PROBLEM SOLVING IN ALGEBRA

Here are various kinds of problems you may encounter.

CONSECUTIVE INTEGER PROBLEMS

Example: The sum of three consecutive integers is 75. Find the first of the three integers.

We could let x, y, and z be the three integers but then there would be an infinite number of solutions for

x+y+z=75.

The trick is to let x be the first integer. Then the number after x is x+1 and the number after that is x+2. Therefore

x+(x+1)+(x+2)=75

3x+3=75

3x=72

x=24

Check: If x=24, x+1=25 and x+2=26

Therefore 24+25+26=75 and the problem checks.

If we had consecutive odd numbers or consecutive even numbers we would list them as x, x+2, x+4, etc.

AGE PROBLEMS

Start looking for relationships. For multiple choice answers it may be best to plug in answers.

Example: If Levi is three times as old as Harmony and if Levi was 10 when Harmony was born, how old is Harmony now?

Let H be Harmony's age now. If Levi was 10 when Harmony was born, Levi will always be 10 years older than Harmony. Therefore, Levi's current age is H+10. Levi is also 3 times as old as Harmony is now, so Levi's current age is 3H. We have two ways of expressing Levi's current age: H+10 and 3H so we can set these expressions equal to each other.

10+H=3H

10=2H

5=H

Sometimes it is expedient to make a chart. Stop when the number in the left column is three times as large as the number in the right column. If the column is very long you may want to skip some values as you hone in on the correct answer.

Levi's Age	Harmony's Age
10	0
11	1
12	2
13	3
14	4
15	5

RATE PROBLEMS

If someone drives 50 mph for 3 hours they will travel 150 miles. We find the distance by multiplying the rate by the time traveled. We use the formula

Rate x Time = Distance

rt = D

Example: Frank travels to work each day along a highway at 60 mph. However, one day construction causes traffic to move at 20 mph. He arrives at work 15 minutes late. How far is it from his home to work?

A. 1.6

B. 7.5

C. 12.5

D. 15

E. 20

Use the formula D = rt.

Note that 15 minutes = of an hour.

Let t be the amont of time, in hours, it takes to get to work if Frank travels at 60 mph.

Then t + is the amount of time, in hours, it takes to travel at 20 mph. We have

60 mph x t = D

20 mph x (t + ) = D

Since the distance traveled is the same,

60t = 20(t + )

60t = 20t + 5

40t = 5

t = hours

= 1/8 hours

= 12.5 minutes

Finally, to find the distance we have to multiply the rate by the time:

60 x 1/8 = 7.5 miles.

WORK PROBLEMS

In many work problems two people work at different rates to complete one job. These are similar to distance problems in that the rate at which a person works is multiplied by the amount of time spent working in order to find the total amount of work done.

Example: If Mark can build a wall of a certain height in 15 hours and if Sean can build the same wall in 22 hours, how long will it take them to build the wall if they work together?

Answer: 330/37 hours or about 8.9 hours

This is a rate problem. Here rate refers to the rate at which one person can do a job. If Mark takes 15 hours to do the job, he does 1/15 of the job each hour, so his rate is 1/15 jobs per hour. Similarly, if Sean takes 22 hours to do the job he does 1/22 of the job each hour so his rate is 1/22 jobs per hour. Now the relationship is:

Rate x Time = Work

Let t = the amount of time it takes to do the job if they work together. The amount of work they do is one job (=1).

Then

Clearing fractions by multiplying both sides by the least common denominator of 15 and 22 we have

22t + 15t = 330

37t = 330

t = 330/37 which is about 8.9 hours.

MIXTURE PROBLEMS

We can mix nuts that cost $2.25 a pound with nuts that cost $4.00 a pound. We can mix solutions that are 20% alcolhol with solutions that are 50% alcohol. We can mix coins that are worth 5 cents each with coins that are worth 10 cents each. In the following problems we are mixing animals that have two feet each with animals that have 4 feet each.

Example: A farmer raises chickens and horses. The animals have a total of 120 heads and 300 feet. How many chickens does he have?

A. 50

B. 60

C. 70

D. 80

E. 90

This is a good problem to use a guessing approach with. You will only need to plug in two answers. Start in the middle with answer choice C. If the answer is too large, try B. If answer B is too large the answer must be A. Similarly, if C is too small, try answer choice D. If D is too small E is the correct answer.

Guess: 70. If there are 70 chickens there must be 50 horses since there are 120 animals altogether.

If each chicken has two feet, there are 70 x 2 = 140 chicken feet.

If each horse has four feet, there are 50 x 4 =200 horse feet.

Therefore, 140 feet and 200 feet= 340 feet, which is too many feet. We have to trade some horses for some chickens to have fewer feet. Therefore, there must be more than 70 chickens and fewer than 50 horses.

Guess: 80. If there are 80 chickens there must be 40 horses.

If each chicken has two feet, there are 80 x 2 = 160 chicken feet.

If each horse has four feet, there are 40 x 4 =160 horse feet.

Now there are 320 feet, which is still too many feet, although we are getting closer. Therefore, there must be more than 80 chickens. In fact, the only larger answer is 90, which is (E).

Algebraic solution:

Let C = the number of chickens and H = the number of horses.

Let 2C = the number of chicken feet and 4H = the number of horse feet.

Then we have two equations and two variables:

C + H = 120

2C+4H=300

Multiply the first equation by 4 and subtract the second equation from the first equation:

4C+4H=480

-2C-4H=-300

We have

2C=180

so that

C=90.

SUMMARY

Actually, the distance problem, the work problem and the mixture problem can all be considered to be rate problems. We have expressed rate in miles per hour, work per hour, and feet per animal. By recognizing similarities among problems you will become better at finding solution methods.