Refresher - MATHEMATICS Quiz 10
PROBLEM 1:
Over the years, a saleswoman has found that the number of vacuum
cleaners she can sell depends on price. The lower the price, the more
she can sell. She has determined that the number of vacuums (x) that
2P
x =- + 28
she can sell at a price (P) is related by the equation 25 . How
much revenue wil be taken in if the vacuums are priced at P250?
Solution:
2P
x=- + 28
25
⎛ 2P ⎞
Revenue = P ⎜ - + 28⎟
⎝ 25 ⎠
⎡ 2(250) ⎤
Revenue = 250 ⎢ - + 28 ⎥
⎣ 25 ⎦
Revenue = P2000
, Refresher - MATHEMATICS Quiz 10
PROBLEM 2:
Coast Guard station A is 200 miles directly east of another station B, A ship is sailing on a
line parallel to and 50 miles north of the line through A and B. Radio signals are sent out
from A and B at the rate of 980 ft/µsec (microsecond). If, at 1:00 P.M. the signal from B
reaches the ship 400 microsecond after the signal from A, locate the position of the ship at
that time.
Solution:
Let us introduce a coordinate system with the stations at points A and B on the x-axis and
the ship at P on the line y = 50. Since at 1:00 P.M. it takes 400 microseconds longer for the
signal to arrive from B than from A, the difference d1 – d2
d1 - d2 = (980)(400)
d1 - d2 = 392,000 ft.
392,000
d1 - d2 =
5280
d1 - d2 = 74.24
At 1:00 P.M. point P is on the right branch of a hyperbola whose equation is
(x2/a2) – (y2/b2) = 1.
x2 y2
- =1
d1 - d2 = 2a a 2 b2
2a = 74.24 x2 y2
- =1
a = 37.12 1378 8622
Let y = 50
a 2 = 1378
2 2
b =c -a 2 x2 (50)2
- =1
1378 8622
b 2 = (100)2 - 1378
x = 42.16 say 42
b 2 = 8622.
The coordinates of P are (42, 50)
PROBLEM 3:
PROBLEM 1:
Over the years, a saleswoman has found that the number of vacuum
cleaners she can sell depends on price. The lower the price, the more
she can sell. She has determined that the number of vacuums (x) that
2P
x =- + 28
she can sell at a price (P) is related by the equation 25 . How
much revenue wil be taken in if the vacuums are priced at P250?
Solution:
2P
x=- + 28
25
⎛ 2P ⎞
Revenue = P ⎜ - + 28⎟
⎝ 25 ⎠
⎡ 2(250) ⎤
Revenue = 250 ⎢ - + 28 ⎥
⎣ 25 ⎦
Revenue = P2000
, Refresher - MATHEMATICS Quiz 10
PROBLEM 2:
Coast Guard station A is 200 miles directly east of another station B, A ship is sailing on a
line parallel to and 50 miles north of the line through A and B. Radio signals are sent out
from A and B at the rate of 980 ft/µsec (microsecond). If, at 1:00 P.M. the signal from B
reaches the ship 400 microsecond after the signal from A, locate the position of the ship at
that time.
Solution:
Let us introduce a coordinate system with the stations at points A and B on the x-axis and
the ship at P on the line y = 50. Since at 1:00 P.M. it takes 400 microseconds longer for the
signal to arrive from B than from A, the difference d1 – d2
d1 - d2 = (980)(400)
d1 - d2 = 392,000 ft.
392,000
d1 - d2 =
5280
d1 - d2 = 74.24
At 1:00 P.M. point P is on the right branch of a hyperbola whose equation is
(x2/a2) – (y2/b2) = 1.
x2 y2
- =1
d1 - d2 = 2a a 2 b2
2a = 74.24 x2 y2
- =1
a = 37.12 1378 8622
Let y = 50
a 2 = 1378
2 2
b =c -a 2 x2 (50)2
- =1
1378 8622
b 2 = (100)2 - 1378
x = 42.16 say 42
b 2 = 8622.
The coordinates of P are (42, 50)
PROBLEM 3:
