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 Post subject: bracket maths help
PostPosted: Wed Oct 24, 2007 6:50 pm 
night before help with....

(3 [8] 4) (3 [14] 7)
(5 [ ? ] 6)

how is answer 21 (from Susan Daughtreys paper 8.


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 Post subject:
PostPosted: Wed Oct 24, 2007 7:02 pm 
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Joined: Mon Jan 30, 2006 4:07 pm
Posts: 2660
Dear Guest

Bright Sparks test 8 q67...my version says answer is 24...

(3 [8] 4) (3 [14] 7)
(5 [ ? ] 6)

3 x 4 = 12 – 4 = 8
3 x 7 = 21 – 7 = 14
5 x 6 = 30 – 6 = 24

Multiplying numbers outside the middle brackets, subtracting number on the right.

Patricia


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 Post subject: 24
PostPosted: Wed Oct 24, 2007 7:31 pm 
Dear Patricia

you are absolutely right answer is 24!
And I can now see how to work it out too!

Every-time I think my son & I have mastered these, a simple combination throws us!

thank you again


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 Post subject:
PostPosted: Sat Nov 10, 2007 6:13 pm 
patricia wrote:
(3 [8] 4) (3 [14] 7)
(5 [ ? ] 6)

3 x 4 = 12 – 4 = 8
3 x 7 = 21 – 7 = 14
5 x 6 = 30 – 6 = 24

Multiplying numbers outside the middle brackets, subtracting number on the right.

Also, 'first number minus 1 times second number' works and is possibly easier as it keeps the numbers smaller. Essentially, we don't need that extra 4 or 7 or 6 in the first place if we just have to deduct it at the end.

(3 - 1) x 4 = 2 x 4 = 8
(3 - 1) x 7 = 2 x 7 = 14
(5 - 1) x 6 = 4 x 6 = 24


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