Care to try this simple(?) algebra problem?

Asked by LostInParadise (32183

) April 28th, 2018

I came across the problem on this link Be warned, the article is a teaser. To read the full article, including the answers to the two problems, you have to open a subscription.

The second problem is fairly easy, but I won’t give it away if anybody wants to to think about it. The first one took a little bit of thought. I had to read through it 2 or 3 times just to figure out what they were saying. See how far you can get.

I would also be interested if anybody could provide a link to the subject of the article without having to take out a subscription.

Observing members: 0

Composing members: 0

4 Answers

Easy. 2+2=10

ragingloli (52231

)“Great Answer” (0

) Flag as…

Full article here.

JeSuisRickSpringfield (8531

)“Great Answer” (2

) Flag as…

Thanks.

Here is the algebra for solving the first problem. Let j, c and z be the current ages of Joey, Chloe and Zoe.

Joey was Chloe’s age j – c years ago. Zoe’s age at that time was z -(j-c)= z-j+c. We are told that Chloe’s current age is twice that number, giving 2( z – j + c) = c. Solving for z gives z = j – c/2.

Chloe will be twice Joey’s age in 2j – c years. At that time, Zoe’s age will be z + 2j – c = (j – c/2) + 2j – c = 3j – 3c/2. We can see that this is three times (j – c/2)

LostInParadise (32183

)“Great Answer” (1

) Flag as…

Here is a simpler way of getting z in terms of j and c. j-c years ago, Zoe was ½ Chloe’s current age of c, so Zoe’s age now is z = c/2 + (j – c) = j – c/2. Thinking of it that way, the problem is almost simple enough to do in your head.

LostInParadise (32183

)“Great Answer” (1

) Flag as…