# Fundamentals of Numeric Computing

## CPAN 112 || Lab 02

• Please provide your answers in the boxes below each question, and do not change the text colour.
• There is no credit if you only state the final answer.
• Please keep the naming conventions requested in this lab and each question.
• Once you complete your lab, rename your word document file to the (CPAN112_LabXX_FirstName_LastName). Replace XX with the lab number (e.g. 01). Replace FirstName and LastName with your first name and last name, respectively.

It will be a 10% mark deduction if you do not follow the guidelines mentioned above

• Set up an Equation for:
• A manufacturer makes two type of products profit on Product A is \$30 per unit and profit on Product B is \$40 per unit. Budgeted monthly profit is \$6000.
• 30x+40y=6000
• A manufacturer processes two type of products. Each unit of the product A needs 20 time units in finishing while product B needs 30 time units. Per day 1200 time units are available. Set up an equation that describes the relationship between the number of units.
• 20x+30y=1200
• If you earn\$ 30000 per year and spend \$29000 per year, write and equation for the amount you save after y years, if you start with nothing.
• X = (30000-29000)*y
• X = 1000y

• Setup the equation and solve:
• Terry invested a total of \$4500. A portion was invested at 4% and the rest was invested at 6%. The amount of Terry’s annual return on each portion is the same. Find the average rate of interest Terry earned on his total investment.
• Total invested 2 Portions i.e. X+Y = 4500 eq(1)
• X*4/100=Y*6/100 => 4X-6Y =0 eq(2)
• Multiply eq(1) with 6 => 6X +6Y = 27000 eq(3) Add eq(2) and eq(3)
• 10X-27000 => X=2700 So, Y = 4500-2700 = 1800
• Total Return => 2700*4/100 + 1800*6/100 = 216 Avg Return = 216/4500 *100 = 4.8%
• Kim invested a total of \$ 24000 in two mutual funds. Her investment in the equity fund is \$4000 less than three times her investment in the Bond value. How much did Kim invested in each value.
• X = equity fund Y = Bond value x+y= 24000
• x= 3y-4000 x+y=24000
• x-3y=-4000 (* -1) 4y=28000
• y= 7000
• x= 3*7000 – 4000 x = 17000
• Kim invested \$7000 in the equity fund and \$17000 in the Bond value.
• Nancy’s sales last week were \$140 less than three times Andrea’s sales. Together they sold \$940. Determine how much each person sold last week?
• X = Nancy sales
• Y = Andrea’s sales X + y = 940
• X = 3y -140 X + y = 940
• X – 3y = -140 (* -1) 4y = 1080
• Y = 270
• X+270 = 940
• X = 670
• Nancy sold \$670 and Andrea sold \$270.

• Solve the system of equations: 2x + 6y = -12   2x – 5y = 10
• 2x + 6y = -12
• 2x – 5y = 10 (* -1)
• 11y = -22
• Y = -2
• 2x + 6*(-2) = -12
• 2x – 12 = -12
• 2x = 0
• X = 0

• Find the slope and y-intercept: (Rewrite the equation in y = mx +b form)
• 4.5x + 9 y = 2 b) x + 2y = 8 c) 4y = 16 d) x = 12
• Y=(-4.5x+2)/9 Y = -0.5x + 0.22
• Slope: -0.5 y-intercept: 0.22
• Y=-0.5x+4
• Slope: -0.5 y-intercept: 4
• y= 4
• slope: 0 y-intercept: 4
• x=12 => 12-x =0
• Infinite slope, no y intercept

• Graph: x + 4y = 8. Rewrite the equation in y = mx +b form Slope m = Rise/ Run =
• Rise = Run =
• • X + 4y = 8 4y = -x + 8
• Y = -0.25x + 2
• Slope m = -0.25/1 = -0.25 Rise = – 0.25
• Run = 1
•  x 0 1 2 3 4 5 6 7 8 y 2 1.75 1.5 1.25 1 0.75 0.5 0.25 0

6. Solve graphically: 2x – y = 1 and 2x + y = 8 ( Find the point of intersection, Where the lines cross each other). • 2x – y = 1
• -y = 1 – 2x
• Y = 2x – 1 (1)
• 2x + y = 8 Y = 8 – 2x
• Y = -2x + 8 (2)

7.

 x 0 1 2 3 4 5 y -1 1 3 5 7 9 x 0 1 2 3 4 5 y 8 6 4 2 0 -2
• Using the first equation into the second:
• Y = 2x -1 (1)
• Y = -2x + 8 (2)
• (1) = (2)
• 2x -1 = -2x +8
• 4x = 9
• X = 9/4 = 2.25
• Using X in the equation (1) Y = 2*(2.25) – 1
• Y = 3.5
•  Intersection between the two lines is A (2.25, 3.5)