Question 28 - HABS Maths paper 2014
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Question 28 - HABS Maths paper 2014
please help is there any trick to working out this question??
Re: Question 28 - HABS Maths paper 2014
Here's the question -
We write s(2,5) as an abbreviation for 2+3+4+5 so that s(2,5) = 14
Similarly
S(6,39) = 6+7+8+9+....+ 38 + 39 = 765
Work out:
s(1,3) =
s(6,40) =
s(7, 38) =
s(1,2) - s(2,3) + S (3,4) - s(4,5)+ ......... - S (18,19) + S (19,20) =
We write s(2,5) as an abbreviation for 2+3+4+5 so that s(2,5) = 14
Similarly
S(6,39) = 6+7+8+9+....+ 38 + 39 = 765
Work out:
s(1,3) =
s(6,40) =
s(7, 38) =
s(1,2) - s(2,3) + S (3,4) - s(4,5)+ ......... - S (18,19) + S (19,20) =
Re: Question 28 - HABS Maths paper 2014
We've had this before - try searching.
Re: Question 28 - HABS Maths paper 2014
The long way round is this:
Total = (number of items) x (mean value)
For s(a,b):
number of items = b-a+1
mean value = (a+b)/2
This gives total = (b-a+1)(a+b)/2
So
s(1,3) = 1+2+3 = 6
Check the formula works: s(1,3)=(3-1+1)*(4/2)=3*2=6
Then
s(6,40)=(40-6+1)*(46/2)=35*23=805
s(7,38)=(38-7+1)*(45/2)=32*22.5=720
or alternatively s(7,38)=s(6,40)-6-39-40=805-85=720
The short way is to use the example they have given of s(6,39)=765, like this:
s(1,3)=1+2+3=6
s(6,40)=s(6,39)+40=765+40=805
s(7,38)=s(6,39)-6-39=765-45=720
Total = (number of items) x (mean value)
For s(a,b):
number of items = b-a+1
mean value = (a+b)/2
This gives total = (b-a+1)(a+b)/2
So
s(1,3) = 1+2+3 = 6
Check the formula works: s(1,3)=(3-1+1)*(4/2)=3*2=6
Then
s(6,40)=(40-6+1)*(46/2)=35*23=805
s(7,38)=(38-7+1)*(45/2)=32*22.5=720
or alternatively s(7,38)=s(6,40)-6-39-40=805-85=720
The short way is to use the example they have given of s(6,39)=765, like this:
s(1,3)=1+2+3=6
s(6,40)=s(6,39)+40=765+40=805
s(7,38)=s(6,39)-6-39=765-45=720
Re: Question 28 - HABS Maths paper 2014
This is one of those "can you spot the short way" type questions. DebsB has wonderfully given the short way to do the second and third questions, using the answer they have already given for s(6,39).
And in the final question, they're looking to see if you spot that most of the terms cancel each other out:
s(1,2) - s(2,3) + s(3,4) - s(4,5)+ ......... - s(18,19) + s(19,20)
= 1 + 2 - 2 - 3 + 3 + 4 - 4 - 5 + ... - 18 - 19 + 19 + 20
= 1 + 20
= 21
And in the final question, they're looking to see if you spot that most of the terms cancel each other out:
s(1,2) - s(2,3) + s(3,4) - s(4,5)+ ......... - s(18,19) + s(19,20)
= 1 + 2 - 2 - 3 + 3 + 4 - 4 - 5 + ... - 18 - 19 + 19 + 20
= 1 + 20
= 21
Re: Question 28 - HABS Maths paper 2014
Nicely done, rah_b92. I hadn't even noticed that the final question was there. Shocking example of failure to read the question properly - let's hope none of our kids have been throwing marks away by missing stuff out in that way!rah_b92 wrote:This is one of those "can you spot the short way" type questions. DebsB has wonderfully given the short way to do the second and third questions, using the answer they have already given for s(6,39).
And in the final question, they're looking to see if you spot that most of the terms cancel each other out:
s(1,2) - s(2,3) + s(3,4) - s(4,5)+ ......... - s(18,19) + s(19,20)
= 1 + 2 - 2 - 3 + 3 + 4 - 4 - 5 + ... - 18 - 19 + 19 + 20
= 1 + 20
= 21
Re: Question 28 - HABS Maths paper 2014
thank you everyone