Tutor MathsGuru: Ask me for your burning Maths questions!

[Dharma]

super star:

HI MATHS GURU

Please help me with this problem.At present,ronald is 3 times as old as his sister.In 22 years time,ronald's age will be 19 years less than twice his sister's age.how old is ronald?

Now
Ron : 3u
Sister : 1u

22 years time
Ron : 3u + 22
Sister : 1u + 22

3u + 22 = 2(1u + 22) – 19
3u + 22 = 2u + 25
1u = 3

Ronald age => 3u = 3 x 3 = 9 yrs old

[Dharma]

kancheongmum:

tianzhu:

Hi clblinym


You're welcome.

Best wishes

Hi tianzhu really like your graphics in your answers. I hope you don't mind explaining to me on the solution of sheep eating grass again. I don't quite get it thank very much.

Hi clblinym and kancheongmum,

This is another way to solve the \"sheep eat grass problem\"

Alternative method
200 sheep can eat up all the grass in 100 days
150 sheep can eat up all the grass in 150 days

Lets say a sheep can eat 1u of grass a day;

200 sheep in 100 days => 200 x 100 x 1u = 20000u
150 sheep in 150 days => 150 x 150 x 1u = 22500u

The difference is due to the fact that the grass in the pastures continues to grow.

Rate of growth of grass = (22500u – 20000u)/(150 – 100) days = 50u/day

Original amt of grass = 20000u – 50u(100 days) = 15000u

The 200 sheep will take 75 days to finish eating the original grass i.e 15000u/200u = 75 days
The balance 5000u (new grass that grew) will be eaten in 5000u/(200u/day)= 25 days

Therefore; 200 sheep will take 100 days (75days + 25days) to complete eating all the grass.

Now lets say that 100 sheep will take N days to eat up all the grass.

No. of days for the 100 sheep to eat up the original amt of grass = 15000u/(100u/day) = 150 days

Amt of new grass grown in N days = 50u(N)

No. of days for the 100 sheep to eat up the original amt of new grass grown
= 50u(N)/(100u/day) = ½ N days

100 sheep takes ½ N days to eat up the new grass; so it must have also taken ½ N days to eat up the original grass.

½ N = 150 days
N = 300 days.

100 sheep will take 300 days to eat up all the grass.

[CJM]

MOE Hater:

CJM:

hi all, pls help with this question :


A large water tank has two inlet pipes ( a large one and a small one ) and one outlet pipe. it takes 3 hours to fill the tank with the large inlet pipe. On the other hand, it takes 4 hours to fill the tank with the small inlet pipe. The outlet pipe allows the full tank to be emptied in 7 hours.

What fraction of the tank ( initially empty ) will be filled in 2.27 hours if all three pipes are in operation? Give your answer to two decimal places.

I don't understand. You asked for a fraction then said to give the answer to 2 decimal places

This is a question given by the teacher.
We do not have the answer and he wants the answer to be converted to decimal and round off to 2 decimal places

[CJM]

maths6a:

MOE Hater:

[quote=\"CJM\"]hi all, pls help with this question :


A large water tank has two inlet pipes ( a large one and a small one ) and one outlet pipe. it takes 3 hours to fill the tank with the large inlet pipe. On the other hand, it takes 4 hours to fill the tank with the small inlet pipe. The outlet pipe allows the full tank to be emptied in 7 hours.

What fraction of the tank ( initially empty ) will be filled in 2.27 hours if all three pipes are in operation? Give your answer to two decimal places.

I don't understand. You asked for a fraction then said to give the answer to 2 decimal places

Sorry I worked answer to be 0.999 so rounded up will be 1.00 which is like whole tank is filled???[/quote]may i know how did u derive the answer?
ds need to show the steps in deriving the answer but I have no clue at all.

thanks

[Dharma]

CJM:

maths6a:

[quote=\"MOE Hater\"]
I don't understand. You asked for a fraction then said to give the answer to 2 decimal places

Sorry I worked answer to be 0.999 so rounded up will be 1.00 which is like whole tank is filled???

may i know how did u derive the answer?
ds need to show the steps in deriving the answer but I have no clue at all.

thanks[/quote]In 1 hr
Inflow : (1/3 + ¼) = 7/12 of tank
Outflow : 1/7 of tank

Net flow : 7/12 – 1/7 = 49/84 – 12/84 = 37/84 of tank filled

In 2.27 hrs (use calculator)
Proportion of of tank filled: (37/84 x 2.27 hrs) = 0.9998 = 1.00 ( to 2 decimal places)

[Kiasu Friend]

Hi Dharma/ Mathsguru,


Please help with this problem (from MGS Prelim).

Study the number pattern below:

5 x 5 = 25
5 x 5 x 5 = 125
5 x 5 x 5 x 5 = 625
5 x 5 x 5 x 5 x 5 = 3,125

(a) In (5 x 5 x 5 x 5 x 5 x 5), i.e., "5 to the power of 6" what is the sum of the digits in the ones and tens place?

(b) What is the sum of the last THREE digits in "5 to the power of 15" ?

What is the sum of the last FOUR digits in "5 to the power of 210"?

Thanks a lot.

[Dharma]

Kiasu Friend:

Hi Dharma/ Mathsguru,


Please help with this problem (from MGS Prelim).

Study the number pattern below:

5 x 5 = 25
5 x 5 x 5 = 125
5 x 5 x 5 x 5 = 625
5 x 5 x 5 x 5 x 5 = 3,125

(a) In (5 x 5 x 5 x 5 x 5 x 5), i.e., \"5 to the power of 6\" what is the sum of the digits in the ones and tens place?

(b) What is the sum of the last THREE digits in \"5 to the power of 15\" ?

(c) What is the sum of the last FOUR digits in \"5 to the power of 210\"?

Thanks a lot.

a)\tSum of digits in ones and tens place = 2 + 5 = 7

b)\t5 to the power of odd number => last 3 digits = 125
5 to the power of even number => last 3 digits = 625
5 to the power of 15 => 15 is odd number ; so last 3 digits = 125
Sum of last 3 digits = 1 + 2 + 5 = 8

c)\t5 to power of 5 = 3125 => Sum of last 4 digits =11
5 to power of 6 = 15,625 => Sum of last 4 digits =18
5 to power of 7 = 78,125 => Sum of last 4 digits = 16
5 to power of 8 = 390,625 => Sum of last 4 digits = 13
5 to power of 9 = 1,953,125=> Sum of last 4 digits = 11

210 divide by 4 =52R2
=> Last 4 digits of 5 to the power of 210 = 5625
5 to power of 210 => Sum of last 4 digits = 18

[CJM]

Dharma:

CJM:

[quote=\"maths6a\"]





Sorry I worked answer to be 0.999 so rounded up will be 1.00 which is like whole tank is filled???

may i know how did u derive the answer?
ds need to show the steps in deriving the answer but I have no clue at all.

thanks

In 1 hr
Inflow : (1/3 + ¼) = 7/12 of tank
Outflow : 1/7 of tank

Net flow : 7/12 – 1/7 = 49/84 – 12/84 = 37/84 of tank filled

In 2.27 hrs (use calculator)
Proportion of of tank filled: (37/84 x 2.27 hrs) = 0.9998 = 1.00 ( to 2 decimal places)[/quote]thank U so much for the solution !

[Kiasu Friend]

Dharma:

Kiasu Friend:

Hi Dharma/ Mathsguru,


Please help with this problem (from MGS Prelim).

Study the number pattern below:

5 x 5 = 25
5 x 5 x 5 = 125
5 x 5 x 5 x 5 = 625
5 x 5 x 5 x 5 x 5 = 3,125

(a) In (5 x 5 x 5 x 5 x 5 x 5), i.e., \"5 to the power of 6\" what is the sum of the digits in the ones and tens place?

(b) What is the sum of the last THREE digits in \"5 to the power of 15\" ?

(c) What is the sum of the last FOUR digits in \"5 to the power of 210\"?

Thanks a lot.

a)\tSum of digits in ones and tens place = 2 + 5 = 7

b)\t5 to the power of odd number => last 3 digits = 125
5 to the power of even number => last 3 digits = 625
5 to the power of 15 => 15 is odd number ; so last 3 digits = 125
Sum of last 3 digits = 1 + 2 + 5 = 8

c)\t5 to power of 5 = 3125 => Sum of last 4 digits =11
5 to power of 6 = 15,625 => Sum of last 4 digits =18
5 to power of 7 = 78,125 => Sum of last 4 digits = 16
5 to power of 8 = 390,625 => Sum of last 4 digits = 13
5 to power of 9 = 1,953,125=> Sum of last 4 digits = 11

210 divide by 4 =52R2
=> Last 4 digits of 5 to the power of 210 = 5625
5 to power of 210 => Sum of last 4 digits = 18

Dear Dharma,

Thanks a lot for giving solution so fast. Very very fast indeed.
But I do not understand the part marked in red. My questions are:

(1) Why divide 210 by 4? Is it because we want the sum of last 4 digits? If we want the sum of last 3 digits, will we divide by 3?

(2) When we divide 210 by 4, we get 52R2. From that, how do we conclude that the last 4 digits of \"5 to the power of 210\" are 5625?

(3) Is there any concept or formula I should learn to understand the above?

Kindly explain these. I am most grateful for sparing your time.
Thank you so much.

[Dharma]

Kiasu Friend:

Dharma:

[quote=\"Kiasu Friend\"]Hi Dharma/ Mathsguru,


Please help with this problem (from MGS Prelim).

Study the number pattern below:

5 x 5 = 25
5 x 5 x 5 = 125
5 x 5 x 5 x 5 = 625
5 x 5 x 5 x 5 x 5 = 3,125

(a) In (5 x 5 x 5 x 5 x 5 x 5), i.e., \"5 to the power of 6\" what is the sum of the digits in the ones and tens place?

(b) What is the sum of the last THREE digits in \"5 to the power of 15\" ?

(c) What is the sum of the last FOUR digits in \"5 to the power of 210\"?

Thanks a lot.

a)\tSum of digits in ones and tens place = 2 + 5 = 7

b)\t5 to the power of odd number => last 3 digits = 125
5 to the power of even number => last 3 digits = 625
5 to the power of 15 => 15 is odd number ; so last 3 digits = 125
Sum of last 3 digits = 1 + 2 + 5 = 8

c)\t5 to power of 5 = 3125 => Sum of last 4 digits =11
5 to power of 6 = 15,625 => Sum of last 4 digits =18
5 to power of 7 = 78,125 => Sum of last 4 digits = 16
5 to power of 8 = 390,625 => Sum of last 4 digits = 13
5 to power of 9 = 1,953,125=> Sum of last 4 digits = 11

210 divide by 4 =52R2
=> Last 4 digits of 5 to the power of 210 = 5625
5 to power of 210 => Sum of last 4 digits = 18

Dear Dharma,

Thanks a lot for giving solution so fast. Very very fast indeed.
But I do not understand the part marked in red. My questions are:

(1) Why divide 210 by 4? Is it because we want the sum of last 4 digits? If we want the sum of last 3 digits, will we divide by 3?

(2) When we divide 210 by 4, we get 52R2. From that, how do we conclude that the last 4 digits of \"5 to the power of 210\" are 5625?

(3) Is there any concept or formula I should learn to understand the above?

Kindly explain these. I am most grateful for sparing your time.
Thank you so much.[/quote]If u notice that sum of the last 4 digits repeats itself every 4 times ( 5 to power of 5) gives 11, (5 to power of 6) gives 18, (5 to power of 7) gives 16 , (5 to power of 8 ) gives 13 and (5 to power of 9) gives 11 again.

5 to power of 5 => 5 divide by 4 = 1R1
5 to power of 6 => 6 divide by 4 = 1R2
5 to power of 7 => 7 divide by 4 = 1R3
5 to power of 8 => 8 divide by 4 = 2
5 to power of 9 => 9 divide by 4 = 2R1

So, from the pattern;
5 to power of 210 => 210 divide by 4 = 52R2 …Remainder 2 (last 4 digits = 5625 => giving a sum of 18 for the last 4 digits)

For sum of last 3 digits; you see the last 3 digits alternate between 125 and 625.
5 to the power of odd number => last 3 digits = 125
5 to the power of even number => last 3 digits = 625

There is no concept or formula….but impt to able to establish pattern.