Week 9 Review: Answers and Solutions

Below are the solutions to the exercises and problems about age problems.

Exercises

1.) Leah is 3 years older than Lanie. The sum of their ages is 29. What are their ages?

Let x = Lanie’s age
x + 3 = Leah’s age

The sum of their ages is 29.

x + (x + 3) = 29
2x + 3 = 29
2x = 29 – 3
2x = 26
x = 26/2
x = 13 (Lanie’s age)

Leah’s age = x + 3 = 13 + 3 = 16

Answer: Lanie 13, Leah 16

2.) Alfred’s thrice as old as Fely. The difference between their ages is 16. What are their ages?

Let x = Fely’s age
3x = Alfred’s age

The difference between their ages is 16.

3x – x = 16
2x = 16
x = 16/2
x = 8(Fely’s age)

3x = 3(8) = 24 (Alfred’s age)

Answer: Alfred 24, Fely 8

3.) Kaye is 4 years younger than Kenneth. The sum of their ages is 42. What are their ages?

Let x = Kenneth’age
x – 4 = Kenneth’s age

The sum of their ages is 42.

x + (x – 4) = 42
2x – 4 = 42
2x = 42 + 4
2x = 46
x = 46/2
x = 23 (Kenneth’s age)
x – 4 = (23)-4 = 19 (Kaye’s age)

Answer: Kaye 19, Kenneth 23.

Problems

1.) Gina is 5 years older than Liezel. In 5 years, the sum of their ages will be 39. What are their ages?

Present ages
Let x = Liezel’s age
x + 5 = Gina’s age.

In 5 years
(x + 5) = Liezel’s age
(x + 5) + 5 = Gina’s age.

The sum of their ages will be 39.

(x + 5) + (x + 5) + 5 = 39
2x + 15 = 39
2x = 39 – 15
2x = 24
x = 24/2
x = 12 (Liezel’s age)
(x + 5) = 12 + 5 = 17 (Gina’s age)

Answer: Gina 17, Liezel 12.

2.) Alex is 7 years older than Ben. Three years ago, the sum of their ages was 29. What are their ages?

Present ages
Let x = Ben’s age
x + 7 = Alex’s age

3 yrs ago
x – 3 = Ben’s age
x + 7 – 3 = Alex’s age

The sum of their ages was 29.

(x – 3) + [(x + 7) – 3] = 29
x – 3 + x + 4 = 29
2x + 1 = 29
2x = 29 – 1
2x = 28
x = 28/2
x = 14 (Ben’s age)
(x + 7) = 14 + 7 = 21 (Alex’s age)

Answer: Alex 21, Ben 14

3.) Yna is 18 years older than Karl. In 8 years, she will be as twice as old as Karl. What are their ages?

Let x = Karl’s age
x + 18 = Yna’s age

In 8 years…
Karl = x + 8
Yna = (x + 18) + 8

…she (Yna) will be as twice as old as Karl

Yna’s age = 2 times Karl’s age

(x + 18) + 8 = 2(x + 8)
x + 26 = 2x + 16
x – 2x = 16 – 26
-x = -10
x = 10 (Kar’s age)
x + 18 = 10 + 18 = 28 (Yna’s age)

4.) Peter’s age is thrice Amaya’s age. In 5 years, his age will be twice Amaya’s age. How old is Peter?

Let x = Amaya’s age
3x = Peter’s age

In 5 years…
Amaya = x + 5
Peter = 3x + 5

…his age will be twice as Amaya’s age

3x + 5 = 2(x + 5)
3x + 5 = 2x + 10
3x – 2x = 10 – 5
x = 5 (Amaya’s age)
3x = 3(5) = 15 (Peter’s age)

5.) Martin is thrice as old as Kaye. If 7 is subtracted from Martin’s age and 5 is added to Kaye’s age, then the sum of their ages is 34. What are their ages?

Let x = Kaye’s age
3x = Martin’s age

If 7 is subtracted from Martin’s age…
3x – 7

…and 5 is added to kaye’s age…
x + 5

…then the sum of their ages is 34.

(3x – 7) + (x + 5) = 34
4x – 2 = 34
4x = 34 + 2
4x = 36
x = 36/4
x = 9 (Kaye’s age)

3x = 3(9) = 27 (Martin’s age)

Answer: Kaye 9, Martin 27.

6.) James is 9 years older than Kevin. Two years ago, his age was twice that of Kevin’s age. How old is James?

Present ages
Let x = Kevin’s age
x + 9 = James’ age

2 years ago

x – 2 = Kevin’s age
(x + 9) – 2 = x + 7 = James’ age

…his age was twice of Kevin
x + 7 = 2(x – 2)
x + 7 = 2x – 4
x – 2x = -4 – 7
-x = -11
x = 11 (Kevin’s age)
x + 9 = 11 + 9 = 20 (James’ age)

Answer: James is 20 years old.

7.) Mark is twice as old as Lorie. Rey is 6 years younger than Mark. Three years ago, the average of the ages of the three of them is 20. What are their present ages?

Present ages
Let x = Lorie’s age
2x = Mark’s age
2x – 6 = Rey’s age

3 years go
x – 3 = Lorie’s age
2x – 3 = Mark’s age)
(2x – 6) -3 = (2x -9) = Rey’s age

The average of their ages was 20.

(Lorie’s age + Mark’s age + Rey’s age ) / 3 = 20
[(x – 3) + (2x – 3) + (2x – 9)]/3 = 20.

Multiplying both sides by 3,

x – 3 + 2x – 3 + 2x – 9 = 20(3)
5x – 15 = 60
5x = 60 + 15
5x = 75
x = 75/5.

x = 15 (Lorie’s age)
2x = 2(15) = 30 (Mark’s age)
2x – 6 = 2(15) – 6 = 30 – 6 = 24 (Rey’s age)

Answer: Lorie 15, Mark 30, Rey 24.

8.) Sam is thrice as old as Vina. Rio is half as old as Vina. The sum of their ages is 54. What are their ages?

Let x – Vina’s age
3x = Sam’s age
x/2 = Rio’s age

The sum of their ages is 54.
x + 3x + x/2 = 54

Multiply both sides by 2.
2(x + 3x + x/2 = 54)2
2(x) + 2(3x) + 2(x/2) = 2(54)
2x + 6x + x = 108
9x = 108
x = 108/9
x = 12(Vina’s age)

3x = 3(12) = 36 (Sam’s age)
x/2 = 12/2 = 6 (Rio’s age)

Answer: Vina 12, Sam 36, Rio 6.

9.) Four years from now, Tina’s age will be equal to Kris’ present age. Two years from now, Kris will be twice as old as Tina. What are their present ages?

Present Ages

x = Kris’age
x – 4 = Tina’s age

2 years from now

x + 2 = Kris’ age
x – 4 + 2 = x – 2 = Tina’s age

4 years from now
x + 4 = Kris’ age
x = Tina’s age

Two years from now, Kris will be twice as old as Tina.
x + 2 = 2(x – 2)
x + 2 = 2x – 4
x – 2x = -4 – 2
-x = -6
x = 6 (Kris’ present age)
x – 4 = 6 – 2 = 4 (Tina’s age)

Answer: Tina 2, Kris 6.