Q&A - PSLE Math

Academic support for Primary 6 and PSLE
Locked
alfretztay
KiasuGrandMaster
KiasuGrandMaster
Posts: 1199
Joined: Sun Sep 12,
Total Likes:5

Re: Q&A - PSLE Math

Post by alfretztay » Mon Jan 25, 2016 9:23 pm

AshleyL wrote:Hi! Can anyone help with this please?

Jim bought some chocolates and gave half of them to Ken.
Ken bought some sweets and gave half of them to Jim.
Jim ate 12 sweets and Ken ate 18 chocolates.
After that, the number of sweets and chocolates Jim had were in the ratio 1:7 and the number of sweets and chocolates Ken had were in the ratio 1:4.
How many sweets did Ken buy?

Thanks!!
Using units and parts,
Chocolates :
Jim : 10u - 5u = 5u
Ken : 5u

Sweets :
Ken : 10p - 5p = 5p
Jim : 5p

For Jim :
5p - 12/5u = 1/7
35p - 84 -------- 5u

For Ken :
5p/5u - 18 = 1/4
20p ------- 5u - 18

20p ------- 35p - 84 - 18
15p ------- 102
1p -------- 102/15 = 6.8
10p ------- 6.8 x 10 = 68

Ans : 68 sweets.

AshleyL
YellowBelt
YellowBelt
Posts: 11
Joined: Tue Jul 06,

Re: Q&A - PSLE Math

Post by AshleyL » Mon Jan 25, 2016 10:08 pm

alfretztay wrote:
AshleyL wrote:Hi! Can anyone help with this please?

Jim bought some chocolates and gave half of them to Ken.
Ken bought some sweets and gave half of them to Jim.
Jim ate 12 sweets and Ken ate 18 chocolates.
After that, the number of sweets and chocolates Jim had were in the ratio 1:7 and the number of sweets and chocolates Ken had were in the ratio 1:4.
How many sweets did Ken buy?

Thanks!!
Using units and parts,
Chocolates :
Jim : 10u - 5u = 5u
Ken : 5u

Sweets :
Ken : 10p - 5p = 5p
Jim : 5p

For Jim :
5p - 12/5u = 1/7
35p - 84 -------- 5u

For Ken :
5p/5u - 18 = 1/4
20p ------- 5u - 18

20p ------- 35p - 84 - 18
15p ------- 102
1p -------- 102/15 = 6.8
10p ------- 6.8 x 10 = 68

Ans : 68 sweets.
Thanks so much! I supposed the 10u and 10p used for chocolates and sweets are based on assumption? I can also use 20u and 20p?

alfretztay
KiasuGrandMaster
KiasuGrandMaster
Posts: 1199
Joined: Sun Sep 12,
Total Likes:5

Re: Q&A - PSLE Math

Post by alfretztay » Mon Jan 25, 2016 11:10 pm

AshleyL wrote:
alfretztay wrote:
AshleyL wrote:Hi! Can anyone help with this please?

Jim bought some chocolates and gave half of them to Ken.
Ken bought some sweets and gave half of them to Jim.
Jim ate 12 sweets and Ken ate 18 chocolates.
After that, the number of sweets and chocolates Jim had were in the ratio 1:7 and the number of sweets and chocolates Ken had were in the ratio 1:4.
How many sweets did Ken buy?

Thanks!!
Using units and parts,
Chocolates :
Jim : 10u - 5u = 5u
Ken : 5u

Sweets :
Ken : 10p - 5p = 5p
Jim : 5p

For Jim :
5p - 12/5u = 1/7
35p - 84 -------- 5u

For Ken :
5p/5u - 18 = 1/4
20p ------- 5u - 18

20p ------- 35p - 84 - 18
15p ------- 102
1p -------- 102/15 = 6.8
10p ------- 6.8 x 10 = 68

Ans : 68 sweets.
Thanks so much! I supposed the 10u and 10p used for chocolates and sweets are based on assumption? I can also use 20u and 20p?
You are most welcome. Yes, the units and parts are based on assumption; keeping the numbers as small as possible, I believe, would help reduce the chances of students making errors in their calculations.

AshleyL
YellowBelt
YellowBelt
Posts: 11
Joined: Tue Jul 06,

Re: Q&A - PSLE Math

Post by AshleyL » Mon Jan 25, 2016 11:17 pm

Using units and parts,
Chocolates :
Jim : 10u - 5u = 5u
Ken : 5u

Sweets :
Ken : 10p - 5p = 5p
Jim : 5p

For Jim :
5p - 12/5u = 1/7
35p - 84 -------- 5u

For Ken :
5p/5u - 18 = 1/4
20p ------- 5u - 18

20p ------- 35p - 84 - 18
15p ------- 102
1p -------- 102/15 = 6.8
10p ------- 6.8 x 10 = 68

Ans : 68 sweets.[/quote]

Thanks so much! I supposed the 10u and 10p used for chocolates and sweets are based on assumption? I can also use 20u and 20p?[/quote]

You are most welcome. Yes, the units and parts are based on assumption; keeping the numbers as small as possible, I believe, would help reduce the chances of students making errors in their calculations.[/quote]

Understand now. Thanks once again for your great help!

S-H
BrownBelt
BrownBelt
Posts: 564
Joined: Mon Feb 08,
Total Likes:1

Re: Q&A - PSLE Math

Post by S-H » Wed Jan 27, 2016 8:44 pm

Hi Can someone help me with the following question, thank you!

5 men are hired to complete a job. If one more man is hired, the job can be completed 8 days earlier. Assuming that all the men work at the same rate, how many more men should be hired so that the job can be completed 28 days earlier?


pirate
KiasuGrandMaster
KiasuGrandMaster
Posts: 4779
Joined: Mon Jul 02,
Total Likes:183

Re: Q&A - PSLE Math

Post by pirate » Wed Jan 27, 2016 10:23 pm

S-H wrote:Hi Can someone help me with the following question, thank you!

5 men are hired to complete a job. If one more man is hired, the job can be completed 8 days earlier. Assuming that all the men work at the same rate, how many more men should be hired so that the job can be completed 28 days earlier?
Let the amount of work 1 man does in 1 day be U.
Amount of work 5 men does in 1 day = 5U
Amount of work 6 men does in 1 day = 6U

Let the no. days required by 5 men to complete the job be D

Amount of work required to complete the job = 5U x D

5U x D = 6U x (D - 8)
5UD = 6UD - 48U
48U = UD
48 = D

No. of days required by 5 men to complete the job = 48 days
Amount of work required to complete the job = 48 x 5U = 240U

To complete the job 28 days earlier = completing the job in 48 - 28 = 20 days

240U / 20 = 12U

No. of men required to complete the job in 20 days is 12
Therefore 12 - 5 = 7 more men should be hired to complete the job 28 days earlier.

alfretztay
KiasuGrandMaster
KiasuGrandMaster
Posts: 1199
Joined: Sun Sep 12,
Total Likes:5

Re: Q&A - PSLE Math

Post by alfretztay » Wed Jan 27, 2016 10:41 pm

AshleyL wrote:Using units and parts,
Chocolates :
Jim : 10u - 5u = 5u
Ken : 5u

Sweets :
Ken : 10p - 5p = 5p
Jim : 5p

For Jim :
5p - 12/5u = 1/7
35p - 84 -------- 5u

For Ken :
5p/5u - 18 = 1/4
20p ------- 5u - 18

20p ------- 35p - 84 - 18
15p ------- 102
1p -------- 102/15 = 6.8
10p ------- 6.8 x 10 = 68

Ans : 68 sweets.
Thanks so much! I supposed the 10u and 10p used for chocolates and sweets are based on assumption? I can also use 20u and 20p?[/quote]

You are most welcome. Yes, the units and parts are based on assumption; keeping the numbers as small as possible, I believe, would help reduce the chances of students making errors in their calculations.[/quote]

Understand now. Thanks once again for your great help![/quote]

It is my pleasure to help.

S-H
BrownBelt
BrownBelt
Posts: 564
Joined: Mon Feb 08,
Total Likes:1

Re: Q&A - PSLE Math

Post by S-H » Wed Jan 27, 2016 10:49 pm

pirate wrote:
S-H wrote:Hi Can someone help me with the following question, thank you!

5 men are hired to complete a job. If one more man is hired, the job can be completed 8 days earlier. Assuming that all the men work at the same rate, how many more men should be hired so that the job can be completed 28 days earlier?
Let the amount of work 1 man does in 1 day be U.
Amount of work 5 men does in 1 day = 5U
Amount of work 6 men does in 1 day = 6U

Let the no. days required by 5 men to complete the job be D

Amount of work required to complete the job = 5U x D

5U x D = 6U x (D - 8)
5UD = 6UD - 48U
48U = UD
48 = D

No. of days required by 5 men to complete the job = 48 days
Amount of work required to complete the job = 48 x 5U = 240U

To complete the job 28 days earlier = completing the job in 48 - 28 = 20 days

240U / 20 = 12U

No. of men required to complete the job in 20 days is 12
Therefore 12 - 5 = 7 more men should be hired to complete the job 28 days earlier.

Hi pirate,

Thank you for your help! :thankyou:

StillThinking
BlueBelt
BlueBelt
Posts: 458
Joined: Sat Oct 03,
Total Likes:7

Re: Q&A - PSLE Math

Post by StillThinking » Thu Jan 28, 2016 8:52 am

:scratchhead: Does anyone realise the questions here mostly all can be solved by very simple algebra?

ozora
BlueBelt
BlueBelt
Posts: 326
Joined: Thu Feb 24,

Re: Q&A - PSLE Math

Post by ozora » Fri Jan 29, 2016 12:30 am

Could someone help in this question? Image
Thanks

Locked