
By: JONAANDR (offline) Tuesday, September 23 2014 @ 11:35 PM CDT (Read 1597 times)



JONAANDR 
Pls help, thanks! (4 questions for P5)
1. Alan had thrice as many stickers as Yvonne. After Alan and Yvonne had given 40% and 20% of their stickers to Wendy respectively, the number of stickers that Wendy had increased by 1/4. Wendy had 1680 stickers in the end. How many stickers did Yvonne have in the end?
2.The number of pupils in Grou0 A was 4 more than the pupils in Group B. Ager a test was given. 72 pupils graduated form Group A to Group B. The number of pupils in Group B became thrice as many as those in Group A. How many pupils were there in Group A at first?
3. Carl, Don and Ed sold donuts at a funfair. Carl sold 1/7 of the total number of donuts. Don sold 5/12 of the remaining donuts and Ed sold the rest. Carl sold 40 donuts less than Ed. How many donuts did the 3 boys sell?
4. Ahmad had 3 albums, P, Q and R. Album P had 4 times as many stamps as Q. He transferred 5 stamps from P to Q and 8 stamps from R to Q. In the end, all the albums had an equal number of stamps. How many stamps did Ahmad have?

Newbie
Registered: 02/27/14 Posts: 6





By: echeewh (offline) Wednesday, September 24 2014 @ 02:02 AM CDT



echeewh 
Hey <JONAANDR>
Appreciate if u can provide <answerkey> , where possible ( for quick verification purpose ). May be good to try and limit your questions to 2 per posting as others may be able to help you solve them as well at the same time. Thanks !!!
Following are my worked solutions for Q1, Q2:
Q1.
A : Y
3 : 1
<Given>
A:
(40/100) × 3u = (6/5)u
Y:
(20/100) × 1u = (1/5)u
Given W stickers increased by 1/4, we have ...
(5/4) W = 1680
(1/4) W = 1680 ÷ 5 = 336
i.e. this increase of 336 stickers comes from (A + Y)
A + Y = (6/5)u + (1/5)u = (7/5)u
(7/5)u > 336
(1/5)u > 336 ÷ 7 = 48
Y (end) = 1u  (1/5)u = (4/5)u
(4/5)u > 4 × 48 = 192
======
Q2. ( can use model method )
<Before>
A: 1u + 4
B: 1u
<After 72 pupils graduated from A to B>
<After>
A: 1u + 4  72 = 1u  68
B: 1u + 72
Given number of pupils in Group B became thrice as many as those in Group A, we have ...
1u + 72 = 3 × (1u  68)
1u + 72 = 3u  204
3u  1u > 72 + 204
2u > 276
1u > 276 ÷ 2 = 138
A (at first) = 1u + 4
1u + 4 > 138 + 4 = 142
========
Trust the above help.
Do let me know again if these are different from your <Answerkey> or if there's further clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 623





By: JONAANDR (offline) Wednesday, September 24 2014 @ 02:29 AM CDT



JONAANDR 
Quote by: echeewhHey <JONAANDR>
Appreciate if u can provide <answerkey> , where possible ( for quick verification purpose ). May be good to try and limit your questions to 2 per posting as others may be able to help you solve them as well at the same time. Thanks !!!
Following are my worked solutions for Q1, Q2:
Q1.
A : Y
3 : 1
<Given>
A:
(40/100) × 3u = (6/5)u
Y:
(20/100) × 1u = (1/5)u
Given W stickers increased by 1/4, we have ...
(5/4) W = 1680
(1/4) W = 1680 ÷ 5 = 336
i.e. this increase of 336 stickers comes from (A + Y)
A + Y = (6/5)u + (1/5)u = (7/5)u
(7/5)u > 336
(1/5)u > 336 ÷ 7 = 48
Y (end) = 1u  (1/5)u = (4/5)u
(4/5)u > 4 × 48 = 192
======
Q2. ( can use model method )
<Before>
A: 1u + 4
B: 1u
<After 72 pupils graduated from A to B>
<After>
A: 1u + 4  72 = 1u  68
B: 1u + 72
Given number of pupils in Group B became thrice as many as those in Group A, we have ...
1u + 72 = 3 × (1u  68)
1u + 72 = 3u  204
3u  1u > 72 + 204
2u > 276
1u > 276 ÷ 2 = 138
A (at first) = 1u + 4
1u + 4 > 138 + 4 = 142
========
Trust the above help.
Do let me know again if these are different from your <Answerkey> or if there's further clarification.
Cheers,
Edward
Q1 : 192
Q2: 142
Q3: 112
Q4: 57
From Singapore Chinese Girls' School 2013 P5 SA2

Newbie
Registered: 02/27/14 Posts: 6





By: echeewh (offline) Wednesday, September 24 2014 @ 07:31 AM CDT



echeewh 
Hey <JONAANDR>
Thanks for the <Answerkey> which help in the verification.
Following are my worked solutions for Q3, Q4 :
Q3.
( Can use model method )
<Sold>
C: (1/7)total
D: (5/12) × (6/7)total = (5/14)total
E: 1  (1/7)total  (5/14)total = (7/14)total
Given Carl (C) sold 40 donuts less than Ed (E), we have ...
E  C = 40
(7/14)total  (1/7)total = 40
(5/14)total = 40
(1/14)total = 40 ÷ 5 = 8
Total donuts sold by the 3 boys = (14/14)total
(14/14)total = 14 × 8 = 112
========
Q4.
<After>
P : Q : R
1 : 1 : 1
Apply <working backwards> method, we have ...
Q: 1u  8
R: 1u + 8
Q: 1u  8  5 = 1u  13
P: 1u + 5
Given Album P had 4 times as many stamps as Q, we have ...
P = 4Q
1u + 5 = 4 × (1u  13)
1u + 5 = 4u  52
4u  1u > 5 + 52
3u > 57
1u > (57/3)
Total stamps A had = 3u
3u > 3 × (57/3) = 57
========
Trust the above help.
Do let me know again if there's further clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 623





By: JONAANDR (offline) Wednesday, September 24 2014 @ 10:37 PM CDT



JONAANDR 
Quote by: echeewhHey <JONAANDR>
Thanks for the <Answerkey> which help in the verification.
Following are my worked solutions for Q3, Q4 :
Q3.
( Can use model method )
<Sold>
C: (1/7)total
D: (5/12) × (6/7)total = (5/14)total
E: 1  (1/7)total  (5/14)total = (7/14)total
Given Carl (C) sold 40 donuts less than Ed (E), we have ...
E  C = 40
(7/14)total  (1/7)total = 40
(5/14)total = 40
(1/14)total = 40 ÷ 5 = 8
Total donuts sold by the 3 boys = (14/14)total
(14/14)total = 14 × 8 = 112
========
Q4.
<After>
P : Q : R
1 : 1 : 1
Apply <working backwards> method, we have ...
Q: 1u  8
R: 1u + 8
Q: 1u  8  5 = 1u  13
P: 1u + 5
Given Album P had 4 times as many stamps as Q, we have ...
P = 4Q
1u + 5 = 4 × (1u  13)
1u + 5 = 4u  52
4u  1u > 5 + 52
3u > 57
1u > (57/3)
Total stamps A had = 3u
3u > 3 × (57/3) = 57
========
Trust the above help.
Do let me know again if there's further clarification.
Cheers,
Edward
Thanks a lot!

Newbie
Registered: 02/27/14 Posts: 6



