# Compilation of my Math mini-contest problems on D.Buzz for January 2021 š

Public domain image from Pixabay š

Last month (January 2021), I hosted a Math mini-contest for D.Buzz. š It was intended to run from the 17th to 24th of the month. š¤ However, because I did not expect that nobody would have answered the Math problem correctly on the last day, I hosted another day on January 26 (to 27) to get a Day 7 winner. š

Listed below are the Math problems I asked, and the solutions for each of them! š

(Each problem and solution is hyperlinked to their original sources.)

# Day 1 Problem

Assuming an onboarded D.Buzz user has a 50% chance of changing their profile picture on their first day and 60% chance of doing so within their first two days, what is the chance that the user will change their profile picture on their second day?

# Day 1 Solution

### Answer cannot be determined due to insufficient given. šÆ

There is not enough data on the given to determine the chance of the user changing their profile picture on Day 2, unless it specifies that the user did not change the profile picture on the first day.

If the user changes his profile picture on the first day, there is nothing on the given that says whether it will affect the chance of the user changing their picture on the second day.

If the user has not changed their picture on the first day, then through conditional probability, we say that the chance the user changes his picture on the second day is 20%. However, if the user did change their picture on the first day, then the chance of the user changing the picture on the second day is unknown.

# Day 2 Problem

A developer can make a Hive dapp alone in 30 days. Another developer can make the same dapp alone in 36 days. If both developers work together, how much time will it take to finish making the dapp, assuming all other variables equal?

# Day 2 Solution

## 16.36 days šÆ

The problem is an algebraic work problem. Let t be the total time required by any developer to finish the Hive dapp alone. Every day, the first developer does 1/30 of the work done when alone, while the second developer does 1/36 of the work done also when alone. The question is how much time the work will take if they work together.

t * [1/30 + 1/36] = 1 day
t = (1 day) / (1/30 + 1/36)
t = (1 day) / [(30+36)[30*36)]
t = (1 day) / (66/1080)
t = (1 day) * 1080 / 66
t = (1080 / 66) days
t = 16.36 days

# Day 3 Problem

A get-together event of 10 D.Buzz members plus @chrisrice and @emafe (for a total of 12 people) is done seated around a round table. If every attendee's seat is random, what is the probability that @chrisrice is seated beside @emafe?

# Day 3 Solution

## 2/11 or 18.18% šÆ

There will be 11 seats remaining after @chrisrice gets seated, and @emafe can be seated beside @chrisrice in either his left or right, which is 2 out of 11 possible chances.

# Day 4 Problem

@chrisrice is 13 years older than @jancharlest. Exactly 9 years ago, @chrisrice was twice as old as @jancharlest. What are their ages at present?

# Day 4 Solution

### @chrisrice is 35 years old and @jancharlest is 22 years old. šÆ

Let x be the present age of @jancharlest in years. Since @chrisrice is 13 years older than @jancharlest, let (x + 13) be the age of @chrisrice.

When we look at 9 years ago, the age of @jancharlest was (x - 9) and the age of @chrisrice was (x + 13 - 9) or simply (x + 4). Since it was stated in the original problem that 9 years ago, the age of @chrisrice was twice the age of @jancharlest, our equation can be worded into: "The age of @chrisrice plus 4 is equal to 2 times the age of @jancharlest after deducting 9 from it", or simply (x + 4) = 2 * (x - 9).

PresentPast (9 years ago)
@jancharlestxx - 9
@chrisricex + 13x + 4

(x + 4) = 2 * (x - 9)
x + 4 = 2x - 18
x - 2x = -18 - 4
-x = -22
x = 22

Since we represented the present age of @jancharlest as x years old, and the value of x is 22, his age in years is 22. Since @chrisrice is 13 years older than @jancharlest, the age of @chrisrice in years is (22 + 13) or simply 35.

# Day 5 Problem

On a poll for Hive users regarding which Hive dapp they use, the following data were obtained:

ā¢ 28% use PeakD
ā¢ 35% use HiveBlog
ā¢ 64% use DBuzz
ā¢ 5% use both PeakD and HiveBlog
ā¢ 15% use both HiveBlog and DBuzz
ā¢ 12% use both PeakD and DBuzz
ā¢ 7% use all three

What percentage of the poll respondents who use DBuzz do not use either HiveBlog or PeakD?

# Day 5 Solution

### The data in the problem are inconsistent with each other. šÆš

From the poll, there are 64% who use DBuzz, and each of them may or may not be using another Hive dapp. There are 34% (obtained from 64%-15%-12%-7%) of all users who are using DBuzz plus at least one other Hive dapp. In total, there are 30% of all users who use DBuzz only, and the percentage of DBuzz-only users from those DBuzz users who use at least one other dapp is 46.875% (obtained from 30% divided by 64% or 0.64).

However, the solution above assumes that we ignore the inconsistency that there are 7% who use all three dapps, but only 5% who use both PeakD and HiveBlog (and may or may not be using DBuzz). That will result in -2% of users who use both PeakD and HiveBlog but not DBuzz, which is impossible.

# Day 6 Problem

In a game of basketball, @chrisrice has an 18% chance of scoring points in one throw while @jancharlest has a 45% chance. If each of them tries to shoot once, what is the chance @jancharlest successfully scores points while @chrisrice does not?

# 36.9% šÆ

@chrisrice@jancharlest
Chance to shoot ball18%45%
Chance not to shoot ball82%55%

The problem is basically the chance of @jancharlest shooting the ball multiplied by the chance of @chrisrice not shooting the ball. The solution is 82% multiplied by 45% (or 0.45), whose result is 36.9%.

# Day 7 (Part 1) Problem

@chrisrice and @jancharlest are playing in a laser tag game with the other D.Buzz admins.

@jancharlest is on top of an elevated platform and is aiming his laser gun directly at @chrisrice below him at an angle of 32.5Ā° downwards from horizontal. The tip of @jancharlest's laser gun is 8 meters away from @chrisrice. The ground is perfectly horizontal.

How high from the ground is the laser gun tip of @jancharlest?

# Day 7 (Part 1) Solution

## 4.30 meters šÆ

Let A be the point where the tip of @jancharlest's laser gun is at. Let B be the point where the target @chrisrice is. Let C be the point of intersection of the vertical line where @jancharlest is at and the horizontal ground line, which form an angle of 90Ā°.

Since the aim of @jancharlest is basically a laser point on the target, the target @chrisrice will be represented as a point. @jancharlest is aiming 32.5Ā° downwards, which is practically the same angle as @chrisrice looking up to @jancharlest's laser gun tip.

We now have a logical triangle ABC. The given data are AC = 8 meters, ACB = 90Ā°, and ABC = 32.5%. By derivation, we get CAB = 57.5% by deducting the two known angles from 180Ā° which is the sum of all angles of a triangle. Since all angles are known, we can use either the Law of Sines, Law of Cosines, or Law of Tangents. For this solution, we use the Law of Sines because it is the most simple among the three formulas.

Law of Sines: a / sin (A) = b / sin (B) = c / sin (C)

For the original problem, we will use b / sin (B) = c / sin (C).

In this solution, [side b] is AC (which is the unknown in this equation) and angle B is CBA or 32.5Ā°, while [side c] is 8 meters and angle C is ACB or 90Ā°.

AC / sin (32.5Ā°) = 8 meters / sin (90Ā°)
AC / 0.5373 = 8 meters / 1
AC = 0.5373 * 8 meters
AC = 4.2984 meters

Therefore, the distance between the ground and @jancharlest's laser gun tip (basically the height) is approximately 4.30 meters. We can have multiple solutions for the problem, but the answer shall be the same.

# Day 7 (Part 2) Problem

In a party game, teams of 2 are to be formed by randomly drawing pieces of paper ("stubs") containing the name of the participants from two separate bags. One stub is to be drawn from each bag, and the two people with their names on the stubs will comprise a new team as partners. This process repeats until either bag has no more stubs.

The stub of @chrisrice is in a bag with 6 other stubs, while @emafe's stub is in another bag with 8 other subs. However, just before the draw, @jancharlest sneaked his stub into one of the bags.

What is the chance that @chrisrice gets partnered with @emafe?

# Day 7 (Part 2) Solution

## 19 out of 180 or 10.56% šÆ

Let C = @chrisrice's stub
Let J = @jancharlest's stub
Let E = @emafe's stub

Since C is in the bag with the less number of stubs among the two bags, there is a 100% chance to pick C in any of the 7 draws from that bag.

(If @jancharlest did not sneak his stub into any of the bags, then we will have gotten the answer right away, which is one out of all the stubs in the bag with @emafe's stub, or 1 out of 9. It does not matter how many draws happened before @chrisrice's stub got picked, since our perspective is from the very start of the drawings.)

@jancharlest inserted J into the bag where either C or E is, where there is an equal (50%) probability of J being in a specific bag. This leads to any the following:

If @jancharlest inserted J where C is, then the bag with C still has 8 stubs while the bag where E is has 9 stubs. In this case, the chance of E getting picked for C is 1 out of 9.

On the other hand, if @jancharlest inserted his stub where @emafe's stub is, then the bag with C still has 7 stubs while the bag with E has 10 stubs. In this case, E getting picked immediately after C is 1 out of 10.

Since there are 50% chances of having either a 1 out of 9 chance or 1 out of 10 chance, we can just average those two chances to arrive at the final answer. (1/9 + 1/10) / 2 = 19/180

There are 19 out of 180 chances or approximately 10.56% chance of @chrisrice getting partnered with @emafe.

### Winner: @jfang003 š

Thanks to all participants!

Mentions: @eturnerx (@eturnerx-dbuzz) and @appukuttan66
Special mentions: @dbuzz, @chrisrice, and @jancharlest
Ā