## PCSR Civil Service Exam Review Guide 10

After learning how to solve number problems, let’s learn how to solve age problems. Watch the videos below and read the articles. Then, we will have exercise later.

PART I: MATH

A. Videos (These videos are in Taglish and the links point to Youtube).

B. Articles

C. Exercises

Part II ENGLISH

A. Vocabulary

Civil Service Exam Vocabulary Review 9

For those who are searching the Review Guides 1 to 9, you can find them here.

## 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

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)

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)

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)

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)

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)

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)

## The Age Problem Solving Series

One of the common types of word problems in mathematics and in many examinations is about Age Problems. This series discusses various problem styles involving age problems and explains in details how they are solved.

The Age Problem Solving Series

How to Solve Age Problems Part 1 discusses simple 2-person problems particularly present-past and present-future age relationships.

How to Solve Age Problems Part 2 discusses a slightly more difficult 2-person problems particularly present-past and present-future age relationships.

How to Solve Age Problems Part 3 discusses age problems that involves fractions.

I am planning to write a fourth part for this series in the near future, but for now, I will focus more on yet uncovered topics.

## How to Solve Age Problems Part 2

This is the second part of the Solving Age Problem Series. We will continue solving age problems that are slightly more complicated that the first part. We have already discussed 3 problems in the first part of this series, so we continue with the fourth problem.

Problem 4

Simon is four years older than Jim. The sum of their ages is 52. How old is Simon?

Scratch Work

This problem is a sort of review of first part of this series. Simon is older than Jim by $4$ years. So, if Jim is $x$ years old, then Simon is $x + 4$ years old. The sum of their ages is $52$. This means that if add $x$ and $x + 4$, then the sum is $52$. That is the equation. Continue Reading

## How to Solve Age Problems Part 1

After a series of tutorials on word problems involving numbers, we now move to learning on how to solve word problems involving age. Age problems are very similar to number problems, so if you have finished reading The Number Word Problem Series, then it will be easier for you to solve the following age problems.

Example 1

Benjie is thrice as old as his son Cedric. The sum of their ages is 64. How old are both of them?

Scratch Work

This is one of those age problems that are very similar to number problems. Let’s take a specific case. If Cedric is say $8$ years old, then Benji is $3(8)$ years old. This means that if Cedric is $x$ years old, then Benjie is $3x$. If we add their ages, the result is $64$. Continue Reading