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PostPosted: Sat Mar 31, 2018 5:27 pm 
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This is from a paper that has at least one question with missing info (typographical error). Is this question also missing information or are we just innumerate?

30 children are going on a trip. The trip is £5 with lunch provided or £3 without. The 30 children pay a total of £110. How many paid £3?


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PostPosted: Sat Mar 31, 2018 5:34 pm 
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I think its complete; I'd use trial and improvement.

If all 30 paid £3 then £90 would be collected, so we know some pay £5. I also noted we need a multiple of 3 that gives a zero in the units. There aren't many to try.

Try 20 paying £3 and 10 paying £5; £60 + £50 = £110


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PostPosted: Sat Mar 31, 2018 5:36 pm 
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I think it’s doable. I’ve been used simultaneous equations to solve it though, which wouldn’t be a method used for 11+, if this is an 11+question, suspect trial and improvement would be the way to go

Cross post with G55


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PostPosted: Sun Apr 01, 2018 1:10 pm 
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Yes, I spoke to a primary school teacher who said that it is a key stage 2, level 5 question. Very few children would solve it and most would use trial and improvement.
It is from a practice paper provided by a company, but is from an old SATS paper.


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PostPosted: Sun Apr 01, 2018 1:47 pm 
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You could use algebra but that is not part of KS2 - noticing that you need a multiple of 3 which gives a unit's digit of 0 means it needs few trials


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PostPosted: Sun Apr 01, 2018 4:48 pm 
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I tackled it a bit differently (I think). If all 30 paid £3 the total would be £90 which is £20 short of the real total, so we know that 10 children paid an extra £2, the difference between £110 and £90, and the other 20 paid £3. No trial and improvement needed?

Could be done in reverse of course. Assume all 30 paid £5, making £150, therefore 20 children must have paid £2 less to bring the £150 down to £110.


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PostPosted: Sun Apr 01, 2018 7:08 pm 
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Location: Buckinghamshire
anotherdad wrote:
I tackled it a bit differently (I think). If all 30 paid £3 the total would be £90 which is £20 short of the real total, so we know that 10 children paid an extra £2, the difference between £110 and £90, and the other 20 paid £3. No trial and improvement needed?

A man after my own heart ... :mrgreen:


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PostPosted: Mon Apr 02, 2018 6:34 pm 
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IrisM65 wrote:
This is from a paper that has at least one question with missing info (typographical error). Is this question also missing information or are we just innumerate?

30 children are going on a trip. The trip is £5 with lunch provided or £3 without. The 30 children pay a total of £110. How many paid £3?


This is a trail and error question. 20 children paid £3, which makes £60. 10 children paid £5 each, which makes £50.


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PostPosted: Wed Apr 04, 2018 6:01 pm 
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I just asked my 10 yr old, and she did it this way:

If all 30 paid £5 each, the total cost would be £150.
The total cost was £110, so £40 less than this.
That £40 means 20 children paid £2 less each (based on £3 being £2 less than £5.)
So 20 paid £3 each and 10 paid £5 each = £110

I did trial and improvement (before asking dd, otherwise I’d have realised her way was much better!)
30 x £5 = £150
30 x £3 = £90
I need to be just under halfway between these totals, so I’ll try over half the children paying £3 and under half paying £5. So:
20 x £3 = £60 and 10 x £5 = £50. Total £110.
That was partly a lucky guess, but if it had been over / under, I’d have adjusted accordingly.

My daughter’s way was more efficient, but it’s doable either way.


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PostPosted: Wed Apr 04, 2018 6:03 pm 
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Only just read anotherdad’s reply properly and realised that dd’s method has already been suggested. Will leave it there in case seeing the steps she used is useful.


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