of course infinite ways of arriving at 1.5% concentration.
Say if we take equal volumes of 1% and 2% and none of the 3% milk, we get 1.5% milk.
or say we take 1 gallon each of 2% and 3% milk to get 2 gallons of 2.5% milk. Now if we add 4 gallons of 1% milk to this, we will get 6 gallons of 1.5% milk.
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
M838TE wrote:
KarishmaB wrote:
MSoS wrote:
Hi, would someone please so kind and explain the question:

Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. If x gallons of the q percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x + y + z gallons of a 1.5 percent grade, what is x in terms of y and z?

(a) y + 3z
(b) (y+z)/4
(c) 2y +3z
(d) 3y + z
(e) 3y + 4.5z

Thanks a lot...

A quick approach:

The question asks you for x in terms of y and z. Whatever values x, y and z can take, this relation should hold.
Since we mix 1%, 2% and 3% milk and get 1.5% milk, one way of mixing them could be that 1% and 2% are mixed in equal quantities (to give 1.5% milk) and 3% milk is not added at all. Which means x = 1, y = 1 and z = 0 should satisfy the relation between x, y and z.
The only relation that satisfies these values is (A).

Note: If multiple options satisfied these values, you could take another set of values e.g. x = 3, y = 0 and z = 1 and check out of the shortlisted options.

KarishmaB
Sorry to revive and old post, but I was hoping you could clarify the way you pick the numbers; I was picking a random set say x =5 ,y=3, and z=2.
However, I realized it came no where close to the posts here when plugging into any of the AC here. 5 =/ 9 in choice A for instance. Then I thought, I guess the numbers have to add up to 1.5% given in the stem, meaning the xyz that I picked have to combine to give 1.5% using the provided concentration for each?
x*1%+y*2% etc / (total gallons)?

Hello M838TE,

You cannot pick any random set of numbers for x, y and z. The numbers x, y and z should satisfy every condition mentioned in the question.

Let’s look again at the necessary conditions given in the question:
x gallons of 1 % milk, y gallons of 2% milk, and z gallons of 3% milk are mixed to make (x + y + z) gallons of 1.5% milk.

We will now translate all this into math:
- Quantity of milk in grade 1 milk: 1% of x gallons = 0.01x
- Quantity of milk in grade 2 milk: 2% of y gallons = 0.02y
- Quantity of milk in grade 3 milk: 3% of z gallons = 0.03z
- Quantity of milk in the mixture created by mixing the three grades of milk above:
• 1.5% of (x + y + z) gallons
• = 0.15(x + y + z)
Hence, 0.01x + 0.02y + 0.03z = 0.015(x + y + z). --- (1)

You did not keep this in consideration while choosing your set of numbers, and hence, went wrong. Understand that you can take ANY three numbers that satisfy equation (1).
For example, you could take x = 4, y = 1, and z = 1 or x = 10, y = 4 and z = 2. Both of these satisfy equation (1).

But you cannot take x = 5, y = 3, and z = 2 since that does not satisfy equation (1).

Now that I’ve answered your core question, I would still suggest that you build methodical approaches that can help you solve not just this one question, but ANY question that the GMAT throws at you. Let me just continue from equation (1) and show you why numbers were not even needed:

• From eq (1), we have 0.01x + 0.02y + 0.03z = 0.015(x + y + z).
• Multiplying throughout by 100, we get:
• 1x + 2y + 3z = 1.5(x + y + z)
• Or x + 2y + 3z = 1.5x + 1.5y + 1.5z
• Now, since we want x in terms of y and z, we’ll just take x on one side and the other two variables to the other side. We get:
• 0.5x = 0.5y + 1.5z
• Multiplying both sides of above equation by 2 to convert 0.5x into x (as we want to find ultimately), we get x = y + 3z

And thus, we land at the correct option: choice A.

Hope this helps!

Best Regards,
Ashish Arora
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
MBAhereIcome wrote:
$$\frac{x+2y+3z}{x+y+z} = \frac{3}{2}$$

$$x = y+3z$$

from where did you get 3/2?
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Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
horrorslive wrote:
MBAhereIcome wrote:
$$\frac{x+2y+3z}{x+y+z} = \frac{3}{2}$$

$$x = y+3z$$

from where did you get 3/2?

The 3/2 in the equation comes from converting the 1.5% to a fraction: 3/2 = 1.5. So, it's basically: $$x+2y+3z = 1.5(x+y+z)$$. Deriving of that is explained here: https://gmatclub.com/forum/three-grades ... ml#p666769
Three grades of milk are 1 percent, 2 percent and 3 percent fat by vol [#permalink]
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