Problem 1 :
Mrs. Saunders can clean the windows of her house in 3 hours. Her daughter can clean the windows in 6 hours. How long will it take them to clean the windows if they work together?
Solution:
Saunders = 1/3 windows per hours
Daughter = 1/6 windows per hours
Working together, they clean
It will take them 2 hours to finish when they work together.
Problem 2 :
A driver can deliver his newspapers in 80 minutes. His friend can take care of the same route in 2 hours. How long would it take them to do the job together?
Solution:
The driver's ratio is of 1/80
His friend's ratio is of 1/120
The together ratio is of 1/x
It would take them 48 minutes to do the job together.
Problem 3 :
One pipe can fill a tank in 8 minutes, a second can fill it in 12 minutes and a third can fill it in 24 minutes. If the tank is empty how long will it take the three pipes, operating together, to fill it?
Solution:
The rate of the first pipe = 1/8
The rate of the second pipe = 1/12
The rate of the third pipe = 1/24
Problem 4 :
Two printing presses, working together, can complete a job in 2 hours. If one press requires 6 hours to do the job alone, how many hours would the second press need to complete the job alone?
Solution:
First press = 1/6 hours
Second press = 1/x hours
working together = 1/2 hours
So, second printing press needs 3 hours by itself.
Problem 5 :
One supplementary angle is 5 more than six times the other angle. What is the measure of each angle?
Solution:
Let one angle = x
other angle = y
x + y = 180° ---> (1)
x - 5 = 6y
x - 6y = 5 ---> (2)
Substitute equation (2) from (1),
x - x + y + 6y = 180 - 5
7y = 175
y = 25°
Put y = 25° in equation (1)
x + y = 180°
x + 25° = 180°
x = 180 - 25
x= 155°
So, one angle = 155°
and other angle = 25°
Problem 6 :
One complementary angles measures 18 less than five times the other angle. What is the measure of each angle?
Solution:
Let one angle = x
Other angle = 5x - 18
x + (5x - 18) = 90°
6x - 18 = 90°
6x = 108
x = 108/6
x = 18
So, one angle = 18°
Other angle = 5(18) - 18
= 90 - 18
= 72°
Problem 7 :
If the perimeter of an equilateral triangle is 24 inches, find a side of the triangle.
Solution:
Perimeter of an equilateral triangle = 3a
24 = 3a
a = 24/3
a = 8 in
Problem 8 :
Each of the equal sides of an isosceles triangle is 4 times the third side. The perimeter of the triangle is 144 inches. Find the sides of the triangle.
Solution:
Let the third side be x.
Each of the equal side is 4x.
Perimeter = 144 in
P = x + 4x + 4x
P = 9x
144 = 5x
x = 16 in
Second side = 4x = 4(16)
= 64 in
Third side = 4x = 4(16)
= 64 in
The sides of the triangle are 16 in, 64 in and 64 in.
Problem 9 :
The perimeter of a triangle is 73 inches. If the second side is 5 inches longer than twice the first side, and the third side is 4 inches less than three times the first side, how long is each side?
Solution:
Let first side be x.
Second side = 2x + 5
Third side = 3x - 4
The sum of the length of the sides is equal to the perimeter.
x + 2x + 5 + 3x - 4 = 73
6x + 1 = 73
6x = 72
x = 72/6
x = 12 in
Second side = 2(12) + 5
= 29 in
Third side = 3(12) - 4
= 32 in
Problem 10 :
The length of a rectangle is three times the difference of the width and two. If the perimeter of the rectangle is 60 cm, what is the length of the rectangle?
Solution:
Let the width be x.
Length = 3(x - 2)
Perimeter of rectangle = 60 cm
Perimeter = 2(length + width)
2(3(x - 2) + x)) = 60
2(3x - 6 + x) = 60
2(4x - 6) = 60
8x - 12 = 60
8x = 72
x = 9 cm
Length = 3(x - 2)
= 3(9 - 2)
= 3(7)
Length = 21 cm
So, width = 9 cm
Length = 21 cm
Problem 11 :
A rectangular playground is enclosed by 440 feet of fencing. If the length of the playground is 20 feet less than 3 times the width, find the dimensions of the playground.
Solution:
Let the width be x feet
The length = 3x - 20 feet
2(x + 3x - 20) = 440
2(4x - 20) = 440
8x - 40 = 440
8x = 480
x = 480/8
x = 60 feet
Length = 3(60) - 20
= 180 - 20
= 160 feet
So, width = 60 feet
Length = 160 feet.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM