Q&A - PSLE Math

[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.

[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

Q2.

http://i56.tinypic.com/2mgwklt.jpg\">

cheers.

[ozora]

i need help on this question.

thanks http://i53.tinypic.com/al3jb4.jpg\">

[ozora]

may i know how to i trace back a solution to the following solution?

Jane and Grace shared a sum of money. If jane gave 1/4 of her share to Grace, Grace would have $400 more than Jane. If Jane gave 1/5 of her share to Grace,
Grace would have $154 more than Jane. How much money did they have altogther?

[tianzhu]

ozora:

i need help on this question.

thanks http://i53.tinypic.com/al3jb4.jpg\">

Hi

Good Morning.

There are a few ways to solve it.You may use equivalent ratios by assigning units, UM or MD.

The numbers in your question are pretty small. Actually in this case, it’s not too tedious to assign units and work systematically to the answer.

Assign 1 unit to the shaded part. Assign 1 unit to the unshaded area of the small square.

Hence, area of big square is 4 units and area of unshaded area of big square is 3 units.This satisfies all the conditions in your question.

1 unit ----- 20
2 units -----40

Area of small square ----- 40 sqcm

Next, I’ll share the UM method.

This is a question on “Common Difference” or “Unchanged Difference”. I believe you’ve used them in answering questions on the topic “Ratio”.

The difference in areas between the two squares and the difference in unshaded areas between the two squares are equal.

Area of small square ----- 1 unit
Area of big square ----- 2 units
Difference -------1 unit

Area of unshaded part of small square ------- 1 unit
Area of unshaded part of big square ------- 3 unit
Difference -------2 units

Make the difference the same

Area of small square ----- 1 unit ----------2units
Area of big square ----- 2 units --------4 units
Difference -------2 unit

1 unit ------20
2 units -------40

Area of small square ----- 40 sqcm

Best wishes