Tutor MathsGuru: Ask me for your burning Maths questions!

[2DMommy]

Yu Xuan:

starlight1968sg:

[quote=\"2DMommy\"]Pls help me with this question.


5 children baked some cookies.
Angela baked 40 more cookies than Josephine but 1/2 as many cookies as Grace. Grace baked 4 times as many cookies as Mary. Mary baked 3/4 as many cookies as Josephine. Christina baked thrice as many cookies as Mary.
How many cookies does each child have :
Angela ?
Josephine ?
Grace ?
Mary ?
Christina ?

What is the ans?
Mine is J:80, M:60, A:120, G:240 and 180.

Hi ... I have got the same answer too.

Using a model ;
J = 4 units
M = 3 units
G = 12 units
A = 4 units + 40

Since Angela should have half as much as Grace, she should have 6 units in total. Hence the 2 units is equal to 40.

1 unit = 20
J - 4 * 20 = 80
M - 3 * 20 = 60
G - 12 * 20 = 240
A - 4 * 20 + 40 = 120
:celebrate:[/quote]Thanks starlight1968sg & Yu Xuan

[Yu Xuan]

Dharma:

trytry:

Hi can anyone help to solve this type of question:


There are 5 boys A. B, C, Dand E. Each time 2 boys will pair up to have their mass taken. The ten readings in kg are 105, 113 116, 117, 118,124,125,130,137 and 139. What is the mass of the third heaviest boy?
TIA.

Mass of all 5 boys
= (105 + 113 + 116 + 117 + 118 + 124 + 125 + 130 + 137 + 139)/4
= 306

Sum of the mass of the 2 lightest boys and 2 heaviest boys
= 105 + 139
= 244

The mass of the 3rd heaviest boy = 306 – 244 = 62

Hi Dharma

You are really very intelligent. I was still cracking my head on how I should proceed after getting 306.

I surely have much to learn from you.

[CJM]

Tang:

CJM:

[quote=\"maths6a\"]i also calculated 20. all the trees that were planted on mulitples of 5 and 10, in this case are the ones on 20m, 40m,... and 20 tress did not be removed.

thank u : )

Hi,

Don't forget to add the one at '0'.[/quote]hi all,

thank you for the quick response and answer.
appreciate it.

cheers

[Dharma]

Yu Xuan:

Dharma:

[quote=\"trytry\"]Hi can anyone help to solve this type of question:


There are 5 boys A. B, C, Dand E. Each time 2 boys will pair up to have their mass taken. The ten readings in kg are 105, 113 116, 117, 118,124,125,130,137 and 139. What is the mass of the third heaviest boy?
TIA.

Mass of all 5 boys
= (105 + 113 + 116 + 117 + 118 + 124 + 125 + 130 + 137 + 139)/4
= 306

Sum of the mass of the 2 lightest boys and 2 heaviest boys
= 105 + 139
= 244

The mass of the 3rd heaviest boy = 306 – 244 = 62

cracking my head on how I should proceed after getting 306.

I surelyyou.[/quote]Hi Yu Xuan,

Thanks but I am actually learning from all of you everyday and I hope that our sharing will be truly beneficial to parents and their P6 kids in this coming PSLE Maths exam.

[firebird]

Dear Maths guru


Good evening.

Please help me on the following:

The sum of 4 numbers is 80.
If you add 3 to the first number, subtract 3 from the second number, multiply the third number by 3 and divide the fourth number by 3, the result will be equal.
What is the difference between the largest and the smallest of the original numbers?

Thank you
firebird

[Yu Xuan]

firebird:

Dear Maths guru


Good evening.

Please help me on the following:

The sum of 4 numbers is 80.
If you add 3 to the first number, subtract 3 from the second number, multiply the third number by 3 and divide the fourth number by 3, the result will be equal.
What is the difference between the largest and the smallest of the original numbers?

Thank you
firebird

Math guru can definitely present this better than me, but may I try it in my primitive way....

My model is;

1st - 2 unit +2
2nd - 3 unit +3
3rd - 1 unit
4th - 9 units
The tricky part is the sum of 1st and 2nd number is 6 units.

Hence 16 unit --> 80
1 unit --> 80/16 = 5
8 unit --> 5 * 8 = 40

Ans: 40

[firebird]

Dear Yu Xuan


Thank you very much for you answer.

With best regards
firebird

[Dharma]

firebird:

Dear Maths guru


Good evening.

Please help me on the following:

The sum of 4 numbers is 80.
If you add 3 to the first number, subtract 3 from the second number, multiply the third number by 3 and divide the fourth number by 3, the result will be equal.
What is the difference between the largest and the smallest of the original numbers?

Thank you
firebird

A + B + C + D = 80
A + 3 = B – 3 = 3C = D/3

A = 3C – 3
B = 3C + 3
D = 9C

(3C – 3) + (3C + 3) + C + 9C = 80
16C = 80
C = 5

A = 15 – 3 = 12
B = 15 + 3 = 18
C = 5
D = 45

Difference between the largest and smallest of the original numbers = D – C = 45 – 5 = 40

[Herbie]

Dear Dharma


I have one qn which i need yr help.

RGPS paper 2 Q 15. is a square embeded into a circle and the area of the square is 400sqm. Qn required us to find the shaded area between the square and circle.

The given answer is 228.32. However if the child uses the a2 + b2 = c2 he will derive at 228.38cm2.

Would you able to advise whether the answer 228.38 cm2 can be marked correct as well.

Tq

[Dharma]

Herbie:

Dear Dharma


I have one qn which i need yr help.

RGPS paper 2 Q 15. is a square embeded into a circle and the area of the square is 400sqm. Qn required us to find the shaded area between the square and circle.

The given answer is 228.32. However if the child uses the a2 + b2 = c2 he will derive at 228.38cm2.

Would you able to advise whether the answer 228.38 cm2 can be marked correct as well.

Tq

Hi Herbie,

I guess it will not make a difference which methed is used to find the diameter of the circle; either by Pythagoras Theorem or just from the Area of the Square (given)

The difference in the answer in the number of decimals places will depend on what one uses as pi.
If qns says pi = 3.142 or 22/7 then do what the question says.
If qn is silent about the value of pi, then use the pi button on the calculator.


http://www.postimage.org/image.php?v=Tst2900