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Tuesday, August 22 2017 @ 02:23 AM CDT
 Forum Index >  Test Paper Related >  Primary 6 Matters
 Help on 3 maths questions
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By: Kuro (offline)  Wednesday, August 14 2013 @ 12:47 PM CDT (Read 4231 times)  
Kuro

1. Refer to attachment.
Ans: 1365.34m2

2. At a swimming meet. School A had 18 more representatives than School B and 6 fewer representatives than School C.
The ratio of the number of boys to girls from the three schools was 1 : 3.
The ratios of the number of boys to the number of girls in Schools A, B and C were 1 : 3. 1 : 5 and 2 : 5 respectively.
How many representatives from the three schools were there in all?
Ans: 96

3. There are two bags of jelly beans labelled A and B. In bag A, there are 470 red jelly beans and 480 green jelly beans.
In Bag B, there are 350 red jelly beans and 200 green jelly beans. Some red and green jelly beans from bag B are transferred
to bag A. As a result, 60% of the jelly beans in Bag A and 20% of those in Bag B are red.
How many red and how many green jelly beans are transferred from Bag B to Bag A?
Ans: Red 310, Green 40

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By: echeewh (offline)  Wednesday, August 14 2013 @ 07:41 PM CDT  
echeewh

Hey <Kuro>.

For Q2, pls refer to solution provided with subject: <p6 2012 nanyang prelim>. This is currently found on pg 2 of this Forum last posted Aug 10, 2013. Thank you.


As for Q3, I remember you posted this question last month and I actually refer you to the concept and method of how to solve this problem in this thread/post listed "P6 Percentage" (date last posted: Apr 8, found in pg 13 as at Jul 8, 2013). Looks like you still dont get it???

Anyway, here's how it works.

Q3.

<Before>
<Bag A>
rA = 470
gA = 480

<Bag B>
rB = 350
gB = 200

<process>
r,g jelly beans from bag B are transferred to bag A. Hence there's a decrease in B, increase in A.

<After>

<bag A>
rA : gA
60 : 40
3 : 2

<bag B>
rB : gB
20 : 80
1 : 4

As this is an internal transfer process, total red (r) jelly beans in both bags A,B (rA + rB ) is the same before and after. Likewise, for the green (g) jelly beans.

Refer the <After> ratio in <bag A> as Part (p) qty and in <bag B> as Unit (u) qty.

3p + 1u = 470 + 350 = 820 -- (1)
2p + 4u = 480 + 200 = 680 -- (2)

Using <Simultaneous> method to solve the unknowns p, u , we eliminate (u) as follows:

Multiply (1) 4x.

(1)x4:
12p + 4u = 3280 -- (3)

(3)-(2):
12p - 2p = 3280 - 680
10p = 2600
1p --> 260

<After>

rA: 3p --> 3 × 260 = 780
gA: 2p --> 2 × 260 = 520

Red(r) jelly beans transferred:
780 - 470 = 310

Green(g) jelly beans transferred:
520 - 480 = 40

=========

Trust this helps.

Do let me know again if there's further clarification.

Cheers,
Edward

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By: echeewh (offline)  Thursday, August 15 2013 @ 05:47 PM CDT  
echeewh

hey <kuro>,

Kindly refer to my attached worked solution for Q1.

FYI, my answer is different from the <Answerkey> of 1365.34 cm2 as I reckon there is some mistake made there. ( with regards to the Unshaded region in Figure A}. Thank you.

Trust this helps.

Do let me know again if there's further clarifications.

Cheers,
Edward

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By: parent603 (offline)  Tuesday, March 25 2014 @ 02:02 AM CDT  
parent603

Hi Edward,

Pls send in answer to Q2 as I am not able to locate the solution. Thanks alot.


Quote:
For Q2, pls refer to solution provided with subject: <p6 2012 nanyang prelim>. This is currently found on pg 2 of this Forum last posted Aug 10, 2013. Thank you.

End of quote

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By: echeewh (offline)  Wednesday, March 26 2014 @ 01:58 AM CDT  
echeewh

Hey <parent603>,

Following is my worked solution for Q2.

A |----|<- 18 ->| --> 1u + 18
B |----| ******** --> 1u
C |---------------|< 6 >| --> 1u + 24

Given the following B:G ratios of each class A, B, C,

we have ...

A --> 4 pA (where p = parts)
B --> 6 pB
C --> 7 pC

Total (A + B + C) (from model) = 3u + 42 --> 18pB + 42 {since B --> 6pB or 1u }

From the model above, we have ...

A - B = 18
C - B = 24
C - A = 6

Transforming to the respective parts total, we have ...

4pA - 6pB = 18 -- (1)
7pC - 6pB = 24 -- (2)
7pC - 4pA = 6 -- (3)

Given that total B:G ratio is 1 : 3, we have ...
<B>
1pA + 1pB + 2pC = 1p -- (4)
<G>
3pA + 5pB + 5pC = 3p -- (5)

Eliminating (p), multiply (4) 3x.

(4)x3:
3pA + 3pB + 6pC = 3p -- (6)

Comparing (6),(5) we can equate them (since =) and we have ...

3pA + 3pB + 6pC = 3pA + 5pB + 5pC
3pB + 6pC = 5pB + 5pC
6pC - 5pC = 5pB - 3pB
1pC = 2pB -- (7)

Replace or substitute (7) in (2) as follows:

7pC - 6pB = 24
14pB - 6pB = 24
8pB --> 24
1pB --> 3

From model above, we have ...
B --> 6 pB (1u)
= 6 × 3 = 18

Total reps (A + B + C) = 3u + 42
--> (3 × 18) + 42
= 54 + 42
= 96

=========

Trust this helps.

Do let me know again if there's further clarification.

Cheers
Edward

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By: Kuro (offline)  Sunday, April 06 2014 @ 03:40 AM CDT  
Kuro

I'm unable to post new topic but need assistance on the following ration question. Thanks.

Class A and Class B have the same number of pupils. The ratio of the number of boys in Class A to the number of boys in Class B is 3:2. The ratio of the number of girls in Class A to the number of girls in Class B is 3:5. Find the ratio of the number of boys in Class A to the number of girls in Class B. Ans: 6:5

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By: echeewh (offline)  Wednesday, April 09 2014 @ 11:52 PM CDT  
echeewh

Hey <Kuro>

Heres my worked solution : ( my sincere apology for missing out ur query)

BA : BB
3 : 2

GA : GB
3 : 5

Given that A : B is 1 : 1, we have ...

BA + GA = BB + GB
3p + 3u = 2p + 5u
3p - 2p -> 5u - 3u
1p -> 2u

Hence ...

BA : BB
3 : 2

-> 6 : 4

BA : GB
6 : 5

=======

Trust this helps.

Do let me know again if there's further clarifications.

Cheers,
Edward

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By: w13158 (offline)  Tuesday, May 06 2014 @ 11:04 PM CDT  
w13158

Let me try to answer q2 as a layman.

2. At a swimming meet. School A had 18 more representatives than School B and 6 fewer representatives than School C.
The ratio of the number of boys to girls from the three schools was 1 : 3.
The ratios of the number of boys to the number of girls in Schools A, B and C were 1 : 3. 1 : 5 and 2 : 5 respectively.
How many representatives from the three schools were there in all?
Ans: 96

The initial part of the question tells you that the ratio of boys to girl is 1:3, so the hint is here that the answer will be a multiples of 4 (use the ratio of 1:3,that is 1+3=4, as there is no remainder). Students are taught how to solve problem using common multiples and factors, thus this problem will be using multiples to solve. For parents who does not know what is multiples of 4, it means the 4 time tables (4, 8, 12, 16, 20, etc)

School A ---> U+18
School B----> U
School C----> U+24 (I hope you know all these)
Thus, the 3U+42= Total students

The first thing is to find out what is U,

School A ratio of boys to girls is 1:3 (1+3=4, this tells you that it is a multiple of 4)
School B is U, so ignore this since B= 1U
School C the ratio of boys to girls is 2:5 (2+5=7, therefore it requires a multiple of 7)

Thus to solve the problem, you need to find the LCM(lowest common multiple) of 4 and 7
The lowest number to start is > 24, the formula is (U+18(24)=multiple of 4(7)>24)

For School A U+18, list the multiples of 4
U 10 14 18 22 26 30 34 38 42 46
+18 18 18 18 18 18 18 18 18 18 18
Multiple of 4 28 32 36 40 44 48 52 56 60 64

For School C U+24, list the multiples of 7

U 4 11 18 25 32 39 46 53 60 67
+24 24 24 24 24 24 24 24 24 24 24
Multiple of 7 28 35 42 49 56 63 70 67 84 91


Look at the U line above to scan for the common number, which is 18. Therefore U is 18.

Thus 3U + 42 = 96,
Check your answer to ensure that it can divide by 4.








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