Tutor MathsGuru: Ask me for your burning Maths questions!
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Vanilla Cake:
Hi Vc,Almighty:
Mrs Lee was 44 years old.Her son was twice her daugther's age.Mrs Lee will be twice her son's age when her daugther is 26 years old.How old will Mrs Lee be when her daughter is 20 years old?
Assume daughter's age to be N.
Mrs Lee's age: 44 years old
Daughter's age: N years old
Son's age : 2N years old
Difference between daughter's age and son's age = N years old
Mrs Lee's age: (52+2N) years old
Daughter's age: 26 years old
Son's age: (26+N) years old
Difference in the age between Mrs Lee and her son is the same.
44-2N = (52+2N)-(26+N)
44-2N=26+N
3N=18
N=6
When daughter is 20 years old, Mrs Lee's age = (52+2N)-6 = (52+12)-6=58 years old
Pls advise the source of this question as this question is almost the same as the 5-mark http://www.wendykoh.com/07/primary6-scgs-maths.pdf from Singapore Chinese Girls' School (Primary) 2007 P6 Prelim.
Sorry , i dont remember from which forum i took this qt. All tricky qts i used to put in my Xel sheet and try to solve them when i get time. So, i dont exactly remember from which website i took this qt.Thanks 4 yr timely help. -
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. -
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/ -
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! -
Hi iFruit
Thank you for the clear explanation.
Best regards
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] -
capricorn28:
Actual workings from DD2.Hi All,
VC's mumcapricorn28:
http://postimage.org/image/s3ym9sis/Paper 2 - Q6 and the ans is 16 degrees
Angle TUV =180⁰-60⁰=120⁰
Angle XWT=180⁰-170⁰=10⁰
360⁰-60⁰=300⁰
Angle XUT =360⁰-300⁰-34⁰-10⁰=16⁰
Angle XUT = 16⁰.capricorn28:
http://postimage.org/image/240e8e9fo/Paper 2 - Q8 and the ans is 72%
Before
L: 4u
F: 6u
After
L: 3.2u
F: 4u
(3.2u+4u)/10ux100% = 72%
72% of the stamps was left.capricorn28:
http://postimage.org/image/jk1hymuc/Paper 2- Q13 and the ans is 80cm
5u=8000 cm²
4u=6400 cm²
Square root of 6400 cm² = 80 cm
Length of the square = 80 cm.capricorn28:
http://postimage.org/image/jk6gkobo/Paper 2 - Q14 and the ans is 40.5cm sq
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 cm²capricorn28:
http://postimage.org/image/2aopcqyys/Paper 2 - Q15 and the ans is $360
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:
http://postimage.org/image/jk4t1btw/Paper 2 - Q17 and the ans is 120
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. -
wkong:
Use guess and check,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/
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 -
kancheongmum Thanks for your help.
Best wishes -
Thanks again VC for your help

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