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Maths Question Consortium 2008 Q30
Posted: Sun Dec 15, 2013 4:13 pm
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.
Re: Maths Question Consortium 2008 Q30
Posted: Sun Dec 15, 2013 4:45 pm
by Guest55
A,B and C are digits .... think about that.
Re: Maths Question Consortium 2008 Q30
Posted: Sun Dec 15, 2013 6:22 pm
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
Re: Maths Question Consortium 2008 Q30
Posted: Sun Dec 15, 2013 6:36 pm
by Guest55
That's why I only gave a hint!
Re: Maths Question Consortium 2008 Q30
Posted: Mon Dec 16, 2013 12:11 pm
by Proud_Dad
10A + B + C = 50
(1)
10B + C + A = 41
(2)
(1) - (2)
9A - 9B = 9
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.
Re: Maths Question Consortium 2008 Q30
Posted: Mon Dec 16, 2013 2:56 pm
by Okanagan
It is simpler than that:
AB + C =50
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
given A is 4, BC must be 37 so B is 3 and C is 7
Re: Maths Question Consortium 2008 Q30
Posted: Mon Dec 16, 2013 3:01 pm
by Proud_Dad
Okanagan wrote:It is simpler than that:
AB + C =50
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
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!