Math Qn on Black and White hexagonal tiles

This is posted by LL♪♫. The following is a question from cbox: A very large floor is tiled with black and white regular hexagonal tiles. Each black tile is surrounded by 6 white tiles and each white tile is surrounded by 3 white tiles and 3 black tiles. Find the approximate ratio of the number of black tiles to the number of white tiles. Solution is below: The picture above is self-explanatory. The approx. ratio of the number of black tiles to the number of white tiles is 1:2.

Math Qn on White, Black and Red Marbles

This is posted by LL ☺

Qn on cbox: There are two containers A and B. Each of them contains 9 white marbles, 9 black marbles and 9 red marbles. If 10 marbles are removed from A and placed into B. how many marbles must be returned from B to A to make sure that there are at least 8 marbles of each colour in A?

Solution:
A: 9W, 9B, 9R
B: 9W, 9B, 9R

10 marbles are removed from A and placed into B. Assume 9W and 1B (any combination will do).
A: 0W, 8B, 9R
B: 18W, 10B, 9R

Now, return marbles from B to A to make sure that there are at least 8 marbles of each colour in A.
The most unlucky picks from B would be: 9R and 10B (A: 0W, 18B, 18R) followed by 8W (A: 8W, 18B, 18R)
Hence for worst case, number of marbles returned is 9+10+8 = 27 marbles.

1, 2, 3, 4, 5 Math Qn

Someone needed help with the following question:
The digits 1, 2, 3, 4 and 5 can form 120 five-digit numbers, each without any repeating digit. What is the sum of these 120 numbers?

Solution by Lim Li:
Each of the digits 1, 2, 3, 4 and 5 appears 24 (i.e. 120/5) times in the Ones, Tens, Hundreds, Thousands and 10 Thousands place.
1+2+3+4+5 = 15
24 x 15 = 360
Hence,
+++++360
++++360
+++360
++360
+360
------------
=3999960 (sum of all 120 numbers)
------------

2000th 7-digit Math Question

Someone posted the following question:
All the 7-digit numbers containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once, and not divisible by 5, are arranged in the increasing order. Find the 2000th number in this list.

Solved by Lim Li ☺☻:
Since the 7-digit number cannot be divisible by 5, the last digit cannot be 5.
If the 1st digit is fixed, there are 1x2x3x4x5x5 = 600 numbers
[1]765432 => 600th number
[2]765431 => 1200th
[3]765421 => 1800th
If the 1st and 2nd digits are fixed, there are 1x2x3x4x4 = 96 numbers
[41]76532 => 1896th
[42]76531 => 1992nd
:
[4312]756 => 1996th
[4315]267 => 1997th
[4315]276 => 1998th
[4315]627 => 1999th
[4315]672 => 2000th
Hence the 2000th number is 4315672

Math Question on ABCDEF

Someone posted the following question on the cbox:

Find the 6-digit number ABCDEF such that 7 x ABCDEF = 6 x DEFABC where DEFABC is another 6-digit number. Different letters stand for different single-digit whole numbers.

Solution:
Let
___
ABC = x
___
DEF = y

7 * (1000x + y) = 6 * (1000y + x)
6994x = 5993y
538x = 461y
Clearly, x = 461 and y = 538
Hence the 6-digit number is 461538.

Math Qn on Red and Blue Marbles

This is posted by LL ☺
Someone posted the following question on the cbox:

Container A contains 250 red marbles and 200 blue marbles. Container B contains 600 red marbles and 150 blue marbles.How many red and blue marbles must be moved from Container A to Container B such that 25% of the marbles in Container A are red and 75% of the marbles in Container B are red?

LL's solution:

before............after
A red 250.........1 unit(u)
..blue 200........3 units(u)

B red 600.........3 parts(p)
..blue 150........1 part(p)

4u+4p=1200 (total of red and blue marbles in containers A and B)
->1u+1p=300
3p+u=850 (total red marbles)
(3p+u)-(1u+1p)=850-300
2p=550, p=275
At first, container B has 150 blue balls.
Now, container B has 1 part=275 which is increased by 275-150=125 blue marbles
At first, container B has 600 red balls.
Now, container B has 3 parts=275x3=825 which is increased by 825-600=225 red marbles

Therefore, 225 red and 125 blue marbles must be moved from Container A to Container B.

Random Math Question

Find the shaded area.
Let the midpoint of line AD be E. (Centre of semi-circle). Let the arc of AC and AD cut at point F.
AE = EF = 3.5 (radius of circle)
BA = BF = 7 (radius of circle)
EB is common. Then,
Triangle ABE is congruent to Triangle FBE. Hence,
angle ABE = angle FBE = tan-1(AE/AB) = tan-1(1/2) (Note: tan-1 means tangent inverse)
angle ABF = angle ABE + angle FBE = 2 tan-1(1/2) (get this angle in degree)
Area of sector ABF = (2 tan-1 (1/2))/360 x 7 x 7 x pi = (49 pi tan-1(1/2))/180

angle AEB = angle FEB = tan-1(AB/AE) = tan-1(2)
angle AEF = angle AEB + angle FEB = 2 tan-1(2) (get this angle in degree too)
Area of sector AEF = (2 tan-1(2))/360 x 3.5 x 3.5 x pi = (49 pi tan-1(2))/720

Area of quadrilateral ABFE = 3.5x7 = 24.5

Area of shaded region = Area of sector ABF + Area of sector AEF - Area of quadrilateral ABFE
= (49 pi tan-1(1/2))/180 + (49 pi tan-1(2))/720 - 24.5
= 11.8cm2 (round off to 1 decimal point)
(exact pi is used, not 22/7)

Math Qn

Someone posted the following question on the cbox:
Let AG be x cm.
Then GB = 12-x
Triangle AGF is similar to Triangle ABC, hence AG = GF = x
Tiangle BGF is similar to Triangle BAE, hence
BG/GF = BA/AE
=> (12-x)/x = 12/6 (since AB = 12, AE = 6)
=> 12-x = 2x
=> x=4
Area of Triangle DEF = 1/2 x ED x AG = 1/2 x 6 x 4 = 12
Area of Triangle DFC = 1/2 x DC x (BC - GF) = 1/2 x 12 x (12-4) = 48
Hence, Area of shaded region = 12 + 48 = 60 cm2.

NMOS 2010 Special Round Qn 5

Question 5: A company could use any of the 3 builder (Anthony, Bob and Charles) to build a wall. Anthony would take A days to build the wall. Likewise, Bob and Charles would take B days and C days to build the wall respectively.
Anthony and Bob take 16.8 days to complete a wall.
Anthony and Charles take 21 days to complete a wall.
Bob and Charles take 28 days to complete a wall.

Solution:
Anthony builds 1 wall in A days. Hence, Anthony builds 1/A wall in 1 day.
Similarly, Bob builds 1/B wall in 1 day, and Charles builds 1/C wall in 1 day.
In 1 day, Anthony and Bob together will build (1/A + 1/B) wall. In 16.8 days, Anthony and Bob will build [16.8x(1/A + 1/B)] wall which is 1 wall since Anthony and Bob complete the wall in 16.8 days. Hence, we get the following equations:

16.8(1/A + 1/B) = 1
21(1/A + 1/C) = 1
28(1/B + 1/C) = 1
Implying..
1/A + 1/B = 1/16.8 -------- (1)
1/A + 1/C = 1/21 ---------- (2)
1/B + 1/C = 1/28 ---------- (3)

(1)+(2)-(3)
2(1/A) = 1/16.8 + 1/21 - 1/28 = 1/14
Therefore, 1/A = 1/28, hence, A = 28.
From (1), 1/B = 1/16.8 - 1/28 = 1/42. Hence, B = 42.
From (2), 1/C = 1/21 - 1/28 = 1/84. Hence, C = 84.

Answer: A+B+C=154

Qn from 2010 SMO (Junior) Special Round

Someone posted the following question on the cbox:
A student divides an integer m by a positive integer n, where n<=100 and claims that m/n=0.167a1a2... show the student must be wrong.

My solution:

Another Math Qn

Another question posted on the cbox:

A travelled from Town P to Q. B travelled from Town Q to P. They started at the same time. They met each other 80km from Q. When A reached Town Q, he immediately travelled back to Town P. When B reached Town P, he also immediately travelled back to Town Q. They met each other 60km from P. What was the distance from Town P to Town Q?

Solution (not drawn to scale) by Lim Li:

Qn from SMOPS 2010

Someone posted the following question from SMOPS 2010 in the cbox:

s=1/((1/2001)+(1/2002)+(1/2003)+(1/2004)+(1/2005)+(1/2006)+(1/2007)+(1/2008)+(1/2009)+(1/2010)). Find the largest whole number smaller than s.

Soln:
Clearly, 10/2001 > (1/2001)+(1/2002)+(1/2003)+(1/2004)+(1/2005)+(1/2006)+(1/2007)+(1/2008)+(1/2009)+(1/2010) > 10/2010
Hence 2001/10 < s < 2010/10
Therefore the largest whole number smaller than s is 200.

Number Patterns

This week's Columbus State University's Math contests are about number patterns.

Problem of the Week:
0, -2, -2, 0, 4, 10, ___, ___

Algebra in Action:
1.5, 1.0, 0.83, 0.75, 0.7, ___, ___

Middle School Madness:
1, 4, 9, 61, 52, 63, 94, ___, ___

Elementary Brain Teaser:
2, 5, 11, 17, 23, 31, 41, 47, ____, ____

Algebra in Action's pattern is more challenging than the rest as only about 1/4 of participants got the answer correct. You should be able to get the other 3 answers quickly. Just in case you can't solve the challenging one, check the solution here by highlighting the text -> 1.5 (3/2), 1.0 (4/4), 0.83 (5/6), 0.75 (6/8), 0.7 (7/10), 0.67 (8/12), 0.64 (9/14)

Question from AMC10A 2010

Someone posted a question (#18) from AMC10A on the cbox. It's easier and less messy to explain it here rather than on the cbox:

Qn: Bernardo randomly picks distinct numbers from the set (1,2,3,4,5,6,7,8,9) and arranges them in descending order to form a 3-digit number. Silvia randomly picks distinct numbers from the set (1,2,3,4,5,6,7,8) and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?

Solution:
Consider two scenarios: (a) Bernardo picks a '9' among the 3 numbers that she picks from her set of 9 numbers (b) Bernardo does not pick any '9'

Bernardo picks ‘9’:
When picking 3 numbers from the set (1,2,3,4,5,6,7,8,9), probability that Bernardo would pick a ‘9’ is 3/9 or 1/3. When this happens, Bernardo’s number (which has to be 9**, since the 3-digit number is to be arranged in descending order) will always be greater than Silvia’s number (obvious, since Silvia's greatest possible 3-digit number can only be 876!).

Bernardo does not pick ‘9’:
Number of different 3-digit numbers: 8C3 = 56
Probability that both Bernardo and Silvia pick the same 3-digit number is 1/56
Probability that both Bernardo and Silvia do not pick the same 3-digit number is 55/56
If both Bernardo and Silvia pick different number, probability that Bernardo’s number is bigger than Silvia’s number is ½ (similarly, probability that Bernardo’s number is smaller than Silvia’s number is also ½)

Therefore the probability that Bernardo’s number is larger than Silvia’s number is:
(Bernardo picks ‘9’) 1/3 + (Bernardo does not pick ‘9’) 2/3 x ½ x (55/56)
= 37/56

Four Maths Problems A Week!

The following is a good website for students who love to do interesting/challenging Math problems.
http://www.colstate.edu/mathcontest/index.php

It is hosted by Columbus State University’s College of Education. There are 4 categories: Problem of the Week, Algebra in Action, Middle School Madness and Elementary Brain Teaser.

Questions in the last 2 categories are usually more elementary. I will usually try to solve the problems in the first 2 categories, while my younger sister will attempt questions in the last 2 categories. If you know the answers, you can submit them. Your name will be published immediately (usually within a day) after your answer has been reviewed and it is correct.

There is a problem a week (every Monday) for each category. I chanced upon this website only recently and I like it! You can also try the past problems, or problems in Challenge Mode (where <50% students got the answers correct).

Have fun!

Another PSLE Math Qn

Another question posted on my blog:
Jim bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Jim. Jim ate 12 sweets and Ken ate 18 chocolates. The ratio of Jim’s sweets to chocolates became 1:7 and the ratio of Ken’s sweets to chocolates became 1:4. How many sweets did Ken buy?

My mother asked my younger sister to solve this question, and her solution (it is correct) is as follows:

PSLE questions

A PSLE student posted the following 2 questions on my cbox. I thought it is easier to explain it here than on the cbox. Here it is:

Geom question:

Given: ABCD is a square, EG=EF=EH. Find angle GFH.

Since ABCD is a square, AB=BC=CD=DA

As EF=BC=DA and AB=GH, hence EG=EH=AB=GH.

Therefore, triangle EGH is equilateral since EG=GH=EH.

angle GEH=60°

angle GEF=60/2=30°

Triangle EGF is isosceles. Hence, angle FGE= angle GFE = (180-30)/2 = 75°

angle GFE = angle HFE = 75°

Therefore angle GFH= 75X2=150°

Balloon question:

Given: String of 2 big balloons is 90cm String of 5 small balloons is is 1.2m If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether?

105 small balloons would have a length of (105/5) X 1.2 = 25.2m = 2520cm

1 big balloon has a length of 90/2=45cm

1 small balloon has a length of 1.2m/5= 0.24m= 24cm

Difference in length of big and small balloon = 45-24= 21 cm

Number of big balloons = 2520/21 = 120

Number of small balloons = 120+105 = 225

Total number of balloons = 120+225 = 345

Fun Math