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PostPosted: Tue Oct 14, 2008 6:35 am 
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Can you help, absolutely perplexed ?

A number 65 bus goes on a circular route shown in the diagram below. The cost of travelling between any two stops is worked out by adding together the fares between consecutive stages. For example, if it costs 40p to travel from A to B and 37p to travel from B to C, then the cost of a ticket from A to C would be 77p.

If it costs 24p to go from C to D and 97p to go from C to E what is the fare from D to E ? I think the answer is 73p

Mavis, Ethel and her friend Dot all get on the bus at E. Mavis buys a ticket from E to G and is charged 82p. Ethel kindly buys a ticket for both herself and her friend. Ethel is going all the way to G, whereas Dot is going to F. Ethel is charged £1.17. What is the fare from F to G ? ________

The total cost of one ticket from G to J and one ticket from G to I is £1
The total cost of one ticket from G to J and one ticket from H to J is £1.20
The total cost of one ticket from G to J, one ticket from G to H and one ticket from I to J, is £1.30

What is the cost of one ticket from G to H ?

I just don't know where to begin and get flooded by all the information, can anyone help?

I can't get the diagram up, but basically letter are in a circle from A to J


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PostPosted: Tue Oct 14, 2008 9:17 am 
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Joined: Wed Mar 08, 2006 8:07 pm
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This is how I would do it.

A to B = 40p
B to C = 37p
C TO D = 24p
C to E = 97p
D TO E = 97p - 24p = 73p
E to G = 82p
E to F to G = 82p
Ethel pays for E to G and F to G
Ethel pays £1.17p
F to G = £1.17p - cost of E to G
F to G = £1.17p - 82p
F to G = 35p

The total cost of one ticket from G to J, one ticket from G to H and one ticket from I to J, is £1.30
(G to H to I to J) + ( G to H) + (I to J)
is the same as from G to J twice
It costs £1.30 for 2 tickets from G to J, so it'll cost 65p for a ticket
from G to J.

Now we can substitute this in the others.

The total cost of one ticket from G to J and one ticket from G to I is £1
65p + a ticket from G to I is £1.00
So a ticket from G to I will be 35p

The total cost of one ticket from G to J and one ticket from H to J is £1.20
65p + a ticket from H to J is £1.20
So a ticket from H to J would be 55p

If G to J is 65p and H to J is 55p,
then G to H should cost 10p.


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PostPosted: Tue Oct 14, 2008 9:26 am 
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I think I might have got the part that says

The total cost of one ticket from G to J, one ticket from G to H and one ticket from I to J, is £1.30
wrong.

It says from G to J and then one from G to H and I to J, H to I is not included in the cost.

You would have to take that into account and re work it but it's along the lines of my earlier post.

Sorry


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PostPosted: Tue Oct 14, 2008 12:32 pm 
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Posts: 139
Let me try to solve the last part of the question. I think the earlier parts are relatively easy.

The total cost of one ticket from G to J and one ticket from G to I is £1
The total cost of one ticket from G to J and one ticket from H to J is £1.20
The total cost of one ticket from G to J, one ticket from G to H and one ticket from I to J, is £1.30

What is the cost of one ticket from G to H ?

First, to reduce the letters, I would assign some single letters to the distance.

GH -> x HI -> y and IJ -> z.
That way GJ it is simply x + y + z

The above statements then become:

1) x + y + z + x + y = 100 (in pence)
2) x + y + z + y + z = 120
3) x + y + z + x + z = 130

Need to reduce the number of unknowns in the above equations.

Take 1 and 2
x + y + z + y is common in both.

1) x + y + z + y = 100 - x
2) x + y + z + y = 120 - z

that is 100 - x = 120 - z
-x = 120 - 100 - z
So x = z - 20


Take 1 and 3
x + y + z + x is common in both.

1) x + y + z + x = 100 - y
3) x + y + z + x = 130 - z

that is 100 - y = 130 - z
So y = z - 30

Now both x and y are represented in z. So substitute the values of x and y in the above equation 1

1) (z - 20) + (z - 30) + z + (z - 20) + (z - 30) = 100
5z - 100 = 100
5z = 200
z = 200 / 5
z = 40

x = z - 20
x = 40 - 20
x = 20

y = z - 30
y = 40 - 30
y = 10


x = 20, y = 10, z = 40

These values must satisfy all the above equations.

Bringing back the sectors...

GH = 20p
HI = 10p
IJ = 40p

Now, I would really like to know if this question is from a 11+ exam paper. Unless I missed a easy trick to solve the above, I honestly believe it is asking too much from a 10/11 year olds to do it and within a short period of time.

Gilly


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PostPosted: Tue Oct 14, 2008 12:52 pm 
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Joined: Tue Oct 14, 2008 6:16 am
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I agree, this is a question from a past haberdasher paper - I am aware that the papers increase in difficulties, this clearly is not a level four or five question, or is it? My son has sat tests for Bexley and Kent (results not yet in), as he is due to sit St Olaves test (which is suppose to be much harder) I wanted to add some complexity and vary the papers, so that he does not lose interest and is prepared for the tests. Thank you for your time, it has been very helpful, now I have to relay this to my son.


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PostPosted: Tue Oct 14, 2008 12:58 pm 
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Joined: Mon Jun 18, 2007 2:32 pm
Posts: 6966
Location: East Kent
sometimes I think that we adults can tend to read too much into a question. I have gone into algebra etc to solve a question where a pupil has got their quicker with a simple diagram


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PostPosted: Tue Oct 14, 2008 1:24 pm 
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Joined: Wed Mar 08, 2006 8:07 pm
Posts: 80
This question has so many tricky parts to it that it's so easy to interprete the question incorrectly.

Correcting the first part of my previous answer again

E to F to G - 82p
Ethel buys a ticket for both herself and her friend. Ethel is going all the way to G, whereas Dot is going to F.

Ethel is charged £1.17
What is the fare from F to G

E to F to G = 82p
E to F to G plus E to F = £1.17

E to F = £1.17 - 82p
E to F = 35p

F to G = 82p - 35p = 47p


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PostPosted: Tue Oct 14, 2008 2:04 pm 
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Joined: Tue Sep 16, 2008 7:30 pm
Posts: 139
angel,

Thanks. Yes, I had a look at the question paper myself and initially thought there may some clue to make out from the diagram. But there is nothing in their picture to draw any idea. Just a yellow bus inside a circle.

At least they had kept this as the last question in that paper thus not upsetting most children until the end. :)


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PostPosted: Tue Oct 14, 2008 2:17 pm 
I have done this paper with a few of my children. So far only one has managed this question and she was hard pushed to explain how she got the right answer. I think it is what I would call a scholarship question--it's there to real sort out the exceptionally bright from the very bright.
I don't think this relates to Sats levels. I think it is testing the ability of a child to keep several factors in their head while processing through logical steps (rather like a chess player would think) rather than expecting a solution through formal algebra.


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PostPosted: Tue Oct 14, 2008 7:54 pm 
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Joined: Tue Sep 16, 2008 7:30 pm
Posts: 139
An update:

As yoyo123 and fm mentioned earlier children sometimes throw simple ideas that adults often miss. So, in the evening I put this question (the last part) to DS without giving any clues. He came up with this idea which simplifies the logic I explained earlier.

To keep in the same tone, I use the same x, y, z here as well.

As per statements 1 and 2, both cover the entire distance (x+y+z) once.
Then both statements have y common as well.
That means z in the second statement must be 20p more than x.

In equation, it is: z = x + 20

If we apply the same logic to statements 2 and 3, we can see x is 10p more than y.

In equation, it is: x = y + 10

Now consider statements 1 and 3 with the same concept above and we can see z is 30p costlier than y.

In equation, it is: z = y + 30

All we then need to do is substitute the value of x and z ( in terms of y ) in statement 1.

It will be something like:

(y + 10) + y + (y + 30) + (y + 10) + y = 100
5y + 50 = 100
5y = 100 - 50
y = 50/5
y = 10

Hope this makes more sense. Sometimes we need to be reminded of the KISS mantra.

Gilly


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