Problem Solving

I understand your solution but this is how i was doing it and for some reason i am getting a wrong answer. Can someone point out to me why i am wrong.

For every 1 unit of Z i would need to add 3 units of X to keep the ratio 1.5 and for every unit of y i would have to add equal amount of x.
converting it to equations,
3x=z
x=y
4x=y+z and hence answer is b.

The way I did it :
ATQ :
1*x + 2*y + 3*z = 1.5*(x+y+z)
or, x + 2y + 3z = 1.5x + 1.5y + 1.5z
or, x - 1.5x = 1.5y - 2y + 1.5z - 3z
or, -0.5x = - 0.5y - 1.5z
or, x = y+3z (divide the above equation by -0.5)
ANS : A

lunarpower wrote:courtesy of user 'karenmeow'

Three Grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 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

1. y+ 3z
2. (y+z)/4
3. 2y+3z
4. 3y + z
5. 3y + 4.5z

I too would plug in values. It helps to recognize that since x=1% and y=2%, if we use the same amounts of x and y and none of z, the result will be a mixture that is 1.5%.

Plug in x=2, y=2, z=0. Since the question is asking for the value of x, our target answer is x=2.

A) y+3z = 2 + 3*0 = 2. Hold onto A.
B) (y+z)/4 = (2+0)/4 = 1/2. Eliminate B.
C) 2y + 3z = 2*2 + 3*0 = 4. Eliminate C.
D) 3y+z = 3*2 + 0 = 6. Eliminate D.
E) 3y + 4.5z = 3*2 + (4.5)*0 = 6. Eliminate E.

The correct answer is A.

The trick to plugging in is to choose numbers that make the math easy.

lunarpower wrote:courtesy of user 'karenmeow'

Three Grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 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

1. y+ 3z
2. y+z/4
3. 2y+3z
4. 3y + z
5. 3y + 4.5z

Ron Please explain........ What's wrong with this. { I am confused. }

Just assume there are
100 ml of y means 2 ml of fat. ( 2 % fat. )
100 ml of z means 3 ml of fat. ( 3% fat. )
400 ml of x means 4 ml of fat. ( 1 % fat )

in all we have 600 ml of solution and 9 ml of fat.

This means 9/600 is 1.5 % of fat,

by this means Both the options are correct option 1 and 4.

1. y+ 3z 100 ml of y + 300 ml of z

4. 3y + z 300 ml of y + 100 ml of z

I find it easier to solve this by doing the equations:

0.01x+0.02y+0.03z = 0.015 (x+y+z) [multiplied by 1000]
10x+20y+30z = 15 (x+y+z) [divided by 5]
2x+4y+6z = 3 (x+y+z)
2x+4y+6z = 3x+3y+3z
4y+6z-3y-3z = 3x-2x
y + 3z = x and voila!

Correct Answer: [spoiler](A)[/spoiler]

Sure this is easy. But I did this algebraically just as MAAJ.

However I have one question:
How do you find the wieghts here? I assumed the percents to be wieghts and got the whole thing wrong.

(x * 1 + y * 2 + z * 3) / (1 + 2 + 3) = 1.5
and I got x = 9 - 2y - 3z (wrong)

But if I did
(x * 1 + y * 2 + z * 3) / (x + y + z) = 1.5
then I get x = y + 3 z (correct)

Could anyone kindly clarify?

yes correct A is the answer...x=y+3z

yes correct A is the answer...x=y+3z

GMATGuruNY wrote:

lunarpower wrote:courtesy of user 'karenmeow'

Three Grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 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

1. y+ 3z
2. (y+z)/4
3. 2y+3z
4. 3y + z
5. 3y + 4.5z

I too would plug in values. It helps to recognize that since x=1% and y=2%, if we use the same amounts of x and y and none of z, the result will be a mixture that is 1.5%.

Plug in x=2, y=2, z=0. Since the question is asking for the value of x, our target answer is x=2.

A) y+3z = 2 + 3*0 = 2. Hold onto A.
B) (y+z)/4 = (2+0)/4 = 1/2. Eliminate B.
C) 2y + 3z = 2*2 + 3*0 = 4. Eliminate C.
D) 3y+z = 3*2 + 0 = 6. Eliminate D.
E) 3y + 4.5z = 3*2 + (4.5)*0 = 6. Eliminate E.

The correct answer is A.

The trick to plugging in is to choose numbers that make the math easy.

Hi Mitch,

What is the reason to choose Z=0 ? and X=Y=2
Missing something here! Please help.

TIA,
GK

GmatKiss wrote:

GMATGuruNY wrote:

lunarpower wrote:courtesy of user 'karenmeow'

Three Grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 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

1. y+ 3z
2. (y+z)/4
3. 2y+3z
4. 3y + z
5. 3y + 4.5z

I too would plug in values. It helps to recognize that since x=1% and y=2%, if we use the same amounts of x and y and none of z, the result will be a mixture that is 1.5%.

Plug in x=2, y=2, z=0. Since the question is asking for the value of x, our target answer is x=2.

A) y+3z = 2 + 3*0 = 2. Hold onto A.
B) (y+z)/4 = (2+0)/4 = 1/2. Eliminate B.
C) 2y + 3z = 2*2 + 3*0 = 4. Eliminate C.
D) 3y+z = 3*2 + 0 = 6. Eliminate D.
E) 3y + 4.5z = 3*2 + (4.5)*0 = 6. Eliminate E.

The correct answer is A.

The trick to plugging in is to choose numbers that make the math easy.

Hi Mitch,

What is the reason to choose Z=0 ? and X=Y=2
Missing something here! Please help.

TIA,
GK

X=1%, Y=2%, Z=3%.
It would involve quite a bit of trial and error to come up with values for X,Y and Z so that X+Y+Z yields a mixture that is 1.5%.
It is FAR easier to come up with values for only TWO of the ingredients.
Thus:
Let Z=0, implying that no Z will be in the mixture.
Thus, the mixture will consist only of X and Y.
Since 1.5% is HALFWAY between X (1%) and Y (2%), combining equal amounts of X and Y will yield a mixture that is 1.5%.
To make the math easy, I chose X=2 and Y=2.