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Bancroft Maths 2017
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Author:  Gee [ Sun Sep 22, 2019 4:58 pm ]
Post subject:  Bancroft Maths 2017

I need help from Maths experts.

I don't know where to start.


Three boxes (A, B and C) contain red balls or yellow balls or both.
Each box contains the same number of balls.
Box A contains all twelve of the red balls and one-ninth of the yellow balls.

How many yellow balls are there altogether?

How many balls are there in each box?

Author:  ToadMum [ Sun Sep 22, 2019 5:43 pm ]
Post subject:  Re: Bancroft Maths 2017

Gee wrote:
I need help from Maths experts.

I don't know where to start.


Three boxes (A, B and C) contain red balls or yellow balls or both.
Each box contains the same number of balls.
Box A contains all twelve of the red balls and one-ninth of the yellow balls.

How many yellow balls are there altogether?

How many balls are there in each box?


Keep upping the number of yellow balls until that number, times 8 - to make the 'added' yellow balls one ninth of the total yellow balls - (or, that number, times 4) gives you the number of balls in total you now have in the first box? Then times the total in the first box by 3?

e.g. 12 red, add 3 yellow, 15 total in that box. 3 x 8, divided by 2, equals 24/2, so 12. Do not that, keep trying...

Author:  Gee [ Sun Sep 22, 2019 6:00 pm ]
Post subject:  Re: Bancroft Maths 2017

Please correct me if my understanding is wrong.

Do one box have both colours, one box only red and one box only yellow?

If so, box A has both colours?

Author:  ToadMum [ Sun Sep 22, 2019 10:15 pm ]
Post subject:  Re: Bancroft Maths 2017

Gee wrote:
Please correct me if my understanding is wrong.

Do one box have both colours, one box only red and one box only yellow?

If so, box A has both colours?


Box A apparently has all the red balls and some yellow balls. So the other two boxes only contain yellow balls.

Author:  Surferfish [ Mon Sep 23, 2019 9:59 am ]
Post subject:  Re: Bancroft Maths 2017

Box A contains all 12 of the red balls and 1/9 of the yellow balls.

The remaining 8/9 of the yellow balls must therefore be in boxes B and C. 4/9 in box B and 4/9 in box C, as each box has the same number of balls.

Because the number of balls in each box is equal, then if y is the total number of yellow balls we can say:

Number of balls in box A = Number of balls in box B
12 + (1/9)y = (4/9)y
12 = (3/9)y

y = 36

Author:  Gee [ Mon Sep 23, 2019 10:15 am ]
Post subject:  Re: Bancroft Maths 2017

Thank you so much for all the replies.

Got the answer too once understood correctly.

Find 1/9 of multiples of 9. It has to be more than 18.

So, the next multiple is 27. But the total of the balls won't be equal with 27 balls.

So, the next multiple of 9 is 36.

1/9 of 36 is 4. 12 + 4 = 16 balls.

36 - 4 = 32 balls

32/2 = 16 balls in each boxes.

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