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By: geethavasavi (offline)  Wednesday, February 10 2016 @ 02:52 AM CST (Read 1822 times)  
geethavasavi

Hi All,

Please help to solve this P6 maths question

Frederic had a candy bag which contained 35 chocolate bars and 50 sweets. Ryan had another candy bag which contained 45 chocolate bars and 10 sweets. After Frederic gave Ryan some chocolate bars and sweets, 40% of Frederic's candy bag contained chocolate bars and 30% of Ryan's candy bag contained sweets. How many chocolate bars and sweets did Frederic give to Ryan altogether?

Thanks & Regards,
geeth

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By: echeewh (offline)  Wednesday, February 10 2016 @ 01:17 PM CST  
echeewh

Hey Geeth,

Following is my worked solution :

<Before>

FC = 35; FS = 50; RC = 45; RS = 10

<After>

FC : FS
40 : 60
2 : 3

RC : RS
70 : 30
7 : 3

Since this is an <Internal Transfer> process, we can apply <Unchanged Total> concept, i.e. total number of chocolate bars (C) at the end <after> is the same as <before>. Likewise, the same applies to total number of sweets (S).

Total C <Before> = FC + RC = 35 + 45 = 80
Total S <Before> = FS + RS = 50 + 10 = 60


Let FC, FS <After> be p, and RC, RS <After> be u

2p + 7u = 80 -- (1)
3p + 3u = 60 -- (2)

(1)x3:
6p + 21u = 240 -- (3)

(2)x2:
6p + 6u = 120 -- (4)

(3)-(4):
21u - 6u --> 240 - 120
15u --> 120
1u --> 8

RC + RS <After> = 10u --> 10 × 8 = 80

RC + RS <Before> = 45 + 10 = 55

Hence, number of C and S given to Ryan (R) is as follows:

80 - 55 = 25

========

Trust this helps.

Do let me know again if this is different from your <anskey> or if there's further clarification.

Cheers,
Edward


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By: geethavasavi (offline)  Wednesday, February 10 2016 @ 09:11 PM CST  
geethavasavi

Hi Edward,

thank you for the reply. the solution is very correct for me but the answer key show "Federic gave 7 chocolate bars and 8 sweets to Ryan altogether" so the total ryan received is "15".
can you help to check if there can be any method to get this answer.

Regards,
Geeth

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By: echeewh (offline)  Thursday, February 11 2016 @ 11:48 AM CST  
echeewh

Hey Geeth,

Further to my worked solution of 25 total number of C and S given to Ryan, we have ...

Ryan <After>
Number of C = 7u --> 7 × 8 = 56
Number of S = 3u --> 3 × 8 = 24

Number of C given to Ryan (R) = 56 - 45 = 11
Number of S given to Ryan (R) = 24 - 10 = 14

In my worked solution,

% of RS = (24 / 80) × 100 = 30% ( this agrees with the question which stated 30% of Ryan's candy bag contained sweets. )

To see if the <anskey> of 7C and 8S given to Ryan is correct, we can always put these values back into the question as follows:

RC = 45 + 7 = 52
RS = 10 + 8 = 18

% of RS = (18 / 70) × 100 ~ 25.7%

However, the question stated 30% of Ryan's candy bag contained sweets.

Hence, I think there must be some mistake in the <Anskey> or somewhere.

Thank you.

Cheers,
Edward

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By: geethavasavi (offline)  Saturday, February 27 2016 @ 05:30 AM CST  
geethavasavi

Hi Edward,


i have an algebra question, not able to solve please help on this.

Sally and Molly shared the total cost of a present.Sally paid $n more than of the cost of the present. Molly paid $5. How much did the present cost?
Give your answer in terms of n in the simplest form.

ans: (50 + n) x 2 =$(100 + 2n)

Thanks & Regards,
Geetha

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By: echeewh (offline)  Monday, February 29 2016 @ 08:41 PM CST  
echeewh

Hey Geeth,

I think there is something wrong with the question itself. Can you please check it again ? Thanks .

Next time pls post a new question if there is one , instead of using an existing post.

Thank you.

Cheers,
Edward

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