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 Forum Index >  Test Paper Related >  Primary 6 Matters P6 Maths SA2 2011 Raffles Girls Q30, Q5, Q7, Q8, Q10 & 14a
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 By: larry.natnat (offline)  Wednesday, July 25 2012 @ 02:00 AM CDT (Read 1859 times)
larry.natnat

pls help to provide step by step workings for the attached questions?

Thanks

Newbie

Registered: 07/01/12
Posts: 2

 By: jo sarah (offline)  Saturday, July 28 2012 @ 10:25 AM CDT
jo sarah

Hi,

No. 30:
Take each solid marking as 1 unit.
The first 3 classes has a total of 10+2/3 units
the last 3 classes has a total of 13+2/3 units
the difference = 3 units, and this difference is the difference in totals
which is 3 x 90 = 270 (because each sub-group has 3 classes).

Since, class 4C has also 3 units.
So, class 4C collected \$270.

No. 5:
ED // AC, so angle DEC = 26 deg (alternate angles)
AB // CD, so angle DCA = 180 deg - angle CAB = 180 deg - 32 deg = 148 deg

Therefore, angle x = 148 deg - 26 deg - 30 deg = 92 deg

No. 7: [constant difference]
If Mary (M) is now 1 u years, and Jimmy is now 1 p years old.
3 years ago: 1 p - 3 = 4 x (1 u - 3)
so, 1 p = 4 u - 9
3 years later: 1 p + 3 = 2 x (1u + 3)
so, 1 p = 2 u + 3
therefore, 2 u = 12
and 1 u = 6

So, Mary is now 6 years old [only ]

No. 8:
BC // AE, so angle CAE = 180 deg - 68 deg - 37 deg (interior angles sum to 180 deg)
= 75 deg
AC // DE, so angle AED = 180 deg - 75 deg (interior angles) = 105 deg

Thus far, friend. It's late now, and it's the Lord's Day tomorrow. God willing, will return with the rest next week.
Cheerio.

Regular Member

Registered: 03/20/12
Posts: 111

 By: jo sarah (offline)  Monday, July 30 2012 @ 10:41 AM CDT
jo sarah

here to continue...

No. 10:
Tabulate as follows:
Pattern no. 1 2 3 4
no. of circles 1 3 5 7
no. of triangles 0 1 4 9

Note: difference in no. of circles from 1 pattern to another is 2, so think in multiples of 2:
and, no. of circles = 2 x pattern no. - 1

Also, no. of triangles are perfect squares = (pattern no. - 1) ^2

therefore,
(a) no. of circles in pattern 5 = (2 x 5) - 1 = 9
(b) no. of triangles in pattern 10 = (10 - 1) x (10 - 1) = 81
(c) no. of triangles in pattern 32 = 31 x 31 = 961 = no. of circles in pattern X
therefore, X = (961 + 1) / 2 = 962 / 2 = 481

No. 14 (a):
2 x circumference = 616 cm
therefore, circumference = 616 / 2 = 308 cm
[the portion marked 150 cm is of no bearing in the question.]

14(b): see attached

Regular Member

Registered: 03/20/12
Posts: 111

 By: larry.natnat (offline)  Monday, July 30 2012 @ 07:22 PM CDT
larry.natnat

Hello Jo Sarah,

Thank you

Newbie

Registered: 07/01/12
Posts: 2

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