Q&A - PSLE Math

[Daddy]

Hi meremortal,


http://i52.tinypic.com/1058s5x.jpg\">

Thanks

[Daddy]

:thankyou: Tianzhu..

[tianzhu]

Daddy:

Hi Tiazhu,

Is my answer corrected?

Jim and his sister shared some money. 1/3 of Jim's amount is 3/4 of his sister's share. If the difference in their share of money was $50, how much money did Jim have?

Jim [][][][][][][][][]
sister [][][][][][][][][][][][]
|------> $50
3u -> $50
1u -> 50/3
Jim -> 9u
= 9 x 50/3
= $150

Hi

1/3 of Jim’s share is equal to of 3/4 his sister.

This means Jim has a total of 9 units and his sister has a total of 4 units

Difference ----- 9 - 4 -------5

5 units -------50
1 unit ------10

Jim ------9 units -------90

Best wishes.

[tianzhu]

Daddy:

:thankyou: Tianzhu..

Hi

You're welcome.

Wishing you the very best for PSLE 2011.

Best wishes

[meremortal]

tianzhu:

Daddy:

Hi Tianzhu,


I still not understand.
Is it the chicken's egg did not do a compare, Stand alone model?
Duck eggs compared with quail eggs, so their model units are the same.
Am I corrected?

Thanks

In this question, there is a link or relationship between duck and quail eggs,. He sold 3 times as many quail eggs as duck eggs.
So, if duck eggs are represented by 2 boxes, then quail eggs are represented by 6 boxes.

The chicken eggs are of different sets of objects; hence the units are of different kind and measure.

Best wishes

To solve out the number of chickens originally, can we resort to algebra?

So from the models we can solve for the Quail and Ducks, and simple deduction will give us the number of Chicken eggs remaining: 205 (all eggs) - 45 (duck) - 90 (Quail) = 70 (Chicken eggs remaining)

So now we know 70 is 40% of all chicken eggs originally. Therefore, 70/0.4 = 175.

Is this allowed in P5/P6?

[small]

Daddy:

Hi Tiazhu,

Is my answer corrected?

Jim and his sister shared some money. 1/3 of Jim's amount is 3/4 of his sister's share. If the difference in their share of money was $50, how much money did Jim have?

Jim [][][][][][][][][]
sister [][][][][][][][][][][][]
|------> $50
3u -> $50
1u -> 50/3
Jim -> 9u
= 9 x 50/3
= $150

Hi,
The answer shouldbe Jimmy had $90

Jimmy - 9 Units
Sister - 4 Units

(9-4) units = $50
5 Units = $50
1U = $10

Jimmy = 9 x $10
= $90

[tianzhu]

meremortal:

Is this allowed in P5/P6?

Hi

If you were talking about the use of algebra in PSLE, here’s the reply from MOE.

http://www.moe.gov.sg/media/forum/2007/20070217.htm

Best wishes.

[ozora]

Any solution to these 2 questions from Rosyths paper?

q1. Abdul , Bern and Chin were all standing in a straight line waiting for a race to start. Chin was 300m ahead of Bern and Bern was 100m ahead of Abdul. At 9 am, they started the race.Abdul overtook Bern in 5 mins and in another 5 mins, Abdul overtook Chin. If Bern’s speed is 150m/min, at what time did Bern overtake Chin?

q2. Alan had some sweets and chocolate. If he ate 1 sweet, the ratio of the number of sweets to the number of chocolate left in the bag would be 2:3.
If he ate 1 chocolate, the ratio of the number of sweets to the number of chocolate left in the bag would be 7:10. What was the ratio of the number of sweets to the number of chocolate he had in the bag ?

[ASK TalkMaths]

ozora:

Any solution to these 2 questions from Rosyths paper?

q1. Abdul , Bern and Chin were all standing in a straight line waiting for a race to start. Chin was 300m ahead of Bern and Bern was 100m ahead of Abdul. At 9 am, they started the race.Abdul overtook Bern in 5 mins and in another 5 mins, Abdul overtook Chin. If Bern's speed is 150m/min, at what time did Bern overtake Chin?

q2. Alan had some sweets and chocolate. If he ate 1 sweet, the ratio of the number of sweets to the number of chocolate left in the bag would be 2:3.
If he ate 1 chocolate, the ratio of the number of sweets to the number of chocolate left in the bag would be 7:10. What was the ratio of the number of sweets to the number of chocolate he had in the bag ?

Hi, i have the solution for q2.

It is in the link https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B81le9JxERBGNjlhYjdhZjctYjIwYi00MTIwLWI5YjctMmE2MzUzNTk2NzQ1&hl=en_GB

[tianzhu]

ozora:

Any solution to these 2 questions from Rosyths paper?

q1. Abdul , Bern and Chin were all standing in a straight line waiting for a race to start. Chin was 300m ahead of Bern and Bern was 100m ahead of Abdul. At 9 am, they started the race.Abdul overtook Bern in 5 mins and in another 5 mins, Abdul overtook Chin. If Bern's speed is 150m/min, at what time did Bern overtake Chin?

Hi

As in any question on “Speed”, it’s always useful to start with a simple diagram showing the relationship between the runners.Once done, things fall into places.

There are two ways to go about it.

Use the familiar Speed triangle learned in school.

Abdul and Bern
In 5 mins, Bern travelled (150*5) or 750m
In 5 mins, Abdul travelled (100+750) or 850m
Hence, Abdul’s speed is 850/5 or 170 m/min

Abdul and Chin
In 10 mins, Abdul travelled (170*10) or 1700m
Chin travelled (1700 - 400) or 1300m in 10 mins.
Chin’s speed is 130m/min

Bern and Chin
Difference between Bern and Chin’s speed ------- 20 m/min.
Chin was 300m ahead of Bern
300/20 ------ 15 mins

Bern overtook Chin at 9.15am

The second way, use “catch up”

Abdul and Bern
Head start ------ 100m
Time to catch up -------5 mins
Speed difference ------100/5 ------20

Abdul’s speed -------- 150+20 ------170m/min

Abdul and Chin
Head start ------ 400m
Time to catch up -------10 mins
Speed difference ------400/10 ------40

Chin’s speed -------- 170 - 40 ------130m/min

Bern and Chin
Head start ------ 300m
Speed difference ------ 20
Time to catch up ------- 300/20 ----- 15 mins

Bern overtook Chin at 9.15am

Best wishes