Algebra I Workbook For Dummies
Page 18
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16. Melissa and Heather drove home for the holidays in separate cars, even though they live in the same place. Melissa’s trip took two hours longer than Heather’s because Heather drove an average of 20 mph faster than Melissa’s 40 mph. How far did they have to drive?
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Answers to Making Formulas Work in Basic Story Problems
The following are the answers (in bold) to the practice problems in this chapter.
1. Jack’s recliner is in the corner of a rectangular room that measures 10 feet by 24 feet. His television is in the opposite corner. How far is Jack (his head, when the recliner is wide open) from the television? (Hint: Assume that the distance across the room is the hypotenuse of a right triangle.) 26 feet
2. Sammy’s house is 1,300 meters from Tammy’s house — straight across the lake. The paths they need to take to get to one another’s houses (and not get wet) form a right angle. If the path from Sammy’s house to the corner is 1,200 meters, then how long is the path from the corner to Tammy’s house? 500 meters
3. A ladder from the ground to a window that’s 24 feet above the ground is placed 7 feet from the base of the building. How long is the ladder, if it just reaches the window? 25 feet. Check out the following figure.
4. Calista flew 400 miles due north and then turned due east and flew another 90 miles. How far is she from where she started? 410 miles. See the following figure.
5. The sum of the measures of the angles of a triangle is 180 degrees. In a certain triangle, one angle is 10 degrees greater than the smallest angle, and the biggest angle is 15 times as large as the smallest. What is the measure of that biggest angle? 150 degrees. Look at the figure for help.
Let x = the measure of the smallest angle in degrees. Then the other two angles measure and 15x. Add them to get . Simplifying on the left, you get . Subtract 10 from each side: . Then, dividing by 17, , and the largest angle, which is 15 times as great, is 150 degrees.
To check, find the measure of the other angle, . Adding up the three angles, you get .
6. A pentagon is a five-sided polygon. In a certain pentagon, one side is twice as long as the smallest side, another side is 6 inches longer than the smallest side, the fourth side is 2 inches longer than two times the smallest side, and the fifth side is half as long as the fourth side. If the perimeter of the pentagon is 65 inches, what are the lengths of the five sides? 8 inches, 16 inches, 14 inches, 18 inches, 9 inches.
Let x represent the length of the smallest side. Then the second side is 2x long, the third side is x + 6, the fourth is 2 + 2x, and the fifth is . Add up all the sides and set the sum equal to 65: x + 2x + x + 6 + 2 + 2x + 1 + x = 65. Simplifying, you get 7x + 9 = 65. Subtract 9 from each side and divide by 7 to get x = 8.
7. The sum of the measures of the angles in a quadrilateral (a polygon with 4 sides) adds up to 360 degrees. If one of the angles is twice as big as the smallest and the other two angles are both three times as big as the smallest, then what is the measure of that smallest angle? 40 degrees. Look at this figure.
Their sum is 360, so , which simplifies to . Dividing by 9, degrees. To check, find the measures of all the angles: and . Adding them all up, you get degrees.
8. The sum of the measures of all the angles in any polygon can be found with the formula , where n is the number of sides that the polygon has. How many sides are there on a polygon where the sum of the measures is 1,080 degrees? 8 sides.
Use the formula , where n is that number you’re trying to find. Replace the A with 1,080 to get . Divide each side by 180 to get 6 = n – 2. Add 2 to each side, and n = 8. So it’s an eight-sided polygon that has that sum for the angles.
9. Two figures are similar when they’re exactly the same shape — their corresponding angles are exactly the same measure, but the corresponding sides don’t have to be the same length. When two figures are similar, their sides are proportional (all have the same ratio to one another). In the figure, triangle ABC is similar to triangle DEF, with AB corresponding to DE and BC corresponding to EF. If side EF is 22 units smaller than side BC, then what is the measure of side BC? 66 units.
A proportion that represents the relationship between sides of the triangle is . Replacing the names of the segments with the respective labels, you get .
Cross-multiply (refer to Chapter 12 if you need more information on dealing with proportions) to get 36(x – 22) = 24x. Distributing on the left, you get 36x – 792 = 24x. Subtract 24x from each side and add 782 to each side for 12x = 792. Dividing by 12, you find that x = 66. (Hint: The numbers would have been smaller if I had reduced the fraction on the left to before cross-multiplying. The answer is the same, of course.)
10. The measure of an exterior angle of a triangle is always equal to the sum of the other two interior angles. If one of the nonadjacent angles in a triangle measures 30 degrees, and if the exterior angle measures 70 degrees less than twice the measure of the other nonadjacent angle, then how big is that exterior angle? 130 degrees. Check out the following figure.
Let x = the measure of a nonadjacent angle. The other nonadjacent interior angle is 30. The exterior angle has measure . Because the measure of that exterior angle is equal to the sum of the measures of the two nonadjacent interior angles, you can write the equation . Solving for x, subtract x from each side and add 70 to each side to get . The two nonadjacent interior angles add up to . This amount is the same as the measure of the exterior angle. Put 100 in for x in .
11. How long will it take you to travel 600 miles if you’re averaging 50 mph? 12 hours.
Use the distance-rate-time formula (d = rt) and replace the d with 600 and the r with 50 to get 600 = 50t. Divide each side by 50 to get 12 = t.
12. What was your average rate of speed (just using the actual driving time) if you left home at noon, drove 200 miles, stopped for an hour to eat, drove another 130 miles, and arrived at your destination at 7 p.m.? 55 mph.
Use the formula d = rt where d = 200 + 330 total miles, and the time is t = 7 – 1 = 6 hours. Substituting into the formula, you get . Divide each side by 6 to get the average rate of speed: .
13. Kelly left school at 4 p.m. traveling at 25 mph. Ken left at 4:30 p.m., traveling at 30 mph, following the same route as Kelly. At what time did Ken catch up with Kelly? 7 p.m.
Let t = time in hours after 4 p.m. Kelly’s distance is 25t, and Ken’s distance is , using because he left a half hour after 4 p.m.
Ken will overtake Kelly when their distances are equal. So the equation to use is . Distributing on the right gives you . Subtract 30t from each side to get . Divide each side by –5, and t = 3. If t is the time in hours after 4 p.m., then , or Ken caught up with Kelly at 7 p.m.
14. A Peoria Charter Coach bus left the bus terminal at 6 a.m. heading due north and traveling at an average of 45 mph. A second bus left the terminal at 7 a.m., heading due south and traveling at an average of 55 mph. When were the buses 645 miles apart? 1 p.m.
Let t = time in hours after 6 a.m. The distance the first bus traveled is 45t, and the distance the second bus traveled is , which represents 1 hour less of travel time. The sum of their distances traveled gives you their distance apart. So . Distribute the 55 to get . Combine the like terms on the left and add 55 to each side: . Divide by 100, . The busses were 645 miles apart 7 hours after the first bus left. Add 7 hours to 6 a.m., and you get 1 p.m.
15. Geoffrey and Grace left home at the same time. Geoffrey walked east at an average rate of 2.5 mph. Grace rode her bicycle due south at 6 mph until they were 65 miles apart. How long did it take them to be 65 miles apart? 10 hours.
Look at the following figure, showing the distances and directions that Geoffrey and Grace traveled.
Geoffrey walks , and Grace rides . Use the Pythagorean theorem:
Multiply each side by 4 to get rid of the decimal:
In 10 hours, Geoffrey will walk , and Grace will ride . Plug these values into the Pythagorean theorem: .
16. Melissa and
Heather drove home for the holidays in separate cars, even though they live in the same place. Melissa’s trip took two hours longer than Heather’s, because Heather drove an average of 20 mph faster than Melissa’s 40 mph. How far did they have to drive? 240 miles.
Because Heather’s time is shorter, let t = Heather’s time in hours. Heather’s distance is 60t, using for her speed. Melissa’s distance is , because she took 2 hours longer. Their distances are equal, so . Distribute the 40 to get . Subtract 40t from each side: . Divide by 20 for t = 4. Heather’s distance is . Melissa’s distance is . It’s the same, of course.
Chapter 19
Relating Values in Story Problems
In This Chapter
Dealing with age problems
Counting up consecutive integers
Working on sharing the workload
Yes, this is another chapter on story problems. Just in case you’re one of those people who are less than excited about the prospect of problems made up of words, I’ve been breaking these problems down into specific types to help deliver the material logically. Chapter 17 deals with formulas, Chapter 18 gets into geometry and distances, and now this chapter focuses on age problems, consecutive integers, and work problems.
Problems dealing with age and consecutive integers have something in common. You use one or more base values or ages, assigning variables, and you keep adding the same number to each base value. As you work through these problems, you find some recurring patterns, and the theme is to use the same format — pick a variable for a number and add something on to it.
As for work problems, they’re a completely different bird. In work problems you need to divvy up the work and add it all together to get the whole job done. The portions usually aren’t equal, and the number of participants can vary.
This chapter offers plenty of opportunities for you to tackle these story problems and overcome any trepidation.
Tackling Age Problems
Age problems in algebra don’t have anything to do with wrinkles or thinning hair. Algebra deals with age problems very systematically and with an eye to the future. A story problem involving ages usually includes something like “in four years” or “ten years ago.” The trick is to have everyone in the problem age the same amount (add four years to each person’s age, if needed for example).
When establishing your equation, make sure you keep track of how you name your variables. The letter x can’t stand for Joe. The letter x can stand for Joe’s age. If you keep in mind that the variables stand for numbers, the problem — and the process — will all make more sense.
Q. Joe’s father is twice as old as Joe is. (He hasn’t always been and won’t be again — think about it: When Joe was born, was his father twice as old as he was?) Twelve years ago, Joe’s father was three years older than three times Joe’s age. How old is Joe’s father now?
A. Joe’s father is 42 years old.
1. Assign a variable to Joe’s age.
Let Joe’s age be x. Joe’s father is twice as old, so Joe’s father’s age is 2x.
2. Continue to read through the problem.
Twelve years ago . . . . Both Joe’s age and his father’s age have to be backed up by 12 years. Their respective ages, 12 years ago, are x – 12 and 2x – 12.
3. Take the rest of the sentence where “Joe’s father was three years . . . ” and change it into an equation, putting an equal sign where the verb is.
“Twelve years ago, Joe’s father” becomes 2x – 12
“was” becomes =
“three years older than” becomes 3 +
“three times Joe’s age (12 years ago)” becomes 3(x – 12)
4. Put this information all together in an equation.
2x – 12 = 3 + 3(x – 12)
5. Solve for x.
2x – 12 = 3 + 3x – 36 = 3x – 33, so x = 21
That’s Joe’s age, so his father is twice that, or 42.
1. Jack is three times as old as Chloe. Ten years ago, Jack was five times as old as Chloe. How old are Jack and Chloe now?
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2. Linda is ten years older than Luke. In ten years, Linda’s age will be thirty years less than twice Luke’s age. How old is Linda now?
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3. Avery is six years older than Patrick. In four years, the sum of their ages will be 26. How old is Patrick now?
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4. Jon is three years older than Jim, and Jim is two years older than Jane. Ten years ago, the sum of their ages was 40. How old is Jim now?
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Tackling Consecutive Integer Problems
When items are consecutive, they follow along one after another. An integer is a positive or negative whole number, or 0. So put these two things together to get consecutive integers. Consecutive integers have patterns — they’re evenly spaced. The following three lists are examples of consecutive integers:
Consecutive integers: 5, 6, 7, 8, 9, . . .
Consecutive odd integers: 11, 13, 15, 17, 19, . . .
Consecutive multiples of 8: 48, 56, 64, 72, 80, . . .
After you get one of the integers in a list and are given the rule, then you can pretty much get all the rest of the integers. When doing consecutive integer problems, let x represent one of the integers (usually the first in your list) and then add on 1, 2, or whatever the spacing is to the next number and then add that amount on again to the new number, and so on until you have as many integers as you need.
Q. The sum of six consecutive integers is 255. What are they?
A. The integers are 40, 41, 42, 43, 44 and 45. The first integer in my list is x. The next is x + 1, the one after that is x + 2, and so on. The equation for this situation reads x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 255. (Note: The parentheses aren’t necessary. I just include them so you can see the separate terms.) Adding up all the x’s and numbers, the equation becomes 6x + 15 = 255. Subtracting 15 from each side and dividing each side by 6, you get x = 40. Fill the 40 into your original equations to get the six consecutive integers. (If x = 40, then x + 1 = 41, x + 2 = 42, and so on.)
Q. The sum of four consecutive odd integers is 8. What are they?
A. The integers are –1, 1, 3, and 5. The equation for this problem is x + (x + 2) + (x + 4) + (x + 6) = 8. It becomes 4x + 12 = 8. Subtracting 12 and then dividing by 4, you get x = –1. You may have questioned using the +2, +4, +6 when dealing with odd integers. The first number is –1. Replace the x with –1 in each case to get the rest of the answers. The problem designates x as an odd integer, and the other integers are all two steps away from one another. It works!
5. The sum of three consecutive integers is 57. What are they?
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6. The sum of four consecutive even integers is 52. What is the largest of the four?
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7. The sum of three consecutive odd integers is 75. What is the middle number?
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8. The sum of five consecutive multiples of 4 is 20. What are they?
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9. The sum of the smallest and largest of three consecutive integers is 126. What is the middle number of those consecutive integers?
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10. The product of two consecutive integers is 89 more than their sum. What are they?
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Working Together on Work Problems
Work problems in algebra involve doing jobs alone and together. Together is usually better, unless the person you’re working with distracts you. I take the positive route and assume that two heads are better than one.
The general format for these problems is to let x represent how long it takes to do the job working together. Follow these steps and you won’t even break a sweat when working work problems: