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Friday, July 21 2017 @ 09:46 PM CDT
 Forum Index >  Test Paper Related >  Primary 6 Matters
 P6 Percentage
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By: Tping (offline)  Monday, April 08 2013 @ 03:46 AM CDT (Read 1303 times)  
Tping

Hi, pls help..

Q1
There were 225 apples and 253 oranges in Box A.
There were 260 apples and 212 oranges in Box B.
Mr Chia moved some apples and oranges from Box B to Box A.
In the end, 40% of the fruits in Box A and 70% of the fruits in Box B were apples.
How many fruits did Mr Chia move from Box B to Box A ?
(Ans: 122)

Q2
Breaker A and Beaker B contain some water.
If 46.5ml of water is drained out from Beaker A,
the volume of the water in Beaker A will be 60% that of the water in Beaker B.
If 35.2ml of water is drained out from Beaker B,
the volume of the water in Beaker B will be 85% that of the water in Beaker A.
What is the total volume of water in Beaker A and Beaker B ?
(Ans: 290.5ml)

Thank you.

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By: echeewh (offline)  Monday, April 08 2013 @ 05:03 AM CDT  
echeewh

Hello Tping,

Following pls find my worked solution:

Q1

Given that apples and oranges were moved from Box B to Box A, this infers an <Internal Transfer - Unchanged Total> concept. In other words, total number of Apples (A) in boxes A,B are the same before and after. Likewise, the total number of Oranges (Or).

<Before>
Total Apples (A) = 225 + 260 = 485
Total Oranges (Or) = 253 + 212 = 465

<process>
A and Or were moved from Box B to Box A.

<After>
Box A:
A : Or
40 : 60
2 : 3

Box B:
A : Or
70 : 30
7 : 3

Apply <Unchanged Total> concept here, we have ...

2p + 7u --> 485 -- (1)
3p + 3u --> 465 -- (2)

Eliminating (p) qty, we do the following:

(1)x3:
6p + 21u --> 1455 -- (3)

(2)x2:
6p + 6u --> 930 -- (4)

(3)-(4):
21u - 6u --> 1455 - 930
15u --> 525
1u --> 35

Box B (after):
7u (A) --> 7 × 35 = 245
3u (Or) --> 3 × 35 = 105

Number of fruits in Box B (before):
260 + 212 = 472

Number of fruits in Box B (after):
245 + 105 = 350

Hence, number of fruits moved from Box B:
472 - 350 = 122

=========

Q2

<after>
A : B
60 : 100
3 : 5

<process>
A: -46.5ml

<before>
A: 3p + 46.5
B: 5p

<process1>
B: -35.2ml

<after1>
A: 3p + 46.5
B: 5p - 35.2

Given that volume of water in Beaker B will be 85% that of the water in Beaker A, we have ...

A : B
100 : 85
20 : 17

From these, we have ...

3p + 46.5 = 20u
5p - 35.2 = 17u

Rearranging these,

20u - 3p --> 46.5 -- (1)
5p - 17u --> 35.2 -- (2)

Eliminating (u) qty, multiply (1) 17x; (2) 20x;

(1)x17:
340u - 51p --> 790.5 -- (3)

(2)x20:
100p - 340u --> 704 -- (4)

(3)+(4):
100p - 51p --> 790.5 + 704
49p --> 1494.5
1p --> 30.5

Total volume in A,B (before):
8p + 46.5 --> [8 × 30.5] + 46.5
= 290.5 ml

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

Trust you find these useful.

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

Cheers,
Edward

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Registered: 04/21/11
Posts: 623

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By: T0116060A (offline)  Saturday, August 10 2013 @ 07:55 AM CDT  
T0116060A

Quote by: echeewh

Hello Tping,

Following pls find my worked solution:

Q1

Given that apples and oranges were moved from Box B to Box A, this infers an <Internal Transfer - Unchanged Total> concept. In other words, total number of Apples (A) in boxes A,B are the same before and after. Likewise, the total number of Oranges (Or).

<Before>
Total Apples (A) = 225 + 260 = 485
Total Oranges (Or) = 253 + 212 = 465

<process>
A and Or were moved from Box B to Box A.

<After>
Box A:
A : Or
40 : 60
2 : 3

Box B:
A : Or
70 : 30
7 : 3

Apply <Unchanged Total> concept here, we have ...

2p + 7u --> 485 -- (1)
3p + 3u --> 465 -- (2)

Eliminating (p) qty, we do the following:

(1)x3:
6p + 21u --> 1455 -- (3)

(2)x2:
6p + 6u --> 930 -- (4)

(3)-(4):
21u - 6u --> 1455 - 930
15u --> 525
1u --> 35

Box B (after):
7u (A) --> 7 × 35 = 245
3u (Or) --> 3 × 35 = 105

Number of fruits in Box B (before):
260 + 212 = 472

Number of fruits in Box B (after):
245 + 105 = 350

Hence, number of fruits moved from Box B:
472 - 350 = 122

=========

Q2

<after>
A : B
60 : 100
3 : 5

<process>
A: -46.5ml

<before>
A: 3p + 46.5
B: 5p

<process1>
B: -35.2ml

<after1>
A: 3p + 46.5
B: 5p - 35.2

Given that volume of water in Beaker B will be 85% that of the water in Beaker A, we have ...

A : B
100 : 85
20 : 17

From these, we have ...

3p + 46.5 = 20u
5p - 35.2 = 17u

Rearranging these,

20u - 3p --> 46.5 -- (1)
5p - 17u --> 35.2 -- (2)

Eliminating (u) qty, multiply (1) 17x; (2) 20x;

(1)x17:
340u - 51p --> 790.5 -- (3)

(2)x20:
100p - 340u --> 704 -- (4)

(3)+(4):
100p - 51p --> 790.5 + 704
49p --> 1494.5
1p --> 30.5

Total volume in A,B (before):
8p + 46.5 --> [8 × 30.5] + 46.5
= 290.5 ml

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

Trust you find these useful.

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

Cheers,
Edward

Cry Wink javascript:emoticon('Mr. Green'Wink


I really loved a lot from these solutions! Thank you Edward! Wink Smile

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