Q&A - PSLE Math
-
:thankyou: Tianzhu..
-
Daddy:
HiHi 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
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. -
Daddy:
:thankyou: Tianzhu..
Hi
You're welcome.
Wishing you the very best for PSLE 2011.
Best wishes -
tianzhu:
To solve out the number of chickens originally, can we resort to algebra?
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.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
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
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? -
Daddy:
Hi,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
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 -
meremortal:
Hi
Is this allowed in P5/P6?
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. -
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 ? -
ozora:
Hi, i have the solution for q2.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 ?
It is in the link https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B81le9JxERBGNjlhYjdhZjctYjIwYi00MTIwLWI5YjctMmE2MzUzNTk2NzQ1&hl=en_GB -
ozora:
HiAny 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?
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 -
ozora:
HiAny 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 ?
Q2 is a variation of the \"total unchanged\" type of questions - just remove the 1 from the total and work the question with initial ratio of 2:3 and final ratio of 7:10. And then add \"1\" back to the total..
If 1 sweet is removed,
Sweets : Chocolate = 2:3 (initial)
If 1 chocolate eaten (assume that now the 1 chocolate is replaced by 1 sweet), Sweets : Chocolate = 7:10 (final)
Make the initial total and final total the same...
Initially, Sweets : Chocolate = 2:3 = 34u : 51u
In the end, Sweets : Chocolate : 7 : 10 = 35u : 50u
So, 1u =1
Number of sweets = 35
Number of chocolates = 51
Ratio of sweets : chocolates = 35 : 51
cheers.
Hello! It looks like you're interested in this conversation, but you don't have an account yet.
Getting fed up of having to scroll through the same posts each visit? When you register for an account, you'll always come back to exactly where you were before, and choose to be notified of new replies (either via email, or push notification). You'll also be able to save bookmarks and upvote posts to show your appreciation to other community members.
With your input, this post could be even better 💗
Register Login