0
$\begingroup$

Write the numbers 3, 4, 5 and 6 in the boxes to give the greatest possible total. You may write each number only once.

enter image description here

This is worth 1 mark in an exam, but the only way I can think of going about it is that 5 and 6 obviously go into the two boxes outside of the fractions, then you have 4 choices, now the only way I can think of doing this is trial and error, and in an exam where you really do need a mark a minute and it's non-calculator, I can't imagine my trial and error taking 1 minute, is there a set way to do this?

I know the answer, but it's is there a better way to get to the answer? Thanks! :)

The way I went about it was: The first number outside each fraction must be 5 or six, well then there's only 4 ways to do it, so I'll test each... $$ 5\frac{1}{3}+6\frac{2}{4}=\frac{142}{12}\\ 5\frac{1}{4}+6\frac{2}{3}=\frac{143}{12}\\ 6\frac{1}{3}+5\frac{2}{4}=\frac{142}{12}\\ 6\frac{1}{4}+5\frac{2}{3}=\frac{143}{12}\\ $$ Therefore the answer is either: $$ 6\frac{1}{4}+5\frac{2}{3}\\or\\5\frac{1}{4}+6\frac{2}{3} $$

$\endgroup$
  • $\begingroup$ It seems an unsatisfactory problem in that the answer is not unique, as posed above. One consideration might be to rule out using $2/4$ on the grounds of not reduced to lowest terms, but there remains the fact that the five and six positions are interchangeable. $\endgroup$ – hardmath May 8 '17 at 1:20
1
$\begingroup$

You are right that you want $5$ and $6$ to be the whole numbers. It doesn't matter which one goes where because you will add them either way. Then you want two big pieces and one little instead of two little and one big, so the $3$ goes under the $2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.