Math in the Modern World Assignment 3.1 ```Exercise 1.1
Use inductive or deductive reasoning to answer the given questions.
1. Predict the next number in each given set of numbers
a. 3, 6, 9, 12, 15,?
b. 1, 3, 6, 10, 15,?
a. 18
b. 21
2. What type of reasoning did you apply in 1a and 1b?
a. Deductive
b. Inductive
3. Tell whether the mathematical statements is true or false.
If false, provide a counterexample. Assume x is any real number
a. If x > 0 then 1/x > 0.
b. The multiplicative inverse 1/x always exists.
c. 𝑥 2 is always positive.
a. True
b. True
c. True
4. In a neighborhood of engineers, it is known that there is a chemical engineer, civil engineer,
mechanical engineer, and an electrical engineer among Tito, Vic, Joey, and Willy. Identify the
correct profession of each engineer given the following clues:
i.
Vic gets home from work after the civil engineer but before the electrical engineer.
ii.
Joey, who is the last to get home from work, is not the electrical engineer.
iii.
The electrical engineer and Joey leave for work at the same time.
iv.
The civil engineer lives next door to Willy.
Willy is the electrical engineer
Joey is the Chemical Engineer
Vic is the Mechanical Engineer
Tito is the Civil Engineer
5. A new bot band composed of four young boys Josh, Jude, Benedict, and Francis was formed by a
recording agency. With one acting as the lead vocalist, each of the three other boys takes cae of
the drum, the keyboard, and the guitar. Based on the clues specified below, identify the role of
each member in the band.
i.
Francis is younger than the vocalist.
ii.
Josh and the guitarist are twins and they are the youngest members of the band.
iii.
Benedict and the guitarist are schoolmates.
Francis is the guitarist
Benedict is the Vocalist
Francis is the drummer
Josh is the pianist
Exercise 2.1
Solve each problem using Polya’s four-step problem solving strategy.
1. The elimination stage of the UAAP’s women’s volleyball competition is played in two rounds
where each competing team plays against each of the other teams once in every round. How
many games are played in the elimination round if there are eight teams in a given season?
i.
ii.
Understand the Problem
Find the total games if there are eight teams.
Device a Plan
We could apply the statistical formula, known as combination, to solve the
problem. To find the total arrangement of each teams. After finding the combination, it
will be the total games played since that in one game there are 2 rounds.
nCr =n!(n-r)!r!
Let:
iii.
iv.
n = teams
r = team competing with another team
Carry out the plan
8P2 =n!/(n-r)! = 8!/(8-2)! = 56 games.
Look back
Team 1 Team 2 Team 3
Team 4
Team 5
Team 6
Team 7
Team 8
Team 1
Game 1 Game 2
Game 3
Game 4
Game 5
Game 6
Game 7
Team 2
Game 8
Game 9
Game 10 Game 11 Game 12 Game 13
Team 3
Game 14 Game 15 Game 16 Game 17 Game 18
Team 4
Game 19 Game 20 Game 21 Game 22
Team 5
Game 23 Game 24 Game 25
Team 6
Game 26 Game 27
Team 7
Game 28
Team 8
Each game consists of 2 rounds and it is not repetitive.
2. Find the digit that is 50 places to the right of the decimal point in the decimal representation of
the rational number.
i.
Understand the problem
Find the digit that is 50 places to the right of the decimal point in the decimal
form of the rational number. By using a sample rational number, 2/27
ii.
iii.
iv.
Device a Plan
First, the rational number should be expressed in decimal form
(0.074074074074074074074…). For instance, if the third digit of a decimal number is the
same as the one in the 6th, 9th, 12th, 15th end etc., it means that any digit where the
place is divisible by 3 is the same digit found in third place.
Carry out the plan
It is to note that the pattern in the decimal sequence repeats after every 3 digits.
In the decimal representation of a rational number, when looking at the 50th digit to the
right of the decimal, it is noticeable that the previous number of it is divisible by 3,
meaning the number after the 3rd digit is the same as the 50th digit.
Look back
A rational number is a number which can be written in the form , where a and b
are both integers and b is not equal to 0.
3. A coffee shop is giving away a signature annual planner. In the mechanics, each customer has to
collect 24 stickers to avail of the said planner, and customers can share stickers. At the end of
the promo period, Tito had the greatest number of stickers, more than enough to get the
planner. Unfortunately, Vic and Joey did not have enough. This is what they did: First, Tito gave
Vic and Joey as many stickers as each had; after this Vic gave Tito and Joey as many stickers as
many as they had. At the end, each of the three friends had exactly enough stickers to get a
planner. How many stickers did each person have at the start?
i.
ii.
iii.
Understand the Problem
Find the number of presents each person had at the start.
Device a Plan
Let x, y, and z be the number of stickers Tito, Vic, and Joey had at the start, respectively.
 Tito gave Vic and Joey as many stickers as each had;
 Vic gave Tito and Joey as many stickers as they had; and
 Joey gave Tito and Vic as many stickers as they had.
After doing these steps, there will be three mathematical expressions, which are the
number of stickers Tito, Vic, and Joey had in the end. Since they all had the same amount
of stickers and were all able to avail the planner, then it means that they all had 24 stickers
in the end. Equate all these mathematical expressions into 24, then solve for x, y, and z.
Carry out the plan
At the start, the three had these amount of stickers:
Tito: x
Vic: y
Joey: z
First, Tito gave Vic and Joey as many stickers as each had.
Tito: x-y-z
Vic: y+y=2y
Joey: z+z=2z
Then, Vic gave Tito and Joey as many stickers as they had.
Tito: (x-y-z)+(x-y-z)=2x-2y-2z
Vic:2y-(x-y-z)-2z= -x+3y-z
Joey: 2z+2z=4z
Lastly, Joey gave Tito and Vic as many stickers as they had
Tito: (2x-2y-2z)+(2x-2y-2z)=4x-4y-4z
Vic: (-x+3y-z)+ (-x+3y-z)= -2x+6y-2z
Joey: 4z-(2x-2y-2z)- (-x+3y-z)= -x-y+7z
Equating these three expressions to 24,
Tito: 4x-4y-4z=24
Vic: -2x+6y-2z=24
Joey: -x-y+7z=24
Solving these using Matlab:
iv.
→
Therefore, the amount of stickers each person had from the start was:
Tito: x=39 stickers
Vic: y=21 stickers
Joey: z=12 stickers
Look back
In the problem, it was stated that at the start, Tito had enough stickers to avail the
planner, but Vic and Joey didn’t. This was satisfied in the results, since Tito got more than
24 stickers, while Vic and Joey got below 24. Moreover, since there were no new stickers
added throughout the entire time, there would be the same total of stickers from the
start and at the end.
At the start:
39 + 21 + 12 = 72 stickers
At the end:
24 + 24 + 24 = 72 stickers
This was also satisfied, which meant that the answers are correct.
4. Iza can earn ₱7,800.00 for doing 3 hours of office work and 2 hours of field work. However, she
gets ₱8,200.00 if she works two hours in the office and 3 hours in the field. What is the rate per
hour of Iza for doing office work and for doing field work?
i.
Understand the problem
The amount per hour she gets when doing office work and doing field work given that
she works 3 office hours and 2 field hours that amounts to ₱7,800; and 2 office hours
and 3 field hours that amounts to ₱8,200.
ii.
Devise a plan
The problem can be solved algebraically. Let x be the amount of office work hours and y
be the amount of field work hours.
𝑥 = 𝑜𝑓𝑓𝑖𝑐𝑒 ℎ𝑜𝑢𝑟𝑠
𝑦 = 𝑤𝑜𝑟𝑘 ℎ𝑜𝑢𝑟𝑠
Make the equations and solve for x and y as this will yield to the rate of how much Iza
earns per office work hours and per field work hours.
a) 3𝑥 + 2𝑦 = 7,800
b) 2𝑥 + 3𝑦 = 8,200
Solve the equation.
iii.
Carry out the plan
3𝑦 + 2𝑥 = 8,200
−2𝑦 − 3𝑥 = 7,800
𝑦 = 400 + 𝑥
3(400 + 𝑥) + 2𝑥 = 8,200
1,200 + 3x + 2x = 8,200
5x = 7,000
𝐱 = 𝟏, 𝟒𝟎𝟎
2y + 3x = 7,800
2y + 3(1,400) = 7,800
2y + 4,200 = 7,800
2y = 3,600
𝐲 = 𝟏, 𝟖𝟎𝟎
Iza gets paid ₱1,400.00 for every office work hour and ₱1,800.00 for every field work
hour.
iv.
Look back
Substituting these values into equations 1 and 2,
1: 3(1400) + 2(1800) = 7800
2: 2(1400) + 3(1800) = 8200
Both equations are satisfied, and thus the answers are correct.
5. Ed’s purse contains some 10-peso, 5-peso, 1 peso, and 25 centavo coins. How many of each type
does he have if the purse has a total of ₱20.50.
i.
Understand the problem
To make a list of the number of type of coins Ed’s purse contains that amounts to
₱20.50.
ii.
Devise a plan
Since the problem did not state how many coins there are in total, a table will be made
to show how many coins there can possibly be which will amount to ₱20.50. To start the
table, we can put the maximum number of coins there will be for each type of coin then
decrease the maximum count by 1 and filling the spaces with the other types of coins.
Assuming Ed’s purse contains at least 1 type of each coin, we will have to cancel out the
parts where a type of coin has zero (0) count on the table.
iii.
Carry out the plan
10.00
2
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
5.00
0
2
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
4
1.00
0
0
5
4
3
2
1
0
10
9
8
7
6
5
4
3
2
1
0
0
0.25
2
2
2
6
10
14
18
22
2
6
10
14
18
22
26
30
34
38
42
2
20.50
…
From here, the 10-peso coin is zero. This table ends here if we were to assume that the
there at least have to be one of each type of coin. The table is narrowed down to rows
which does not have a zero.
10.00
5.00
1.00
0.25
1
1
5
2
1
1
4
6
20.50
1
1
3
10
1
1
2
14
1
1
1
18
Answer: these are all combinations to which there is at least one type of coin there are
in Ed’s purse.
iv.
Look back
By making an assumption of Ed’s purse having at least one for each type of coin, the
results gathered has been limited to 5 possible outcomes.
Exercise 3.1 (p. 105-107)
Solve the following problems using Polya’s four-step problem-solving strategy.
1. In the complex number system, i1=i; i2=-1; i3=-i; i4=1; i5=i, … Find i173.
i. Understand the Problem
Find i173from the given examples.
ii.
Device a Plan
Referring to the table below:
iii.
iv.
2.
Exponent of i
Equivalent
1
i
2
-1
3
-i
4
1
5
i
6
-1
There can be seen a pattern, where the same value repeats after a period of 4. From this,
divide 173 by 4, then compare the remainder to the table. Or by trial and error, we know
that 172 is 1 because, we all knew that the same value repeats after a period of 4,
therefore i173 is I which is the value after 1.
Carry out the plan
173 / 4 = 43 remainder 1. This means that i173=i1=I or
172 is 1 therefore 173 is i.
Look back
Using modulo operations,
173 mod 4 = 1 mod 4
Find the last digit of the sum: 32018+42018
i.
Understand the Problem
Compute for the last digit of the sum of two numbers raised to an exponent.
ii.
Device a Plan
Referring to the table below:
n
Last digit of 3n
Last digit of 4n
1
3
4
2
9
6
3
7
4
4
1
6
5
3
iii.
iv.
4
There can be seen a pattern, where the last digit of 3 repeats after a period of 4, while
the last digit of 4nrepeats after a period of 2. From this, divide the exponent by 4 and 2,
then compare them to the table of 3n and 4n, respectively.
Carry out the plan
2018 / 4 = 504 remainder 2. This means that 32018=32=9.
2018 / 2 = 1009. Since there is no remainder, the exponent is equal to the period for 4n.
42018=42=16. The last digit is 6.
Finding the sum, 9 + 6 = 15.
Look back
Using modulo operations,
2018 mod 4 = 2 mod 4
2018 mod 2 = 0 mod 2 = 2 mod 2
3.
Yan was born exactly 78 days before Dong was born. If Dong was born on a Monday, what day
was Yan born?
i.
Understand the Problem
Find the day Yan was born.
ii.
Device a Plan
If today is Monday, then exactly 7 days ago was also Monday knowing that there are 7
days a week. From this, find the nearest multiple of 7 from 78, then work backwards from
there, until the exact day was computed.
iii.
Carry out the plan
Let today be a Monday. The nearest multiple of 7 from 78 is 77, which meant that it was
also a Monday 77 days ago. Referring to the table below:
Days from Dong’s Birth
Day
77
Monday
78
Sunday
79
Saturday
80
Friday
81
Thursday
82
Wednesday
83
Tuesday
It was a Sunday, 78 days from when Dong was born. Therefore, Yan was born on a Sunday.
iv.
Look back
It would always be a Sunday 78 days before a Monday. For example, if today is June 3,
which is a Monday, 78 days ago was March 17, which was a Sunday. Therefore, the answer
is correct.
4.
The set {0,1} forms the binary system. There are exactly two 1-digit binary numbers, namely 0 and
1; there are four 2-digit binary numbers, namely 00, 01, 10, and 11; and there are eight 3-digit binary
numbers, namely 000, 001, 010, 100, 011, 101, 011, and 111.
a. List all 4-digit binary numbers.
b. How many 4-digit binary numbers are there?
c. How many 5-digit binary numbers are possible?
d. How many n-digit binary numbers are possible?
i.
Understand the Problem
Find the number of n-digit numbers possible, and list the 4-digit binary numbers
possible.
ii.
Device a Plan
(a) Listing all the 4-digit binary numbers
















0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Listing the number of binary numbers for 1 to 4 digits,
iii.
iv.
Number of Digits
Number of Binary Numbers
1
2
2
4
3
8
4
16
Find the formula for the pattern, then solve for the number of 5-digit binary
numbers.
Carry out the plan
From the table above, the formula for how many n-digit binary numbers possible
is 2n (d). Substituting n = 5, there are 25=325-digit binary numbers (c).
Look back
Listing all the 5-digit binary number
































00000
00001
00010
00011
00100
00101
00110
00111
01000
01001
01010
01011
01100
01101
01110
01111
10000
10001
10010
10011
10100
10101
10110
10111
11000
11001
11010
11011
11100
11101
11110
11111
In total, there are 32 numbers, which means that the answer is correct.
5.
Consider a rectangle with length l and width w. Now fold it into two, draw a line on the fold, and
unfold the rectangle. Observe that there are now three rectangles, two small rectangles, and the original
rectangle.
a. Fold the rectangle into three, draw lines on the folds, then unfold. How many rectangles
do you see?
b. Fold the rectangle into four, draw lines on the folds, then unfold. How many rectangles
do you see?
c. From your observations in (a) and (b), how many rectangles are formed if the original
rectangle is folded into n pieces?
i.
Understand the Problem
Find the pattern of all possible rectangle when it is folded
ii.
Device a Plan
By using the data above. We could see that the pattern follow the
summation of n, as shown below:
an = x = x(x+1)/2
iii.
Carry out the plan
a.
a3 = x = 3(3+1)/2 = 6
b.
a4 = x = 4(4+1)/2 = 10
c.
an = x = n(n+1)/2
iv.
Look back
# of Rectangle
Total combined Rectangle
1
1
2
3
3
6
4
10
n
n(n+1)/2
6.
Finding the odd. In a grocery store, Perla was challenged by a promo girl to identify which pack in
a box of 25 packs of liquid detergent is heavier than the rest. According to the promo girl, all packs in the
box weigh the same except for the one which is heavier. Using only a balance scale, Perla was challenged
to identify the single heavier pack for a maximum of three (3) weighing attempts only. If she succeeds,
she gets the box for free. How should Perla weigh the 25 packs to successfully identify the heavier pack?
i.
Understand the Problem
Find if Perla can identify the heavier pack for a maximum of three weighing attempts only.
ii.
Device a Plan
To find the heavier one, we can compare any two box, leaving the third out. If the
two box weigh the same, then the heavier box must be one of those not on the balance.
Otherwise, it is the one indicated as heavier by the balance.
iii.
Carry out the plan
iv.
Look back
Since that the can succeed in finding the heavier box with the use of only 3 trials
by using the method above.
7.
Can Perla succeed in the challenge if there are 30 packs in the box with 29 having the same weight
i.
Understand the Problem
Identify if Perla can succeed in finding the one pack that is heavier out of 30 packs.
ii.
iii.
iv.
Device a Plan
Identify the maximum amount of packs that can give a 100% chance of winning.
Then check whether the given packs is within its range.
3n = Maximum packs for weighing
Wherein, n is the number of trials given.
Carry out the plan
Perla cannot 100% succeed since that the total pack is 30. By using the formula,
3
3 = 27, which is the total amount of packs that can give a 100% chance of winning. Thus,
30 packs is out of range of achieving a 100% chance of winning.
Look back
33 = 27
30 > 27
Thus, Perla cannot 100% succeed in the challenge.
8.
John bought ten (10) bottles of vitamin C tablets (same brand) from a certain pharmacy. After
getting home, the saleslady in the pharmacy called and informed him that one of the bottles is a complete
counterfeit (meaning, 9 bottles contain all real medicines while one bottle contains all counterfeit items).
He was told that the real medicines were 10 mg each tablet while the counterfeits are heavier at 11 mg
each table. Using a digital weighing machine, can John identify which bottle contain a counterfeit in just
one (1) weighing attempt?
i.
Understand the Problem
Can John identify which bottle of Vitamin C tablets contain a counterfeit in just one
attempt?
ii.
Device a Plan
Identify the chance of getting the counterfeit bottle out of the ten bottles.
iii.
Carry out the plan
The possibility of John identifying which bottle of Vitamin C tablets that contains
the counterfeit in just one attempt has a probability of 1/10. A digital weighing machine
is used, therefore, only one bottle can be weighed at a time. If by luck, he got the bottle
that contains the counterfeit first, he can definitely identify it in just one attempt.
However, there are 9 bottles more, thus, if he got all those before the counterfeited one,
it would be a total of ten attempts.
iv.
Look back
John only has a 10% chance of getting the counterfeit bottle.
9.
While waiting for their parents to arrive, Sansa and Arya decided to play a game. They collected
25 stones and placed it in an urn. They are to take turns taking away 1, 2, or 3 stones from the urn. The
person who will take the last stone loses. Sansa took the first move. What are her chances of winning the
game?
i.
Understand the Problem – What are the chances of Sansa winning the game?
ii.
Device a Plan
25 initial stone
2 players
Can get to 1, 2, or 3
The only sure chance that Sansa will win the game if she is able to leave five stones in the end
for Arya to choose from. Arya could get 3 stones from the five, then Sansa could get 2, leaving Arya with
the last one; or, Arya could get 2, Sansa would get 2 as well, still leaving Arya with the last one. Another
scenario would be that Arya would only get 1, Sansa would get 3, and Arya would still be left with the
last stone.
iii.
Carry out the plan
Use permutation
3!+3!+3!+3!=24
is the chance of losing.
iv.
Look back
Therefore there is a very minimal chance of Sansa to win the game.
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