Math 111 Final Exam Hints and Answers - Autumn 2007, Exams of Mathematics

Hints and answers for the final exam of math 111 course in autumn 2007. It covers various problems related to economics, kinematics, and finance, including calculating the cost, revenue, and height of different quantities, as well as finding the time and interest rate for certain financial situations.

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Pre 2010

Uploaded on 03/09/2009

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MATH 111 FINAL EXAM Hints and Answers
Autumn 2007
1. (a) HINT: Draw the least steep diagonal line that intersects TC and compute its slope.
ANSWER: approximately $6.67 per Krumpette
(b) HINT: Draw a diagonal line with slope 11 and find the quantity at which it intersects
AC.
ANSWER: approximately 5.8 Krumpettes
(c) HINT: Draw the line tangent to TC (or V C) at q= 15 and compute its slope.
ANSWER: approximately 12 dollars or dollars per Krumpette
(d) HINT: Sketch the graph of TR , a diagonal line with slope 7.5, and find the quantities
at which the T R graph is above the T C graph.
ANSWER: from about 10.6 to about 16.3 hundred Krumpettes
(e) HINT: Slide your ruler to find the quantities at which the tangent line to T C is parallel
to T R.
ANSWER: approximately 3.7 and 13.7 hundred Krumpettes
(f) HINT: Find the largest vertical gap between T R and T C (when T R is above TC ).
ANSWER: approximately 12.5 hundred dollars (or 1250 dollars)
2. (a) TRANSLATION: The average trip speed of the red car at t= 14 seconds is 1.29 feet
per second.
(b) TRANSLATION: R(5) R(1) > R(6) R(2)
(c) ANSWERS: F; F; T
3. (a) T C(q) = cq +f,T R(q) = pq
(b) HINT: Set T R(q)T C(q) = 100 and solve for q.
ANSWER: q=100 + f
pcPies
(c) HINT: Use your answer to part (b).
ANSWER: q= 140 Pies
4. (a) HINT: Evaluate H(0).
ANSWER: 15 feet
(b) HINT: The height of the rock is a quadratic whose graph is a parabola that opens down.
The greatest height the rock reaches is the y”-coordinate of the vertex.
ANSWER: 24.77 feet
(c) HINT: The rock hits sea level when H(t) = 0.
ANSWER: 2.03 seconds
(d) HINT: The height of the second rock at time tis H(t10).
ANSWER: 16t2+ 345t1835
5. (a) HINT: The balance changes from an OLD value, A(3) = P(1.02)3, to a NEW value,
A(3.2) = P(1.02)3.2. Use the formula for proportionate change and simplify.
ANSWER: 0.004
(b) HINT: You have two options. Either use the fact that A(5) = P(1.02)5= 2345 to
compute your principal now (P= 2123.94) and then use that value to compute A(5.5).
Or just compute what a principal of $2345 would become after 0.5 years: 2345(1.02)0.5.
ANSWER: $2368.33
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MATH 111 – FINAL EXAM Hints and Answers Autumn 2007

  1. (a) HINT: Draw the least steep diagonal line that intersects T C and compute its slope. ANSWER: approximately $6.67 per Krumpette (b) HINT: Draw a diagonal line with slope 11 and find the quantity at which it intersects AC. ANSWER: approximately 5.8 Krumpettes (c) HINT: Draw the line tangent to T C (or V C) at q = 15 and compute its slope. ANSWER: approximately 12 dollars or dollars per Krumpette (d) HINT: Sketch the graph of T R, a diagonal line with slope 7.5, and find the quantities at which the T R graph is above the T C graph. ANSWER: from about 10.6 to about 16.3 hundred Krumpettes (e) HINT: Slide your ruler to find the quantities at which the tangent line to T C is parallel to T R. ANSWER: approximately 3.7 and 13.7 hundred Krumpettes (f) HINT: Find the largest vertical gap between T R and T C (when T R is above T C). ANSWER: approximately 12.5 hundred dollars (or 1250 dollars)
  2. (a) TRANSLATION: The average trip speed of the red car at t = 14 seconds is 1.29 feet per second. (b) TRANSLATION: R(5) − R(1) > R(6) − R(2) (c) ANSWERS: F; F; T
  3. (a) T C(q) = cq + f , T R(q) = pq

(b) HINT: Set T R(q) − T C(q) = 100 and solve for q. ANSWER: q =

100 + f p − c Pies (c) HINT: Use your answer to part (b). ANSWER: q = 140 Pies

  1. (a) HINT: Evaluate H(0). ANSWER: 15 feet (b) HINT: The height of the rock is a quadratic whose graph is a parabola that opens down. The greatest height the rock reaches is the “y”-coordinate of the vertex. ANSWER: 24.77 feet (c) HINT: The rock hits sea level when H(t) = 0. ANSWER: 2.03 seconds (d) HINT: The height of the second rock at time t is H(t − 10). ANSWER: − 16 t^2 + 345t − 1835
  2. (a) HINT: The balance changes from an OLD value, A(3) = P (1.02)^3 , to a NEW value, A(3.2) = P (1.02)^3.^2. Use the formula for proportionate change and simplify. ANSWER: 0. (b) HINT: You have two options. Either use the fact that A(5) = P (1.02)^5 = 2345 to compute your principal now (P = 2123.94) and then use that value to compute A(5.5). Or just compute what a principal of $2345 would become after 0.5 years: 2345(1.02)^0.^5. ANSWER: $2368.

(c) HINT: Use the fact that A(10) = P (1.02)^10 = 1234 to compute P. Then use the value of P that you find to compute A(8.5) and A(7) and subtract those two balences. ANSWER: $35.

  1. (a) HINT: Solve for r: 295000(1 + r)^7 = 367000. ANSWER: r = 0. 0317 (b) HINT: Either let B(k) = B(0) · 2 k, where k is the number of 10-minute periods that have elapsed, set B(4.2) = 260 and solve for B(0); or let B(t) = B(0) · Bt/^10 , where t is time in minutes, set B(42) = 260 and solve for B(0). ANSWER: 14.15 million bacteria (c) HINT: A(t) = 3000(1.25)t, where t is time in years. You want A(4.75). ANSWER: $8658.
  2. (a) A: $11,600; B: $11,410.95; C: $11,399.68; A is best

(b) A: $18,000; B: $19,346.77; C: $19,251.43; B is best (c) ANSWER: 6.77%

  1. (a) HINT: Solve for t: 50000e^0.^098 t^ = 1500000. ANSWER: 34.71 years (b) HINT: Solve for t: 50000

( 1 + 0.^0984

) 4 t = 1500000. ANSWER: 35.13 years (c) HINT: Solve for r: 50000

( 1 + 12 r

) 12 · 40 = 2000000. ANSWER: 9.26%