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 2011 SA1 RGPS Maths Paper 2 Q10
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By: geniuskids (offline)  Sunday, May 06 2012 @ 10:20 PM CDT (Read 1171 times)  
geniuskids

Pls help with solutions!

(Q10) Mr Lim had some stickers and stamps for his class. The number of stickers was half as many as the number of stamps. After each pupil received 3 stickers and 8 stamps from Mr Lim. he had 39 stickers and 2 stamps left. What was the total number of stickers and stamps Mr Lim had at first?

Thank you

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By: geniuskids (offline)  Sunday, May 06 2012 @ 10:53 PM CDT  
geniuskids

Also pls help with Questions 16 -Pattern

Don't understand the answer given as well. ..

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By: echeewh (offline)  Monday, May 07 2012 @ 11:03 PM CDT  
echeewh

hey there,

Following pls find worked solution for your Q10.

Method 1 ( Simultaneous Eqn method )
----------------------------------------------------

<Before>
S : P
1 : 2

<Process>
3 S and 8 P were given out to each pupil; hence the proportion given out is 3 : 8

<After>
39 : 2

Using Simultaneous Eqn method, we have ...

1p - 3u --> 39 --- (1)
2p - 8u --> 2 --- (2)

As question is asking for total number of both at first, we shall eliminate (u).

(1) x 8: 8p - 24u --> 312 --- (3)
(2) x 3: 6p - 24u --> 6 --- (4)

Subtracting (4) from (3) will eliminate (u).

8p - 6p --> 312 - 6
2p --> 306
1p --> 153

Therefore, total number of stickers and stamps = 3p --> 3 x 153 = 459


Method 2 ( Cross-Multiply method )
-----------------------------------------------

Using the above <Before> <After> models shown above, we have <Before> - <After> = <Process> / <Given Away> since the <After> model is using whole numbers ( actual qty ). Rewriting this, we have ...

S : P
1 : 2

-39 -2

3 : 8

Applying Cross-Multiply technique,

( 1u - 39 ) / 3 = ( 2u - 2 ) / 8
8 x ( 1u - 39 ) = 3 x ( 2u - 2 )
8u - 312 = 6u - 6
8u - 6u --> 312 - 6
2u --> 306
1u --> 153

Hence, total number of stickers and stamps = 3u --> 3 x 153 = 459


*NB. If these 2 methods / techniques dont work for you , then one can also use the standard Before and After model concept ( Drawing ) to derive the solution.

======================================

Trust the above helps.

Anything pls do clarify.

Regret to say this, I can do one question at a time and there are numbers of questions to attend to Smile. Will do Q16 soon Smile.

Cheers,
Edward

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By: echeewh (offline)  Tuesday, May 08 2012 @ 09:31 PM CDT  
echeewh

Q16 ( Pattern question )

Refer to Pattern shown on Paper.

Following is my worked solution:

Where n = Pattern Number,
Shaded Squares = (n + 1) x (n + 2) + 4
Unshaded Squares = (n x 2) + (n + 2) = 3n + 2

(a) Shaded Squares in Pattern 55
= (56 x 57) + 4 = 3196

(b) Number of Squares in Pattern 250
= Shaded Squares + Unshaded Squares
= ((251 x 252) + 4) + ((3 x 250) + 2)
= 63256 + 752
= 64008

* Solution provided for (b) requires one to merge the above 2 equations together ( in which case I doubt a primary student can do ).
Merging the 2 gives (n + 2) x (n + 4), that is why the solution provided as ( 252 x 254 )

====================================================

Trust this helps.

Cheers,
Edward

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