Tutor MathsGuru: Ask me for your burning Maths questions!

[Herbie]

Hi I have one qn


Candle A and Candle B are of the same length. Candle A which is broader can burn for 5 hours while Candle B the thinner candle can burn for 4 hour, 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?

Kindly advise.

Tx

[Vanilla Cake]

Herbie:

Candle A and Candle B are of the same length. Candle A which is broader can burn for 5 hours while Candle B the thinner candle can burn for 4 hour, 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?

This question is from River Valley Primary School 2010 P6 Prelim Paper 2 Q17 [5 marks].

Refer to http://psle2010a.blogspot.com/2010/09/rvps-p6-prelim-2010-p2-q17.html
A similar question solved by http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=280&postdays=0&postorder=asc&start=1530.

[Herbie]

Hi! Thanks Vc! I hv another qn. A fruitseller had some oranges, pears n durians. After selling some of them, there were 6 pears for every 5 oranges left and 7 oranges for every 3 durians left. He sold 168 oranges. He had 456 pears and oranges left. The no of pears sold was the Same as the no of durians sold. How many oranges had he at first? If the no of pears was twice the ö ofDurian at firSt, what the total no of pears and durians sold

[Daddy]

Hi ,


I have two questions to ask.

1) The total cost of 4 similar pouches and 7 similar lipsticks is $205. If 3 such pouches and 1 lipstick cost $90, find the cost of each pouch?


2) Alan and Jenny had an equal amount of money. Alan spent $20 each day and Jenny spent $36 each day. When Jenny had $40 left, Alan had five times as much money left as she. How much money did Alan have at first?

Thanks
Daddy

[Herbie]

Hi daddy

4p + 7L = 205
3p +1L= 90

12p + 21L = 615
12p + 4l = 360

21L -4L= 615-360
!7L=225
L=15
insert L=15 into the earlier eqn n you will get p= 25
each pouch = $25

[corneyAmber]

Daddy:

Hi ,


2) Alan and Jenny had an equal amount of money. Alan spent $20 each day and Jenny spent $36 each day. When Jenny had $40 left, Alan had five times as much money left as she. How much money did Alan have at first?

Thanks
Daddy

Money Jenny had at first = 40 + 36(# of days)
Money Alan had at first = (40x5) + 20(# of days)
40 + 36(# of days) = 200 + 20(# of days)
16(# of days) = 160
# of days = 10

Hence money Alan had at first = 200 + 20(10) = $400

[corneyAmber]

Daddy:

1) The total cost of 4 similar pouches and 7 similar lipsticks is $205. If 3 such pouches and 1 lipstick cost $90, find the cost of each pouch?

Thanks
Daddy

Alternative method:

3P + 1L = 90 => 1L = 90-3P
4P + 7L = 205
4P + 7(90-3P) = 205
4P - 21P = 205 -630
17P = 425
P = $25

[corneyAmber]

kancheongmum:

Hi ksi and ganchiong

I don't think the answer is 77. You cannot divide the volume of the container by the volume of the cuboid. You have to approach this type of question this way. Minimize the remainder. Divide each dimension by each dimension. Try not to have remainder.

6/3 -- 2 cuboids
11/2 -- 5 cuboids remain 1 cm
14/2 -- 7 cuboids

Max no. 2 x 5 x 7 = 70 cuboids

Yes the original answer is too simplistic.

[ganchiong]

kancheongmum:

ksi:

[quote=\"ganchiong\"]
yes

container = 6x11x14 = 924
cuboid = 2x2x3 = 12
max. Number of cuboids = 924/12 = 77

Hi ksi and ganchiong
I don't think the answer is 77. You cannot divide the volume of the container by the volume of the cuboid. You have to approach this type of question this way. Minimize the remainder. Divide each dimension by each dimension. Try not to have remainder.

6/3 -- 2 cuboids
11/2 -- 5 cuboids remain 1 cm
14/2 -- 7 cuboids

Max no. 2 x 5 x 7 = 70 cuboids[/quote]Thank you all for your reply.

I am really confused... :? now.
Actually i tried to draw it out and the answer is 77. Any Maths Guru out there can clarify?

[YLH88]

Dharma:

[Moderator's note: Article edited & selected for http://www.kiasuparents.com/kiasu/content/some-questions-and-answers-primary-6-math]

YLH88:

Hi,

Need help with the below 2 questions :

1) Nanyang 2010 P6 Prelim paper 1 Qn 21
The cost of 3 mangoes is the same as the cost of 5 grapefruits.
The cost of 3 mangoes is also the same as the cost of 10 pears.
Find the ratio of the cost of a mango to the cost of a grapefruit to the cost of a pear.
(ans : 10 : 6 : 3)

2) Nanyang 2010 P6 Prelim paper 1 Qn 30
The figure below is formed by stacking 4 pieces of square paper one on top of another. The papers have different prints and sizes (of sides 3cm, 4cm, 5cm and 6cm).
The 6-cm piece is placed at the bottom of the stack, followed by the 5-cm piece, then the 4-cm piece and the 3-cm piece is placed right on top.
Find the sum of the area A and B.
(ans : 8 cm2)

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


Thank you very much!

Nanyang 2010 P6 Prelim paper 1 Qn 21
3M = 5G
1M = 5/3 G
1M : 1G = 5 : 3 = 10 : 6

3M = 10P
1M = 10/3 P
1M : 1P = 10 : 3

1M : 1G : 1P = 10 : 6 : 3

Nanyang 2010 P6 Prelim paper 1 Qn 30
Area of large square – Area of yellow square = 6^2 – 3^2 = 36 – 9 = 27
Yellow square => 3 x 3
Orange square => 4 x 4
White square => 5 x 5
Blue square => 6 x 6

Length of A = Width of A = 6 - 4 = 2
Area of A = 2 X 2 = 4
Area of A = Area of B [Symmetry]
Area of A + Area of B = 4cm2 x 2 = 8cm2

Hi Dharma,

Thank you for the solution!