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Mathematics of Modern World., Exercises of Mathematics

Problem Solving & Reasoning: Puzzles & Riddles.

Typology: Exercises

2021/2022

Available from 05/17/2022

lirry-mercury
lirry-mercury 🇵🇭

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Problem Solving & Reasoning
Puzzles & Riddles
1. I am an odd number. Take away a letter and I become even. What number am I?
Answer: SEVEN
Conclusion: Using Polya’s Problem solving we can prove that the answer is SEVEN.
a. Understand the problem: Odd number, take away a letter then becomes even.
b. Devise a plan: Let’s assume numbers from (1-10). Let all the numbers change into word numbers,
then find the odd numbers.
c. Carry out the plan: From the odd numbers one, three, five & seven (1,3,5 & 7), find the word that
can be change into the word even.
d. Review the solution: Now we can see from the respective numbers that only the odd number (7)
is suitable from the word SEVEN itself by taking away the letter “S” it will now become EVEN,
therefore the answer is SEVEN.
2. Two fair coins are tossed simultaneously. In how many different chances It can have the two heads
(2) and two tails (2)?
Answer: Two heads have (1) chance & Two tails have chances (1).
Conclusion: Using Polya’s Problem solving we can prove that the answer has one chances (1) for
each, two heads and two tails.
a. Understand the problem: There are many different orders by flipping two fair coins
simultaneously, it could have two heads (HH) also it could also have two tails (TT) and of course
there are other possibilities like a head and a tails (HT).
b. Devise a plan: I’ll make a list of all possible orders and organized list to ensure that each of the
different chances or possibilities will be listed once and once only.
c. Carry out the plan: By tossing two fair coins simultaneously, we can see that there are 4 different
chances shown below:
HH (Heads & Heads)
HT (Heads & Tails)
TH (Tails & Heads)
TT (Tails & Tails)
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Problem Solving & Reasoning

Puzzles & Riddles

  1. I am an odd number. Take away a letter and I become even. What number am I? Answer: SEVEN Conclusion: Using Polya’s Problem solving we can prove that the answer is SEVEN. a. Understand the problem: Odd number, take away a letter then becomes even. b. Devise a plan: Let’s assume numbers from (1-10). Let all the numbers change into word numbers, then find the odd numbers. c. Carry out the plan: From the odd numbers one, three, five & seven (1,3,5 & 7), find the word that can be change into the word even. d. Review the solution: Now we can see from the respective numbers that only the odd number (7) is suitable from the word SEVEN itself by taking away the letter “S” it will now become EVEN, therefore the answer is SEVEN.
  2. Two fair coins are tossed simultaneously. In how many different chances It can have the two heads (2) and two tails (2)? Answer: Two heads have (1) chance & Two tails have chances (1). Conclusion: Using Polya’s Problem solving we can prove that the answer has one chances (1) for each, two heads and two tails. a. Understand the problem: There are many different orders by flipping two fair coins simultaneously, it could have two heads (HH) also it could also have two tails (TT) and of course there are other possibilities like a head and a tails (HT). b. Devise a plan: I’ll make a list of all possible orders and organized list to ensure that each of the different chances or possibilities will be listed once and once only. c. Carry out the plan: By tossing two fair coins simultaneously, we can see that there are 4 different chances shown below:  HH (Heads & Heads)  HT (Heads & Tails)  TH (Tails & Heads)  TT (Tails & Tails)

d. Review the solution: As we organized the list, it shows us the different possibilities and we can conclude, there are (4) different chances in which we can obtain two heads (1) and two tails (1) in out of four (4) by flipping two coins simultaneously.

  1. What goes up must come down. It went up. It will come down. Answer: Deductive Reasoning. Conclusion: Because of the conclusion is a specific case of a general assumption, this argument is deductive reasoning. The conclusion is What goes up must come down.
  2. In the sequence 1, 2, 4, 7, 11, 16 the next most probable number is 22. Answer: Inductive Reasoning. Conclusion: The argument reaches the conclusion based on specific sentence, so it is an inductive reasoning. The conclusion is the next most probable number is 22.
  3. Mary has four daughters, and each of her daughters has a brother. How many children does Mary have? Answer: Five, each daughter has the same brother. Conclusion: Using Polya’s Problem solving we can prove Five children on which each daughter has the same brother. a. Understand the problem: Mary has four (4) daughters were each of the daughters has a brother. b. Devise a plan: As the sentence said that each of the daughter has a brother, this means that the word BROTHER is singular or only one (1). c. Carry out the plan: Since the Mary has a (4) daughter and each of them has a brother we can assume that the brother is Mary’s son, it is the only one that they have and all her daughter is telling the same person, their brother therefore Mary have 5 children. d. Review thee solution: We can now conclude that Mary have four (4) daughter and one (1) son, where each of her daughter has only one brother that they have and thus Mary have 5 children.