Mathematics 1090 PRACTICE EXAM I-Solutions Spring 2005

```Mathematics 1090
2005
PRACTICE EXAM I-Solutions
Spring
1. Find the slope-intercept equation for the following lines:
(a) The line of equation 2y + 3x + 2 = 0
Solution: We solve for y in the general form of the equation
3
2y + 3x + 2 = 0 ⇐⇒ 2y = −3x − 2 ⇐⇒ y = − x − 1
2
This is then the slope intercept form of the equation.
(b) The line passing through the points (1, 2) and (−1, 3)
Solution: We find the slope:
m=
1
3−2
=−
−1 − 1
2
1
Point slope form y − 2 = − (x − 1) and we arrange it to find the slope-intercept from:
2
1
1
1
1
5
y − 2 = − (x − 1) ⇐⇒ y = − x + + 2 ⇐⇒ y = − x +
2
2
2
2
2
(c) The line of slope 2 passing through the point (1, 1)
Solution: Point slope form: y − 1 = 2(x − 1) and we arrange it to find the slopeintercept from:
y − 1 = 2(x − 1) ⇐⇒ y = 2x − 2 + 1 ⇐⇒ y = 2x − 1
(d) The line parallel to y = 2x + 3 and passing through (0, 1)
Solution: As the line must be parallel to the line of equation y = 2x + 3 it must
have the same slope m = 2. Moreover the line must pass through (0, 1) which on the
y-axis, then the y-intercept is 1, so the slope-intercept equation is:
y = 2x + 1
(e) The line perpendicular to y = x + 2 and passing through (1, 0)
Solution: As the line must be perpendicular to the line of equation y = x + 2 its
slope is m = − 11 = −1. Moreover the line must pass through (1, 0) so the point-slope
equation is y − 0 = −(x − 1). Then the slope intercept form is
y = −x + 1
2. Suppose that the cost of a business property is \$720000. The owning company wants
to use a straight-line depreciation schedule for a period of 120 months. If y is the
value of the property after x months, write the linear equation of the depreciation
schedule.
Solution: Here x is the time in months, and y is the value of the building in dollars.
The depreciation schedule is represented by a line passing through the point (0, 720000)
which is the point corresponding to the purchase (time=0 and value=720000) and the
point (120, 0) which is the end of the depreciation after 120 months, that is when the
building is worth nothing. So the slope is
m=
0 − 720000
= −6000
120 − 0
Moreover the y-intercept is 720000 so the equation of the depreciation schedule is
y = −6000x + 720000
3. The total amount of a simple investment rate is given by
A(t) = P + P rt
where P is the principal investment, r is the yearly rate and t is the time (in years,
i.e. t = 0 when you invest the money and for instance the amount of money after 10
years is A(10), i.e. when t = 10).
You want to invest \$10000, and your goal is to have \$14000 after 8 years. For what
rate r can you achieve this goal?
Solution: We setup the equation representing our goal, that is having \$14000 on the
account after 8 years:
A(8) = 14000
Now we solve it for r, using the fact that P = 10000:
A(8) = 14000 ⇐⇒ 10000 + 10000r &times; 8 = 14000 ⇐⇒ 80000r = 4000 ⇐⇒ r = 5%
So to achieve the goal we need to have a yearly rate of 5%.
4. You want to invest money so that if you decide to cash it out after 4 years you would
have \$8000 and if you finally decide to cash it out after 10 years you would have
\$15000. Find the equation of the line representing the amount of money A in function
of the elapsed time t (since the investment) that would match your expectations.
Then compare with the formula of the preceding exercise to determine the rate and
the principal investment to achieve your goals.
Solution: The amount of money is given by
A(t) = P + P rt
Then A is a linear function of t (the time). Our goal here is to have \$8000 after 4 years,
that is we want A(4) = 8000 and also \$15000 after 10 years, that is A(10) = 15000.
So that means that the graph of the linear function A(t) is a line passing through the
points (4, 8000) and (10, 15000). We the slope of this line:
m=
7000
15000 − 8000
=
10 − 4
6
Then we find the point-slope form of the equation (here A is y-coordinate t is xcoordinate):
A − 8000 =
7000
14000
7000
7000
(t − 4) ⇐⇒ A = 8000 +
t−
=
t + 10000/3
6
6
3
6
Then as we know that A(t) = P + P rt then we must have
P = 10000/3 = \$3333.33 and P r =
3
7000
7
7000
⇐⇒ r =
&times;
=
= 0.35 = 35%
6
10000
6
20
5. You want to repaint your house. You calculated that you need 15 gallons of paint. At
the paint store they only sell a thick paint that you must dilute with a special solvent.
The salesman tells you that you must prepare a solution containing 10% of solvent
to have a good result. How many gallons of pure paint and solvent must you buy to
prepare the 15 gallons of diluted paint?
Solution: We let x be the quantity of paint and y the quantoty of slovent, both
in gallons. Then first we must prepare a total of 15 gallons with the paint and the
solvent so we have the equation
x + y = 15
Moreover the quantity of solvent must be 10% of the paint, so we have
y = 0.1x ⇐⇒ x = 10y
So we must solve the following system
(
x = 10y
x + y = 15
by substitution the second equation becomes:
10y + y = 15 ⇐⇒ 11y = 15 ⇐⇒ y = 15/11 = 1.4
and the x = 10y = 13.6. So you must buy 13.6 gallons of paint and 1.4 gallons of
solvent.
6. Find
( the market equilibrium for the following laws of supply and demand.
p = − 51 q + 200
, calculate also the M.E. in the case of a tax. of \$6 per sold unit.
a)
p = q + 80
Solution: To find the market equilibrium before tax by solving:
6
1
− q + 200 = q + 80 ⇐⇒ 120 = q ⇐⇒ q = 100
5
5
then the corresponding price is p = 100 + 80 = 180. So the Market Equilibrium is
reached for the price of \$180 and a quantity if 100.
With taxation, the new law of supply is p = q + 80 + 6 = q + 86. We find the market
equilibrium after tax by solving:
1
6
− q + 200 = q + 86 ⇐⇒ 114 = q ⇐⇒ q = 95
5
5
then the corresponding price is p = 95 + 86 = 181. So the Market Equilibrium is
reached
for the price of \$181 and a quantity if 95.
(
2p − q = 50
b)
pq = 100 + 2q
Solution: To find the market equilibrium we solve the system by substitution: from
the first equation we get
2p − q = 50 ⇐⇒ q = 2p − 50
and we put this in the second equation:
p(2p − 50) = 100 + 2(2p − 50) ⇐⇒ 2p2 − 50p = 100 + 4p − 100 ⇐⇒ 2p2 − 54p = 0
⇐⇒ 2p(p − 27) = 0
⇐⇒ p = 0 or p = 27
Then we compute the corresponding quantities: for p = 0 we find that q = −50 &lt; 0,
so we discard this solution. For p = 27 we find q = 54 − 50 ⇐⇒ q = 4. So the
market equilibrium is for p = 27 and q = 4.
7. A company selling dryers has fixed costs of \$5250, and variable costs of 100 + 0.1x
per unit.
a) Write the formula for the cost function C(x).
Solution: We have C(x) = 5250 + x(100 + 0.1x) = 5250 + 100x + 0.1x2 .
b) If the selling price for a dryer is \$210, calculate the profit function P (x) and find
the break-even point(s).
Solution: As the revenue function is R(x) = 210x, we have
P (x) = R(x) − C(x) = 210x − 5250 − 100x − 0.1x2 = −0.1x2 + 110x − 5250
The break-even points are given by solving the equation P (x) = 0 that is by solving
−0.1x2 + 110x − 5250 = 0
We compute the discriminant ∆ = 1102 −4&times;(−0.1)&times;(−5250) = 12100−2100 = 10000.
This is positive so we have 2 solutions
√
√
−110 + 10000
−10
−110 − 10000
x1 =
=
= 50 and x2 =
= 1050
2 &times; (−0.1)
−0.2
−0.2
So the break-even points occur when selling 50 or 1050 dryers.
```