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 By: t.h.seah (offline)  Monday, September 22 2014 @ 09:53 PM CDT (Read 3320 times)
t.h.seah

1. If 25% of boys left the computer club, the ratio of the number of boys remain to the number of girls would be 5 :8. On children day celebration each girl received 4 sweet fewer than each boy. The
boys received 294 more sweet than the girl and a total of 1806 sweets were given out.
a. what was the ratio of boy to the number of girl in the computer club?
b. what was the total number of boys and girl in the club.

2. Arifin, Bobby and Calvin each had some stickers. If Ariffin gave 1/3 of his stickers to Bobby and Bobby gave 40 of his stickers to Calvin, the 3 boys would have the same amount of stickers.
During the school Value-In-Action event, Ariffin gave 20% of his stickers to a younger boy while Bobby gave 10% of his to another boy. As a result, Bobby had 12 stickers fewer than Calvin .
How many stickers had Ariffin than Calvin in the end?

Newbie

Registered: 07/28/11
Posts: 3

 By: echeewh (offline)  Tuesday, September 23 2014 @ 02:01 AM CDT
echeewh

Hey <t.h.seah>

Following is my worked solution to Q1 :

Q1.

<After>
B : G
5 : 8

If 25% (1/4) boys left, it will be left with 5p of boys in this <After> comparison.

(3/4) --> 5p
(4/4) --> 5p × (4/3) = (20/3)p

<Before>
B: (20/3)p
G: 8p

x3:

B: 20u
G: 24u

(a)
B : G
20 : 24
5 : 6

<On Childrens day - Sweets>
B |----<-4->| --> (1u + 4)
G |----| --> 1u

Given boys received 294 more sweets than the girls and a total of 1806 sweets were given out, we have ...

5p(1u + 4) - 6p(1u) = 294
5pu + 20p - 6pu = 294
20p - pu = 294
pu --> 20p - 294 --> (1)

5p(1u + 4) + 6p(1u) = 1806
5pu + 20p + 6pu = 1806
11pu + 20p = 1806 --> (2)

Sub (1) in (2), we have ...
11(20p - 294) + 20p = 1806
220p - 3234 + 20p = 1806
240p --> 1806 + 3234 = 5040
1p --> 21

(b)
Total number of Boys and Girls in club (B + G)
= 11p --> 11 × 21 = 231

========

Trust this helps.

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

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: echeewh (offline)  Tuesday, September 23 2014 @ 05:52 AM CDT
echeewh

Hey <t.h.seah>

One clarification is needed for Q2.

Appreciate if you can recheck this question as I think there could be some possible typo errors or missing info that may be necessary for solving the problem.

Thank you.

Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: t.h.seah (offline)  Tuesday, September 23 2014 @ 09:41 AM CDT
t.h.seah

thank you Edward for yr help!

Newbie

Registered: 07/28/11
Posts: 3

 By: echeewh (offline)  Saturday, November 15 2014 @ 12:07 AM CST
echeewh

Hey <t.h.seah>,

Following is my worked solution for Q2:

*** - for alignment purpose

<After>
A: 1p ***** B: 1p ***** C: 1p

<Work backwards>
B: +40
C: -40
A: (2/3) = 1p {If Ariffin gave 1/3 of his stickers to Bobby}
(1/3) = (1/2)p
(3/3) = (3/2)p
B: -(1/2)p + 40

<Before>
A: (3/2)p
B: 1p - (1/2)p + 40 = (1/2)p + 40
C: 1p - 40

<During event>
A: -20% (-1/5)
B: -10% (-1/10)

<After1>
A: (4/5) × (3/2)p = (6/5)p
B: (9/10) × [(1/2)p + 40] = (9/20)p + 36
C: 1p - 40

Given Bobby (B ) had 12 stickers fewer than Calvin (C), we have ...

B = C - 12
(9/20)p + 36 = 1p - 40 - 12
1p - (9/20)p = 36 + 40 + 12
(11/20)p --> 88
1p --> 88 ÷ (11/20) = 88 × (20/11) = 160

A - C (end) = (6/5)p - (1p - 40)
= (1/5)p + 40
--> [(1/5) × 160] + 40
--> 32 + 40 = 72

Ariffin had 72 more stickers than Calvin in the end.

======

Trust this helps.

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

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: echeewh (offline)  Saturday, November 15 2014 @ 12:13 AM CST
echeewh

Hey <t.h.seah>,

Following is my worked solution for Q2:

*** - for alignment purpose

<After>
A: 1p ***** B: 1p ***** C: 1p

<Work backwards>
B: +40
C: -40
A: (2/3) = 1p
(1/3) = (1/2)p
(3/3) = (3/2)p
B: -(1/2)p + 40

<Before>
A: (3/2)p
B: 1p - (1/2)p + 40 = (1/2)p + 40
C: 1p - 40

<During event>
A: -20% (-1/5)
B: -10% (-1/10)

<After1>
A: (4/5) × (3/2)p = (6/5)p
B: (9/10) × [(1/2)p + 40] = (9/20)p + 36
C: 1p - 40

Given Bobby ( had 12 stickers fewer than Calvin (C), we have ...

B = C - 12
(9/20)p + 36 = 1p - 40 - 12
1p - (9/20)p = 36 + 40 + 12
(11/20)p --> 88
1p --> 88 ÷ (11/20) = 88 × (20/11) = 160

A - C (end) = (6/5)p - (1p - 40)
= (1/5)p + 40
--> [(1/5) × 160] + 40
--> 32 + 40 = 72

Ariffin had 72 more stickers than Calvin in the end.

======

Trust this helps.

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

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

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