Transcript TEST TAKING TIPS Presented by Janice Levasseur

MATH 90 CHAPTER 4
PART I
MSJC ~ San Jacinto Campus
Math Center Workshop Series
Janice Levasseur
Polya’s 4 Steps to Problem Solving
1.
2.
3.
4.
•
Understand the problem
Devise a plan to solve the problem
Carry out and monitor your plan
Look back at your work and check your
results
We will keep these steps in mind as we
tackle the application problems from the
infamous Chapter 4!
1. Understand the problem
• Read the problem carefully at least twice.
– In the first reading, get a general overview of the
problem.
– In the second reading, determine (a) exactly what you
are being asked to find and (b) what information the
problem provides.
• Try to make a sketch to illustrate the problem.
Label the information given.
• Make a list of the given facts.
Are they all pertinent to the problem?
• Determine if the information you are given is
sufficient to solve the problem.
2. Devise a Plan
to Solve the Problem
• Have you seen the problem or a similar problem
before?
• Are the procedures you used to solve the similar
problem applicable to the new problem?
• Can you express the problem in terms of an
algebraic equation?
• Look for patterns or relationships in the problem
that may help in solving it.
• Can you express the problem more simply?
• Will listing the information in a table help?
2. continued.
• Can you substitute smaller or simpler
numbers to make the problem more
understandable?
• Can you make an educated guess at the
solution? Sometimes if you know an
approximate solution, you can work
backwards and eventually determine the
correct procedure to solve the problem.
3. Carry Out and Monitor Your
Plan
• Use the plan you devised in step 2 to
solve the problem.
• Check frequently to see whether it is
productive or is going down a dead-end
street. If unproductive, revisit Step 2.
4. Look Back at Your Work and
Check Your Results
• Ask yourself, “Does the answer make sense?”
and “Is the answer reasonable?” If the answer is
not reasonable, recheck your method for solving
the problem and your calculations.
• Can you check the solution using the original
statement?
• Is there an alternative method to arrive at the
same conclusion?
• Can the results of this problem be used to solve
other problems?
When given a total amount, we can use a single
variable to represent two unknowns.
For example, I have $20 and a buy something.
If you do not know how much I spent, there would
be two unknowns, how much I spent and
how much I have left.
• What if I spent $6, what would I have left? $14
• What if I spent $12, what would I have left? $8
• You have a total of $20.
• If s represents the amount you spent,
• then 20 – s represents the amount you have.
Coin Problem
Ex: A coin bank contains 27 coins in
dimes and quarters. The coins have a
total value of $4.95. Find the number
of dimes and quarters in the bank.
1. What are we being asked to find?
# of dimes and # of quarters in the bank
2. Can you express the problem in terms of an
algebraic equation? Can you list the information
given in a table to help solve the problem?
Coin Problem
Ex: A coin bank contains 27 coins in
dimes and quarters. The coins have a
total value of $4.95. Find the number
of dimes and quarters in the bank.
1. What are we being asked to find?
# of dimes and # of quarters in the bank
2. Can you express the problem in terms of an
algebraic equation? Can you list the information
given in a table to help solve the problem?
A coin bank contains 27 coins in dimes and
quarters. The coins have a total value of $4.95.
Find the number of dimes and quarters in the bank.
N
# of coins
dimes
quarters
total
d
27 – d
27
Let d = # of dimes
*
V
=
T
coin value
total value
0.10
0.10d
0.25
0.25(27 – d)
4.95
 27 – d = # of quarters
Use the information in the table to write an algebraic equation:
Total value in dimes + total value in quarters = total value
0.10d + 0.25(27 – d) = 4.95
3. Using the plan devised in Step 2,
solve the (algebraic) problem
0.10d + 0.25(27 – d) = 4.95
0.10d + 6.75 – 0.25d = 4.95
6.75 – 0.15d = 4.95
– 0.15 d = – 1.8
d = 12
Quarters: 27 - d
27 – d = 27 – 12 = 15
Solution: 12 dimes and
 d = 12
15 quarters
4. Did we answer the question being asked?
Is our answer complete? Check the solutions.
Yes! Check: 12 dimes + 15 quarters = 27 coins
Yes!
12(.10) + 15(.25) = 1.20 + 3.75 = 4.95
Ex: A total of $7000 is deposited into two simple
interest accounts. One account pays a simple
interest of 10% annually. The other pays an
annual interest rate of 15% simple interest. How
much should be invested in each account so that
the total interest earned is $800 ?
1. What are we being asked to find?
How much should be invested in each account.
2. Can you express the problem
in terms of an algebraic equation?
Is there an applicable formula?
Can you list the information given in a table
to help solve the problem?
The annual simple interest earned formula is:
I = Pr
I = simple interest, P = principal (initial investment),
and r = interest rate (%)
We have two accounts:
Account #1 and Account #2.
We want to know how much to invest in each
account.
Let p = principal for account #1.
A total of $7000 is deposited into two simple interest accounts. One
account pays a simple interest of 10% annually. The other pays an
annual interest rate of 15% simple interest. How much should be
invested in each account so that the total interest earned is $800.
P
principal
Acct #1
*
p
Acct #2 7000 – p
total
7000
Since p = Principal Acct #1
r
interest rate
0.10
0.15
=
I
Simple interest
0.10p
0.15(7000 – p)
800
 7000 – p = Principal Acct #2
Use the information in the table to write an algebraic equation:
Interest Acct #1 + Interest Acct #2 = total Interest
0.10p + 0.15(7000 – p) = 800
3. Using the plan devised in Step 2, solve the
(algebraic) problem
0.10p + 0.15(7000 – p) = 800
0.10p + 1050 – 0.15p = 800
1050 – 0.05p = 800
– 0.05 p = – 250
 p = 5000
p = $5000
Principal Acct #2
7000–p = 7000 – 5000 = 2000
Solution: $5000 in Acct #1 and
$2000 in Acct #2
4. Did we answer the question being asked? Is our
answer complete? Check the solutions.
Yes! Check: $5000 + $2000 = $7000 principal
Yes!
.10(5000) = 500 and .15(2000) = 300
$500 + $300 = $800 interest
Investment word problem,
set up a system of equations &
solve it.
• 1st equation – amounts: what was
invested.
• 2nd equation – value: sum of
percentages time amount equals the
interest.
• Solve by one of the above methods.
Investment word problem,
set up a system of equations & solve it.
• Bret has $50,000 to invest, some at 12% and the rest at
8%. If his annual interest earned is $4,740, how much
did he invest at each rate?
m ultiply by   8 
m ultiply by  100 
x  y  50000
.12 x  .08 y  4740
 8 x  8 y   400000
12 x  8 y 
474000
4x 
74000
x  $18500 invested at 12%
y  $31500 invested at 8%