General Question

LostInParadise's avatar

Can you solve this different kind of algebra problem?

Asked by LostInParadise (30211points) February 2nd, 2012
5 responses
“Great Question” (1points)

I found this problem on that other Q&A site (Yahoo! Answers) and thought it was kind of fun to solve. It requires nothing more than the most basic algebra plus a little bit of reasoning and a little bit of effort. The story behind the problem is a bit lame, but I could not think of anything better. If you have an improvement let me know.

Five children are weighed in pairs. Their weights in kg are:
58 60 61 64 64 65 67 68 70 71

What are the five weights?

Observing members: 0
Composing members: 0

Answers

6rant6's avatar

The two heaviest together are 71 pound, the two lightest together are 58. From this we can find the middle one is 33.

Let’s try 36 for the heaviest (at least one must weigh more than 35).

71–36 = 35

so one must be 35

But then the 70 pair is problematic…

so let’s try 37.

71–37 = 34
70–37 = 33
So the middle one is right.

Now for the lower 2….

30, and 28….
Can’t get 60.
31 and 27

27, 31, 33, 34, 37

I believe that’s it.

flutherother's avatar

If we call the weights of the five children in ascending order a,b,c,d and e
Then it follows that…

a+b=58
a+c=60
a+d=61
a+e=64
b+c=64
b+d=65
b+e=67
c+d=68
c+e=70
d+e=71

Adding both sides together 4(a+b+c+d+e) =648
As we know from above what any four of these weights add up to we can then calculate the 5th

The weights of the children are therefore 27,31,33,35 and 36Kg

LostInParadise's avatar

@flutherother , How did you determine b+c and a+e. How do you know that a+e is less than c+d?

LostInParadise's avatar

Order the weights a<=b<=c<=d<=e.
If you think about it, you will see that the smallest sum is a+b and the next smallest is a+c. Similarly the largest sum is d+e and the next largest is c+e. That is the only immediate conclusion we can come to.

a+b = 58
a+c=60
c+e=70
d+e=71

That is four equations for five variables. We need one more equation. I did not see @flutherother ‘s trick of using the sum of all the pairs of weight sums. I used a much more convoluted approach. Getting 4(a+b+c+c+e)=648 provides the extra equation.
a+b+c+d+e = 162.
We then have c =(a+b+c+d+e) – (a+b) – (d+e) = 162 – 129 = 33. Having c, it is easy to use the equations to find the remaining values.

flutherother's avatar

I assumed too much and got the wrong answer. @6rant6 got the right answer and @LostInParadise has the right method.

Answer this question

Login

or

Join

to answer.

Mobile | Desktop


Send Feedback   

`