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m_a_m_a [10]
2 years ago
12

Mrs Wright spent 2/9 of her paycheck on food and 1/3 on rent. She spent 1/4 of the remainder on transportation. She had $210 lef

t. How much was Mrs. Wright's paycheck ?
Mathematics
1 answer:
Mashutka [201]2 years ago
3 0

Answer:

Mrs. Wright's paycheck is $630

Step-by-step explanation:

Let x = Mrs. Wright's paycheck.

Mrs. Wright spent 2/9 of her paycheck on food. This means that the amount of money spent on food is 2/9 × x = 2x/9

She spent 1/3 on rent. This means that the amount spent on rent is

1/3 ×x x = x/3

Amount she spent on food and rent is x/3 + 2x/9 =3x + 2x /9

= 5x/9

The remainder is her pay check - the amount that she spent on food and rent. It becomes

x - 5x/9 = (9x- 5x)/9 = 4x/9

She spent 1/4 of the remainder on transportation. It means that she spent 1/4 × 4x/9 = x /9 on transportation.

Amount left = 4x /9 - x/9= 3x/9

She had $210 left. Therefore,

210 = 3x/9

3x = 9×210 = 1890

x = 1890/3

x = $630

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Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

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Then, we need:

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  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

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  • 2 must be related with 2,
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  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

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