Tutor MathsGuru: Ask me for your burning Maths questions!

[Vanilla Cake]

2DMommy:

In a box, there are yellow, red and green beads, 1/3 of the beads in the box are yellow. There are 30 more red beads than yellow beads. The rest of the 15 beads are green. How many beads are there altogether?

Hi 2DMommy,
Pls advise the source of your question so that we (KSP members) know the level of this question in order to apply the appropriate method to solve.
Not sure whether this method is suitable or not?

Yellow: u
Red: u+30
Green: 15

Total number of beads = 3u.
Number of green beads = 3u-u-(u+30)=3u-u-u-30 = u-30

u-30=15
u=45
3u= 3x45=135

Number of beads altogether = 135

If this is really a P3 question then model solutions are easy to understand than the above method. Pls kindly wait for other KSP members to post their model solutions.

[wkong]

Thanks VC for your help.



Another 1 more question that needs help:

The figure is not drawn to scale. It shows a rectangle divided into 6 parts. Each part has a different area.

a) Find the total area of P and Q.

b) Find the perimeter of the figure.



http://www.postimage.org/

[YLH88]

Hi all,


Need help in the below question :

Sam and Harry started running at the same time from the opposite ends of the diameter of a circular track.
Both of them were running in the clockwise direction.
Sam took 12 minutes to run round the track once while Harry took 15 minutes to do the same.
How long did Sam take to run past Harry for the first time ?
(ans : 30 minutes)

http://postimage.org/image/1z70q6hqc/

Thank you!

[capricorn28]

Hi iFruit


Thank you for the clear explanation.

Best regards
Capricorn28

[capricorn28]

Hi All,


Need help on Catholic High School P5 SA2 - Year 2009
Paper 2 - Q6 and the ans is 16 degrees
Paper 2 - Q8 and the ans is 72%
Paper 2- Q13 and the ans is 80cm
Paper 2 - Q14 and the ans is 40.5cm sq
Paper 2 - Q15 and the ans is $360
Paper 2 - Q17 and the ans is 120

http://postimage.org/image/jk1hymuc/

http://postimage.org/image/s3ym9sis/

http://postimage.org/image/240e8e9fo/

http://postimage.org/image/jk4t1btw/

http://postimage.org/image/jk6gkobo/

http://postimage.org/image/2aopcqyys/
[/img][/list]

[Vanilla Cake]

capricorn28:

Hi All,

Actual workings from DD2.


VC's mum

capricorn28:

Paper 2 - Q6 and the ans is 16 degrees

http://postimage.org/image/s3ym9sis/
Angle TUV =180⁰-60⁰=120⁰
Angle XWT=180⁰-170⁰=10⁰
360⁰-60⁰=300⁰
Angle XUT =360⁰-300⁰-34⁰-10⁰=16⁰

Angle XUT = 16⁰.

capricorn28:

Paper 2 - Q8 and the ans is 72%

http://postimage.org/image/240e8e9fo/
Before
L: 4u
F: 6u

After
L: 3.2u
F: 4u

(3.2u+4u)/10ux100% = 72%

72% of the stamps was left.

capricorn28:

Paper 2- Q13 and the ans is 80cm

http://postimage.org/image/jk1hymuc/
5u=8000 cm²
4u=6400 cm²
Square root of 6400 cm² = 80 cm

Length of the square = 80 cm.

capricorn28:

Paper 2 - Q14 and the ans is 40.5cm sq

http://postimage.org/image/jk6gkobo/
24÷8x5=15 cm
24-15 = 9 cm
40-9 = 31 cm
½x40x15 = 300 cm²
½x31x9 = 139.5 cm²
½x24x40 = 480 cm²
480-139.5-300=40.5 cm²

Shaded area = 40.5 cm²

capricorn28:

Paper 2 - Q15 and the ans is $360

http://postimage.org/image/2aopcqyys/
Before
J: 6u
K: 5u

After
J: 3.6u
K: 7.4u

In the end
J: 7.3u
K: 3.7u

7.3u-3.7u=3.6u
3.6u=$216
6u = $360

At first, Jude had $360.

capricorn28:

Paper 2 - Q17 and the ans is 120

http://postimage.org/image/jk4t1btw/
Hall (2n)
B : G
3u/4 : u/4
3 : 1
15 : 5

Canteen (1n)
B : G
2p/5 : 3p/5
2 : 3
4 : 6

1u=6
20u=120

Number of pupils in the hall = 120.

[Vanilla Cake]

wkong:

The figure is not drawn to scale. It shows a rectangle divided into 6 parts. Each part has a different area.

a) Find the total area of P and Q.
b) Find the perimeter of the figure.

http://www.postimage.org/

Use guess and check,
Area of P = 8 cmx 6 cm = 48 cm²
Area of Q = 5 cm x 4 cm = 20 cm²
(a) Total area of P and Q = (48+20) cm² = 68 cm²

Length of rectangle = (5+6+4) cm = 15 cm
Breadth of rectangle = (8+5) cm = 13 cm
(15x2)+(13x2) = 30+26 = 56 cm
(b) Perimeter of the figure = 56 cm

[youngtay]

kancheongmum Thanks for your help.


Best wishes

[wkong]

Thanks again VC for your help

[Vanilla Cake]

YLH88:

Hi all,

Need help in the below question :

This is an interesting question but nobody has forward his/her solution yet.I am not please with my algebra method and let's hope that others will forward with better methods for sharing and learning.Could you pls advise the source of this question?

VC's mum

YLH88:

Sam and Harry started running at the same time from the opposite ends of the diameter of a circular track.
Both of them were running in the clockwise direction.
Sam took 12 minutes to run round the track once while Harry took 15 minutes to do the same.
How long did Sam take to run past Harry for the first time ?
(ans : 30 minutes)

http://postimage.org/image/1z70q6hqc/

LCM of 12 and 15 is 60.
Work out 60m as the length (circumference) of the circular track even though 60m is not a realistic number.
Sam's average speed = 60m÷12 = 5 m/min
Harry's average speed = 60m÷15 = 4 m/min

Note that both of them started running at the same time from the opposite ends of the diameter of a circular track means the difference in their distance = ½x60m = 30 m.
For them to meet for the first time, Sam must run more than Harry by 30 m in the same amount of time.

Assume the time taken to be T,
5xT= (4xT)+30
5T=4T+30
T=30

Sam took 30 min to run past Harry for the first time.