Tutor MathsGuru: Ask me for your burning Maths questions!

[Vanilla Cake]

Herbie:

Hi Vc, wrt the RV qn 17 on candle A n B, i hv looked at uncle tianzhu model on james ang post but i still can't figure out how to do the RV QN can help to post the solution? Tx

Hi Herbie,

As school reopens tomorrow, I will be busy preparing for my year 2 (Sec 2) EOY exam.
Pls kindly refer to http://psle2010a.blogspot.com/2010/09/rvps-p6-prelim-2010-p2-q17.html which I believe are clear and well explained.

Question 17
Candle A and candle B are of the same length. Candle A, which is broader, can burn for 5 h while Candle B, the thinner candle, can burn for 4 h. If both candles are lighted at the same time, how long does it take for Candle A to be twice as long left as Candle B?

My suggested solution
Using algebra :

Assume the number of hours burnt be n
1 on the left side is 5/5 and 1 on the right hand side is 4/4

1-n/5=2(1-n/4)
1-n/5=2-n/2
(5-n)/5=(4-n)/2
2(5-n)=5(4-n)
10-2n=20-5n
3n=10
n= 3⅓ hours

It takes 3⅓ hours for Candle A to be twice as long left as Candle B.

Hope that my solution does not cause more confusion for you.

[YLH88]

MOE Hater:

(20-8)cm/2 = 6cm


6cm x 20cm x 20cm = 2400cm3 (cubic centimetre)

4 litres = 4000cm3

4000cm3 - 2400cm3 = 1600cm3

1cm x 8cm x 8cm + 1cm x 20cm x 20cm = 464cm3

(1600cm3 / 464cm3/cm) + 6cm = 9.45cm (correct to 2 d.p.)

Hi MOE Hater,

Thank you for the solution.

[YLH88]

Vanilla Cake:

firebird:

Dear Vanilla cake


Good afternoon.

Thank you very much for your solution for Henry Park Q18 paper 2, SA 2010 question.

With best regards
Firebird

Dear Firebird,

As school reopens tomorrow, I will be busy preparing for my year 2 (Sec 2) EOY exam. Hope that you are able to understand my solution posted earlier this morning.

Hi Vanilla Cake,

All the best to your year 2 EOY exam

[YLH88]

Hi all,


Sorry need help again with the below questions :

1) Veronica bought some tulips and lilies in the ratio of 4 : 7
Each tulip cost $1.20 more than each lily.
If she spent $305.50 altogether and the amount she spent on the lilies was $3.90 more than the amount she spent on the tulips, how many tulips did she buy ?


2) Nanyang 2010 P6 Prelim Paper 2 Qn 15
The table below shows the day of the week that 1st January falls on from the year 1981 to 2000.
Note that Day 1 of a week is a Monday and Day 7 is Sunday.
There are 31 days in the month of January.

a) What number does the letter C represent ? (ans : 3)
b) What number does the letter F represent ? (ans : 7)
c) Which day of the week will 1st February 2012 fall on ? (ans : Wednesday)

http://www.postimage.org/image.php?v=Pqcy_EA


Thank you !

[Vanilla Cake]

YLH88:

2) Nanyang 2010 P6 Prelim Paper 2 Qn 15

The table below shows the day of the week that 1st January falls on from the year 1981 to 2000.
Note that Day 1 of a week is a Monday and Day 7 is Sunday.
There are 31 days in the month of January.

a) What number does the letter C represent ? (ans : 3)
b) What number does the letter F represent ? (ans : 7)
c) Which day of the week will 1st February 2012 fall on ? (ans : Wednesday)

http://www.postimage.org/image.php?v=Pqcy_EA

Vanilla Cake:

Q2 is easy. You need to know something about leap year.OK, briefly leap year is the year where the number can be divided by 4 without any remainder. After every leap year, you jump by 2 days.

Look at Q2 again, you will find that a leap year starts from 1984 onwards:
1984
1988
1992
1996
2000
2004
2008

by understanding how leap years work, you will get:

A: 1
B: 2
3 - 2003
4
E: 6
F: 7 - 2006
G: 1
H: 2
I: 4
J: 5
2011 : 6
2012 : 7

(a) Letter C (2003) represents 3.

(b) Letter F (2006) represents 7

1st Jan 2012 -> Sunday
29 Jan 2012 -> Sunday ( add 4 weeks ie 4x7 = 28 days )
30 Jan 2012 -> Monday
31 Jan 2012 -> Tuesday
1 Feb 2012 -> Wednesday
(c) The day of the week which 1st February 2012 will fall on - Wednesday.

Hi YLH88,

I am the mum of Vanilla Cake.Pls take a look at http://psle2010a.blogspot.com/2010/09/challenging-p6-2010-sa2-nanyang-p2-q15.html which is detailed and better than Vanilla Cake's explanation.
You may wish to browse through http://psle2010a.blogspot.com/search?updated-min=2010-01-01T00%3A00%3A00%2B08%3A00&updated-max=2011-01-01T00%3A00%3A00%2B08%3A00&max-results=50 in which he had olved and posted most of the 2010 P6 Prelim Maths questions from top schools.
Thanks.

VC's mum

[firebird]

Dear Vanilla cake


Good morning.

Many thanks for your help and blessing given to my son.

God bless you.

With best regards
Firebird

[Dharma]

YLH88:

Hi all,


Sorry need help again with the below questions :

1) Veronica bought some tulips and lilies in the ratio of 4 : 7
Each tulip cost $1.20 more than each lily.
If she spent $305.50 altogether and the amount she spent on the lilies was $3.90 more than the amount she spent on the tulips, how many tulips did she buy ?

Thank you !

Q1)

Amt spent on tulips = (4u)(1p + $1.20)
Amt spent on lilies = (7u)(1p)

Total spending = (11u)(1p) + ($4.80)(1u) = $305.50
Difference in spending = (3u)(1p) – ($4.80)(1u) = $3.90

(14u)(1p) = $309.40
(1u)(1p) = $22.10

($4.80)(1u) = $305.50 – 11($22.10) = $62.40
1u = 13

No. of tulips bought = 4u = 4 x 13 = 52

[corneyAmber]

YLH88:

Hi all,


Sorry need help again with the below questions :

1) Veronica bought some tulips and lilies in the ratio of 4 : 7
Each tulip cost $1.20 more than each lily.
If she spent $305.50 altogether and the amount she spent on the lilies was $3.90 more than the amount she spent on the tulips, how many tulips did she buy ?

Thank you !

Method = Algebra
Answer = 52 Tulips

Total = 305.50
Cost of tulips = 150.80
Cost of lilies = 154.70

Let price for lilies be p, then price of tulips will be (p+1.20)
Let total flowers be T, # of tulips = 4/11T, # of lilies = 7/11T

(1) 4/11T(p+1.2) = 150.80
(2) 7/11T(p) = 154.70

Substitute (2) into (1) and solving, T = 143.

Hence # of tulips 4/11 x 143 = 52

Vanilla cake, 91 is the number of lilies....so probably your method is correct but in a rush you mixed up the flowers.

[Hifive]

Hi


Need help with the following questions:

Source : RGS P5 2009 SA2 PAPER 2

1. Mr Muhamad has 3 cubical fish tanks of different size. The length of Tank A is twice the length of Tank B and the length of Tank B is twice the length of Tank C. Mr Muhamad filled 20% of Tank A, 35% of Tank B and 40% of Tank C with water at first. He then decided to pour all the water from Tank B and Tank C into Tank A. What percentage of Tank A is filled with water?

2. The average height of a group of boys is 1.64m. When 2 new boys each of height of 1.70m joined the group, the average height became 1.65m. How many boys were in the group at the end?

Thanks in advance for your help.

[Vanilla Cake]

Hifive:

Source : RGS P5 2009 SA2 PAPER 2

Q18. Mr Muhamad has 3 cubical fish tanks of different size. The length of Tank A is twice the length of Tank B and the length of Tank B is twice the length of Tank C. Mr Muhamad filled 20% of Tank A, 35% of Tank B and 40% of Tank C with water at first. He then decided to pour all the water from Tank B and Tank C into Tank A. What percentage of Tank A is filled with water?

A: 4u
B: 2u
u

20/100x4ux4ux4u = 12.8u³
35/100x2ux2ux2u = 2.8u³
40/100xuxuxu = 0.4u³

(12.8+2.8+0.4)u³ = 16u³
16 u³/(4ux4ux4u) = 16 u³/64 u³=0.25
0.25x100% = 25%
The percentage is 25%

Hifive:

Source : RGS P5 2009 SA2 PAPER 2
Q16. The average height of a group of boys is 1.64m. When 2 new boys each of height of 1.70m joined the group, the average height became 1.65m. How many boys were in the group at the end?

1.64xu =1.64u
1.64u+1.7x2=1.64u+3.4
1.65x(u+2)=1.65u+3.3
1.64u+3.4=1.65u+3.3
0.01u=0.1
u=10
10+2=12

There were 12 boys in the group at the end.

Above workings from my DD2.
Hope this helps.

VC's mum