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. \(33\frac13\%\)
C. \(40\%\)
D. \(50\%\)
E. \(66\frac23\%\)
Answer: B
Source: Official Guide
Seed mixture \(X\) is \(40\) percent ryegrass and \(60\) percent bluegrass by weight; seed mixture \(Y\) is \(25\) perce
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This looks like a job for weighted averages!Vincen wrote: ↑Fri Aug 06, 2021 7:38 amSeed 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. \(33\frac13\%\)
C. \(40\%\)
D. \(50\%\)
E. \(66\frac23\%\)
Answer: B
Source: Official Guide
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
Let \(M = x + y\)Vincen wrote: ↑Fri Aug 06, 2021 7:38 amSeed 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. \(33\frac13\%\)
C. \(40\%\)
D. \(50\%\)
E. \(66\frac23\%\)
Answer: B
Source: Official Guide
\(M =\) New mixture
\(x =\) Mixture \(X\)
\(y =\) Mixture \(Y\)
What do we need to find? \(\Rightarrow \dfrac{x}{M}\ast 100\)
Equating Ryegrass in the mixture
\(0.4x + 0.25y = 0.3M\)
\(0.4x + 0.25(M-x) = 0.3M\)
\(0.4x + 0.25M - 0.25x = 0.3M\)
\(0.15x = 0.05M\)
\(\dfrac{x}{M} = \dfrac{1}{3}\)
Hence, \(33.33\%.\)