Q&A - PSLE Math

[MathIzzzFun]

isetan:

Hi, need help on this question:--



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Two candles of different thickness were placed on a table. Candle B was placed on a stand of height 8 cm to reach the same height as Candle A. At 0630, Candle A was lit. One hour later, Candle B was lit. Both candles burnt down to the height marked 'X' at 1100. At 1600, Candle B burnt out while Candle A only burnt out at 2000. The rate of burning for each candle was constant throughtout.

(a) Find the ratio of the time taken for Candle A to burn to \"X\" to the time taken for Candle B to burn to \"X\".

(b) Find the original height of Candle A.

TIA.

Hi

I presume you are able to work out (a)

Time taken for A to burn to X --> 4.5h
Time taken for B to burn to X --> 3.5h
Total time taken for A to burn out --> 13.5h
Total time taken for B to burn out --> 8.5h
4.5/13.5 A = 3.5/8.5 B
A/B = 21/17
A --> 21 units, B --> 17 units
4 units --> 8 cm
So, original height of A --> 21/4 x 8 = 42 cm

cheers.

[cimman]

brastilava:

Hi, please help in this question:


Tin A, Tin B and Tin C are required to paint a complete wall. If only Tin A is used, we are short of 7 litres. If only Tin B is used, we are short of 5 litres. If only tin C is used, we are short of 3 litres. How many litres of paint is required to paint a complete wall?

here's another solution using equations:

let w = paint needed to paint one wall

A + B + C = w
A = w - 7 -- (short of 7 litres to paint the wall)
B = w - 5
C = w - 3

(w - 7) + (w - 5) + (w - 3) = w
3w - 15 = w
2w = 15
w = 7.5

[Neat]

tianzhu:

Neat:

Hi! Good evening,


Please help with this 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?

Thanks

Hi

There are a few ways to arrive at the solution.

1/3 of Jason and Edward’s money ------- 837 – 648 ------- 189

Hence, their total money ------ 3*189 ------ 567

Sam’s money@first ------- 837 – 567 ------- 270

270/3 ------ 90

Therefore, Edward’s money@first ------- 4*90 ------ 360

837 – 360 – 270 ------ 207 (Jason’s money@first)

Best wishes

[quote]Thanks a lot, tianzhu.

If unit method is to be used, which item is to be made constant?

Regards[/quote]

[tianzhu]

Neat:

Thanks a lot, tianzhu.

If unit method is to be used, which item is to be made constant?

Regards

Hi

Good Morning.

I am trying to understand “which item is to be made constant”. Could you help to elaborate on it?

Alternatively, you may also present the solution in “Units and Parts”.

The relationship between Edward and Sam is given.

Edward ------ 4 units
Sam ------- 3 units

We differentiate between Edward/Sam and Jason by using parts for Jason.

1 unit + 1 part ------- 837 – 648 ------- 189

3 units + 3 parts ------ 3*189 ------ 567 (total amount)

Sam’s money@first ------- 837 – 567 ------- 270

3 units ------- 270
1 unit ------- 90

Therefore, Edward’s money@first ------- 4*90 ------ 360

837 – 360 – 270 ------ 207 (Jason’s money@first)

Best wishes

[Neat]

tianzhu:

Neat:

Thanks a lot, tianzhu.

If unit method is to be used, which item is to be made constant?

Regards

Hi

Good Morning.

I am trying to understand “which item is to be made constant”. Could you help to elaborate on it?

Alternatively, you may also present the solution in “Units and Parts”.

The relationship between Edward and Sam is given.

Edward ------ 4 units
Sam ------- 3 units

We differentiate between Edward/Sam and Jason by using parts for Jason.

1 unit + 1 part ------- 837 – 648 ------- 189

3 units + 3 parts ------ 3*189 ------ 567 (total amount)

Sam’s money@first ------- 837 – 567 ------- 270

3 units ------- 270
1 unit ------- 90

Therefore, Edward’s money@first ------- 4*90 ------ 360

837 – 360 – 270 ------ 207 (Jason’s money@first)

Best wishes

[quote]Hi tianzhu!

Good morning.

Yes, this is the method that I am looking for.

As I understand, if unit transfer method is used, we must always equalized the units and parts.

I used the following to solve but was stucked. Now, I understand.


J E S

before ? 4u 3u
change -1p -1p -
after 2p ? 3u

Thanks a lot, tianzhu.
[/quote]

[isetan]

MathIzzzFun:

isetan:

Hi, need help on this question:--



http://i50.tinypic.com/1g65hi.jpg\">

Two candles of different thickness were placed on a table. Candle B was placed on a stand of height 8 cm to reach the same height as Candle A. At 0630, Candle A was lit. One hour later, Candle B was lit. Both candles burnt down to the height marked 'X' at 1100. At 1600, Candle B burnt out while Candle A only burnt out at 2000. The rate of burning for each candle was constant throughtout.

(a) Find the ratio of the time taken for Candle A to burn to \"X\" to the time taken for Candle B to burn to \"X\".


Thanks Tianzhu!
(b) Find the original height of Candle A.

TIA.

Hi

I presume you are able to work out (a)

Time taken for A to burn to X --> 4.5h
Time taken for B to burn to X --> 3.5h
Total time taken for A to burn out --> 13.5h
Total time taken for B to burn out --> 8.5h
4.5/13.5 A = 3.5/8.5 B
A/B = 21/17
A --> 21 units, B --> 17 units
4 units --> 8 cm
So, original height of A --> 21/4 x 8 = 42 cm

cheers.

[ksoh3000]

Hi


Can anyone help me with this qn ?

-Grace cycled from home to the park at 7.05 a.m. at a constant speed of 300m/min. She passed Colin at 7.15 a.m. as he started cycling at a constant speed from his house to the park. At 7.30 am, Colin was 1.5km ahead of Grace. Colin reached the park at 7.40 am.
(a) Find Colin’s speed, giving yr answer in kilometres per hour .
(b)How far is Grace’s house from the park? Give yr answer in kilometres.

Tks

[tianzhu]

Hi


Here’s another way to look at the candle question.

Life is just like a candle in the wind.

As I get older, I start to appreciate more the fragility and uncertainty in life.

We are just a flame atop a candle.

A draft from an unseen corner may put an end to it.

Best wishes

http://farm9.staticflickr.com/8151/7158587127_c44efa4cf1_z.jpg\">

[Spongy]

Hi I have 2 questions to ask, thanks in advance.


1.Aunt May baked some square, round and oval cookies. For every 3 square cookies that were baked, 2 round cookies were baked. For every 4 square cookies that were baked, 3 oval cookes were baked.
(a) Find the ratio of the number of square cookies to round cookies to oval cookies baked.
(b) After Aunt May baked another 120 round cookies, 30% of the cookies now were round cookies. How many square cookies did Aunt May Bake?

2.Mr Lim earned $3000 monthly,which was 20% more than the monthly salary of Mr Tan
(a) What was Mr Tan’s monthly salary?
(b) When both Mr Lim’s and Mr Tan’s monthly salaries were increased by the same percentage, Mr Lim earned $590 more than Mr Tan. What was the percentage increase in their salaries?

[tianzhu]

Spongy:

Hi I have 2 questions to ask, thanks in advance.


1.Aunt May baked some square, round and oval cookies. For every 3 square cookies that were baked, 2 round cookies were baked. For every 4 square cookies that were baked, 3 oval cookes were baked.
(a) Find the ratio of the number of square cookies to round cookies to oval cookies baked.
(b) After Aunt May baked another 120 round cookies, 30% of the cookies now were round cookies. How many square cookies did Aunt May Bake?

Hi

Square:Round ------ 3:2 -------- 12:8

Square:Oval ------- 4:3 ------- 12:9

The units for “Square cookies” are repeated in both set of ratios, so make them the same.

Ratio of the number of square cookies to round cookies to oval cookies baked ------- 12:8:9

Squre cookies ------- 12 units
Round cookies ------- 8 units + 120
Oval cookies ------- 9 units

8 units + 120 ------- 30%

21 units ------- 70%

3 units ------- 10%

Hence, 8 units + 120 ------- 9 units

This gives 1 unit ------- 120

Number of square cookies --------12*120 ------1440

Best wishes