Problem Solving

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 the mixture is x?


a. 10 %
b.33.3%
c.40%
d.50%
e.66.67%

qa is b


although the answer explanation is nice in the back of the og 11 book, I am looking for a faster and less algebraic approach. ANy help?

well, i dont know what approach did the book use but here is mine..

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%

0.4x+0.25y = 0.3(x+y)

Mutiply by 100

40x+25y = 30x+30y

10x=5y

Once u get x/y = 1/2

This is part to part ratio

so part/whole ratio for x would be 1/3 = 33.3 %

Aha! Now I get it! One step closer. I am beginning to see that most of these higher end questions are simply a matter of framing the question correctly and then moving on!

Simple Approach

X : Y
40 25
30

5 : 10

1 : 2

Therefore

X / X+Y = 1 / 3 = 33 1/3rd

I missed the question the 1st time but got it right the 2nd time, however, I still don't understand the concept... since originally X = 40% ryegrass & Y = 25% ryegrass then shouldn't X have more weight in the 30% mixture?

If X = 33.33% of the new mixture, then does that make Y = 66.66% of the new mixture? It just doesn't make sense to me why X has a lower percentage of the new mixture when X originally carried 40% ryegrass vs. Y which originally carried 25% ryegrass.

Any insight into this conundrum would be appreciated

Hi BlindVision,

This is a weighted average concept. In a weighted average between 2 values, the weighed average will always be closer to the average of the larger group. For instance if in a class girls average 8yrs old and boys average 10 years old, if the weigthed average of all kids is 8.1yrs, it's because there are a lot more girls than boys. The logic behind this is simple. If the kids were all girls, the average would be 8. If the kids were all boys, the average would be 10. The larger group (boys vs girls) will pull the weighted average (like gravity)

In this case, X is 40% rye and Y is 25% rye. Since the overall mixture is closer to 25% than to 40% rye, there must be more Y than X in the overall mixture.

Actually, we can be even more specific. The weighted average is twice as far from X (30 is 10 away from 40) as it is from Y (30 is 5 away from 25), so the larger group (in this case Y) must be twice the size of the smaller group. Thus without doing any math I can guarantee that Y:X is 2:1, or the mix is 66% Y and 33% X.

In case the solutions above are not sufficient, two detailed solutions and a 30-second hack (a simplified version of my reasoning above) are attached. This question is #223 in the OG12. Those who cannot see the attachment can read it here.

Hope this helps,
-Patrick

simba12123 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 the mixture is x?


a. 10 %
b.33.3%
c.40%
d.50%
e.66.67%

qa is b

although the answer explanation is nice in the back of the og 11 book, I am looking for a faster and less algebraic approach. ANy help?

A less algebraic approach would be to use the answer choices.

The answers represent the percentage of the mixture that will be X. Let's start with C, which says the mixture will be 40% X, 60% Y.

Let's say we use 40 of X and 60 of Y, so our total mixture is 40 + 60 = 100.
X is 40% ryegrass, giving us .4 * 40 = 16 ryegrass.
Y is 25% ryegrass, giving us .25 * 60 = 15 ryegrass
So we have 16 + 15 = 31 ryegrass in the total mixture.
What percent of the total mixture is ryegrass? 31/100 = 31%

Since we need the mixture to be 30% ryegrass, we need to use just a little less X, which has a higher percentage of ryegrass.

So the correct answer must be B.

I love the explanation Patrick_GMATFix... I like the intuitive solution the most. However, I have not fully grasped the concept of the weighted avg. I understand it in the case of lets say school grades of kids, i.e.: 20 girls had an avg. grade of 2.1 and 10 boys had an avg. of 2.5, then you would calculate it the following way: ((20*2.1)+(10*2.5))/30). What keeps me up at night is the weighted average of average speed, i.e. question 149 in the OG 12 on page 173: "During a trip, Francine traveled x percent of the total distance at an average speed of 40 mpg and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?". What is the concept behind this question? May you help me with that... Thanks lots in advance.

cheers

GMATGuruNY wrote:
A less algebraic approach would be to use the answer choices.

The answers represent the percentage of the mixture that will be X. Let's start with C, which says the mixture will be 40% X, 60% Y.

Let's say we use 40 of X and 60 of Y, so our total mixture is 40 + 60 = 100.
X is 40% ryegrass, giving us .4 * 40 = 16 ryegrass.
Y is 25% ryegrass, giving us .25 * 60 = 15 ryegrass
So we have 16 + 15 = 31 ryegrass in the total mixture.
What percent of the total mixture is ryegrass? 31/100 = 31%

Since we need the mixture to be 30% ryegrass, we need to use just a little less X, which has a higher percentage of ryegrass.

So the correct answer must be B.

My question is, shouldn't we apply the theory of alligation in such mixture problems? Its quite straight forward because it considers only one component of the mixture. I am not questioning the other methods, but I found the alligation method to be more suitable here.

I know in theory this is what everyone is saying but I just want to make sure I am doing it right algebraically. I haven't seen anyone create an equation linking x and y and substituting it. Does this take too much time?

1=X+Y --> Y=1-X

(4/10)X+(1/4)Y = (3/10)(X+Y)

(4/10)X+(1/4)(1-X) = (3/10)(x+(1-X))

get common denominator

(16x+(10-10x))/40 = (3/10)

6x+10=12

6x=2

x=2/6 = 1/3 => 33.3%

OK here is what I did ..Please tell me where I am going wrong
I assumed 1 gm of both X and Y
So I will have a mixture of X and Y as 2 gm
and the ryegrass in mixture is 2*.3
and hence the percent of weight of the mixture=(1*.4/2*.3)*100

What am I doing wrong????

cramya wrote:0.4x+0.25y = 0.3(x+y)

Mutiply by 100

40x+25y = 30x+30y

10x=5y

Once u get x/y = 1/2

This is part to part ratio

so part/whole ratio for x would be 1/3 = 33.3 %

I didnt get the question so I followed this approach and ended up with:

10x = 5y

2x=y

Ratio: x: y: x+y = 2:1:3

so x/total = 2/3 and got it wrong. But HOW?