Tutor MathsGuru: Ask me for your burning Maths questions!

[benorito]

This is from Red Swastika Prelim 2010 Q 8:


Raju and Gywneth shared some sweets. If Raju gave 1/2 of his share to Gywneth, Gywneth will have 8 more sweets from Raju. If Raju gave 1/4 of his share to Gywneth, Gywneth would have 2 more sweets than Raju. How many sweets did Raju have at first ?

The given answer was 12. Is this correct ?

[kancheongmum]

benorito:

This is from Red Swastika Prelim 2010 Q 8:


Raju and Gywneth shared some sweets. If Raju gave 1/2 of his share to Gywneth, Gywneth will have 8 more sweets from Raju. If Raju gave 1/4 of his share to Gywneth, Gywneth would have 2 more sweets than Raju. How many sweets did Raju have at first ?

The given answer was 12. Is this correct ?

Hi this is how my ds solve the problem as he is not good with model:

get the common denominator for both fraction 1/2 --- 2/4
Raju has a total of 4u at first
this is internal transfer so total for both cases are equal

1st case --- Raju gave Gywneth 2u --- total --- 2u + 2u +8 = 4u +8
2nd case ---Raju gave Gywneth 1u --- total --- 3u + 3u +2 = 6u +2

4u + 8 = 6u + 2
2u --- 6
1u --- 3

4u --- 3 x 4 = 12 sweets

[corneyAmber]

benorito:

This is from Red Swastika Prelim 2010 Q 8:


Raju and Gywneth shared some sweets. If Raju gave 1/2 of his share to Gywneth, Gywneth will have 8 more sweets from Raju. If Raju gave 1/4 of his share to Gywneth, Gywneth would have 2 more sweets than Raju. How many sweets did Raju have at first ?

The given answer was 12. Is this correct ?

The answer is correct.

Raju [ ][ ][ ][ ] -> 4 parts (before giving away)
After for Raju if he gave 1/2 [ ][ ]
After for Gyw if she recd 1/2 [ ][ ]<8>

After for Raju if he gave 1/4 [ ][ ][ ]
After for Gyw if she recd 1/2 [ ][ ][ ] <2>

So (Raju)2 parts + (Gyw)2 parts + 8 = (Raju)3 parts + (Gyw)3 parts + 2
4 parts + 8 = 6 parts + 2
2 parts = 6
1 part = 3

So Raju has 4 parts initially => (4*3) = 12 sweets

[Dharma]

benorito:

This is from Red Swastika Prelim 2010 Q 8:


Raju and Gywneth shared some sweets. If Raju gave 1/2 of his share to Gywneth, Gywneth will have 8 more sweets from Raju. If Raju gave 1/4 of his share to Gywneth, Gywneth would have 2 more sweets than Raju. How many sweets did Raju have at first ?

The given answer was 12. Is this correct ?

1/2 - (-1/2) = 1 whole => 8
1/4 - (-1/4) = 1/2 => 2

Diff = 1/2 => 8 - 2 = 6
1 whole => 12 ( total number of sweets Raju had at first )

[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!

[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

[little_twin_stars]

Dear all,


May I know is there any good books to learn and practice Mathematics Solving? I think I really need it in order to understand and prepare myself when I teach my daughter.

TIA.

[ganchiong]

Need help in the following question:


Jeffrey wants to pack some identical cuboids into a container. What is the maximum number of cuboids Jeffrey can pack into the container?

Dimension for cuboid:2cmx2cmx3cm and container: 6cmx11cmx14cm

TIA

[corneyAmber]

ganchiong:

Need help in the following question:


Jeffrey wants to pack some identical cuboids into a container. What is the maximum number of cuboids Jeffrey can pack into the container?

Dimension for cuboid:2cmx2cmx3cm and container: 6cmx11cmx14cm

TIA

is the answer 77?

[ganchiong]

ksi:

ganchiong:

Need help in the following question:


Jeffrey wants to pack some identical cuboids into a container. What is the maximum number of cuboids Jeffrey can pack into the container?

Dimension for cuboid:2cmx2cmx3cm and container: 6cmx11cmx14cm

TIA

is the answer 77?

yes