Tutor MathsGuru: Ask me for your burning Maths questions!

[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.

[Kiasu Friend]

Dear Dharma,


I understand perfectly now. Thank you very much.

In my view, even a personal 1-to-1 Tutor visiting home for tutoring may not teach as well as you (no offense meant for tutors, please). You are truly great!

[Dharma]

Kiasu Friend:

Dear Dharma,


I understand perfectly now. Thank you very much.

In my view, even a personal 1-to-1 Tutor visiting home for tutoring may not teach as well as you (no offense meant for tutors, please). You are truly great!

Thanks. I'm glad u understand.

[liketoeat]

Hi, I need help with the following questions…


1) Jane has some money. After spending $100 on 3 bags and 7 belts, she is short of $4 if she wants to buy another bag. However, if she buys 1 more belt, she will have $6 left.
a) How much does one bag cost? (Ans: $13)
b) How much does Jane have? (Ans: $109)

2) Car A left Town X for Town Y at the same time when Car B left Town Y for Town X. The average speed of Car A was 56 km/h and the average speed for Car B was 72 km/h. The two cars passed each other at a point 24 km from the midway of the two towns. What is the distance between Town X and Town Y? (Ans: 384km)

Thank you.

[Brenda10]

liketoeat:

1) Jane has some money. After spending $100 on 3 bags and 7 belts, she is short of $4 if she wants to buy another bag. However, if she buys 1 more belt, she will have $6 left.

a) How much does one bag cost? (Ans: $13)
b) How much does Jane have? (Ans: $109)

Hi, I have come across the similar question. However now I try to solve this question by following the previous sample I couldn’t get the answer same as you provided. I can only get (a) $17, (b) $113

That’s mean that I kept the wrong sample!! :!: Sorry not able to help and I also need to seek help from the rest.

[tianzhu]

Brenda10:

Hi, I have come across the similar question. However now I try to solve this question by following the previous sample I couldn’t get the answer same as you provided. I can only get (a) $17, (b) $113

Hi Brenda10.

Your answers are correct.

Best wishes

[tianzhu]

liketoeat:

2) Car A left Town X for Town Y at the same time when Car B left Town Y for Town X. The average speed of Car A was 56 km/h and the average speed for Car B was 72 km/h. The two cars passed each other at a point 24 km from the midway of the two towns. What is the distance between Town X and Town Y? (Ans: 384km)

Hi

One way to solve this question is to use the concept that distance varies in direct proportion to speed. The faster the speed, the more distance is covered.

Car A ----- Speed ---- 56(7) ------- Distance -------7u
Car B ------ Speed ----- 72(9) -------Distance -------9u

The midpoint is at 8u (16/2)
1u ------24 km
16u ------384 km

Best wishes

[Cheerfuldad]

Hi all,


Please help!

Q. 400 people took part in a camp. 73 of them were adults and the rest were children. 2/5 of the boys and 1/4 of the girls were lower primary pupils while the rest of them were upper primary pupils. There were 9 more upper primary girls than upper primary boys.

a) How many upper primary pupils were there?

b) What percentage of the people who took part in the camp were girls?


TIA