Tutor MathsGuru: Ask me for your burning Maths questions!

[Vanilla Cake]

http://postimage.org/image/32rskcgh0/

Pls assist to provide worked solution for this problem sum.
Thks in advance.

[Dharma]

Vanilla Cake:

http://postimage.org/image/32rskcgh0/

Pls assist to provide worked solution for this problem sum.
Thks in advance.

Hi Vanilla Cake,


http://postimage.org/image/ttml6zd0/

[wahwah]

Hi all,

Can someone help me with the following question? Many thanks in advance.

One thousand "2007" are joined together to form a new number 200720072007…20072007(repeated 1000times).
The first m digits and the last n digits of this new number are then removed such that the sum of digits of the resulting number is 7999. Given that both m and n are at least 4, find the greatest value of m-n.

(1) 436
(2) 438
(3) 439
(4) 443
(5) None one the above

[Vanilla Cake]

http://postimage.org/image/vm18njpg/

http://postimage.org/image/18g81qxw/
Thks in advance for your help.

[Dharma]

Vanilla Cake:

http://postimage.org/image/vm18njpg/

http://postimage.org/image/18g81qxw/
Thks in advance for your help.

Dear Vanilla Cake,

Q3.

Area of AOD : Area of COD = OA : OC = 2 : 3
Area of COD : Area of BOC = OD : OB = 3 : 2

Area of AOD/ Area of BOC = 2/3 x 3/2 = 1

Therefore, area of AOD : Area of BOC = 1 : 1

Q28

If you draw the symmetrical lines of the shifted vertical and horizontal diameters to the vertical and horizontal diameters that pass through origin O.

You will find the shaded and unshaded portions cancel off each other leaving a rectangular shaded portion of length 6cm and width 4cm.

Nett area = 6cm x 4cm = 24cm2

[babyluvv]

Hi blessedami,


Thank you for helping!!
:lol: :lol: :lol: :lol:

[Dharma]

wahwah:

Hi all,

Can someone help me with the following question? Many thanks in advance.

One thousand \"2007\" are joined together to form a new number 200720072007...20072007(repeated 1000times).
The first m digits and the last n digits of this new number are then removed such that the sum of digits of the resulting number is 7999. Given that both m and n are at least 4, find the greatest value of m-n.

(1) 436
(2) 438
(3) 439
(4) 443
(5) None one the above

7999 divided by 9 = 888R7
The resulting number left has 888 sets of 7200 + a single digit “7”.
Since the last digit of the number ends with “7”, the remaining number will start with “2”
and the last 4 digits (n = 4) is 2007
Total number of digits = 4000 (“2007” is repeated 1000 times)
Total digits of the resulting number = 888x4 +1 = 3553
m = 4000 – 3553 – 4 = 443

The greatest value of m – n = 443 – 4 = 439

[wahwah]

Hi Dharma,

Thank you so much! We didn’t understand what the question means when it says " m and n are at least 4". My son only managed to divide 7999 by 9 but didn’t know what to do with 888R7. Now he understands, thanks!

Can I trouble you with another Math question?

Divide 0.123456789 by 0.9192939495 (do not use a calculator). The first 3 digits after the decimal point are

(1)132
(2)133
(3)134
(4)135
(5)None of the above

[Dharma]

wahwah:

Hi Dharma,

Thank you so much! We didn't understand what the question means when it says \" m and n are at least 4\". My son only managed to divide 7999 by 9 but didn't know what to do with 888R7. Now he understands, thanks!

Can I trouble you with another Math question?

Divide 0.123456789 by 0.9192939495 (do not use a calculator). The first 3 digits after the decimal point are

(1)132
(2)133
(3)134
(4)135
(5)None of the above

I think the simplest way is to do long division.

1234567890 divided by 9192939495 ....you will get 0.134...

[MathIzzzFun]

wahwah:

Hi Dharma,

Thank you so much! We didn't understand what the question means when it says \" m and n are at least 4\". My son only managed to divide 7999 by 9 but didn't know what to do with 888R7. Now he understands, thanks!

Can I trouble you with another Math question?

Divide 0.123456789 by 0.9192939495 (do not use a calculator). The first 3 digits after the decimal point are

(1)132
(2)133
(3)134
(4)135
(5)None of the above

Hi

It might be quicker to do bracketing

0.123456789 / 0.92 < 0.123456789 / 0.9192939495 < 0.123456789 / 0.91
9
(ie faster to divide by a 2/3 digit number and need only find the first 4 digits after decimal point)

0.1341.... < 0.123456789 / 0.9192939495 < 0.1343....

so the first 3 digits after the decimal point is 134

cheers