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Difficult Math Problem #119 - Arithmetic

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by 800guy » Mon Apr 16, 2007 7:56 am
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 1/3%

(C) 40%

(D) 50%

(E) 66 2/3%

from diff math doc, oa coming when people respond with answers
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Re: Difficult Math Problem #119 - Arithmetic

by ijinusean » Mon Apr 16, 2007 8:48 am
X(0.4)+Y(0.25) = 0.3(X+Y)

0.1X = 0.05Y

Therefore 2X=Y

Total solution is X+Y and we want to calculate X/(X+Y) which is

X/(X+2X) = 1/3 so answer is B
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by Cybermusings » Mon Apr 16, 2007 9:33 am
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 1/3%

(C) 40%

(D) 50%

(E) 66 2/3%

4:6 (or 2:3) & 25:75 (or 1:3)

(40x/100+25y/100)/x+y = 30/100
Solving it gives us 10x=5y
or x/y=1/2
Thus answer should be B
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OA

by 800guy » Wed Apr 18, 2007 8:15 am
OA:

(x+y)30/100 = x*40/100 + y*25/100
30x + 30y = 40x + 25y
y = 2x or y/x = 2/1 or y:x = 2:1 hence x = 33 1/3%
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by BTGmoderatorRO » Sat Dec 09, 2017 8:50 am
Let the weight of the mixture X be x
and the weight of the mixture Y be y

let r, b and f represent the weight of ryegrass,bluegrass and fescue respectively
In mixture X;
weight of r in X, i.e $$_{r_x}$$ = 40% of x = 0.4x
weight of b in X i.e $$_{b_x}$$ = 60% of x =0.6x

In mixture Y
weight of r in y i.e $$_{r_y}$$ = 25% of y =0.25y
weight of f in y, i.e $$_{f_y}$$ = 75% of y = 0.75%
In mixture of X and Y i.e x+y :
weight of r i n x + y i.e $$_{r_{x+y}}$$ = 30% of x+y = 0.3(x+y)
But $$_{r_{x+y}}$$ = $$_{r_x}$$ + $$_{r_y}$$
= 0.3(x+y)= 0.4x + 0.25y
0.3x+0.3y = 0.4x +0.25y
0.1x=0.05y
x=0.5y
Thertfoe percentage of x in x+y is
x $$_{\frac{x}{x+y}}$$ * 100% = $$_{\frac{0.5y}{0.5y+y}}$$ * 100%
Therefore the % of in x+y is $$_{\frac{0.5y}{1.5y}}$$ * 100% = 33.1/3%
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by BTGmoderatorRO » Sat Dec 09, 2017 8:53 am
Let the weight of the mixture X be x
and the weight of the mixture Y be y

let r, b and f represent the weight of ryegrass,bluegrass and fescue respectively
In mixture X;
weight of r in X, i.e $$_{r_x}$$ = 40% of x = 0.4x
weight of b in X i.e $$_{b_x}$$ = 60% of x =0.6x

In mixture Y
weight of r in y i.e $$_{r_y}$$ = 25% of y =0.25y
weight of f in y, i.e $$_{f_y}$$ = 75% of y = 0.75%
In mixture of X and Y i.e x+y :
weight of r i n x + y i.e $$_{r_{x+y}}$$ = 30% of x+y = 0.3(x+y)
But $$_{r_{x+y}}$$ = $$_{r_x}$$ + $$_{r_y}$$
= 0.3(x+y)= 0.4x + 0.25y
0.3x+0.3y = 0.4x +0.25y
0.1x=0.05y
x=0.5y
Thertfoe percentage of x in x+y is
x $$_{\frac{x}{x+y}}$$ * 100% = $$_{\frac{0.5y}{0.5y+y}}$$ * 100%
Therefore the % of in x+y is $$_{\frac{0.5y}{1.5y}}$$ * 100% = 33.1/3%
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by Brent@GMATPrepNow » Sat Dec 09, 2017 12:40 pm
800guy wrote: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 1/3%
(C) 40%
(D) 50%
(E) 66 2/3%
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

For more information on weighted averages, you can watch this video: https://www.gmatprepnow.com/module/gmat- ... ics?id=805
Here are some additional practice questions related to weighted averages:
- https://www.beatthegmat.com/weighted-ave ... 17237.html
- https://www.beatthegmat.com/weighted-ave ... 14506.html
- https://www.beatthegmat.com/average-weig ... 57853.html
- https://www.beatthegmat.com/averages-que ... 87118.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Sun Sep 29, 2019 6:03 pm
800guy wrote: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 1/3%

(C) 40%

(D) 50%

(E) 66 2/3%

from diff math doc, oa coming when people respond with answers
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 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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