PCSR REVIEW SERIES WEEK 11: Work Problems

This is the 11th week in our review series. After learning about motion problems which discuss distance, rate and time, let’s now learn how to solve work problems. Below are the articles and videos that you can read and watch. Later, we will have practice problems.

ARTICLES

VIDEOS

Enjoy learning!

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.

Week 8 Review: Answers and Solutions

These are the solutions and answers to the problems in Week 8 Review on Number Problems.

Problem 1

One number is 3 more than the other. Their sum is 27. What are the numbers?

Let x – smaller number
x + 3 – larger number

Their sum is 27, so
x + (x + 3) = 27
2x + 3 = 27
2x = 27 – 3
2x = 24
x = 24/2
x = 12 (smaller number)
x + 3 = 15 (larger number).

Problem 2
One number is 5 less than the other. Their sum is 51. What are the numbers?

Let x – larger number
x – 5 –  smaller number

And their sum is 51. So,

x + (x – 5) = 51
2x – 5 = 51
2x = 51 + 5
2x = 56
x = 56/2
x = 28 (larger number)
x – 5 = 28 – 5 = 23 (smaller number).

Answer: 23 and 28

Problem 3

One number is 3 times the other number. Their sum is 48. What are the numbers?

Let x – smaller number
3x – larger number

And their sum is 48. So,

x + 3x = 48
4x = 48
x = 48/4
x = 12(1st number)

2nd number = 3x
3(12) = 36

Answer: 12 and 36

Problem 4
One number is 5 times the other number. Their difference is 52. What are the numbers?

Let x – smaller number
5x – larger number

And their difference is 52. So,

5x – x = 52
4x = 52
x = 52/4
x = 13 (smaller number)
5x = 5(13) = 65.

Checking: -13 – (-65)
-13 + (65) = 52

Answer: 13 and 65

Problem 5
The sum of three numbers is 36. The second number is 5 more than the first number and the third number is 8 less than the first number. What are the three numbers?

Let x – 1st number
x + 5 – 2nd number
x – 8 – 3rd number

Their sum is 36. So,
x + (x + 5) + (x – 8) = 36
3x – 3 = 36
3x = 36 + 3
3x = 39
x = 13 (1st number)
2nd number = x + 5 => (13) + 5 => 18
3rd number = x – 8 => (13) – 8 => 5

Checking: 13 + 18 + 5 = 36

Problem 6

The sum of three numbers is 98. The second number is twice the first number and the third number twice the second number. What are the three numbers?

14, 28 & 56

Let x = 1st number
2x = 2nd number (twice the first)
2(2x)=3rd number (twice the second)

And their sum is 98. So,

x + (2x) + 2(2x) = 98
x + 2x + 4x =98
7x = 98
x = 98/7
x = 14 (1st number)

2nd number = 2x => 2(14) => 28

3rd number = 2(2x) => 2(2(14)) => 2(28) => 56

Problem 7

One number is two more than thrice the other. Their sum is 26. What are the two numbers?

Let x – 1st number
3x + 2 = 2nd number (two more than thrice the other)

And their sum is 26.

x + (3x + 2) = 26
4x + 2 = 26
4x = 26 – 2
4x = 24
x = 24/4
x = 6 (1st number)

2nd number = (3x + 2) => 3(6) + 2 => 18 + 2 => 20

Answer: 6 and 20

Problem 8

One number is thrice the other. When 3 is added to the larger and 7 is subtracted from the smaller, their sum becomes 32. What are the two numbers?

Let x – smaller number
3x – larger number (thrice the other)

When 3 is added to larger number… = 3x + 3

…and 7 is subtracted to smaller = x – 7

Their sum becomes 32. So,

(3x + 3) + (x – 7) = 32
4x – 4 = 32
4x = 32 + 4
4x = 36
x = 36/4
x = 9(smaller number)

Larger number = 3x = 3(9) = 27

Checking:
When 3 is added to larger number = 27 + 3 = 30
And 7 is subtracted to smaller number = 9-7 = 2

Their sum is 32 = 30 + 2 = 32

Answer: 9 and 27

Problem 9

The sum of two consecutive numbers is 91. What are the two numbers?

Let x – first number
x + 1 – 2nd number

x + (x + 1) = 91
2x + 1 = 91
2x = 91 – 1
2x = 90
x = 90/2
x = 45 (1st number)

2nd number => x + 1 => 45 + 1 => 46

Answer: 45 and 46

Problem 10
The sum of two positive consecutive EVEN integers is 66. What are the two numbers?

Let x – 1st number
x + 2 = 2nd number

x + (x + 2) = 66
2x + 2 = 66
2x = 66 – 2
2x = 64
x = 64/2
x = 32 (1st number)

2nd number => x + 2 => 32 + 2 => 34

Answer: 32 and 34

PCSR Problem 11

The sum of two positive consecutive ODD integers is 36. What are the two numbers?

Let x – 1st odd number
x + 2 – 2nd odd number

And their sum is 36.

x + (x + 2) = 36
2x + 2 = 36
2x = 36-2
2x = 34
x = 34/2
x = 17(1st number)
x + 2 = 17 + 2 = 19 (2nd number)

Checking:

17 + 19 = 36
And 17 and 19 are both odd numbers

Answer: 17 and 19

Problem 12

The sum of three positive consecutive ODD integers is 81. What are the three integers?

Let x – 1st odd integer
x + 2 – 2nd odd integer
x + 4 – 3rd odd integer

Their sum is 81.

x + (x + 2) + (x + 4) = 81
3x + 6 = 81
3x = 81 – 6
3x = 75
x = 75/3
x = 25 (1st int)

2nd int = (x + 2) => 25 + 2 => 27
3rd int = (x + 4) => 25 + 4 => 29

Checking:

25 + 27 + 29 = 81
They are consecutive ODD integers.

Answer: 25, 27 & 29

Problem 13

The sum of the smallest and the largest of five positive consecutive integers is 108. What is the third integer?

Let x – 1st integer
x + 1 = 2nd integer
x + 2 = 3rd integer
x + 3 = 4th integer
x + 4 = 5th integer

Since the sum of the first and the fifth is 108,

x + (x + 4) = 108
2x + 4 = 108
2x = 108 – 4
2x = 104
x = 104/2
x = 52 (smallest number).

2nd int. => (x + 1) => 52 + 1 => 53
3rd int => ( x + 2) => 52 + 2 => 54
4th int. => (x + 3) => 52 + 3 => 55
5th int. => (x + 4) => 52 + 4 => 56

Since we are looking for the third integer, the answer is 54.

Problem 14
The average of four positive consecutive EVEN integers is 19. What is the largest integer?

Let x – 1st even integer
x + 2 = 2nd even integer
x + 4 = 3rd even integer
x + 6 = 4th even integer

Their average is 19.

(x + (x + 2) + (x + 4) + (x + 6))/4 = 19
(4x + 12)/4 = 19

Multiplying both sides of the equation by 4,

4x + 12 = 19(4)
4x + 12 = 76
4x = 76 – 12
4x = 64
x = 64/4
x = 16(1st even int).

2nd even int. = x + 2 => 16 + 2 => 18
3rd even int. = x + 4 => 16 + 4 => 20
4th even int. = x + 6 => 16 + 22 => 22

Checking:

(16 + 18 + 20 + 22)/4 = 19
(76)/4 = 19
19 = 19

Answer: 22(largest number)

PCSR Problem 15
The average of seven positive consecutive integers is 31. What is the smallest integer?

Let x – 1st integer
x + 1 = 2nd integer
x + 2 = 3rd integer
x + 3 = 4th integer
x + 4 = 5th integer
x + 5 = 6th integer
x + 6 = 7th integer

Their average is 31.

(x + (x + 1) + (x + 2) + (x +3) + (x + 4) + (x + 5) + (x + 6))/7 = 31
(7x + 21)/7 = 31
7x + 21 = 31(7)
7x + 21 = 217
7x = 217 – 21
7x = 196
x = 196/7
x = 28(1st integer)

2nd int. = x + 1 => 28 + 1 => 29
3rd int. = x + 2 => 28 + 2 => 30
4th int. = x + 3 => 28 + +3 => 31
5th int. = x + 4 => 28 + 4 => 32
6th int. = x + 5 => 28 + 5 => 33
7th int. = x + 6 => 29 + 6 => 34

Checking:
(28 + 29 + 30 + 31 + 32 + 33 + 34)/7 = 31
217/7 = 31
31 = 31

Since we are looking for the smallest integer, the answer is 28.

Week 8 Review: Practice Exercises and Problems

After learning how to solve number problems, let’s have some practice exercises.

Week 8 Review: Practice Exercises and Problems

1.) One number is 3 more than the other. Their sum is 27. What are the numbers?

2.) One number is 5 less than the other. Their sum is 51. What are the numbers?

3.) One number is 3 times the other number. Their sum is 48. What are the numbers?

4.) One number is 5 times the other number. Their difference is 52. What are the numbers?

5.) The sum of three numbers is 36. The second number is 5 more than the first number and the third number is 8 less than the first number. What are the three numbers?

6.) The sum of three numbers is 98. The second number is twice the first number and the third number twice the second number. What are the three numbers?
7.) One number is two more than thrice the other. Their sum is 26. What are the two numbers?

8.) One number is thrice the other. When 3 is added to the larger and 7 is subtracted from the smaller, their sum becomes 32. What are the two numbers?

9.) The sum of two positive consecutive numbers is 91. What are the two numbers?

10.) The sum of two positive consecutive EVEN integers is 66. What are the two numbers?

11. ) The sum of two positive consecutive ODD integers is 36. What are the two numbers?

12.) The sum of three positive consecutive ODD integers is 81. What are the three integers?

13.) The sum of the smallest and the largest of five positive consecutive integers is 108. What is the third integer?

14.) The average of four positive consecutive EVEN integers is 19. What is the largest integer?

15.) The average of seven positive consecutive integers is 31. What is the smallest integer?

Enjoy solving!

Solving Quadratic Word Problems in Algebra

Quadratic Equations are equations of the form ax^2 + bx + c = 0 where a, b and c are real numbers and a \neq 0. Depending on the form of the equation, you can solve for x by extracting the quare root, factoring, or using the quadratic formula.This type of equation appears in various problems that involves multiplication and usually appears in the Civil Service Exams.

The following series details the method and strategies in solving problems involving quadratic equations.

How to Solve Quadratic Word Problems Part 1 is about solving problems involving consecutive integers. In this problem, the product of consecutive numbers is given and factoring was used to solve the problem.  » Read more

Area of Rectangle: Worked Examples

In the previous post, we have learned some problems on how to solve area of squares. In this post, we are going to solve problems involving area of rectangles. The area of rectangle is one of the most common questions in the Geometry part of the Civil Service Examination. In this post, we are going to learn how to solve problems involving area of rectangles. The formula for the Area (A) can be calculated by multiplying the length (l) and the width (w). That is,

A = l \times w.

Problem 1

Find the area of a rectangle with length 8cm and width 5 cm.

Solution

A = l \times w

A = 8 \times 5

A = 40

So, the area of the rectangle is 40 cm^2

Problem 2

The area of a rectangle is 12 cm^2 and its width is 3 cm. What is its length?

Solution

A = l \times w

Substituting the given, we have

12 = l \times 3.

To get l, we have to divide both sides by 3.

\dfrac{12}{3} = \dfrac{l \times 3}{3}

4 = l

Therefore, the length is equal to 3.

Problem 3

The length of a rectangle is twice its width. Its area is 32 square units. What are the dimensions of the rectangle?

Solution

Let x = width of the rectangle

Since the length is twice, it is 2x.

So,

w = x
l = 2x

A = l \times w

A = 2w \times w

32 = 2w^2

Dividing both sides by 2, we have

16 = w^2.

Getting the square root of both sides, we have

4 = w

So, l = 2w = 2(4) = 8.

So, the dimension of the rectangle is 4 by 8.

Problem 4

The length of a rectangle is two more than its width. Its area is 48 square units. What are the dimensions of the rectangle?

Solution

Let length be equals x and width be equals x + 2. That is

w = x
l = x + 2

A = lw
48 = x(x + 2)
48 = x^2 + 2x

Subtracting 48 from both sides, we have
x^2 + 2x - 48 = 0.

Factoring, we have

(x + 8)(x - 6) = 0

x = -8, x = 6

So, the width is 6 units and the length is 6 + 2 = 8 units.

Check:
Is the length two more than 6?
Is 6 \times 8 equals 48?

The Ratio Word Problems Tutorial Series

This is a series of tutorials regarding ratio word problems. Ratio is defined as the relationship between two numbers where the second number is how many times the first number is contained. In this series of problems, we will learn about the different types of ratio word problems.

How to Solve Word Problems Involving Ratio Part 1 details the intuitive meaning of ratio.  It uses arithmetic calculations in order to explain its meaning. After the explanation, the algebraic solution to the problem is also discussed.

How to Solve Word Problems Involving Ratio Part 2 is a continuation of the first part. In this part, the ratio of three quantities are described. Algebraic methods is used as a solution to solve the problem.

How to Solve Word Problems Involving Ratio Part 3 in this post, the ratio of two quantities are given. Then, both quantities are increased resulting to another ratio.

How to Solve Word Problems Involving Ratio Part 4 involves the difference of two numbers whose ratio is given.

If you have more math word problems involving ratio that are different from the ones mention above, feel free to comment below and let us see if we can solve them.

The Solving Number Word Problems Series

Word Problems are difficult to many. The Solving Number Word Problems Series is the first series of detailed tutorials on how to solve various number problems. Here are the posts.

(1) How to Solve Number Problems Mentally

This introduction discusses various strategies used to solve easy number word problems. Before you solve a problem using paper and pencil, you should try to solve it first mentally.

(2) How to Solve Number Problems Part 1 

This part solves the same numbers in (1) but using algebra. The objective of this part is to introduce how to set up equations based on “word phrases.”

(3) How to Solve Number Problems Part 2

This part introduces more problems that are slightly more complicated than in (2). It also introduces “number problems in disguise.”

(4) How to Solve Number Problems Part 3

This part of the series focus on how to solve consecutive numbers. Problem of consecutive numbers are very common in math tests.

(5) How to Solve Number Problems Part 4

This post discusses more complicated problems and also introduces how to set up solutions to number problems with fractions.

What’s more to come?

Maybe, I’ll have one more post for this series in the future. But for now, I will focus on the next topic which is about age problems.

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