Q&A - PSLE Math

[tianzhu]

Daddy:

Thanks MathIzzzfun for ur explaination.

Thanks tianzhu for ur help.

:thankyou:

Hi

You’re welcome.

Best wishes

[verykiasu2010]

tianzhu:

michyms:

In today's Forum, someone wrote in about a maths question in a prelim paper: Three halls contained 9,876 chairs altogether. One-fifth of the chairs were transferred from the first hall to the second hall. Then, one-third of the chairs were transferred from the second hall to the third hall and the number of chairs in the third hall doubled. In the end, the number of chairs in the three halls became the same. How many chairs were in the second hall at first?


Can anyone enlighten how this is done?

Hi

The total number of chairs in the three halls remains the same. Use a strategy called “Working Backwards”. It's helpful to start with a simple diagram.

In the end, the number of chairs in the three halls became the same..

9876/3 -------- 3292

Second hall
2 units ------ 3292
1 unit ------ 1646
3 units –----- 4938

First hall
4 parts ------ 3292
1 part ------ 823

Number of chairs in second hall at first ------- 4938 - 823 ------ 4115

An alternative way, use MD.

Best wishes

http://farm7.static.flickr.com/6064/6123038262_409ab05ff4_z.jpg\">

well done!

I also tested out my DS just now. Still able to solve it within regulation time. phew !

[MathIzzzFun]

Cheerfuldad:

Hi MathIzzzFun,


Thank you for your help!

Cheers!

MathIzzzFun:

[quote=\"Cheerfuldad\"]Hi all,

Please help on the following question:

Jason, Edward and Sam had a total of $837. Jason had the least amount of money. The ratio of Edward's money to Sam's money was 4:3 at first. Jason and Edward each spent 1/3 of their money. Given that the three boys had $648 left, how much did Jason have at first?

TIA

Hi

$837 - $648 = $189
So 1/3 of Jason & Edward's money = $189
Jason and Edward had a total of 3 x $ 189 = $ 567
Amount that Sam had = $837 - $ 567 = $ 270
Total amount Edward and Sam had = 7/3 x $ 270 = $630

Amount Jason had at first = $ 837 - $ 630 = $207

cheers.

[/quote]u r welcome

cheers.

[MathIzzzFun]

PiggyLalala:

MathIzzzFun:

[quote=\"nanosphere\"]pls help me with this question :?:


Andy, billy and charlie cycled at the same time from point X towards another cyclist ahead of them. Andy, billy and charlie took 6 minutes,10 minutes and 12 minutes respectively to overtake the cyclist. Andy's speed was 25km/h and billy's speed was 21km/h.
(a) How far was billy from point X when he overtook the cyclist?

(b) What was Charlie's cycling speed?

:thankyou:

Hi

this one would be quite easily solved using \"area\" method.. here's the usual method ..

a) Billy's distance from X = 10min x 21km/h = 3.5km

When Andy caught up with the cyclist, distance travelled by Andy = 6min x 25km/h = 2.5km
At this time, Billy travelled 6min x 21 km/h = 2.1km.
So, 6min after they started cycling, Billy is 2.5km - 2.1km = 0.4km behind the cyclist.

Billy caught up with the cyclist after cycling for 10 min ie 4 min after Andy caught up with the cyclist.
0.4km/4min= 6km/h so Billy is cycling 6km/h faster than the cyclist ie
cyclist's speed = 21km/h - 6km/h =15km/h

Billy took 10min to catch up with the cyclist, 10min x 6km/h = 1km ie the cyclist was 1km ahead when Andy, Billy and Charlie started cycling.

Charlie took 12min to make up 1km, 1km/12min = 5km/h ie Charlie was cycling 5km/h faster than the cyclist.

Charlie's speed = 15km/h + 5km/h = 20 km/h

cheers.

Hi MathIzzzFun,
I am interested in the easy 'area' method to solve this question. Would you mind posting the solution here too? Thank you very much.[/quote]Hi

there are different approaches to solve this particular problem .. tianzhu has provided another approach by examing the cyclist journey.

personally, i prefer the area method for this because it allows one to \"track\" each individual in the problem sum and it also allows one to extract other information from the diagram if there is a need to. The area method is also handy in tackling \"average\" problem sums.

http://i53.tinypic.com/2w6gb2h.jpg\">

cheers.

[PiggyLalala]

MathIzzzFun:

MathIzzzFun:

[quote=\"nanosphere\"]pls help me with this question :?:


Andy, billy and charlie cycled at the same time from point X towards another cyclist ahead of them. Andy, billy and charlie took 6 minutes,10 minutes and 12 minutes respectively to overtake the cyclist. Andy's speed was 25km/h and billy's speed was 21km/h.
(a) How far was billy from point X when he overtook the cyclist?

(b) What was Charlie's cycling speed?

:thankyou:

Hi

this one would be quite easily solved using \"area\" method.. here's the usual method ..

a) Billy's distance from X = 10min x 21km/h = 3.5km

When Andy caught up with the cyclist, distance travelled by Andy = 6min x 25km/h = 2.5km
At this time, Billy travelled 6min x 21 km/h = 2.1km.
So, 6min after they started cycling, Billy is 2.5km - 2.1km = 0.4km behind the cyclist.

Billy caught up with the cyclist after cycling for 10 min ie 4 min after Andy caught up with the cyclist.
0.4km/4min= 6km/h so Billy is cycling 6km/h faster than the cyclist ie
cyclist's speed = 21km/h - 6km/h =15km/h

Billy took 10min to catch up with the cyclist, 10min x 6km/h = 1km ie the cyclist was 1km ahead when Andy, Billy and Charlie started cycling.

Charlie took 12min to make up 1km, 1km/12min = 5km/h ie Charlie was cycling 5km/h faster than the cyclist.

Charlie's speed = 15km/h + 5km/h = 20 km/h

cheers.

Hi

there are different approaches to solve this particular problem .. tianzhu has provided another approach by examing the cyclist journey.

personally, i prefer the area method for this because it allows one to \"track\" each individual in the problem sum and it also allows one to extract other information from the diagram if there is a need to. The area method is also handy in tackling \"average\" problem sums.

http://i53.tinypic.com/33nu539.jpg\">

cheers.[/quote] :thankyou: very much. Will look through this area method.

As a kiasuparent, I am interested in all different approaches to solve a sum. From there, I can then pick the approach that best suit my son.

[michyms]

Thank you tianzhu!

[MathIzzzFun]

PiggyLalala:

MathIzzzFun:

[quote=\"MathIzzzFun\"]

Hi

this one would be quite easily solved using \"area\" method.. here's the usual method ..

a) Billy's distance from X = 10min x 21km/h = 3.5km

When Andy caught up with the cyclist, distance travelled by Andy = 6min x 25km/h = 2.5km
At this time, Billy travelled 6min x 21 km/h = 2.1km.
So, 6min after they started cycling, Billy is 2.5km - 2.1km = 0.4km behind the cyclist.

Billy caught up with the cyclist after cycling for 10 min ie 4 min after Andy caught up with the cyclist.
0.4km/4min= 6km/h so Billy is cycling 6km/h faster than the cyclist ie
cyclist's speed = 21km/h - 6km/h =15km/h

Billy took 10min to catch up with the cyclist, 10min x 6km/h = 1km ie the cyclist was 1km ahead when Andy, Billy and Charlie started cycling.

Charlie took 12min to make up 1km, 1km/12min = 5km/h ie Charlie was cycling 5km/h faster than the cyclist.

Charlie's speed = 15km/h + 5km/h = 20 km/h

cheers.

Hi

there are different approaches to solve this particular problem .. tianzhu has provided another approach by examing the cyclist journey.

personally, i prefer the area method for this because it allows one to \"track\" each individual in the problem sum and it also allows one to extract other information from the diagram if there is a need to. The area method is also handy in tackling \"average\" problem sums.

http://i53.tinypic.com/33nu539.jpg\">

cheers.

:thankyou: very much. Will look through this area method.

As a kiasuparent, I am interested in all different approaches to solve a sum. From there, I can then pick the approach that best suit my son. :)[/quote]
u r welcome.. I just noted a typo error in the working.. I have amended it.

cheers.

[Daddy]

Hi all,


Need help on these questions. Thanks…

1. The average number of books owned by a group of boys and girls in Pr. 6J is 35.
The average number of books owned by the boys is 20 while the average number of books owned by the girls is 60.
If there are 40 pupils in the class, how many boys are there?

2. Mary and john bought some clips and ribbons.
John bought 40% of the total number of clips and ribbons.
Altogether, they bought 30 more ribbons than clips.
Mary bought 2/3 of the ribbons and 50% of the clips.
What was the total number of clips and ribbons that John bought?

Thanks

[nanosphere]

:thankyou: for all yr help :grphug:

[MathIzzzFun]

Daddy:

Hi all,


Need help on these questions. Thanks..

1. The average number of books owned by a group of boys and girls in Pr. 6J is 35.
The average number of books owned by the boys is 20 while the average number of books owned by the girls is 60.
If there are 40 pupils in the class, how many boys are there?

2. Mary and john bought some clips and ribbons.
John bought 40% of the total number of clips and ribbons.
Altogether, they bought 30 more ribbons than clips.
Mary bought 2/3 of the ribbons and 50% of the clips.
What was the total number of clips and ribbons that John bought?

Thanks

Hi

Q1 is a typical \"chicken & goat\" problem in disguise.
total 40 x 35 = 1400 books.
Total 40 pupils - each boy owned 20 books, each girl own 60 books.

Assume all girls --> 60 x 40 = 2400 books, ie 1000 extra books.
60 - 20 = 40, each boy own 40 less books.
1000/40 = 25 boys
So, 15 girls & 25 boys.

Q2. 90 ribbons & 60 clips, John = 30 ribbon + 30 clips, Mary = 60 ribbon + 30 clips... I'll post the solution later.

cheers.