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 Post subject: Help with maths questionPosted: Fri Dec 11, 2020 10:53 am

Joined: Wed Mar 22, 2017 2:52 pm
Posts: 151
Hi. This is a maths question both my DS and DH could not solve. Maybe they didn't spend enough time on it but if anyone has a good solution, please let me know (thanks!):

"What 5-digit number has the following features:
If we put the numeral 1 at the beginning, we get a number three times smaller than if we put the numeral 1 at the end of the number?

In other words, if you think the answer is the number 34567, then you want the number 134567 to be one third of 345671, but it isn't, so what's the number?"

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 Post subject: Re: Help with maths questionPosted: Fri Dec 11, 2020 5:58 pm

Joined: Fri Mar 17, 2006 6:12 pm
Posts: 1491
Location: Birmingham
Best solved by Algebra

If n is the 5 digit number then putting 1 in front means you are adding 100,000

Similarly, if you add 1 at the end you are multiplying n by 10 and then adding 1.

So if the former is 1/3 of the latter therefore

3(n+100,000) = 10n + 1

Hence 3n + 300,000 = 10n + 1

Or 7n = 299,999

Hence n = 42,857

Required some serious thinking, is this really an 11+ exam question?
Maybe the last Maths question for Independent School 11+ entrance which are often used as a scholarship type question

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 Post subject: Re: Help with maths questionPosted: Fri Dec 11, 2020 10:43 pm

Joined: Sun Oct 04, 2020 2:15 pm
Posts: 41
Location: Trafford
Don't know if I'm reading it wrong. But I think both numbers need to be 5 digits in total.

I get (a) 19997 and (b) 59991
So they've both got 999 in the middle and (a) multiplied by 3 equals (b).

I solved it like this....
(a) is A B C D E
(b) is F B C D A
We know A is 1 from the question.
E times 3 must make A as 1. So E can only be 7 because there's no other way to make a 1 in the A position in number (b). E times 3 is 21
E is 7 carries 2 over in to the D column. So D is a number, when multiplied by 3 and then add 2 still is D in that column. 4 and 9 both fit this requirement.
If D is 4, you can't find C.
So D must be 9. 3 times 9 is 27 plus the 2 carried over is 29.
But we're going to carry a 2 over again. So the C is 9 again because it worked for me last time.
B 9 again for the same reason.

Edit clarity hopefully

Edit again. If it is 6 digits, stick another 9 in the middle

Edit again. B=4 does work because the carry problem isn't an issue in the last column.

So 14997 and 44991 also works. Throw in another 9 if you want 6 digits

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 Post subject: Re: Help with maths questionPosted: Sat Dec 12, 2020 12:10 pm

Joined: Wed Mar 22, 2017 2:52 pm
Posts: 151
Yes, this was the last question on an independent 11+ maths paper. Thanks for your replies!

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 Post subject: Re: Help with maths questionPosted: Sat Dec 12, 2020 12:15 pm

Joined: Wed Mar 22, 2017 2:52 pm
Posts: 151
The sheet gave the answer as 43,857! It may of course be wrong but, regardless, do independents really expect this level of maths problem solving in 10/11-year-olds even for a scholarship question? I'd argue that very few would be exposed to this level unless they have parents who are teachers/maths focused especially if you're in the state sector.

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 Post subject: Re: Help with maths questionPosted: Sat Dec 12, 2020 3:15 pm

Joined: Fri Mar 17, 2006 6:12 pm
Posts: 1491
Location: Birmingham
turnip08 wrote:
The sheet gave the answer as 43,857! It may of course be wrong but, regardless, do independents really expect this level of maths problem solving in 10/11-year-olds even for a scholarship question? I'd argue that very few would be exposed to this level unless they have parents who are teachers/maths focused especially if you're in the state sector.

As I said, Independent schools often set a very difficult final Maths question. The vast majority of candidates would not be able or expected to answer, or indeed may not have the time to answer. However, schools are on the lookout for children with outstanding mathematical or logic abilities who have probably been taught topics at home way beyond the National Curriculum.

If they can answer questions like that they will have a good chance of being offered a fee reduction scholarship.

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 Post subject: Re: Help with maths questionPosted: Sat Dec 12, 2020 3:26 pm

Joined: Fri Dec 11, 2020 12:48 pm
Posts: 10
Private schools dont expect students to know out of the syllabus. If they know its good.
The main motive is to select the students who thinks any complicated question in a simple logical way.
They want to select the students with very good logical skills.
Any Maths question can be solved with different methods..using advanced maths or simple school maths.

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 Post subject: Re: Help with maths questionPosted: Sat Dec 12, 2020 3:36 pm

Joined: Fri Dec 11, 2020 12:48 pm
Posts: 10
When students are appearing for Independent school or grammar school exams, they should be taught how to think and simply any complicated question into a simple question.
If they learn that approach then they try to simply any hard question into a simple one..they dont have to know the advanced maths to solve a difficult question.

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 Post subject: Re: Help with maths questionPosted: Sat Dec 12, 2020 4:00 pm

Joined: Wed Mar 22, 2017 2:52 pm
Posts: 151
Thanks for all your replies. So, for this particular question, would you say that a highly gifted child in Y6 would be able to solve this particular question by using their existing up-to-year 6 maths syllabus knowledge if they simply applied their thinking/logical skills (assuming no maths focused parents/teachers in the family)?

Very curious as I have no clue in how to solve this!

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 Post subject: Re: Help with maths questionPosted: Sat Dec 12, 2020 4:07 pm

Joined: Fri Dec 11, 2020 12:48 pm
Posts: 10
Sorry, it didnt post properly and the letters got moved. I am editing to make sure it posts properly

Let the 5 digits number be ABCDE

1ABCDE
------X 3
---------
ABCDE1
3XE should give *1 means E =7 and 3X7=21
------2 <----carried
1ABCD7
------X3
---------
ABCD71

3XD+2 = *7, which means D should be 5 and 3X5+2=17

-----12<------------ carried
1ABC57
------X3
------------
ABC571

3XC+1=*5, that means C=8 and 3X8+1=25

---212 <-------------carried
1AB857
------X3
---------
AB8571

3XB+2=8 which means B = 2 and 3X2+2=8

---212
1A2857
------X3
----------
A28571

3XA=*2 which means A=4, 3X4=12 and 3X14=42

1-212
142857
------X3
--------
428571

Therefore the answer is ABCDE=42857 is a 5 digits number which satisfies the given conditions.
This can be also solved with Algebra. But, I dont think students of year5 can think in Algebra in the exam hall.
If they learn how to solve any question in a simple way then they can answer any question easily.

Last edited by Mathsatish on Sat Dec 12, 2020 4:41 pm, edited 2 times in total.

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