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Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 % fescue. If a mixture of X and Y contains 30% ryegrass, what percent of the weight of the mixture is X?
A. 10%
B. \(33\frac{1}{3}\)%
C. 40%
D. 50%
E. \(66\frac{2}{3}\)%
A. 10%
B. \(33\frac{1}{3}\)%
C. 40%
D. 50%
E. \(66\frac{2}{3}\)%
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Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of this mixture is X ?
(A) 10%
(B) 33.33 %
(C) 40%
(D) 50%
(E) 66.66 %
METHOD 1:
Assuming the weight of the mixture to be 100g**, then the weight of ryegrass in the mixture would be 30g.
Also, assume the weight mixture X used in the mixture is Xg, then the weight of mixture Y used in the mixture would be (100-X)g.
So we can now equate the parts of the ryegrass in the mixture as:
0.4X + 0.25(100-X) = 30
0.4X + 25 - 0.25X = 30
0.15X = 5
X = 5/0.15 = 500/15 = 100/3
So the weight of mixture X as a percentage of the weight of the mixture
= (weight of X/weight of mixture) * 100%
= (100/3)/100 * 100%
= 33%
METHOD 2
wt. of 1st mixture = x
therefore concentration of ryegrass in 1st mix = 0.4x
wt. of 2nd mixture = y
therefore concentration of ryegrass in 2nd mix = 0.25y
0.4x+0.25y = 0.3(x+y)
0.4x-0.3x = 0.3y - 0.25y
0.1x=0.05y
or
2x=y
so if weight of x = 50grams
weight of y = 100 grams
total weight of mix = 150 grams
percentage of x in 150 grams of mix is 150*x/100 = 50
x = 50*100/150
x = 100/3
x = 33.3%
(A) 10%
(B) 33.33 %
(C) 40%
(D) 50%
(E) 66.66 %
METHOD 1:
Assuming the weight of the mixture to be 100g**, then the weight of ryegrass in the mixture would be 30g.
Also, assume the weight mixture X used in the mixture is Xg, then the weight of mixture Y used in the mixture would be (100-X)g.
So we can now equate the parts of the ryegrass in the mixture as:
0.4X + 0.25(100-X) = 30
0.4X + 25 - 0.25X = 30
0.15X = 5
X = 5/0.15 = 500/15 = 100/3
So the weight of mixture X as a percentage of the weight of the mixture
= (weight of X/weight of mixture) * 100%
= (100/3)/100 * 100%
= 33%
METHOD 2
wt. of 1st mixture = x
therefore concentration of ryegrass in 1st mix = 0.4x
wt. of 2nd mixture = y
therefore concentration of ryegrass in 2nd mix = 0.25y
0.4x+0.25y = 0.3(x+y)
0.4x-0.3x = 0.3y - 0.25y
0.1x=0.05y
or
2x=y
so if weight of x = 50grams
weight of y = 100 grams
total weight of mix = 150 grams
percentage of x in 150 grams of mix is 150*x/100 = 50
x = 50*100/150
x = 100/3
x = 33.3%
Baten80
Just focus on ryegrass. X has 40% ryegrass and Y has 25% ryegrass to give 30% ryegrass in mixture. Therefore, X:Y = 1:2 (because distance of X and Y from average is in the ratio 10:5 i.e. 2:1. If this in unclear, check out https://gmatclub.com/forum/tough-ds-105651.html#p828579)
Hence X is 33.33% in the mixture.
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JeffTargetTestPrep
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Joined: 04 Mar 2011
Last visit: 05 Jan 2024
Posts: 2,973
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Posts: 2,973
carollu
We are given information about ryegrass, bluegrass, and fescue. However, we are told the mixture of X and Y is 30 percent ryegrass. Thus, we only care about the ryegrass, so we can ignore fescue and bluegrass.
We are given that seed mixture X is 40% ryegrass and that seed mixture Y is 25% ryegrass. We are also given that the weight of the combined mixture is 30% ryegrass. With x representing the total weight of mixture X, and y representing the total weight of mixture Y, we can create the following equation:
0.4x + 0.25y = 0.3(x + y)
We can multiply the entire equation by 100:
40x + 25y = 30x + 30y
10x = 5y
2x = y
The question asks what percentage of the weight of the mixture is x.
We can create an expression for this:
x/(x+y) * 100 = ?
Since we know that 2x = y, we can substitute in 2x for y in our expression. So, we have:
x/(x+2x) * 100 = x/(3x) * 100 = 1/3 * 100 = 33.33%
Answer: B
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Let M = x + y
M = New mixture
x = Mixture X
y = Mixture Y
What do we need to find ?? => (x/M)*100
Equating Ryegrass in the mixture -
.4x + 0.25y = 0.3M
.4x + 0.25(M-x) = 0.3M
.4x + 0.25M - 0.25x = 0.3M
.15x = .05M
x/M = 1/3
Hence ans = 33.33%.
M = New mixture
x = Mixture X
y = Mixture Y
What do we need to find ?? => (x/M)*100
Equating Ryegrass in the mixture -
.4x + 0.25y = 0.3M
.4x + 0.25(M-x) = 0.3M
.4x + 0.25M - 0.25x = 0.3M
.15x = .05M
x/M = 1/3
Hence ans = 33.33%.
Posts: 6,727
sondenso
This looks like a job for weighted averages!
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...
Mixture X is 40 percent ryegrass
Mixture Y is 25 percent ryegrass
Let x = the PERCENT of mixture X needed (in other words, x/100 = the proportion of mixture X needed)
So, 100-x = the PERCENT of mixture Y needed (in other words, (100-x)/100 = the proportion of mixture Y needed)
Weighted average of groups combined = 30%
Now take the above formula and plug in the values to get:
30 = (x/100)(40) + [(100-x)/100](25)
Multiply both sides by 100 to get: 30 = 40x + (100-x)(25)
Expand: 3000 = 40x + 2500 - 25x
Simplify: 3000 = 15x + 2500
So: 500 = 15x
Solve: x = 500/15
= 100/3
= 33 1/3
So, mixture X is 33 1/3 % of the COMBINED mix.
Answer: B
RELATED VIDEO
Posts: 1,012
sondenso
B. x = weight of x
40 x + 25 (1-x) = 30
solve for x, x = 33.33%
RG detail table:
X----------------------------------Y
40%------------------------------25%
-----------------30%--------------
30-25 = 5------------------40-30 = 10
RG ratio 5:10 = 1:2
Therefore, % of X in new mixture = 1/3 = 33.3%
X----------------------------------Y
40%------------------------------25%
-----------------30%--------------
30-25 = 5------------------40-30 = 10
RG ratio 5:10 = 1:2
Therefore, % of X in new mixture = 1/3 = 33.3%
General Discussion
Posts: 8
the answer is
E.66 2/3%
solve for 25x + 40 [1-x] = 30
so, x = 2/3 = 66 2/3%
E.66 2/3%
solve for 25x + 40 [1-x] = 30
so, x = 2/3 = 66 2/3%
Hi Tejal777
we can make the method 2 more simpler
METHOD 2
wt. of 1st mixture = x
therefore concentration of ryegrass in 1st mix = 0.4x
wt. of 2nd mixture = y
therefore concentration of ryegrass in 2nd mix = 0.25y
0.4x+0.25y = 0.3(x+y)
0.4x-0.3x = 0.3y - 0.25y
0.1x=0.05y
or
2x=y
[b]so if weight of x = 50grams
weight of y = 100 grams
total weight of mix = 150 grams
percentage of x in 150 grams of mix is 150*x/100 = 50
x = 50*100/150
x = 100/3
x = 33.3%[/b].
what we need is x/x+y *100
we know x and y so solving it we get 33.33%
we can make the method 2 more simpler
METHOD 2
wt. of 1st mixture = x
therefore concentration of ryegrass in 1st mix = 0.4x
wt. of 2nd mixture = y
therefore concentration of ryegrass in 2nd mix = 0.25y
0.4x+0.25y = 0.3(x+y)
0.4x-0.3x = 0.3y - 0.25y
0.1x=0.05y
or
2x=y
[b]so if weight of x = 50grams
weight of y = 100 grams
total weight of mix = 150 grams
percentage of x in 150 grams of mix is 150*x/100 = 50
x = 50*100/150
x = 100/3
x = 33.3%[/b].
what we need is x/x+y *100
we know x and y so solving it we get 33.33%
IMHO this task is better to solve with absolute numbers.
Say, there are 100 liters of X (40+60) and Y (25+75) that add to 200 liters in total.
The weight of reygrass in the united mixture is 0,3*200=60.
The weight of reygrass in X is 40.
40/60=2/3=33,33%
Say, there are 100 liters of X (40+60) and Y (25+75) that add to 200 liters in total.
The weight of reygrass in the united mixture is 0,3*200=60.
The weight of reygrass in X is 40.
40/60=2/3=33,33%
take
X=100
In it:
Ryegrass=40
Others = 60
take
Y=y
Ryegrass=0.25y
Others = 0.75y
Mix both;
Total weight:
X+Y=100+y
In it;
Weight of Ryegrass = 40+0.25y
Given:
40+0.25y=0.3(100+y)
40+0.25y=30+0.3y
0.05y=10
y=200
So; total X+Y = 100+200=300
Percentage of X in it: (X/X+Y)*100 = (100/300)*100=100/3=33.33%
Ans: B
X=100
In it:
Ryegrass=40
Others = 60
take
Y=y
Ryegrass=0.25y
Others = 0.75y
Mix both;
Total weight:
X+Y=100+y
In it;
Weight of Ryegrass = 40+0.25y
Given:
40+0.25y=0.3(100+y)
40+0.25y=30+0.3y
0.05y=10
y=200
So; total X+Y = 100+200=300
Percentage of X in it: (X/X+Y)*100 = (100/300)*100=100/3=33.33%
Ans: B
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x - wt of Mix X added
y - wt of Mix Y added
0.4x + 0.25y = 0.3(x+y)
0.1x = 0.05y
x/y = 5/10 = 1/2
y/x = 2
(y+x)/x = 3
=> x/(y+x) * 100 = 100/3 = 33.33%
y - wt of Mix Y added
0.4x + 0.25y = 0.3(x+y)
0.1x = 0.05y
x/y = 5/10 = 1/2
y/x = 2
(y+x)/x = 3
=> x/(y+x) * 100 = 100/3 = 33.33%
Final Requirement is to calculate "X/(X+Y) * 100"
We have the information about the quantity by weight of ryegrass in X,Y and also the quantity by weight of ryegrass after mixing i.e in X+Y.
So the equation satisfying the above mentioned criteria would be :
(40/100)*X+(25/100)*Y=(30/100)*(X+Y)
Solving this will lead us to the result...
2X=Y
So, the weight percent of X :
(X/X+Y)*100 = (X/X+2X)*100 = 33.33%
We have the information about the quantity by weight of ryegrass in X,Y and also the quantity by weight of ryegrass after mixing i.e in X+Y.
So the equation satisfying the above mentioned criteria would be :
(40/100)*X+(25/100)*Y=(30/100)*(X+Y)
Solving this will lead us to the result...
2X=Y
So, the weight percent of X :
(X/X+Y)*100 = (X/X+2X)*100 = 33.33%
Posts: 49
It might be easier to think of things in terms of their weight. Assume that you're holding 100kg of the mixture. Of this 100kg, you have a certain amount of kilograms of X, and a certain amount of kilograms of Y. That's where you begin with the X + Y = 1. Note that you also know that there is 30kg of ryegrass in your mixture.
Each kilogram of X would consist of 400g of ryegrass and 600g of bluegrass, while each kilogram of Y would comprise 250g of ryegrass and 750g of fescue. If you want to know how many kilograms of X the mixture contains, you need to work out how you can arrive at 30kg of ryegrass, with exactly 70kg of other stuff (bluegrass + fescue) in the mix. That's what the rest of the working is referring to.
Each kilogram of X would consist of 400g of ryegrass and 600g of bluegrass, while each kilogram of Y would comprise 250g of ryegrass and 750g of fescue. If you want to know how many kilograms of X the mixture contains, you need to work out how you can arrive at 30kg of ryegrass, with exactly 70kg of other stuff (bluegrass + fescue) in the mix. That's what the rest of the working is referring to.
Sachin9
Yes it is. The question tells us that mixture X and mixture Y (both of which are themselves mixtures of ryegrass and bluegrass) are mixed together to give a super mixture. We need to find the weight of mixture X in this super mixture.
We know the % of ryegrass in X and in Y and we also know the % of ryegrass in the super mixture (i.e. the average % of ryegrass when you mix X and Y). This tells us the ratio of X and Y in the super mixture i.e. in what ratio were X and Y mixed together (using our standard weighted averages method). When you have the ratio of X and Y, you can say what the % of mixture X is in the super mixture.
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sondenso
Quote:
Your first assumption is incorrect here: In this my approach was that let weight of X be 100 and weight of Y be 100
If you are assuming that the weights are 100 each and the combined weight is 200, what is left to find? X would be 1/2 of the mixture 100/200. What you have to find is the ratio of the weights in which they were mixed. The ratio may not be 1:1 as you have assumed here.
X has 40% ryegrass and Y has 25% ryegrass. Total mix has 30% ryegrass.
So X/Y = (25 - 30)/(30 - 40) = 1/2
So X by weight is 1 part (say 100 gms) and Y is 2 parts (say 200 gms). So X is 1/3 of the total mix i.e. 33.33%
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Using Alligation method
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VeritasPrepKarishma
Quote:
Highlighted part is wrong. Use the formula as it is.
X has 40% ryegrass and Y has 25% ryegrass to give 30% ryegrass in mixture
w1/ w2=(A2–Aavg)(Aavg–A1)
wX/wY = (AY - Aavg)/(Aavg - AX)
wX/wY = (25 - 30)/(30 - 40) = 1/2
So wX/wY = 1/2. You get X:Y as 1:2
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JeffTargetTestPrep
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sondenso
This is a great problem because we can solve it using equations and substitution, or we can employ some estimation because the equation we create is a weighted average. Let’s first do the equation-substitution method.
Equation-Substitution Method
We are given information about ryegrass, bluegrass, and fescue. However, we are told the mixture of X and Y is 30 percent ryegrass. Thus, we only care about the ryegrass, so we can ignore fescue or bluegrass.
We are given that seed mixture X is 40% ryegrass and that seed mixture Y is 25% ryegrass. We are also given that the weight of the combined mixture is 30% ryegrass. With x representing the total weight of mixture X, and y representing the total weight of mixture Y, we can create the following equation:
0.4x + 0.25y = 0.3(x + y)
We can multiply the entire equation by 100, and we have:
40x + 25y = 30x + 30y
10x = 5y
2x = y
The question asks what percent of the weight of the mixture is x.
We can create an expression for this:
x/(x+y) * 100 = ?
Since we know that 2x = y, we can substitute in 2x for y in our expression. So we have:
x/(x+2x) * 100 = x/(3x) * 100 = 1/3 * 100 = 33.33%
The answer is B.
Estimation Method
This is a method I would suggest using only if you are short on time.
We are given that seed mixture X is 40% ryegrass and that seed mixture Y is 25% ryegrass. We are also given that the overall weight of the mixture is 30% ryegrass.
Since 30% is closer to 25% than it is to 40%, we know that there is more of mixture Y than there is of mixture X, but not by a large amount. So let’s analyze our answer choices.
A) 10%
This tells us that the mixture is made up of 10% x and 90% y. That discrepancy is much too large to be correct, based on the overall weight of the mixture.
B) 33 1/3%
This tells us that the mixture is made up of 33 1/3% x and 66 2/3% y. This could be the answer.
C) 40%
This tells us that the mixture is made up of 40% x and 60% y. This could be the answer.
D) 50%
This tells us that the mixture is made up of 50% x and 50% y. Since we know that y is weighted more heavily in the mixture, we know this cannot be the answer.
E) 66.67%
This tells us that the mixture is made up of 66 2/3% x and 33 1/3% y. Since we know that y is weighted more heavily in the mixture, we know this cannot be the answer.
So now we have our answer choices narrowed down to B and C. However, there is something interesting about answer choice B. Notice that if we were to add together answer choice B (33 1/3%) and answer choice E (66 2/3%), our sum would be 100%. These are strategic answer choices, because the GMAT is hoping that if we make a mistake in our calculations we will determine that x represents 66 2/3% of the mixture rather than 33 1/3% of the mixture. Since answer choice C (40%) does not have a corresponding trap answer there is a higher likelihood that the answer is B (33 1/3%) than it is C (40%).
As mentioned at the beginning, the surest solution technique is to create an equation to solve it and actually complete the math. However, if you are short on time or have trouble setting up the equation, estimation is a method you can employ.
The answer is B.
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