Solutions to Age Problems in Algebra

Solutions to http://denviant.blogspot.com/2012/04/engineering-quiz-1-age-problems-in.html

1. Solution:

x = Mark’s age

y = his son’s age

2x = 6y + 8 (Eq. 1)

(x – 10) + (y – 10) = 44 (Eq. 2)

x + y = 64

x = 64 – y (substitute to eq. 1)

2(64 – y) = 6y + 8

128 – 2y = 6y + 8

8y = 120

y = 15

Mark’s son is 15 years old.

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2. Solution:

x = John’s age

x – 13 = (x + 7)/3

3(x – 13) = x + 7

3x – 39 = x + 7

2x = 46

x = 23

John is 23 years old.

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3. Solution:

x = age of JM’s son

4(x + 7) = 41 + 7

4x + 28 = 48

4x = 20

x = 5

JM’s son is 5 years old.

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4. Solution:

x = Revned’s age

y = his son’s age

x = 3y (eq. 1) substitute to eq. 2

x – 4 = 4(y – 4) (eq. 2)

3y – 4 = 4y – 16

y = 12

Revned’s son is 12 years old.

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5. Solution:

x = number of years needed for Maria to be three times as old as her daughter

45 + x = 3(5 + x)

45 + x = 15 + 3x

2x = 30

x = 15

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6. Solution:

x = Verna’s present age

	Past	Present
Mary Ann	x	24
Verna	24/2	x

x – 12 = 24 – x

2x = 36

x = 18

Their gap is 24 – 18 = 6. So when Mary Ann was 18 years old, Verna was 12, and 24 (Mary Ann’s age now) is twice of 12 (Ana’s past age).

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7. Solution:

x = number of children in the family

y = sum of parent’s ages

z = sum of children’s ages

y = 2z (eq. 1) substitute to eq. 2 and 3

y – 5(2) = 4(z – 5x) (eq. 2)

y – 10 = 4z – 20x

2z – 10 = 4z – 20x

20x – 2z = 10 (eq. 4)

y + 15(2) = z + 15x (eq. 3)

y + 30 = z + 15x

2z + 30 = z + 15x

15x – z = 30 (eq. 5)

Eliminating from Equations 4 and 5

20x – 2z = 10

30x – 2z = 60

-10x = -50

x = 5

There are 5 children in the family.

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8. Solution:

x = Bobadilla’s age

y = Bobin’s age

x = y + 5 (eq. 1) substitute eq. 1 to 2

(x + 5)(y + 5) = (1.5)(x)(y) (eq. 2)

(y + 5 + 5)(y + 5) = (1.5)(y + 5)(y)

y² + 15y + 50 = 1.5y² + 7.5y

0.5y² – 7.5y – 50 = 0

             .5

y² – 15y – 100 = 0

(y – 20)(y + 5) = 0

y = 20

Bobin’s present age is 20.

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9. Solution:

x = number of years needed for Kevin to be three times as old as Gabriel

41 + x = 3(9 + x)

41 + x = 27 + 3x

2x = 14

x = 7

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10. Solution:

x = Ianne’s present age

y = Paul’s present age

z = Ronnel’s present age

x = 2y (eq. 1) substitute to eq. 3

y = x/2

z = 2x (eq. 2) substitute to eq. 3

x + y + z + 30 = 58 (eq. 3)

x + y + z = 28

2(x + x/2 + 2x = 28)

2x + x + 4x = 56

7x = 56

x = 8

Ianne’s present age is 8.

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