A rabbit on a controlled diet is fed daily 300 grams of a

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.

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

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.

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

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.

10 % of 300 = 30 gms
15% 0f 300 = 45 gms

using allegation
45- 38/ 38- 30

= 7/ 8
therefore x :y is in the ratio 7: 8

7/ 15 X 300 = 140 gms

We can create two equations:

x + y = 300

y = 300 - x

and

0.1x + 0.15y = 38

Substituting, we have:

0.1x + 0.15(300 - x) = 38

0.1x + 45 - 0.15x = 38

7 = 0.05x

140 = x

Answer: B

Looking at the solutions, it seems pretty straightforward to come up with the 2 equations. But my first attempt at this problem was to derive the following:
1. X + Y = 300

X: 0.1P, Y: 0.15P (from "Food X contains 10% protein and food Y contains 15% protein")

2. 0.1P + 0.15P = 38

.. and just got stuck with 2 equations and 3 unknowns. Is there an approach I could use to immediately see that "10% protein" is referring to 10%(X)?

VeritasKarishma

Thanks in advance.

Whenever you use variables, ensure that you know exactly what it stands for.

X - Amount of food X in grams
Y - Amount of food Y in grams

What is P?

When you say X has 10% protein, you mean to say that 10% of X is protein.

10% of X + 15% of Y = Total protein = 38 gms

Anyway, such questions can be done easily using weighted averages discussed here:
https://www.youtube.com/watch?v=_GOAU7moZ2Q
https://anaprep.com/arithmetic-weighted-averages/

KarishmaB

300 grams = x + y

10% of x = protein in X
15% of y = protein in y

Q1) why can we do the following:

10% of 300 = 30gms of protein in X
15% of 300 = 45gms of protein in Y

If we follow the allegation process, we will land up with a ratio of 7:8.

Q2) why can we do, 7a+8a = 300gms?

You can average out the percentage of protein or the amount of protein. Both will give you the same result as long as you ensure that everything else suits the situation you choose.

X has 10% protein. (concentration of protein in X is 10%)
Y has 15% protein. (concentration of protein in Y is 15%)
Avg concentration of protein in Mixture is 38/300 * 100 = 38/3% protein (which is 12.67% but since it is a rounded value, we can keep 38/3 to retain accuracy)

Now use
\(\frac{w1}{ w2} = \frac{(15 - 38/3)}{(38/3 - 10)} = \frac{7}{8}\)

So X and Y are in the ratio 7:8.
If there is 300 gms of mixture, X will be 140 gms and Y will be 160 gms.


Alternatively, you are saying that 300 gms of X has 30 gms of protein. (Mind you, 300 gms of X, not 300 gms of mixture)
300 gms of Y has 45 gms of protein. (concentration of protein in X is 30/300 and concentration of protein in Y is 45/300)
300 gms of mixture has 38 gms of protein (avg concentration of protein in mixture is 38/300)

So in what ratio were X and Y mixed to give the mixture?

\(\frac{w1}{w2} = \frac{(45 - 38)}{(38 - 30)} = \frac{7}{8}\)

We have the denominator same so that they become comparable.
Essentially, the second method is the same as

\(\frac{w1}{w2} = \frac{(45/300 - 38/300)}{(38/300 - 30/300)}\)

All 300s get cancelled. Note that in both the methods, we are averaging the concentrations.

Just like the first method is the same as
\(\frac{w1}{w2} = \frac{15/100 - (38/3)/100}{(38/3)/100 - 10/100}\)

If you plug in options in this question, particularly option d, you match the daily requirement of 38g of protein, why is this option wrong then?

No, it does not fit in.

X is 10% and Y is 15%.

Be careful, you have to fit in for X.
Thus 10% of 160 + 15% of 140 = 16+21 = 37. But it should be 38.

10% of 140 + 15% of 160 = 14+24 = 38. So, option B.

I believe you filled up opposite values of X and Y in option D.

What others have said is completely fine, i'd avoid equation in this case since there is a proper spread between options.

Knowing that (38/300)*100 is actually more than 12,5% means that you will have an asymmetric weight of components (we are told that X is 10% and Y is 15% so basically 15% is "closer" to the result coming from the formula above that is about 12,7%. So using either the scale/line method to visualize it you can quickly conclude that 140 is the proper answer for X (because Y needs to be more than 150, in this case 160) (since the 12,7% is closer to 15% than it is to 10%, the part at 15% will "weight more" into the final mixture; conversely the part at 10% will "weigth less" into the final mixture)

Of course, in other cases in which you have a narrow margin between options you will have to do the formula.

I think the slight difficulty in this problem is how to start. Once you understand how to start then you need only to pay attention to arithmetic.