Problem Solving

Poisson wrote:Hi Mitch,

Could you please explain how to apply this approach to the following question? Each time I try it, I get an incorrect answer.

Mix A is 35% red and 65 % blue.
Mix B is 50% red and 50% blue.
If you begin with 20 pounds of mix A, how many pounds of mix B do you need to end up with a combined mixture that is 40% red and 60% blue?

A. 5
B. 10
C. 15
D. 20
E. 25

I keep ending up with an answer of 5 once I take the reciprocal of the distances.

Thanks very much.

A = 35% red, B = 50% red, the mixture = 40% red.

A(35%)-----5-----40%-----10-----B(50%)

A:B = 10:5 = 20:10.
Thus, the 20 pounds of A must be mixed with 10 pounds of B.

The correct answer is B.

GMATGuruNY wrote:

Poisson wrote:Hi Mitch,

Could you please explain how to apply this approach to the following question? Each time I try it, I get an incorrect answer.

Mix A is 35% red and 65 % blue.
Mix B is 50% red and 50% blue.
If you begin with 20 pounds of mix A, how many pounds of mix B do you need to end up with a combined mixture that is 40% red and 60% blue?

A. 5
B. 10
C. 15
D. 20
E. 25

I keep ending up with an answer of 5 once I take the reciprocal of the distances.

Thanks very much.

A = 35% red, B = 50% red, the mixture = 40% red.

A(35%)-----5-----40%-----10-----B(50%)

A:B = 10:5 = 20:10.
Thus, the 20 pounds of A must be mixed with 10 pounds of B.

The correct answer is B.

Thank you so much!! I have one more mixture problem that I can't seem to figure out.


Mix A which is 20% chocolate is blended with Mix B which is 30% chocolate.

The initial blend is 5 pounds of Mix A and 10 pounds of Mix B. If the final blend is 22% chocolate, how many pounds of mix A were added to the initial blend?

A. 40
B. 35
C. 30
D. 25
E. 20


A(20%)-----2-----22%-----8-----B(30%)


A:B = 8:2 = 4:1
This is the ratio for the final blend

The ratio for the initial blend:
Mix A: (0.2)(5) = 1 pound chocolate
Mix B: (0.3)(10) = 3 pounds chocolate

A:B = 1:3

I'm not sure where to go from here. I can see that the part for Mix A increased from 1 to 4. But I'm struggling to get to a final answer. Please advise

Thanks so much

Poisson wrote:

GMATGuruNY wrote:

Poisson wrote:Hi Mitch,

Could you please explain how to apply this approach to the following question? Each time I try it, I get an incorrect answer.

Mix A is 35% red and 65 % blue.
Mix B is 50% red and 50% blue.
If you begin with 20 pounds of mix A, how many pounds of mix B do you need to end up with a combined mixture that is 40% red and 60% blue?

A. 5
B. 10
C. 15
D. 20
E. 25

I keep ending up with an answer of 5 once I take the reciprocal of the distances.

Thanks very much.

A = 35% red, B = 50% red, the mixture = 40% red.

A(35%)-----5-----40%-----10-----B(50%)

A:B = 10:5 = 20:10.
Thus, the 20 pounds of A must be mixed with 10 pounds of B.

The correct answer is B.

Thank you so much!! I have one more mixture problem that I can't seem to figure out.


Mix A which is 20% chocolate is blended with Mix B which is 30% chocolate.

The initial blend is 5 pounds of Mix A and 10 pounds of Mix B. If the final blend is 22% chocolate, how many pounds of mix A were added to the initial blend?

A. 40
B. 35
C. 30
D. 25
E. 20


A(20%)-----2-----22%-----8-----B(30%)


A:B = 8:2 = 4:1
This is the ratio for the final blend

The ratio for the initial blend:
Mix A: (0.2)(5) = 1 pound chocolate
Mix B: (0.3)(10) = 3 pounds chocolate

A:B = 1:3

I'm not sure where to go from here. I can see that the part for Mix A increased from 1 to 4. But I'm struggling to get to a final answer. Please advise

Thanks so much

To start you've got 5 pounds of mixture A and 10 pounds of mixture B. We're going from this initial composition to a final composition solely by adding more of mixture A (this would need to be stated explicitly in the question.) So if we're not adding more of mixture B, we know that there will also be 10 pounds of mixture B in the final mixture, right?

You correctly calculated that the final composition will be 4:1 in favor of mixture A. So set it up like this

Mixture A: 4x
Mixture B: x

If there are 10 pounds of mixture B, then x = 10, and 4x = 40.

The final composition will be
Mixture A: 40
Mixture B: 10

We started with 5 pounds of mix A and ended with 40 pounds of mix A. Thus, we must have added 40-5 = 35 pounds.

Poisson wrote: Mix A is 35% red and 65 % blue.
Mix B is 50% red and 50% blue.
If you begin with 20 pounds of mix A, how many pounds of mix B do you need to end up with a combined mixture that is 40% red and 60% blue?

Mix A = 35% red, so .35a is red
Mix B = 50% red, so .5b is red

Together, we want Mix A + Mix B = 40% red, or for .4*(a + b) to be red. That gives us

.35a + .5b = .4*(a + b)

or

.1b = .05a

or 2b = a

So we need twice as much of Mix A as we do of Mix B, or 10 pounds of Mix B.

Poisson wrote:Mix A which is 20% chocolate is blended with Mix B which is 30% chocolate.

The initial blend is 5 pounds of Mix A and 10 pounds of Mix B. If the final blend is 22% chocolate, how many pounds of mix A were added to the initial blend?

Be careful with overusing Mitch's alligation method; as often as not, you're trying to squeeze a square peg into a round hole. On this question, for instance, I think it's easier just to work with the ratio (pounds of chocolate / total pounds).

Initially, we've got

Chocolate = (20% of 5 + 30% of 10) => 4 pounds
Total = 5 + 10 => 15 pounds

Then, when we add x pounds of Mix A, we have

Chocolate = (4 + 20% of x)
Total = (15 + x)

We know that Chocolate/Total => (4 + .2x)/(15 + x) = 22% = 22/100, so just cross multiply and solve:

100*(4 + .2x) = 22*(15 + x)

400 + 20x = 330 + 22x

70 = 2x

35 = x

One other benefit of my last approach: it allows you to solve for x directly without having to interpret a result, which should help decrease confusion and uncertainty on test day!