Q&A - PSLE Math

[tianzhu]

Beikiasu:

Why you up the ratio to \"40\" ?

Hi

You can use any equivalent ratios and still arrive at the same answer.

I use 40 so as to avoid working with decimals.

Best wishes

[Beikiasu]

Oldschool :thankyou:


How you get the \"105%\"?

[tianzhu]

Easy-going:

Hi

Can anyone please help me with these problem sums taken from My Pals are Here Textbook 6A Pg 100
Qn 29
Zac and Sally share some stickers. If Zac gives Sally 1/2 of his stickers, Sally will have 48 more stickers than Zac. If Zac gives 1/4 of his share to Sally, Sally will have 28 more stickers. How many stickers does Zac have at first?

Hi

You may use MD or work with units.

Scenario 1

If Zac gives Sally 1/2 of his stickers, Sally will have 48 more stickers than Zac.

At first

Zac ----- 4 units
Sally ------ 48

In the end

Zac ----- 2 units
Sally ------ 2 units + 48

Scenario 2

If Zac gives 1/4 of his share to Sally, Sally will have 28 more stickers.

At first
Zac ----- 4 units
Sally ------ 48

In the end
Zac ----- 3 units
Sally ------ 1 unit + 48

1 unit + 48 ----- 3units +28

2 units ------ 20
1 unit ------ 10

Zac@first ------ 40

Best wishes

[Oldschool]

Beikiasu:

Oldschool :thankyou:


How you get the \"105%\"?

Hi,

25% dropped from April to May

25% of April
= 1/4 of 140%
= 35%

So, remained is 140%-35% = 105%

Or in other words, 3/4 of April remained
=3/4 of 140%
=105%

Regards

[Beikiasu]

tianzhu, Oldschool :thankyou:

[altp10]

PapayaDad:

altp10:

Hi. I need help with the following Speed question.


Joel & Cliff ran a 100m race in which Joel crossed the finishing line just as Cliff reached the 90m mark.
a) Since Joel won the first race by 10m, he decided to start the second race 10m behind the starting line. Both boys ran at the same speed as in the first race. Who won the second race?
b) In the third race, Cliff started 10m ahead of the starting line. Again assuming that both boys ran at the same speed as in the first race, who was the winner of this race?

Thank you in advance.

I assume Joel finish the race in 10 seconds. That means his speed is 10 meters per second. At that time (10 second) Cliff only reach 90m ; so Cliff speed is 9 meters per second.

Joel start 10 meters behind, meaning he need to cover 110 meters, his speend is 10 meters per second, so he took 11 seconds.

Cliff speed is 9 meters per second, at 11 second he is at 99meter only, still behind Joel.

Cliff then start 10m in front, need to cover 90 meters, his speed is 9 meter per second, so complete in 10 seconds. Joel also took 10 second to reach. so they reach together.

Hi PapayaDad,

Thank u for the advise. However, in my solution in Pg 885, I seemed to have derived a different answer.. Could u kindly advise me on my solution please?

Thanks alot.

[tianzhu]

altp10:

altp10:

Hi. I think I managed to solve this myself. However would still like to have some advice to confirm my answer.


According to first two sentences, J wins by 10m.
So breaking the 100m into 10 segments of 10m each, J would be faster than C by 1m in every segment. This also mean J runs 11m while C runs 10m.

a) Since J runs 11m (J adds 11) & C runs 10m (C adds 10):
J C
-10 0
1 10
12 20
23 30
34 40
45 50
56 60
67 70
78 80
89 90
100 100
Answer to a) Tie between the 2 boys.

b) Since J runs 11m (J adds 11) & C runs 10m (C adds 10):
J C
0 10
11 20
22 30
33 40
44 50
55 60
66 70
77 80
88 90
99 100
Answer to b) C won.

Please do advise if I'm correct.
Thank you.

(Sorry I can't load any diagrams as I'm typing these using iPhone)

Hi

You may solve this question by listing, but a more efficient way is to make use of speed ratio.

Joel & Cliff ran a 100m race in which Joel crossed the finishing line just as Cliff reached the 90m mark.

This means for every 10m covered by Joel, Cliff ran 9m.

(a)Joel is the winner

Joel is required to cover 110m.He covered 10m at one interval; hence he needed 11 intervals to finish 110m. At 11 intervals, Cliff covered 99m, therefore short of the finishing point of 100m.

(b)They finish at the same time.

Joel needed to run 100m and Cliff was required to run 90m.At an interval of 10m, Joel completed the race (100m) at 10 intervals. At 10 intervals, Cliff completed 90m, hence arriving at the same time as Joel.

Best wishes

[altp10]

tianzhu:

altp10:

[quote=\"altp10\"]

Hi. I think I managed to solve this myself. However would still like to have some advice to confirm my answer.


According to first two sentences, J wins by 10m.
So breaking the 100m into 10 segments of 10m each, J would be faster than C by 1m in every segment. This also mean J runs 11m while C runs 10m.

a) Since J runs 11m (J adds 11) & C runs 10m (C adds 10):
J C
-10 0
1 10
12 20
23 30
34 40
45 50
56 60
67 70
78 80
89 90
100 100
Answer to a) Tie between the 2 boys.

b) Since J runs 11m (J adds 11) & C runs 10m (C adds 10):
J C
0 10
11 20
22 30
33 40
44 50
55 60
66 70
77 80
88 90
99 100
Answer to b) C won.

Please do advise if I'm correct.
Thank you.

(Sorry I can't load any diagrams as I'm typing these using iPhone)

Hi

You may solve this question by listing, but a more efficient way is to make use of speed ratio.

Joel & Cliff ran a 100m race in which Joel crossed the finishing line just as Cliff reached the 90m mark.

This means for every 10m covered by Joel, Cliff ran 9m.

(a)Joel is the winner

Joel is required to cover 110m.He covered 10m at one interval; hence he needed 11 intervals to finish 110m. At 11 intervals, Cliff covered 99m, therefore short of the finishing point of 100m.

(b)They finish at the same time.

Joel needed to run 100m and Cliff was required to run 90m.At an interval of 10m, Joel completed the race (100m) at 10 intervals. At 10 intervals, Cliff completed 90m, hence arriving at the same time as Joel.

Best wishes[/quote]
Hi Tianzhu,

Alright now I get it..
I had taken J=11m & C=10m
It should be J=10m & C=9m instead.

Thanks Tianzhu & PapayaDad.
Regards

[cimman]

Beikiasu:

3) The ratio of Jamal's mass to Katty's mass is 4:5.Jamal's mass is increased by 40% and Kathy's mass is decreased. What percentage of kathy's mass must be decreased so that thier total mass remains the same?TaoNan, 2011, Question 12, 4 mark, Ans:32%

the key thing about percentage problem is to know where to put the 100% or the base. It is not immediately obvious where the base is. Language skills and general knowledge here will help. Luckily, the vocabulary used in percentage problems are not that wide, so it is possible to memorize some key words and infer where is the 100%. In the above problem, the word \"by\" is commonly used in problem sums. ie. increased by.... the value that comes after this is the Transfer value, not the After value. Just memorize \"by\" = Transfer value.

Black fonts denote values transferred directly from the problem sum. Red fonts denote derived values.
http://i46.tinypic.com/35bba5y.png\">
since the Total value remains the same, then the Total Transfer value = 0.
To get Total Transfer value = 0, Kathy Transfer value must be the same as Jamal's Transfer value, but with the opposite sign.

[mrsoon]

Fiona is 3 kg lighter than Eliza and 7 kg heavier than Tina.

If Fiona is j kg, find the total weight of the 3 girls.

This problem sum is taken from last year CA1 paper from a school.
The working j + (j + 3) + (j - 7) = 3j + 3 - 7
This will result in dealing with negative number.
Is this in the PSLE syllabus?