Category: Math Word Problems

LCM and GCD Exercises Set 2

Math Word Problems, Numerical Reasoning

Here are some Civil Service exam exercises on GCD and LCM.

1.) What is the GCD of 8, 20, and 28?

2.) What is the GCD of 21, 35, and 56?

3.) What is 18/54 in lowest terms?

4.) What is 38/95 in lowest terms?

5.) What is the LCM of 6 and 8?

6.) What is the LCM of 5, 6, and 12?

7.) What is the product of the LCM and the GCD of 4, 8, and 20?

8.) There are 18 red marbles and 27 blue marbles to be distributed among children. What is the maximum number of children that can receive the marbles if each kid receives the same number of marbles for each color and no marble is to be left over?

9.) In a school sportsfest, there are 60 Grade 4 pupils, 48 Grade 5 pupils and 36 Grade 6 pupils. What is the largest number of teams that can be formed if the pupils in each Grade level are equally distributed and no pupil is left without a team?

10.) In a disco, the red lights blink every 3 seconds and the blue lights blink every 5 seconds. If the two colored lights blink at the same time if you turn them on, they will blink at the same time every ___ seconds.

Answers:

1.) Answer: 4
Solution:
Divisors of 8 – 1, 2, 4, 8
Divisors of 20 – 1, 2, 4, 5, 10, 20
Divisors of 28 – 1, 2, 4, 7, 14, 28

2.) Answer: 7
Solution:
Divisors of 21 – 1, 3, 7, 21
Divisors of 35 -1, 3, 5, 7, 35
Divisors of 56 -1, 2, 7, 8, 14, 28, 56

3.) Answer: 1/3
Note: Reducing fractions to lowest terms is one of the applications of GCD.
Solution:
Divisors of 18 – 1, 2, 6, 9, 18
Divisors of 54 – 1, 2, 3, 9, 18, 27, 54
Numerator = 18 divided by 18 (GCD) = 1
Denominator = 54 divided by 18 (GCD) = 3
Final answer: 1/3

4.) Answer: 2/5
Solution:
Divisors of 38 – 1, 2, 19, 38
Divisors of 95 – 1, 2, 3, 19, 95

5.) Answer: 24.
Solution:
Multiples of 6 – 6, 12, 18, 24
Multiples of 8 – 8, 16, 24

6.) Answer: 60.
Solution:
Multiples of 5 – 5, 10, 15, 20, 25, 30 … 55, 60
Multiples of 6 – 6, 12, 18, 24, 30, 36, 42, 48, 54, 60
Multiples of 12 – 12, 24, 36, 48, 60

7.) Answer: 160.
Solution
GCD of 4, 6, and 20 is 4.
LCM of 4, 6, and 20 is 40.
4 x 40 = 160.

8.) Answer: 9
Solution:
Divisors of 18 – 1, 2, 3, 6, 9, 18
Divisors of 27 – 1, 2, 3, 9, 27

9.) Answer: 12.
Solution:
Divisors of 60 – 1, 2, 3, 4, 5, 6, 10, 12, 20, 30, 60
Divisors of 36 – 1, 2, 3, 4, 6, 9, 12, 18, 36
Divisors of 48 – 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

10.) Answer: 15 seconds.
Solution
LCM of 3 and 5 is 15

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Week 11 Review: Answers and Solutions

Math Word Problems

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

Problem 1
Aria can do a job in 7 days. What part of the job is finished after she worked for 3 days?

Aria can do a job in 7 days. So meaning, she can do 1/7 job each day.

1st day – 1/7
2nd day – 1/7
3rd day – 1/7

So, 3(1/7) = 3/7

Answer: 3/7

Problem 2
Katya can do a job in 5 days. Marie can do the same job in 6 days. If they both worked for 1 day, what part of the job is finished?

Katya can do a job in 5 days which means each day 1/5 of the job.
Marie can do a job in 6 days which means each day 1/6 of the job.

Let x = part of the job finished for 1 day

1/5 + 1/6 = 1/x
LCD: 30x

Multiplying both sides of the equation by 30, we have
(30x)(1/5 + 1/6) = (30x)(1/x)
(30x/5) + 30x(1/6) = (30x/x)
6x + 5x = 30
11x = 30
x = 30/11 or 2 8/11
Answer: 2 and 8/11 days.

Problem 3
Ramon can paint a house in 6 days. Ralph can do the same job in 10 days. If they both worked for 2 days, what part of the job is done if they were to work together the whole time?

Ramon = 6 days (or 1/6 part each day)
Ralph = 10 days (or 1/10 part each day)

Let w – part of the job done for 2 days

w = 2(1/6) + 2(1/10)
w = 2/6 + 2/10

LCD: 30

30(2/6 + 2/10)
((30/6)*2)/30 + ((30/10)*2)/30
(5*2)/30 + (3*2)/30
10/30 + 6/30
16/30 or 8/15

8/15 – work done for 2 days.

If they were to worked together, they will finish the work in x days using the equation

1/6 + 1/10 = 1/x

We multiply both sides of the equation which is 30x giving us

30x(1/6) + 30x(1/10) = 30x(1/x)
(30x)/6 + (30x)/10 = (30x/x)
5x + 3x = 30
8x = 30
x = 30/8
x = 15/4

This means that both of them will finish the work in 15/4 days. This means that the amount of worked finished is (8/15)/(15/4) = (8/15)(4/15) = 32/225.

Problem 4
One hose can fill a pool in 3 hours and a smaller hose can fill the same pool in 4 hours. How long will it take the two hoses to fill the entire pool?

Let x – total time to fill the pool

1/3 + 1/4 = 1/x
LCD: 12x

Multiplying both sides by 12x, we have

12x(1/3 + 1/4 = 1/x)12
(12x)/3 + (12x)/4 = 12
4x + 3x = 12
7x = 12
x = 12/7 or 1 5/7 hours

Answer: 1 5/7 hours

Problem 5
Marco can dig a ditch in 5 hours and he and Jimmy can do it in 2 hours. How long would it take Jimmy to dig the same ditch alone?

Let 1/x – Jimmy’s time alone
1/5 – Marco’s time

1/x + 1/5 = 1/2
LCD: 10x
10x(1/x + 1/5) = (10x)(1/2)
10 + 2x = 5x
2x – 5x = -10
-3x = -10
x = -10/-3 or 3 1/3 hrs

Problem 6
Maria can paint a fence in 6 days and Leonora can do the same job in 7 days. They start to paint it together, but after two days, Leonora left, and Maria finishes the job alone. How many days will it take Leonora to finish the job?

Job done by Maria in 2 days = 2/6
Job done by Leonora in 2 days = 2/7
Job done by Maria in the remaining days = x/6

2/6 + 2/7 + x/6 = 1

LCD:42x

Multiplying both sides of the equation by 42, we have

42(2/6 + 2/7 + x/6) = (42)(1)
84/6 + 84/7 + (42x)/6 = 42
14 + 12 + 7x = 42
26 + 7x = 42
7x = 42 – 26
7x = 16
x = 16/7 or 2 2/7 days

Problem 7
An inlet pipe can fill a pool in 4 hours. An outlet pipe can fill the same pool in 6 hours. One day, the pool was empty. The owner opened the inlet pipe but forgot to close the outlet pipe. How long will it take to fill the pool?

Let x – hours to fill the pool
1/4 – 1/6 = 1/x
LCD: 12x

(12x)(1/4) – (12x)(1/6) = (12x)(1/x)
(12x)/4 – (12x)/6 = (12x)/x
3x – 2x = 12
x = 12 hours

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Week 10 Review: Answers and Solutions

Math Word Problems, Mathematics, Numerical Reasoning

After learning how to solve motion problems, let’s answer some exercises and problems. In the solutions, we let d = distance, r = rate, and t = time.

Exercises

1.)   A car travels and average speed of 75 kph. If it traveled for 3.5 hours, what is the total distance traveled?

d = rt
d = (75 kph)(3.5 hrs) =
d = 262.5 km

2.) A bus traveled 4 hours from City A to City B which is 450 kilometers apart. What is its average speed?

d = rt
450 km = (4 hrs)(r)
r = (450 kph)/(4 hrs)
r = 112.5 km

Problem

1.) Two cars left City A at 8:00 am going to City B using the same route. Car 1 traveled at the average speed of 60 kph while Car 2 traveled at an average speed of 50kph. At what time were the two cars 25 kilometers apart?

Let x = time traveled by the two cars
60x – 50x = 25
10x = 25
x = 25/10
x = 2.5 hours

2.5 hours = 2hours and 30 mins
2 hours and 30 minutes after 8:00 am is 10:30 am.

Answer: 10:30 AM

2.) The road distance from Sapiro City to Lireo City is 195 km. Car 1 left Sapiro City going to Lireo City at an average speed of 70kph. Car 2 left City Lireo City going to Sapiro City at an average speed of 60 kph. If both cars left the two cities at the same time and use the same road, after how many hours will the two cars meet?

Car 1
Rate = 70kph
Time = x
Distance = 70x

Car 2
Rate = 60kph
Time = x
Distance = 60x

Total distance = 195kph

70x + 60x = 195
130x = 195
x = 195/130

x = 1.5hrs

3.) A red car left Vigan at 9:00 AM and traveled to Manila at an average speed of 45 kph. After one hour, a white car left the same place for Manila using the same route at an average speed of 60 kph. At what time will the white car overtake the red car?

RED CAR
Rate = 45kph
Time = x+1
Distance = 45(x+1)

WHITE CAR
Rage = 60kph
Time = x
Distance = 60x

60x = 45(x+1)
60x = 45x+45
60x – 45x = 45
15x = 45
x = 45/15
x = 3hours

3 hours after 10:00am is 1:00 p.m.

Note: We add 3 hours to 10:00 am because the second car left at 10:00 am.

Answer: 1:00pm noon

4.) Two cars started from the same point, at 12nn, traveling to opposite directions at 50 and 60 kph, respectively. What is the distance between them at exactly 3:30 PM?

CAR 1
Rate = 50kph
Time = 3.5 hrs
Distance = (50 kph)(3.5 hrs) = 175

CAR 2
Rate = 60kph
Time = 3.5 hrs
Distance = (60 kph)( 3.5 hrs) = 210

What is the distance between them at exactly 3:30 PM?
175 km + 210 km = 385 km

5.) Two cars from the same point traveling to opposite directions at 75 and 85 kph, respectively. After how many hours will they be 240 kilometers apart?

75x + 85x = 240
160x = 240
240/160 = x
x = 1.5 hours

6.) A blue car leaves City A for City B at exactly 8:00 AM traveling at average speed of 55 kph. A gray car leaves City B for City A at the same time traveling at an average speed of 45 kph. The distance between the two cities is 75 kilometers.

If the two cars use the same route, what time will they pass each other?

let x be the time
d = r x t

sum of the distance traveled by 2 cars is equal to 75km
so

55x + 45x = 75
100x = 75
x = 75/100 or 0.75 hours
0.75 hours (45 mins)

They will pass each other 45 mins after 8:00am so the answer is 8:45 am.

Answer: 8:45 am

Enjoy!

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Week 10 Review: Practice Exercises and Problems

Math Word Problems, Mathematics, Numerical Reasoning

After learning how to solve motion problems, let’s answer some exercises and problems.

Exercises

1.)   A car travels and average speed of 75 kph. If it traveled for 3.5 hours, what is the total distance traveled?

2.) A bus traveled 4 hours from City A to City B which is 450 kilometers apart. What is its average speed?

Problem

1.) Two cars left City A at 8:00 am going to City B using the same route. Car 1 traveled at the average speed of 60 kph while Car 2 traveled at an average speed of 50kph. At what time were the two cars 25 kilometers apart?

2.) The road distance from Sapiro City to Lireo City is 195 km. Car 1 left Sapiro City going to Lireo City at an average speed of 70kph. Car 2 left City Lireo City going to Sapiro City at an average speed of 60 kph. If both cars left the two cities at the same time and uses the same road, after how many hours will the two cars meet?

3.) A red car left Vigan at 9:00 AM and traveled to Manila at an average speed of 45 kph. After one hour, a white car left the same place for Manila using the same route at an average speed of 60 kph. At what time will the white car overtake the red car?

4.) Two cars started from the same point, at 12nn, traveling to opposite directions at 50 and 60 kph, respectively. What is the distance between them at exactly 3:30 PM?

5.) Two cars from the same point traveling to opposite directions at 75 and 85 kph, respectively. After how many hours will they be 240 kilometers apart?

6.) A blue car leaves City A for City B at exactly 8:00 AM traveling at average speed of 55 kph. A gray car leaves City B for City A at the same time traveling at an average speed of 45 kph. The distance between the two cities is 75 kilometers.

If the two cars use the same route, what time will they pass each other?

7.) A gray car leaves Batangas City for Quezon City at 11:00 am at an average speed of 50 kph. After one hour, a black car leaves Quezon City for Batangas at an average speed of 60 kph. If the two cars use the same route,  at exactly what time will the two cars pass each other?

Enjoy!

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PCSR REVIEW SERIES WEEK 10: Motion Problems

Math Word Problems, Mathematics, Numerical Reasoning

We have learned number problems and age problems. Now, we learn how to solve motion problems or problems involving distance, rate, and time. Below are the articles and videos that you can watch to learn about motion problems.

ARTICLES

VIDEOS

Enjoy!

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Week 9 Review: Practice Exercises and Problems

Math Word Problems, Mathematics, Numerical Reasoning

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

Exercises

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

2.) Nico is 3 times as old as Jimmy. The sum of their ages is 48. How old are they?

3.) Mia’s age is 1 more than twice Lito’s age. The sum of their ages is 25. How old are they?

4.) Alfred’s thrice as old as Fely. The difference between their ages is 16. How old are they?

Problems

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

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

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

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

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?

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

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?

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?

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?

In the next post, I am going to post the solutions to these problems.

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PCSR REVIEW SERIES WEEK 9: Age Problems

Math Word Problems, Mathematics

After learning number problems, let’s learn about age problems. Below are the articles and videos that you can use for studying age problems.

ARTICLES

VIDEOS

Good luck!

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Week 8 Review: Answers and Solutions

Math Word Problems, Mathematics, Numerical Reasoning

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.

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