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 Posted: Sun Mar 30, 2014 12:51 am
Hi everyone,
I have couple of question from Magdalen private school maths past paper, is there any trick or simple way to solve it rather than Trial and error?

Q1) Place one of each of the numbers 1,2,3,4,5,7 in a region of the diagram below so that the
total in each of the circles is the same. The number ‘6’ is correctly placed as a hint.

I don't know how to insert the diagram in here so i put the link to the past paper (Q22)
The link to the paper is:
http://www.mcsoxford.org/resource.aspx?id=8878

Q2) Insert the symbols +, +, - , × , × into the boxes(dotted line) below
to make a total of 33. You must use all of the symbols in the list.
1...2...3...4...5...6

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 Posted: Sun Mar 30, 2014 1:09 am

Joined: Mon Aug 12, 2013 9:13 am
Posts: 452
Q1 is Venn diagram to come up with same totals

Q2 Insert the symbols +, +, - , × , × into the boxes below
to make a total of 33. You must use all of the symbols in the list.

I worked backwards to see how can I get 33 which is 30 + 3. You can get 30 from 5 x 6 AND get 1 from others.

1 + 2 x 3 - 4 + 5 x 6
1 + 6 - 4 + 30 [BODMAS]
33

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 Posted: Sun Mar 30, 2014 12:08 pm

Joined: Wed Aug 01, 2007 11:19 am
Posts: 888
I think q1 is trial and error - although trying to be at least a bit systematic about it and starting with e.g. 1, 2, 3 in the other 3 overlaps (which isn't the right answer, lol), rather than just putting in random numbers. I don't know whether there is more than one answer.

q2 I did 1x2-3+4+5x6 so that's at least two answers for that one! I guess (like in Countdown), go for the bigger chunks first e.g. 5x6 and then adjust.

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 Posted: Sun Mar 30, 2014 5:18 pm

Joined: Sat Dec 03, 2011 3:14 pm
Posts: 688
for the Venn diagram put the smallest numbers in the overlaps and then try jugglling the larger numbers

The left hand circle will have 7, 2, 6, 1
Right hand will have 3, 2, 6, 5
Bottom circle will have 5, 6, 1, 4

Sorry can't copy the pattern so just giving the numbers

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 Posted: Sun Mar 30, 2014 9:10 pm

Joined: Thu Jun 14, 2012 9:28 am
Posts: 121
Q1) I believe there is a logical/mathematical approach to this question. We are told that the numbers in each circle add up the same amount. Thus we know that the total of numbers appearing in each circle will be a multiple of three.

We also know that the total of all numbers appearing in each circle will contain all the numbers from 1 to 7 once, the number 6 a further two times and a set of three numbers that overlap. The total of all numbers 1 to 7 is 28 plus 12 (6 appearing twice) is 40. The set of three overlapping numbers will take this total to a multiple of 3. The lowest set of the three overlapping numbers is 1,2 and 3, totalling 6. This takes the total to 46 as a minimum. The first number after 46, divisible by 3 is 48. Therefore, the set of three overlapping numbers add up to 8 (48-40). The set could be 1,2 and 5 or 1,3 and 4.

Each circle would have 4 numbers totalling 16. Removing the constant number 6, the remaining three numbers need to add to ten. The only way we would produce a total of 10, if one of the numbers is 7, is if the set of three numbers is 1,2 and 5. The numbers would appear as follows (shared numbers in red):

Circle 1 7 1 2
Circle 2 2 3 5
Circle 3 5 4 1
I hope this is easy to follow. Thank god I am not a teacher!

Tagore

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 Posted: Tue Apr 01, 2014 8:01 am

Joined: Sat Aug 10, 2013 11:46 am
Posts: 216
7 1 4
7 2 3
5 4 3

Also works

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