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
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Difficult Math Problem #119 - Arithmetic
This topic has expert replies
-
Cybermusings
- Legendary Member
- Posts: 559
- Joined: Tue Mar 27, 2007 1:29 am
- Thanked: 5 times
- Followed by:2 members
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
(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
-
BTGmoderatorRO
- Moderator
- Posts: 772
- Joined: Wed Aug 30, 2017 6:29 pm
- Followed by:6 members
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%
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%
-
BTGmoderatorRO
- Moderator
- Posts: 772
- Joined: Wed Aug 30, 2017 6:29 pm
- Followed by:6 members
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%
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%
GMAT/MBA Expert
-
Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
This looks like a job for weighted averages!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%
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
GMAT/MBA Expert
-
Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7247
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
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: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
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
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
