Refresher - MATHEMATICS Quiz 18
PROBLEM 1:
Rick has two part-time jobs, one paying $7 per hour and the other paying
$5 per hour. He must earn at least $140 per week to pay his expenses
while attending college. How much is his maximum income per week?
Solution:
x = the number of hours per week he works on the first job
7x = per week on the first job
y = the number of hours per week he works on the second job
5y = per week on the second job
7x + 5y ≥ 140
when x = 10 y = 15
7(10) + 5(15) = $145
when x = 5 y = 25
7(5) + 5(25) = $160
, Refresher - MATHEMATICS Quiz 18
PROBLEM 2:
A corner lot of land is 122.5 m. on one street and 150 m. on the other
street, the angle between the two streets being 75˚. The other two lines of
the lot are respectively perpendicular to the lines of the streets. What is
the perimeter of the boundary of the lot?
Solution:
122.5
Sin 15˚ =
h
c
h = 473.30 m a
x = 473.30 – 150 122.5
x = 323.30 m y
b
y 75ß 15ß
tan 15˚ =
323.30 150 x
y = 86.63 m. h
122.5
tan 15˚ =
a
a = 457.18 m
86.63
Sin 15˚ =
b
b = 334.71 m
c = 457.18 – 334.71
c = 122.47 m.
Perimeter = 122.5 + 150 + 86.63 + 122.47
Perimeter = 481.60 m.
PROBLEM 1:
Rick has two part-time jobs, one paying $7 per hour and the other paying
$5 per hour. He must earn at least $140 per week to pay his expenses
while attending college. How much is his maximum income per week?
Solution:
x = the number of hours per week he works on the first job
7x = per week on the first job
y = the number of hours per week he works on the second job
5y = per week on the second job
7x + 5y ≥ 140
when x = 10 y = 15
7(10) + 5(15) = $145
when x = 5 y = 25
7(5) + 5(25) = $160
, Refresher - MATHEMATICS Quiz 18
PROBLEM 2:
A corner lot of land is 122.5 m. on one street and 150 m. on the other
street, the angle between the two streets being 75˚. The other two lines of
the lot are respectively perpendicular to the lines of the streets. What is
the perimeter of the boundary of the lot?
Solution:
122.5
Sin 15˚ =
h
c
h = 473.30 m a
x = 473.30 – 150 122.5
x = 323.30 m y
b
y 75ß 15ß
tan 15˚ =
323.30 150 x
y = 86.63 m. h
122.5
tan 15˚ =
a
a = 457.18 m
86.63
Sin 15˚ =
b
b = 334.71 m
c = 457.18 – 334.71
c = 122.47 m.
Perimeter = 122.5 + 150 + 86.63 + 122.47
Perimeter = 481.60 m.
