Question

1.    Solve the following simultaneous equations (i) graphically and (ii) using the elimination method.

      (a)    2x + 3y = 12.5       (y on the vertical axis)       (b)     4P – 3Q = 3      (p on the vertical axis)                    -x +2y =6                                                                            P +2Q = 20             

2.    Suppose the demand and supply of a good are given as

            P = 80 – 2Q    and P=20 + 4Q

(a) Calculate the equilibrium price and quantity, algebraically.

(b)   Suppose a per unit tax of $12.00 is levied on sellers,   show graphically the effect of this per unit tax on the equilibrium price and quantity if any in the market.

3. Suppose the population grows (in thousands) according to the following relationship,   P=125e^0.012t , where t is the number of years.   Using logarithms, find the number of years t, it would take for the population P, to double, that is equal 250.   Also comment on the meaning of the exponent.

                                   

4.   Suppose you are given the two quadratic equations, (i) x²+6x+5=0,   and (ii) 2x2-7x-9=0 . Solve each of these equations, classifying the types of solutions (roots) found.

5.   Suppose you are given the following inverse demand function, p=200Q+1   and the inverse supply function, p=5+0.5Q.   With p on the vertical axis and Q on the horizontal, draw these two functions.    Also solve for the equilibrium Q* and equilibrium price p*.

6.    Suppose the labour demand function is given as w = 18 – 1.6L  and the labour supply function is given as w=6+0.4L.    Determine the equilibrium wage and equilibrium number of workers algebraically.   Draw the above labour demand and labour supply functions on a diagram with w on the vertical axis and illustrate the effect of a binding minimum wage.    

7.     Solve the following system of linear equations using Cramer’s Rule

1. 2y-20x=24                                     (b) 2p=100-4Q

8x+4y=72                                             2p=10+6Q

Answer 1

--x + y = 6 2 then adding with Given equation is 2x+342 12.5 for eliminiting n, multiplying the equin by 0 we have 22437 = 12.5 - +4y= 12 75 = 34.5 24.5 = 3.5 Put y= 24.3 in eqan @ we get 27-6 Hence, the solution of the system is 249-6 = 1 7 X=1 Yz24:5 -305 48.-38-3 P +28 = 20 For eliminating p we multiply the equn by-4 and then adding with o de get AP-38=8 -4P -80:-80 Pur 8 = 7 in equi- -118--77 .:8= 7 P=20-2x7 Hence, the solution of the system is P = hs 7 S ae gel 6 --

a. Y -10 -x+2y=6 5 2x+3y=12.5 (1,3.5) Х -10 -5 0 5 10 --5 --10

b. P -10 P+2Q=20 4P-3Q=3 (7,6) 5 -5 0 5 10 Q --5