20 ANSWERS
1. If all earnings above the €4000 are
taxed at 25%. Then the linear function
between take-home-pay, THP, and
earnings, E, can be written as:
a) THP = 3000 + 0.75E
b) THP = 0.25E + 4000
c) THP = 0.75E + 1000
d) None of the above
THP =E – 0.25 (E – 4000 ) = E – 0.25E +
1000 = 1000 + 0.75E
1
2. If a linear function can be written as y
= a + bx, which of the following
statements are true:
a) the intercept of the function = a,
and the slope of the function is = b
b) the slope of the function = a, and
the intercept of the function is = b
c) the intercept of the function is a,
and the slop of the function is bx
d) None of the above
intercept of function: value of y when x = 0
and hence = a
slope of the function: change in y due to a
change in x = b (may be positive of
negative)
2
3. A Charity Organisation gets €3000
each year from an anonymous patron. The
remaining income must be raised, and is a
linear function of the number of events
organised E. The slope of the function
relating total income to the number of
events organised is 2. Last year a total of
300 events were organised countrywide.
As a result the Charity had a total income
of
a) €3,300
b) €3,600
c) €6,600
d) None of the above
Income Y = a + bE
Y = 3000 + 2(300) = 3600
3
4. If a student Demands 6 pints of
Guinness when the price per pint P= €2,
and only demands 4 pints when the price
per pint P = €3 per pint. Then the demand
function D = a + bP can be written as
a) D = 6 + 2P
b) D = -10 – P
c) D = 10 – 2P
d) None of the above
Eq1: 6 = a + 2b
Eq.2 4 = a + 3b
solving: 6 – 2b = 4 – 3b and so b = -2
subbing b=-2 into either eq1 or eq2, we
find a = 10
Thus, D = 10 – 2P
4
5. The solution(s) to the
2
equation  2 x  x  10  0 are:
a) there are no solutions
b)
8
c) -2 and +2.5
d) None of the above
quadratic
solve using formula
a=-2, b=1, c=10
b2  4ac  12  4 210  81  0
 b  b2  4ac  1  81  1  9


2a
2 2
4
1 9
1 9
x
 2
x
 2 .5
4
4
x
5
6. Using the rules of indices,
4 3

 can be written as
2x
expression
a)
8 x12
b)
8 x7
c)
8x
d) None of the above
solution: 2 x 
4 3
7.
 
2 .x
3
4 3
the
 8 x 4.3  8 x12
Using the rules of logs, the expression
 xy 
log b  
 z  can be re-written as
log b x  log b y  log b z
a)
log b x  log b y  log b z
b)
log b  log b y  log b z
c)
d)
None of the above
rule: log x + log y =log xy
and log xy – log z = log (xy/z)
6
8. If the demand function for a good is Q
= 25 – 5P, then the inverse demand
function can be written as:
a)
Q=5–P
b)
P = 5 – 1/5 Q
c) P = Q – 25
d) None of the above
re-express, putting P on the left hand
side… so 5P = 25 – Q and hence P = 5 – 1/5
Q
7
9. The equilibrium Price and Quantity
levels in a market with a demand function
QD = 20 – ½ P and supply function QS = –
6 + ½ P is:
a) P = 26 and Q = 7
b) P = 0 and Q = 0
c) P = 14 and Q = 10
d) None of the above
Equilibrium is where QD = QS
So 20 – ½ P = - 6 + ½P
Hence P = 20 + 6 = 26
Solve for equilibrium Q by substituting P
into either QD or QS:
QD = 20 – ½ P = 20 – 13 = 7 so
equilibrium Q = 7
8
10. If a sales tax of amount T per unit is
imposed on suppliers in the market
described in question 9 above, what is the
new equilibrium quantity and price?
a) P = 26 + ½ T and Q = 7 – ¼ T
b) P = 14 + ½ T and Q = 10 + ½ T
c) P = 26 – T and Q = 7 + T
d) None of the above
Effect of sales tax: New Supply curve given
as Qs = - 6 + ½ (P-T) = -6 + ½ P - ½ T
Solving for new equilibrium: set Qd = new
Qs and hence
20 – ½ P = -6 + ½ P - ½ T
Rearranging gives us: 26 + ½ T = P
Subbing in P = 26 + ½ T to the demand
function we get:
Q = 20 – ½ (26 + ½ T) = 20 – 13 – ¼ T = 7
-¼T
(i.e. tax has effect of increasing equilibrium
P and reducing equilibrium Q)
9
11. Differentiating the function y = f (x) =
(3x3+2x2 + x + 5)4 with respect to x gives
the derivative:
a) 4((3x3+2x2 + x + 5)3
b) 4(3x3+2x2 + x + 5)3 ( 9x2+4x + 1)
c) (9x2+4x + 1)3
d) None of the above
Apply chain (or implicit function) rule: let
v = 3x3+2x2 + x + 5 and y = v4
So dy/dx = dy/dv . dv/dx
So dy/dx = (4v3 ) (9x2+4x + 1) and hence
dy/dx = 4(3x3+2x2 + x + 5)3 ( 9x2+4x + 1)
10
12. The first derivative, dy/dq, of the
function y = ln (5q+1 ) is
5
a) 5q  1
1
b) 5q  1
c) 5q + 1
d) None of the above
Apply the chain rule: y = ln(v) where v =
f(q) = 5q + 1
dy/dq = dy/dv . dv/dq = (1/v) (5) = 5/ (5q +
1)
11
13. The first derivative of the function y =
x2 e2x can be written as:
a) e2x ( 2x2 + 2x)
b) e2x + 2x
c) 2x (e2x )
d) None of the above
Apply product rule, and note derivative
of e2x with respect to x is given as 2e2x
So (x2 2e2x) + (e2x. 2x) which can be
written as e2x ( 2x2 + 2x)
14. The slope of the function y = x3 – 4x2
+ 12x + 3 can be written as :
a) + 12
b) 3x2 – 8x + 12
c)
6x – 4
d)
None of the above
The slope of the function is given by the
first derivative:
dy/dx = 3x2 – 8x + 12
12
15. The second derivative, d2y/dx2, of the
function y = (3x +2) (4x +1) can be
written as:
a)
d2y
 24 x  11
2
dx
d2y
b) dx 2  7 x
d2y
0
c) dx 2
d) None of the above
first derivative = dy/dx = 24x + 11
second derivative = d2y/dx2 = 24
13
16. If the Total Cost function is TC = 3Q2
+ 9Q +3 then the Average Cost can be
written as
a) 3Q3 + 9Q2 + 3Q
b) 6Q + 9
c)
3Q  9 
3
Q
d) None of the above
Average Costs = Total Cost/Q = 3Q + 9 +
3/Q
14
17. If the Total Cost function is TC = 3Q2
+ 9Q +3 then the Marginal Cost can be
written as
a) 3Q3 + 9Q2 + 3Q
b) 6Q + 9
c)
6Q  9 
3
Q
d) None of the above
Marginal Cost = change in TC from a
small change in Q = dTC / dQ = 6Q + 9
15
18. Given the demand function: q = -2p +
100, the elasticity of demand when p = 10
is given as
a) 0
b) +0.25
c) - 0.25
d) None of the above
Elasticity of demand = dQ/dP . P /Q
p = 10, so q  2p  100  80 , and
dq
 2
dp
Therefore the point elasticity of demand is:
ED 
p dq 10

(2)  0.25
q dp 80
16
19. If the demand curve for a firm is given
by the function P=f(Q), then the value of
Q that minimises total revenue must
satisfy:
a) f (Q) =0 and f (Q) = 0
b) f (Q) =0 and f (Q) > 0
c) f (Q) =0 and f (Q) < 0
d) None of the above
need to satisfy first order condition: the
first derivative (or slope) must = 0 for a
stationary point so f’(Q) = 0 AND
For that stationary point to be a minimum,
we need to satisfy second order condition:
the change in the slope around the
stationary point must be positive (i.e.
moving from 0 to a positive slope, hence
‘uphill’) so f’’(Q) > 0
17
20. Find and classify any stationary points
of the following function y  x  6x  15 in order
to determine which of the following
statements are true.
a) max at x = –3 and min at x = 3
b) minimum at x = –3
c) maximum at x = –3
d) None of the above
2
first order condition: dy/dx = 2x+6 = 0 and
hence x = -3 at stationary point
second order condition: d2y/dx2 = 2 > 0
hence a min at x = -3
18