Word Problems

Many math students have problems with word problems. Perhaps that's why they are called word problems. To start with it is helpful to know how to translate English words into mathematical language. See http://www.purplemath.com/modules/translat.htm for some help with this.

Below is a three step approach that works for solving just about any word problem. Practice these steps and hopefully you won't have any more problems with word problems.

1. Translate the problem from English into math, using as many variables as it takes. Use variables that make sense. For example, use 't' for time and 'r' for rate. However, avoid variables that can be confused with numbers (e.g., s can be confused with 5).
2. Substitute from one equation into another. Once the problem has been translated into algebraic language, you will likely find you have too many variables. Take what one variable equals and substitute it into another equation.
3. Solve the equation you substituted into and then use that result in the other equation.

If you find the examples below, click on Basic Word Problems for some simpler examples.

Example 1 -- Hockey Points

Hockey teams receive two points when they win and one point when they tie. One season a team won the championship when they received 66 points. They won 12 more games than they tied. How many wins and how many ties did they team have?

Step 1 -- Translate

Use w = wins and t = ties. The appropriate translation is below each relevant part of the word problem.

Hockey teams receive two points when they win and one point when they tie. One season a team won the championship when they received 66 points.

2w + t = 66

They won 12 more games than they tied.

w = t + 12

Step 2 -- Substitute

The second equation gets substituted into the first replacing the 'w' with what it equals (t + 12)

2(t + 12) + t = 66

Step 3 -- Solve

How many wins and how many ties did they team have?

2t + 24 + t = 66 3t + 24 = 66 3t = 66 -24 3t = 42 t = 42/3 t = 14

Substitute this number of ties (14) into the other equation.

w = (14) + 12 w = 26

So there were 26 wins and 14 ties.

Example 2 -- Trains running on parallel tracks

Here's another example. Two trains are traveling in the same direction along parallel tracks. The first train, which left first, is moving at 100 miles per hour while the second train is traveling at 110 miles per hour. The first train passes a station at 11:10 pm. The second train, which left later, passes the same station at 11:40 pm. What time will the faster second train catch up to the slower first train?

To solve this one you need to know the formula for distance, rate and time which is d = rt. As there are two trains we'll use d₁, r₁ and t₁ for the first train and d₁, r₁ and t₁ for the second.

Step 1 -- Translate

Two trains are traveling in the same direction along parallel tracks. The first train, which left first, is moving at 100 miles per hour while the second train is traveling at 110 miles per hour.

r₁ = 100

r₂ = 110

The first train passes a station at 11:10 pm. The second train, which left later, passes the same station at 11:40 pm.

Recognizing that the first, slower train is on the tracks for 30 minutes more or half an hour less than the second faster one, we get.

t₁ = t₂ + 1/2

What time will the faster second train catch up to the slower first train?

We need to recognized that the distance from the starting point for both trains where the second train catches up to the first is the same. Therefore

d₂ = d₁

Step 2 -- Substitute

We can the last translation about and substitute in the formula for distance to write

r₂t₂ = r₁t₁

As we know both rates, we can substitute those values in.

110 t₂ = 100 t₁

Finally we substitute in what t₂ equals to get

110 t₂ = 100 (t₂ + 1/2)

Step 3 -- Solve

So now that we have an equation with one unknown, thanks to our substitution above (copied below), we can solve for t₁.

110 t₂ = 100 (t₂ + 1/2)

110t₂ = 100t₂ + 50

10t₂ = 50

t₂ = 5 hours

Therefore the second train meets up with the first 5 hours after the 11:40 PM station stop which is 4:40 AM.

For more of this type, see http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Distance%20Rate%20Time%20Word%20Problems.pdf

Example 3 -- Consecutive integers

Here is a more complex example.

One leg of a right triangle has a length of 7 m. The other sides have lengths that are consecutive integers. Find these lengths.

To solve this one you need to know the Pythagorean theorem which applies to right triangles. It a² + b² = c² where c is the side (hypotenuse) across from the right angle and a and b are the other two sides. We also need to understand what consecutive integers are. If we call the first integer x then the second one (y) will be x + 1.

Step 1 -- Translate

Pythagoras' theorem tells us that

a² + b² = c²

We know one of the sides that we will call a is 7 m. That translates to

a = 7

Our translation of consecutive integers tells us that if the next side is b then the c must be b + 1. That translates to

c = b + 1

Step 2 -- Substitute

We know that a = 7 so we can substitute that into Pythagoras' theorem and get

7² + b² = c²

Now we substitute b + 1 in for c and get

7² + b² = (b + 1)²

Step 3 -- Solve

Now we are ready to solve. The final translation is

7² + b² = (b + 1)²

49 + b² = b² + 2 b + 1

49 = 2 b + 1

2 b = 48

b = 24

And c being the next consecutive integer must be 25.

Example 4 -- Inequalities and Dental Bills

An insurance company has two dental plans. Under plan A you have to pay the first $190 of your dental bills plus 25% of the rest. Under plan B you pay the first $200, but only 20% of the rest. What would your dental bills have to be for plan B to save you money. Assume you have over $200 in bills.

Step 1 -- Translate

Let n = the the total number of dollars in dental bills you have.

Plan A translates as follows.

Pay the first $190 of your dental bills plus 25% of the remainder owed.

$190 + 25% of (t - $190)

or

$190 + 0.25(t - $190)

Plan B translates as follows.

Pay the first $200 of your dental bills plus 20% of the remainder owed.

$200 + 20% of (t - $200)

or

$200 + 0.2(t - $200)

Finally the cost of plan B must be less than plan A translates to

Cost of plan B < Cost of plan A

Step 2 -- Substitute

From above we know that Plan A = $190 + 0.25(t - 190)

and Plan B = $200 + 0.2(t - $200)

So substituting into

Cost of plan B < Cost of plan A we get $200 + 0.2(t - $200) < $190 + 0.25(t - $190)

Step 3 -- Solve

$200 + 0.2(t - $200) < $190 + 0.25(t - $190)

$200 + 0.2t - $40 < $190 + 0.25t - $47.50

Collect like terms

$160 + 0.2t < $142.50 + 0.25t

Subtract $142.50 from both sides gets you

$17.50 + 0.2t < 0.25t

To isolate the t, we subtract 0.2t from both sides to get

$17.50 < 0.05t

Finally divide both sides by 0.05 to get

$350 < t

Therefore you would save with plan B if the total number of dollars in dental bills was greater than $350.

Example 5 -- Making Change

A student makes a $10.50 purchase at the bookstore with a $20 bill. The store has no bills and gives the change in quarters and fifty cent pieces. There are 30 coins in all. How many of each kind are there?

Step 1 -- Translate

Let the number of quarter's equal Q and the number of $0.50 pieces equal F.

Let's focus on the number of coins first. So the number of quarters and the number of $.50 pieces equals 30 coins in all. This translates to

Q + F = 30

For the second equation we also need to recognize that the amount of change student gets back from the $20 bill is $9.50. That $9.50 is made up of quarters and $.50 pieces. Remembering the quarter is $0.25, this translates to

0.25Q + 0.5F = 9.50

Step 2 -- Substitute

If Q + F = 30 then

F = 30 - Q

We can now substitute F into the second equation and get

$0.25Q +$0.5(30 - Q) = $9.50

Step 3 -- Solve

0.25Q + 0.5(30 - Q) = 9.50

0.25Q + 15 - 0.5Q = 9.50

0.25Q - 0.5Q = 9.50 - 15

-0.25Q = -5.5

Q = 22

So we now know there are 22 quarters and given that the total number of coins was 30 there must be a eight $0.50 pieces.

F = 30 - Q

F = 30 - 22

F = 8

We can verify our answers by substituting in Q and F into the original equation.

Example 6 -- A boat's speed in still water

This example is distance, rate and time one similar to the train problem above. However, the algebra at the solution end is more complex.

The current in a stream moves at a speed of 2 km/h. A boat travels 40 km upstream and 40 km downstream in a total of 14 hours. What is the speed (rounded to the nearest tenth) of the boat in still water?

Step 1 -- Translate

To solve this one, like the train problem, you need to know the formula for calculating distance, rate and time. That is d = rt or its equivalent variants r = d/t and t = d/r. Using those three variables, we translate. While the speed or rate of the boat is constant, the current does affect its progress. When it goes upstream, the rate of the boat is slowed by 2 km/h. We translate that to

r - 2

On its return trip, the boat goes downstream, the rate of the boat is increased by 2 km/h. We translate that to

r + 2

We know that the distance up the stream is 40 km. The boat travels that same distance on its downstream return trip. That translates to

d = 40

However there are two different times, as the trip upstream against the current will take longer than the one downstream with the current. We'll use t₂ for the time it took to get up the river and t₂ for the time to get down again. We are told that the combined upstream and downstream trips took a total of 14 hours. That translates to

t₁ + t₂ = 14

This last translation is the cornerstone to the problem.

Step 2 -- Substitute

We know now we need to focus on the formula which solves for time t = d/r. That focuses us on what equivalent values we can substitute in for the two different times.

The upstream time (t₁) is the distance (40 km) divided by the upstream rate (r-2).

The downstream time (t₁) is the same distance (40 km) divided by the downstream rate (r+2).

Substituting these values into

t₁ + t₂ = 14

gives us

40/(r-2) + 40/(r+2) = 14

Step 3 -- Solve

So now that we have an equation with one unknown, thanks to our substitution above (copied below). So, we can solve for r, the boat's speed/rate in still water.

40/(r-2) + 40/(r+2) = 14

You will need to remember the steps for adding fractions (i.e., finding the common denominator which is (r-2)(r+2). You will end up with a quadratic equation that requires either the quadratic formula or completing the square skills to solve it. If you are unfamiliar with those concepts, Google either term.

The solution rounded to the nearest tenth is that the boat in still water travels at 6.3 km/hr.

For more distance, rate and time word problems; see http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Distance%20Rate%20Time%20Word%20Problems.pdf

Example 7 -- The Power of Synergy

We know that when humans co-operate and work together some pretty amazing things result. This word problem demonstrates that power.

Person A can paint the neighbour's house six times as fast as Person B. The year A and B worked together, it took them two days. How long would it take each to paint the house?

Step 1 -- Translate

To solve this one, like the problems above, you need to know the appropriate formula. The formula to calculate the power of co-operation on a single task is

1/A + 1/B = 1/T

where A is the time it takes person A, B the time for person B and T the total time it takes when they work together.

We also need to translate the first sentence of the problem -- Person A can paint the neighbour's house six times as fast as Person B.

B = 6A

The above translation can be a bit tricky, so try some real number to test out your equation. We know person A is faster than B so if it take A one day, it will take B 6 days.

For the next translation, we know that when they work together, the total time it takes them to paint the house is two days. That translates to

T = 2

Step 2 -- Substitute

We start with the formula for synergy.

1/A + 1/B = 1/T

We substitute into that our translations from above, A = 6B and T = 2 to get

1/A + 1/6A = 1/2

Step 3 -- Solve

You will need to remember the steps for adding fractions (i.e., finding the common denominator which is 6B. Applying the common denominator to each term gets us

6/6A + 1/6A = 3A/6A

Collecting like terms gets us

7/6A = 3A/6A

Multiplying both sides by 6A gets us

7 = 3A

Therefore

A = 7/3 or 2 1/3 days.

As B is six time slower

B = 7/3 times 6 or 14 days.