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 Question from Lucy, a student: Hi, For the following question: Ten pounds of mixed nuts contain 50 percent peanuts. How many pounds of peanuts must be added so that the final mixture has 60 percent peanuts?(A) 2.5 (B) 5 (C) 6 (D) 10 (E) 12.5, my math book gives my the following explanation: Try choice (C): 6 pounds of peanuts. The original 10-pound mixture contained 50 percent peanuts, or 5 pounds. Now you have 11 pounds of peanuts in a 16-pound mixture, more than 60 percent. This result automatically eliminates choices (D) and (E) as well.Why would this result automatically rule out D and E? How can you tell? Thank you for taking the time to answer my question.

We have two responses for you

Lucy,

I have to say I don't like this way of setting the question very much; in the real world guessing and checking is not a great way to do arithmetic, and in exams it's often penalized as "not showing your work."

The idea is that there are "too many peanuts" with the six-pound attempt, so more peanuts will make things worse. And that's true, but it depends on knowing that the function that gives the proportion y in terms of the pounds of peanuts x

y = (5+x)/(10+x)

is "monotone increasing", meaning that as x gets bigger y gets bigger. This needs either clever algebra or easy calculus to show properly.

Good Hunting!
RD

Hi Lucy,

I completely agree with RD that the guess and check method is far from the best way to approach this problem. I would use some logic and basic arithmetic.

The 10 pound mixture is 50% peanuts so it is 50% not peanuts, that is there are 5 pounds of non-peanuts in the mixture. When you add more peanuts the number of pounds of non-peanuts doesn't change. Thus in the final mixture the 5 pounds of non-peanuts is 40% of the mixture. To get from 40% to 60% you multiply by $\frac64$ and hence 60% of the new mixture is $5 \times \frac64 = 7.5 \mbox{ pounds.}$ Hence you need to add 2.5 pounds of peanuts to the 5 pounds you already have to have the new mixture contain 60% peanuts.

Penny

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