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 Posted: Mon Aug 06, 2012 3:14 pm

Joined: Mon Jul 30, 2012 5:04 am
Posts: 14
HELLO after invalueable help from you guys earlier now im stuck on these questions on GL paper 7 nv
GL PAPER7 NV sec 4 q#12 sec6 q#11

TEST 11A SEC 3 Q#12 SEC5 Q#6,8

TEST C SEC5 Q#6,7

TEST D SEC 3 Q#8

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 Posted: Mon Aug 06, 2012 10:11 pm

Joined: Thu Nov 12, 2009 2:36 pm
Posts: 192
Location: East Kent
Pack 2, Test 7

Section 4, Q12 - each of the triangle is symettrical on all 3 sides which rules out all but D

11A - there is a whole 'sticky' thread at the top of this forum with lots of 11A answers on, your answer may be in there.

11C - section 5

Q6 - the shape in each row gets larger as it goes along so answer needs to have a big circle (rules out b & e), the last shape on the row has the same 'markings' as the first two shapes so only leaves d

Q7 - on each row the location of the shading/dark area alternates, so top row it is top, bottom, top. Bottom row is bottom, top, bottom. So the ones on the middle row that you have are left, right so the missing one must be a 'right' - rules out B & C. The pattern of shading is on a diagonal pattern, so the missing shape has to have diaganol shading so that rules out E. Each verticle row is made up a small, medium and large shape - the row with the missing shape already has a small and medium size shape so that only leaves D as the possible answer.

Test D, Section 3 Q8

The black dot moves around 3 positions each time (so from the first square it moves from west, through south west, south and lands on south east), so the missing square moves 3 positions again so needs to move from SE through East, North East and lands on North. This only leaves E as the answer. Also the after the black dot has landed in a circle, on the next in pattern, the circle the black dot had been in moves to the middle of the pattern.

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 Posted: Sat Aug 11, 2012 9:25 am

Joined: Mon Jul 30, 2012 5:04 am
Posts: 14

only couple of more weeks to go now before real test .

ill be glad when its all over.

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 Posted: Sat Aug 11, 2012 10:15 am

Joined: Fri Nov 17, 2006 9:54 pm
Posts: 1564
Location: caversham
dottie1974 wrote:
Pack 2, Test 7

Section 4, Q12 - each of the triangle is symettrical on all 3 sides which rules out all but D

By coincidence we struggled with this question last night and so I was going to ask for help.

Even with the help I still can't see it, I wonder if I have a version with a printing error? Or it could be my eyes!

Paper says on back Code MCNR7 2(9.11)

Also struggling with the previous question Q11, too much going on looks like an optical illusion.

Edit: As you've got paper 7 out, section 6 Q11 the answer is (e)? I can't explain it to DD?

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 Posted: Sat Aug 11, 2012 11:14 am

Joined: Thu Nov 12, 2009 2:36 pm
Posts: 192
Location: East Kent
Steve - I have the same pack as you!

pack 2 Test 7, Section 4 question 12 - it is tricky trying to explain but I will give it another go.

If you hold the question paper in front of you the right way up and look at the positioning of the inside shape, on the examples given, if you rotate the paper so that one of the other two sides of the triangle is at the bottom running horizontally, the inside shape stays the same. But if you do the same with:

a) Holding the question paper properly, the inside square's edge runs parallell to the bottom of the triange, if you then rotate the page clockwise so the next edge of triangle is running horizontally, the square inside no longer runs parallell to the edge of the square so it can't be this one.

b) Same as above - in the first position there are two points of the star at the bottom but after rotating the paper to the next position the star only has one point at the bottom

c) Same as above - in the first position the oval inside is vertical but after rotating the paper it becomes horizontal.

d) the hexagon stays has an edge parallel to the triangle whichever way the paper is rotated

e) in the first position the pentagons edge is horizontal to the bottom of the triange, after rotating the page it doesn't.

I hope this explains it a bit better? Its all to do with rotational symettry I guess? It is hard to try and explain it.

I can't work out the other one though (Paper 7, section 6, q11) - sorry

Two years ago when dd1 did the 11+ I couldn't get my head around NVR at all, it does appear that I am improving with practice but I am still a bit pants at it lol

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 Posted: Sat Aug 11, 2012 12:39 pm

Joined: Fri Nov 17, 2006 9:54 pm
Posts: 1564
Location: caversham
Many thanks I can see that now. It's a kind of symmetry?

GL2 does contain a number of different (to GL1) question types, all good practice, sometimes the easiest (with hindsight) ones are the hardest as instinctively you start looking for a tricky solution.........

DS1 (awaiting GCSEs) offered to help, he used to be very good at NVR, but he struggled. Suggests you do need to have your NVR brain in gear and running.

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 Posted: Thu Sep 13, 2012 12:12 am

Joined: Thu Sep 13, 2012 12:05 am
Posts: 29
"I can't work out the other one though (Paper 7, section 6, q11) - sorry"

im new so please be gentle

i think i have it and im going to give it a go????

pp7 sec 6 qu11
when there is a bold line, then the dot is in the middle of the shape
and when the shape has a dotted line then the dot is outside the shape
therefore in (e) the shape has a bold line and the dot is outside it therefore it is the odd one out

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