Maths Question Consortium 2008 Q30

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Salcombe
Posts: 22
Joined: Sat Dec 08, 2012 7:17 pm

Maths Question Consortium 2008 Q30

Post by Salcombe »

Completely stumped. From the Godolphin and Latymer website:

AB and C represent different digits.

AB + C =50

BC + A = 41

What are the values of A, B and C?


Yikes. This is tricky. Any suggestions how to tackle.
Guest55
Posts: 16254
Joined: Mon Feb 12, 2007 2:21 pm

Re: Maths Question Consortium 2008 Q30

Post by Guest55 »

A,B and C are digits .... think about that.
Salcombe
Posts: 22
Joined: Sat Dec 08, 2012 7:17 pm

Re: Maths Question Consortium 2008 Q30

Post by Salcombe »

I have just realised how easy this is. I thought the first two 'digits' were to be multiplied first. Doh!

Thanks anyhow for looking
Guest55
Posts: 16254
Joined: Mon Feb 12, 2007 2:21 pm

Re: Maths Question Consortium 2008 Q30

Post by Guest55 »

That's why I only gave a hint!
Proud_Dad
Posts: 500
Joined: Fri Oct 11, 2013 9:55 am

Re: Maths Question Consortium 2008 Q30

Post by Proud_Dad »

10A + B + C = 50 :arrow: (1)
10B + C + A = 41 :arrow: (2)

(1) - (2) :arrow: 9A - 9B = 9

:arrow: A = B + 1

So substituting this in (2) gives:

10B + C + B + 1 = 41
11B + C = 40

Since C is a single digit the only value B can have is 3.

Therefore, C = 7

And A = 4.
Okanagan
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Location: Warwickshire

Re: Maths Question Consortium 2008 Q30

Post by Okanagan »

It is simpler than that:
AB + C =50 :arrow: C can't be more than 9 so A must be 4 (and B+C must be 10 although you don't need that fact)

BC + A = 41 :arrow: given A is 4, BC must be 37 so B is 3 and C is 7
Proud_Dad
Posts: 500
Joined: Fri Oct 11, 2013 9:55 am

Re: Maths Question Consortium 2008 Q30

Post by Proud_Dad »

Okanagan wrote:It is simpler than that:
AB + C =50 :arrow: C can't be more than 9 so A must be 4 (and B+C must be 10 although you don't need that fact)

BC + A = 41 :arrow: given A is 4, BC must be 37 so B is 3 and C is 7
Yeah, you're right. I over complicated it a bit!
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