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PostPosted: Wed Apr 27, 2011 6:25 am 
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Do not panic anyone this is not an 11 plus question.

DS has this for homework. We have eventually by trial and error got what we think is the right answer but can anyone explain how we should have mathematically worked it out.

Amrita has written down four whole numbers. If she chooses three of her numbers at a time and adds up each triple, she obtains totals of 115, 153, 169 and 181.

What is the largest of Amrita's numbers?

A 66
B 53
C 91
D 121
E 72


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PostPosted: Wed Apr 27, 2011 6:29 am 
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Location: East Kent
I would tend to work down from the 121 and eliminate.
can't be 121 as there are no others small enough to make 181

So trial and error really. Sorry not much help!


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PostPosted: Wed Apr 27, 2011 9:32 am 
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This is how I did it - but it seems a bit convoluted, there may be a quicker way:

Calling the numbers a, b, c, d in ascending order,

(1) a + b + c = 115
(2) a + b + d = 153
(3) a + c + d = 169
(4) b + c + d = 181

subtract (1) from (4)
d - a = 181 - 115 = 66
a = d + 66

If d = 121, a = 55 and b + c = 60:
a + d = 176 so (b + c = 60) doesn't work because b and c would both have to be negative for eqns (2) and (3) to work out

If d = 91, a = 25 and b + c = 90:
a + d = 116
b = 153 - 116 = 37
c = 169 - 116 = 53
53 + 37 = 90 so that works.

Takes a long time tho', and since they're looking for people with a natural flair for maths the official solution may have something more brisk and not very intuitive to the rest of us.

Mike


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PostPosted: Wed Apr 27, 2011 10:30 am 
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Location: Watford, Herts
Minesatea wrote:
Amrita has written down four whole numbers. If she chooses three of her numbers at a time and adds up each triple, she obtains totals of 115, 153, 169 and 181.

What is the largest of Amrita's numbers?

Each of the original numbers occurs in three of the totals, so the total of the four numbers is (115+153+169+181)/3 = 206, and thus the largest number was 206 - 115 = 91.

Checking the answer (always a good idea): similarly the other numbers are

206 - 153 = 53
206 - 169 = 37
206 - 181 = 25

and we can reconstruct the totals

25 + 37 + 53 = 115
25 + 37 + 91 = 153
25 + 53 + 91 = 169
37 + 53 + 91 = 181


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PostPosted: Wed Apr 27, 2011 10:44 am 
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Location: Watford, Herts
A simpler variant from the 1988 challenge:
Quote:
Weighing the baby at the clinic was a problem. The baby would not keep still and caused the scales to wobble. So I held the baby and stood on the scales while the nurse read off 78 kg. Then the nurse held the baby while I read off 69 kg. Finally I held the nurse while the baby read off 137 kg. What was the combined weight of all three (in kg)?

A 142 B 147 C 206 D 215 E 284


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PostPosted: Wed Apr 27, 2011 11:48 am 
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Thanks WP. That was the answer he got but it took 20 mins so I was sure there had to be a simple way!


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PostPosted: Wed Apr 27, 2011 11:51 am 
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I reckon for a junior maths challenge they are looking for someone to think around the problem, rather than just applying a learnt formula.

That's my excuse anyway


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PostPosted: Sun May 29, 2011 7:00 pm 
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Are these questions typical of 11+ (I know the OP's question wasn't, but the rest?) They are beyond me.


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PostPosted: Tue May 31, 2011 1:53 pm 
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menagerie wrote:
Are these questions typical of 11+ (I know the OP's question wasn't, but the rest?) They are beyond me.


JMC questions are aimed at the top third, of year 8 children.
http://www.mathcomp.leeds.ac.uk/individ ... challenge/


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PostPosted: Tue May 31, 2011 5:57 pm 
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Thanks Mitasol. That's some comfort. :shock: I'm sure my husband could do them, but not convinced he could explain how to our sons...


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