
By: lisatang (offline) Wednesday, March 28 2012 @ 11:51 PM CDT (Read 3116 times)



lisatang 
Dear all pls assist in the following questions. Thanks Lisa
1)The ratio of the number of longans Eugwnw had to the number that Sumei had was 6:7. After Eugene ate 1/3 of his longans and Sumei ate half of hers,
Eugene was left with 12 more longans than Sumei. How many longans did both of them have altogether at first?
2) Pls refer to attachment.

Junior
Registered: 01/26/12 Posts: 19





By: echeewh (offline) Thursday, March 29 2012 @ 03:49 AM CDT



echeewh 
Hey Lisa,
Q1.
E : S
6 : 7
Since S ate 1/2 of hers, we can split the 7 parts of S into smaller units of 2 each, giving 14units (u). In other words, we take an equivalent ratio of the above by 2x.
E : S
12 : 14
E ate 1/3 of his 12 units while S ate 1/2 of her 14 units.
Therefore, the remaining units left by E and S are as follows:
E : S
8 : 7
Given that E was left with 12 more longans than S, we have ...
1 u > 12
Total (at first) = 26u > 26 x 12
= 312
===================================================================
Q2. (refer to attachment Q16)
Shaded Area = 1 u > 2 x 2 = 4 cm2
Area ABCD = 9 u > 9 x 4 = 36
Area PQRS = 12 + 36 = 48
(a) Length of PQRS = 48 / 6 = 8 cm
(b) Unshaded ABCD = 36  4 = 32
Unshaded PQRS = 48  4 = 44
Therefore, total unshaded = 32 + 44 = 76 cm2
===================================================================
Trust the above solutions help. Appreciate that answerkey can be provided so that its easier to confirm our answer.
Cheers.

Active Member
Registered: 04/21/11 Posts: 627





By: echeewh (offline) Thursday, March 29 2012 @ 03:50 AM CDT



echeewh 
Hey Lisa,
Q1.
E : S
6 : 7
Since S ate 1/2 of hers, we can split the 7 parts of S into smaller units of 2 each, giving 14units (u). In other words, we take an equivalent ratio of the above by 2x.
E : S
12 : 14
E ate 1/3 of his 12 units while S ate 1/2 of her 14 units.
Therefore, the remaining units left by E and S are as follows:
E : S
8 : 7
Given that E was left with 12 more longans than S, we have ...
1 u > 12
Total (at first) = 26u > 26 x 12
= 312

Active Member
Registered: 04/21/11 Posts: 627





By: lisatang (offline) Friday, March 30 2012 @ 05:13 AM CDT



lisatang 
Dear Echeewh
Thanks for your help. However, I dont quite get Q1. Can You explain again.
Lisa

Junior
Registered: 01/26/12 Posts: 19





By: echeewh (offline) Friday, March 30 2012 @ 07:17 PM CDT



echeewh 
Hey Lisa,
Ratio is 6 : 7 parts. Its the same as when you draw out the model. Since S ate 1/2 of hers, the 7 parts that S originally had is not divisible by 2 (without remainder). Hence, it wil be easier to subdivide or breakdown 1 part into 2 smaller units. So 7 parts = 7 x 2 = 14 units. Once we do this for S, we must do the same for E. Hence, the 6 parts of E = 6 x 2 = 12 units. So with the 14 units of S, it can nowbe divided equally by 2 (since S took 1/2). This has the same principles in PartWhole modelling.
Hope this helps to clarify.
Rgds,
Edward

Active Member
Registered: 04/21/11 Posts: 627





By: lisatang (offline) Tuesday, April 03 2012 @ 10:36 PM CDT



lisatang 
Thanks got a better idea now. Btw do u know where can I learn more about P5 maths.

Junior
Registered: 01/26/12 Posts: 19





By: echeewh (offline) Wednesday, April 04 2012 @ 12:33 AM CDT



echeewh 
Hi Lisa
The splitting of parts into equal units in Partwhole modelling is similar to adding 'unlike' fractions in which denominators of the fractions in the sum/expression are not similar. Hence, you need to find a common denominator / common multiple based on the given denominators. Technique and idea are very similar to splitting into equal units or subunits in math models.
Where to learn?? Well, there are some good reference books around and one of them being Conquer Problem Sums (aka Thinking Math @OnSponge). This book is available at www.onSponge.com as well as in certain schools' bookshops; but not available at Popular. Look out for authors like Ammiel Wan, Kelvin Ong, Andrew Er. Some have written books on concepts n methods (techniques) to help solving problem sums, etc and you can get them at Popular. Again, like any other thing else, it all comes with constant practice and challenging oneself to the most difficult / challenging questions.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 627



