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 By: JONAANDR (offline)  Tuesday, September 23 2014 @ 11:35 PM CDT (Read 1737 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: 627

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

Quote by: 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

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: 627

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

Quote by: 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

Thanks a lot!

Newbie

Registered: 02/27/14
Posts: 6

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