in fact one eq is enough: .1x + .15(300-x) = 38

solve for x, x = 140

Set up two equations.

We know that 10%x+15%y=38 grams or .1x+.15y=38

We also know that x+y=300 grams.

2 eq. with 2 variables. Solve them and you should get it.

Much simpler method is there...
38 gms means approx. 12.66%. Then use the mixture alligation formula to calculate the ratio of X to Y, ie,

X/Y= (15-12.66)/(12.66-10)=2.33/2.66.
Hence amt of X in total mixture is 300*2.33/(2.33+2.66)=139.2~140
Answer B.

D

Assume x grams is of mixture A and 300-x is of mixture B
Eqn is:

(300-x)*.1 + .15x = 38 (10% and 15% protein in Mixtures A and B respectively)

Solve for x =160

0.1x + 0.15y = 38

x+y = 300
x = 300-y

so 0.1(300-y) + 0.15y = 38
30-0.1y + 0.15y = 38
0.05y = 8
y = 160 grams

then x = 140 grams.

Ans B

Its wat you feel comfortable with or the one which you do faster and accurately. You have to develop that approach by practice.
Atleast I think so. I solved the problem using Mixtures & Alligation formula coz I feel comfortable with that.

Sunil

MGMAT has a great method that has really helped me to do these problems quickly.

For mixture problems, set up a chart:

the two different mixtures go on the right hand side, the quantities of each go on the top:

Protein Other
X

Y

The problem says the two mixtures are added together to produce a new mixture totaling 300 grams, so add to the chart:

Protein Other Total
X X

Y Y

X+Y 300

Now add the information about how much protein is in each mixture

Protein Other Total
X .1X X

Y .15Y Y

X+Y 38 300


Now there are two equations with 2 variables:

.1X + .15Y=38
X + Y = 300

Simple method, but it really helps break down harder questions.

can someone explain the logic of the mixture and allegation formula?

the ratio of (difference between the high value and new percentage)
over
(diff b/t high and new) + (diff b/t new and low value)

what does this ratio represent and why, when multiplying the whole to it does it give you the amount in final.....

We know that
1.00x+1.00y=300

and that
0.10x+0.15y=38

so
0.10x+0.15y=38
*10=
1.00x+1.50y=380

Align them under eachother
1.00x+1.50y=380
1.00x+1.00y=300

Means 0.50y=80
80*2=160
y=160
x=300-y
x=300-160
x=140

let x= mixture A ,, then 300-x is Mixture B

.0.1 x + (300-x)0.15=38

after the solving the equation ,,x=140

This problem is relatively easy once you break it up the way it's supposed to be broken up...

they basically say X lbs +Y lbs = 300 lbs

Now, equate the protein levels together:

0.10 X + 0.15 Y = 38

Solve

just calculation:

0.1*X+0.15*(300-X)=38
X=140

we can solve by testing answers
lets start with B...
let qty of X be 140 gms. so Y will be 160 gms.
proteins from X = 14 gms
proteins from Y = 24 gms
add both, we get 38gms.

so X must be 140 gms.

30 sec reasoning out: 38/300 - is a bit less than a 13% protein content. Hence 13% is closer to 15%Y than to 10%X - this means that there is just sligthly more Y component because should those 2 be equal the protein content would have been sharply 12.5% which is not the case.

Immediately we kick out 3 answer choices >= 150 and are left with A and B. A is too low for the aboe slight deviation - so B is much closer to truth.

Hi,
this weighted approach,if mastered, can actually make a difference in your Quant score..
Good approach and you should try this in almost all Q that asks for an average in a mixture, or for a quantity, when an average is given..
+1 kudos
_________________

I am going through word problem types of questions because I think i bombed this type of q in my last gmat both time-wise and accuracy wise. And what i see from OG about 30% of all questions comprise the word problem type.

During these 2 days I solved numerous word problems and here are my observations:

- very time consuming because the intuitive approach is to apply algebra and form equations, on average easy question take almost the same time as hard ones
- given my incline to make silly mistakes in algebraic calculations I should rely more on reasoning out the easy questions
- majority of easy word problems employ weighted average or mean concepts which can be done by reasoning much faster than by doing algebra - ultimately this will save at least 30 secs on each such question
-

Here is a visual that should help.

_________________

The answer choices are wide enough to narrow down. Food X < average, therefore options C to D are out. Can't be 100 because it is to far away from the average, hence B.

\(.1x+.15(300-x)=38\)

\(10x+4500-15x=3800\)

\(5x=700\)

\(x=140\)