Q&A - PSLE Math

[Maths Hub]

PiscesLeo:

Posted this in P5 thread but there's no reply so hope to get help here.


Tank A and tank B are identical. 5/6 of tank A and 1/4 of tank B are filled with water. The mass of tank B with its contents is 7/13 of the mass of tank A with it's contents. The total mass of the two tanks with their contents is 1400g. Find the mass of the empty tank.

Thank you very much

(i)Difference (Fraction):
5/6 - 1/4 = 7/12

(ii)Difference (Weight):
20 units --> 1400
1 unit--> 1400/20 = 70
13-7 = 6
6 X 70 = 420

(iii) Putting both (i) and (ii) together, we will get:
7/12--> 420 g
1/12--> 420/7 = 60 g
1/4--> 180 g

(iv) We can also get the weight of the tank with 1/4 water from (ii):
Tank+1/4 water--> 70 X 7 = 490

(v) Now we find the answer using (iii) and (iv):
Empty tank--> 490 - 180 = 310 g

Hope that helps.

[CDE]

Anyone can help with this question?


A and B has some sweet. If A eat 2 a day, and B eat 1 a day - A will have 68 left while B has none. If A eat 1 a day and B eat 2 a day - A will have 122 left while B has none.
How many sweet do they have at first?

Thanks

[Maths Hub]

Chalupa:

Please help me in this question


There were 1600 pupils in a primary school.When the number of boys increased by 76 and the number of girls decreased by 4%,the total enrollment in the school increased by 4%.How many boys were there in the school at first?

104% X 1600 = 1664
1600 + 76 = 1676
So there was a drop of 1676-1664 = 12
4% of girls--> 12
100% of girls --> 12 X 25 = 300
Boys at first --> 1600 - 300 = 1300

Hope that helps.

[PiscesLeo]

:thankyou: MathIzzzFun and Maths Hub :thankyou:

[iCreative Math]

A and B has some sweet. If A eat 2 a day, and B eat 1 a day - A will have 68 left while B has none. If A eat 1 a day and B eat 2 a day - A will have 122 left while B has none. How many sweet do they have at first?


The answer is A has 140 sweets and B has 36 sweets at first.

I have coincidentally recorded two YouTube videos recently explaining a similar question. The questions have different figures but the concept tested and hence the problem-solving method is identical.

You might want to get your child to view the YouTube videos.

I have recorded two different videos, one using the \"normal\" method and another using the \"Guess & Check\" method.

The generic simplified problem-solving methods taught in the videos will apply to all similar questions of this nature.

Hope it is helpful.


Best Regards
iCreative Math

YouTube Lesson Videos:
http://www.youtube.com/watch?v=JXpXuz2Tqf4 (PSLE “Double-IF” Question)

http://www.youtube.com/watch?v=jcU2hv-GTBE (PSLE “Double-IF” Question using “Guess & Check” Method)

[Mary Joy]

Please help me with this question!Thank you


In an auditorium,30% of the seats are reserved for the platinum class.The ratio of the number of seats for silver class to the number of seats for gold class is 12:13. After the auditorium underwent a renovation,the number of seats in the auditorium is increased by 20%. There are now 1440 seats reserved for the platinum class.

a) how many seats are reserved for the silver and gold class after the renovation?

b) how many seats are reserved for the silver class before the renovation?

[iCreative Math]

Dear David59, you might want to consider another method to solve the question below:-


1) Gerald had 60 sweets more than Herlina. After eating half of what they had, the ratio of the number of Gerald’s sweets to the number of Herlina’s sweets became 3:2.

a) How many sweets did Gerald eat?

b) Gerald gave Herlina some sweets and the ratio became 8:7. How many sweets did Gerald give Herlina?

If Gerald has 60 sweets more than Herlina, after both of them have eaten half of what they have, Gerald will have 30 sweets more than Herlina.

Since the ratio then becomes 3:2, we can deduce that 1 unit = 30 sweets.

i.e. (a) Gerald has 90 sweets; Herlina has 60 sweets.

(b) since the ratio becomes 8:7, we can deduce Gerald has given 10 sweets to Herlina, i.e. Gerald has 80 sweets, Herlina has 70 sweets.

The model approach is fine but if students can grasp this concept, this method will save them some time.

Good to learn this concept and be aware because it can be used in several other types of problem sums as well including speed questions.

Basically, two steps:-

(a) first, recognise that the difference between A & B becomes half when both A & B are reduced by half, i.e. if A is 100 more than B, when A & B are reduced by half, then A becomes 50 more than B.

(b) and if the question then state that the ratio becomes e.g. 5:3, then we know straight away that 2 units = 50.

Hope this helps.


Best Regards,
iCreative Math

[Mary Joy]

Please help me to answer this question


David and his brother John decided to cycle from his home to the library using the same route. They started cycling at the same time. David cycled at a speed of 15km/hr. Both of them did not change their speed throughout the race . When John covered 1/3 the distance , David was 4.5 km ahead of him.David reached the library at 4.45pm . What time did John reach the library?

[david59]

Mary Joy:

please help me with these questions thank you


2.Jim bought some chocolates and gave half of them to ken.ken bought some sweets and gave half of them to Jim.Jim ate 12 sweets and ken ate 18 chocolates.after that the Number of sweets and chocolates Jim had were in the ratio 1:7 and the number of sweets and chocolates ken had were in the ratio. 1:4

How many sweets did ken buy?

This is another method of solving this problem sum. Hope it is explained clearly to you. God bless.


http://i43.tinypic.com/2z3uc6u.jpg\">

[MathIzzzFun]

Mary Joy:

Please help me to answer this question


David and his brother John decided to cycle from his home to the library using the same route. They started cycling at the same time. David cycled at a speed of 15km/hr. Both of them did not change their speed throughout the race . When John covered 1/3 the distance , David was 4.5 km ahead of him.David reached the library at 4.45pm . What time did John reach the library?

assuming that David continues to cycle after reaching the library ...

for every 1/3 distance John cycles, David cycles 1/3 distance + 4.5km
--> John cycles 3/3 distance, David cycles 3/3 distance + 3 x 4.5 km

3 x 4.5 km = 13.5 km --> David would have cycled extra 13.5km when John reached the library

Time taken for David to cycle 13.5 km = 13.5/15 h = 0.9 h = 54 min

4.45 pm + 54 min --> 5.39 pm

John reached the library at 5.39pm

cheers.