Problem Solving Techniques
I. General suggestions for working word problems
(1) Read the problem carefully. Make sure that you understand all of the terms
involved in the problem.
(2) Make a sketch or table to illustrate the problem if appropriate.
(3) Guess the answer and then test your guess. This will help you to understand the
problem better, it will give you an idea of what the answer might be, it will
help you in writing an appropriate equation or function to solve the problem,
and it will help you in testing this equation or function.
(4) In some problems, there seems to be inadequate information. Provide your own
data for the problem and then work the problem. Try another set of data and
work the problem. If you get the same answer with different data, then the
answer is probably independent of this data and the answer you get is the correct
one. Then you might try to prove that this answer is correct.
II. Value problems
Use the following: (value/item)(no.of items) = total value
For example: (5¢/nickel)(7 nickels) = 35¢
($2.00/kg)(3.5 kg) = $7.00
III. Uniform motion problems
If distance = d , rate = r , and time = t , use the following:
d = rt
or
r=
d
t
or
t=
d
r
Note: Be sure that the time units in the rate match up with time units in time.
For example, (60mi/hr)(3 hr) = 180 mi, but
(60mi/hr)(30 min) = (60mi/hr)(1/2 hr) = 30 mi.
IV. Simple interest problems
Use the following: I = Prt, where I is the amount of interest ($), P is the
interest ($) (amount borrowed or invested), r is the interest
rate (%) per time unit (usually years), and t is the time
measured in the same time unit in which the rate is given.
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V. Mixture problems
Use the following: (% pure substance in solution)(amt of solution)=amt of pure
substance
Also, use the following if solution A is combined with solution B to give solution C:
amt of pure substance in A + amt of pure substance in B = amt of pure substance in C
Example: 20 ml of a 5% acid solution combines with 30 ml of a 10% solution to give
50 ml of an 8% solution because (20)(.05)+(30)(.10) = (50)(.08)
An interesting and reliable procedure for solving mixture problems involves the
following tic-tac-toe set-up:
*
*
*
*
Weak
*
* Strong – Intermediate
Amt of
Solution % *
*
%
Weak Solution
*
*
******************************************************************
*
*
* Intermediate *
* Solution % *
*
*
******************************************************************
*
*
Strong
*
* Intermediate – Weak
Amt of
Solution % *
*
%
Strong Solution
*
*
*
*
To solve, set up the following proportion:
Strong – Intermediate %
-----------------------------------
Amt of Weak Solution
---------------------------------
=
Intermediate – Weak %
Amt of Strong Solution
2
Example: How much pure acid must be added to 20 ml of a 30% acid solution to get
a 50% acid solution?
*
*
30% *
*
50%
20 ml
*******************************************************
*
50% *
*******************************************************
x
100% *
*
20%
*
*
50% 20ml
⇒
=
x
20%
5 20ml
⇒ 5 x = 40ml ⇒ x = 8ml
=
x
2
VI. Work problems
When two or more people (or machines) work together, the individual times for
completing a job do not add up. The individual rates of work do add together to
give the single combined rate.
Example: John can paint a fence in 3 hours. Pete can paint the same fence in 6
1
1
hours. Thus, John’s rate = fence / hr and Pete’s rate = fence / hr .
3
6
1
1
Therefore, the rate if they work together is fence / hr + fence / hr =
3
6
1
fence / hr. Note that it would take 2 hours to paint the fence if they
2
⎛1
⎞
work together since ⎜ fence / hr ⎟ (2 hours) = 1 fence.
⎝2
⎠
A technique that will make work problems easier to manage is to let “the job” be a
number that is divisible by all of the “times” involved in the problem. This will give
the “rates” whole number values.
Example: John can paint a fence in 6 hours and Jim can paint the same fence in 8
hours. How long would it take to paint the fence if they work together?
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Let Fence ≡ 24 Sections (24 is divisible by 6 and 8)
John’s rate =
24 Sections
= 4 Sections / Hr
6 Hours
24 Sections
= 3Sections / Hr
8 Hours
+____________________________________________
Jim’s rate =
= 7 Sections/ Hr
Together’s rate
7 Sections/ Hr
×
? Hr
= 24 Sections
7 x = 24
⇒
3
x = 3 Hr
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VII. Variation
A. Direct
Let x and y denote two nonnegative quantities. Then y varies directly as x ,
or y is directly proportional to x , if there is a nonzero constant k such that
y = kx . k is called the constant of proportionality. Suppose y varies directly
as x ; when x = x1 , then y = y1 ; when x = x2 , then y = y 2 . Thus, y1 = kx1
y1 y 2
y
y
=
. This is another way
and y 2 = kx2 ⇒ 1 = k and 2 = k ⇒
x1 x2
x1
x2
of representing direct variation. Sometimes, y varies directly as a power of
x . Let n be a positive real number. Then y varies directly as the nth power of
x , or y is directly proportional to the nth power of x , if there exists a nonzero
constant k such that y = kx n . Similarly, this can also be represented by
y1
y
= 2 .
x1n x 2 n
Example: The distance a free-falling object falls from rest varies directly as the
square of the time it falls (disregarding air resistance). If an object falls 1024
feet in 8 seconds, how far will it fall in 12 seconds?
Solution 1: d = kt 2 ⇒ 1024 = k (8 2 ) ⇒ 1024=64k ⇒ k = 16.
Thus, d = 16t 2 ⇒ d = 16( 12 2 ) = 2304 feet.
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Solution 2:
d1
2
d2
=
t1
t2
d 2 = 2304 feet.
2
⇒
1024
8
2
=
d2
12 2
⇒
64d 2 = 147456 ⇒
B. Inverse
Let x and y denote two nonnegative quantities. Then y varies inversely as
x , or y is inversely proportional to x , if there exists a nonzero constant k
k
such that y = . Suppose y varies inversely as x ; when x = x1 , y = y1 ;
x
k
k
when x = x2 , y = y 2 . Thus, y1 =
and y 2 =
⇒ x1 y1 = k and
x1
x2
x2 y 2 = k ⇒ x1 y1 = x2 y 2 . This formula is another way of representing
k
inverse variation. If y varies inversely as the nth power of x, then y =
xn
or x1n y1 = x 2 n y 2 .
Example: The illumination produced by a light source varies inversely as the
square of the distance from the source. The illumination of a light source at
5 meters is 70 candela. What is the illumination 12 meters from the source?
Solution 1: I =
k
d
I=
1750
12 2
2
⇒ 70 =
k
5
2
⇒ k = 1750. Thus, I =
1750
d2
⇒
≈ 12.15 candela.
Solution 2: I1d12 = I 2 d 2 2 ⇒ (70)(5 2 ) = I 2 (12 2 ) ⇒ 1750 = I 2 (144) ⇒
I 2 ≈ 12.15 candela.
C. Combined or joint
Suppose y varies directly as x and z 2 , and inversely as w and v 3 . Then there
k
k
exist constants k1 , k 2 , k 3 , k 4 such that y = k1 x, y = k 2 z 2 , y = 3 , and y = 4 .
w
v3
These four equations can be written as one equation which represents the
5
⎡ xz 2 ⎤
=
combined variation: y k ⎢
⎥ . Note that the variables which vary directly as
3
⎣⎢ wv ⎦⎥
y are multiplied and are placed in the numerator of the fraction, and the variables
which vary inversely as y are multiplied and are placed in the denominator.
Example: The volume V of a gas varies directly as the Kelvin temperature T and
inversely as the pressure P. If V = 10 when T = 280 and P = 6, find V if
T = 300 and P = 10.
(10)(6) 3
⎛ 280 ⎞
⎡T ⎤
Solution: V = k ⎢ ⎥ ⇒ 10 = k ⎜
= . Thus,
⎟ ⇒ k=
280
14
⎝ 6 ⎠
⎣P⎦
3
3 ⎛ T ⎞ 3 ⎛ 300 ⎞
V = ⎜ ⎟= ⎜
⎟=6 .
7
14 ⎝ P ⎠ 14 ⎝ 10 ⎠
D. Notes
1. In arithmetic, when we say that 2 quantities are proportional, we really mean
“directly proportional.”
2. Usually, direct variation between the variables x and y is implied under the
following conditions: (i) if x increases, so does y (and vice versa);
(ii) x = 0 ⇔ y = 0.
3. Usually, inverse variation between the variables x and y is implied under the
following conditions: (i) if x increases, then y decreases (and vice versa);
(ii) x ≠ 0 and y ≠ 0.
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