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 Post subject: Stumped! how do you work out this questionPosted: Fri Aug 31, 2012 12:50 pm

Joined: Mon Mar 05, 2012 7:55 pm
Posts: 148
Hi have difficulty in answering this question I found on a test paper:

Problem:

In November 2010 Becky is two years older than Mary who is three times Anna's age.

In November 2011 Anna is one third of Becky's age. Mary is 12 years older than Anna.

How old are the three girls in November 2010?

- Thanks in advance for any help with this one

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 Post subject: Re: Stumped! how do you work out this questionPosted: Fri Aug 31, 2012 1:16 pm

Joined: Thu May 10, 2012 8:30 am
Posts: 247
yummycool wrote:
- Thanks in advance for any help with this one

Let the girls' ages be B for Becky, M for Mary, A for Anna. Make sure you take them all as being in 2010.

We have B=M+2, and M=3A from the first line.

We have 3(A+1) = (B+1) and M=A+12 from the second line.

Substituting for M we have A+12=3A, so Anna is 6. M=3A so Mary is 18, and B=M+2 so Becky is 20. Checking that everything else is consistent, M=3A is true (18=3x6) and 3(A+1)=(B+1) is true, as 3(6+1) = 20+1.

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 Post subject: Re: Stumped! how do you work out this questionPosted: Fri Aug 31, 2012 1:17 pm

Joined: Mon Aug 22, 2011 8:20 pm
Posts: 1706
Location: Warwickshire
In November 2010 Becky is two years older than Mary who is three times Anna's age.

In November 2011 Anna is one third of Becky's age. Mary is 12 years older than Anna.

How old are the three girls in November 2010?

Regardless of when you measure it Mary will always be 12 years older than Anna, and Becky will always be 2 years older than Mary

M = A+12
B = A+14
B = M+2

In Nov 2010:
M = 3A

one year later Anna is one third of Becky's age.
so (in Nov 2010) A+1 = (B+1)/3

substituting gives A+1 = (A+14+1) / 3
multiply both sides by 3 gives 3A+3 = A+15
subtract A+3 from both sides gives 2A = 12
therefore A = 6

Putting 6 back into the equations
M = A+12
B = A+14
gives
M = 6+12 = 18
B = 6+14 = 20

Anna is 6
Mary is 18
Becky is 20

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 Post subject: Re: Stumped! how do you work out this questionPosted: Fri Aug 31, 2012 5:29 pm

Joined: Thu Jun 14, 2012 9:28 am
Posts: 121
If I may offer another way involving basic algebra, whihc appears in more ways then one at this stage.

From the question, it is clear that Anna is the youngest. Let her age in Nov 2010 be X. It follows that Mary is 3X and Becky (3X + 2).

In Nov 2011, Becky wil be (3X + 3), Mary will be (3X + 1) and since Anna is a third of Mary's age, she will be (3X + 3)/3. this looks much simpler when written down systematically:

Becky Mary Anna
Nov 2010 3X + 2 3X X
Nov 2011 3X + 3 3X + 1 (3X + 3)/3 = X + 1

We already know that in Nov 2011, Mary - Anna = 12

Therefore, (3X + 1) - (X+1) = 12

Solve this equation and you get X = 6. The rest follows.
Tagore

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 Post subject: Re: Stumped! how do you work out this questionPosted: Fri Aug 31, 2012 9:41 pm

Joined: Mon Mar 05, 2012 7:55 pm
Posts: 148
Thanks to everyone who has responded to this question...the support is on this forum is appreciated

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 Post subject: Re: Stumped! how do you work out this questionPosted: Fri Aug 31, 2012 10:04 pm

Joined: Thu Nov 08, 2007 9:57 pm
Posts: 1167
.

Last edited by Belinda on Sat Nov 03, 2012 6:59 pm, edited 1 time in total.

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 Post subject: Re: Stumped! how do you work out this questionPosted: Sat Sep 01, 2012 7:41 am

Joined: Mon Jul 04, 2011 1:47 pm
Posts: 2151
Location: Warwickshire
My dd would just skip the question! Isn't that what you're supposed to do if you can't do it?! Mind you, that's probably why she has a very slim chance of passing ... but I still thought you weren't supposed to spend ages on one question if it looks too difficult?

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 Post subject: Re: Stumped! how do you work out this questionPosted: Sat Sep 01, 2012 9:39 am

Joined: Tue Jul 21, 2009 9:56 pm
Posts: 8228
I think in Kent you could get away without doing a question like that and still get a superselective score. Yes even if you are great at algebra that question takes a little while relative to some others.

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 Post subject: Re: Stumped! how do you work out this questionPosted: Sat Sep 01, 2012 2:38 pm

Joined: Mon Mar 05, 2012 7:55 pm
Posts: 148
ginx mystery - I agree with you. I thought the same when dd looked at the question as if it had from outer space!!

We have many other areas to reinforce for the next while

(btw I still have no clue as to how to work out that last sum - but from the responses there are many who do). My dd was ok when I showed her the answers.

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 Post subject: Re: Stumped! how do you work out this questionPosted: Sat Sep 01, 2012 7:31 pm

Joined: Thu Jun 14, 2012 9:28 am
Posts: 121
You are quite right - at 11+ DC's are not expected to solve such problems with algebra. The examiner maybe looking for a more analytical approach. This is another possible solution.

Nov 2011 Mary is 12 years older than Anna. Write down the numbers 2 to 9 and the corresponding numbers, adding twelve each time. Something like this:

2 14
3 15

and so on. Now look for the pair, which when reduced by 1, satisfies the criteria of Nov 2010 when Mary is 3 times Anna's age. The pair 7 and 19 should result. Becky's age in Nov 2010 is a simple calculation.

I have had the luxury of time to think about this. Our DC's have a minute at the most to come up with a solution. Where was this from? Hope above is more helpful.

and....The Best of Luck to all the DC's taking their exams this month.

Tagore

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