Problem
Santa is on his way to the North Lumber
Company to buy some more supplies to finish the square pen he is building to hold his famous reindeer. He has already completed
three sides of the pen and has spent exactly $250. The fence posts that he used cost $10 each. How much will he have to spend
to finish the pen?
Solution
$70.
Remove all the fencing except for
the four corner posts, which cost $10 each, or $40. The three completed sides must have cost $210, or $70 a side. The fourth
side will also cost $70.
Problem
At this time of year, many postal workers
have to work overtime, so a supervisor at one large post office planned a late-night snack for the employees. She ordered
1 extra large pizza for every two workers, 1 large bag of potato chips for every three workers, and 1 two-liter bottle of
cola for every four workers. When the order arrived, 26 items were delivered. How many employees were working that evening?
Solution
24 workers.
Suppose there are x workers. The
number of pizzas delivered will be (x workers)/( 2 workers/pizza) and similarly for the other two items. The total number
of items is then x/2 + x/3 + x/4 = 26. Solving for x, we find that there must be 24 workers that evening. We could also reason
that the number of workers must be a common multiple of 2, 3, and 4. These multiples are 12, 24, 36, 48, 60, 72 . . . . Set
up a ratio table, giving the number of items for each possible number of workers, to find that 24 workers result in the given
numbers of items.
Problem
In a group of 24 tree decorations (ornaments
and candy canes), 12 are gold. (a) If you know that none of the ornaments is gold, what is the greatest number of ornaments
possible in the group? (b) If there are 9 ornaments in the group, what is the fewest possible number of candy canes that are
gold? What is the greatest possible number? (c) Suppose that there are 2 ornaments for every candy cane and that no candy
cane is gold. How many ornaments are not gold?
Solution
(a) 12 ornaments; (b) 3 candy canes, 12
candy canes; (c) 4 ornaments.
(a)From the fact that 12 decorations
are gold, 12 must not be gold. If none of the ornaments is gold, there can be at most 12 ornaments. (b) If there are 9 ornaments,
then there are 15 candy canes. If you assume that all 9 ornaments are gold, then 3 candy canes must be gold. As fewer ornaments
are gold, more candy canes must be gold, up to the maximum of 12 candy canes. (c) If there are 2 ornaments for every 1 candy
cane, then there are 16 ornaments and 8 candy canes. Since none of the candy canes is gold, 12 of the ornaments must be gold,
leaving 4 ornaments, which are not gold.
Problem
Ricardo is wrapping a gift. He wants to
cover an 8 in. × 10 in. × 5 in. box with wrapping paper. How many square inches of paper will he need if the paper fits perfectly
over the box? How much ribbon will he need if the bow itself adds 15 inches?
Solution
340 in.2;71 in.
To find the surface area of the
box, use the formula SA = 2l × w + 2l × h + 2w × h = 2(10 in.)(8 in.) + 2(10 in.)(5 in.) + 2(8 in.)(5 in.) = 160 in.2 + 100
in.2 + 80 in.2 = 340 in.2. To find the length of ribbon needed, add the top and bottom lengths (10 in. + 10 in. = 20 in.),
the top and bottom widths (8 in. + 8 in. = 16 in.), the heights of the four sides (4 × 5 in. = 20 in.) and the extra 15 inches
required for the bow. The total length is 20 in. + 16 in. + 20 in. + 15 in., or 71 in.
Problem
Your school has a telephone chain to notify
people of weather-related school cancellations. In the first round of calls, the school secretary calls two different homes.
In the second round, each of those families calls two other families. Each family has to make only two calls, but the chain
is still a fast way to notify a lot of people. How many families are called in the fifth round of calls? How many rounds are
needed to call all 123 families at your school?
Solution
32, 6.
In each round, the number of families
called is twice the number called in the previous round. The total number of families called in round R is the sum of each
of the numbers of families called in all previous rounds. In round 1, two families are called. In round 2, four more families
are called, which means that six families have been informed by the end of the round. Continuing, we see that by round 3,
6 + 8 or 14, families have been informed; by round 4, 14 + 16, or 30, families have been informed; by round 5, 30 + 32, or
62, families have been informed; and by round 6, the remaining 61 families will have been informed.
Problem
A recipe for cooking a turkey states to
allow 20 minutes for each pound of turkey and to add 5 minutes per pound if the turkey is stuffed. Our turkey weighs 16 pounds
and will be stuffed. To have dinner at two o’clock in the afternoon, what time does the turkey need to go in the oven?
Allow 1/2 hour for cooling and carving. Explain your answers and your thinking.
Solution
No later than 6:50 a.m. Since the
turkey will be stuffed, it will need to roast for 25 minutes per pound for 400 minutes, or 6 hours, 40 minutes. Adding 1/2
hour (30 minutes) for cooling and carving, the total amount of lead time necessary is 7 hours, 10 minutes. The latest time
to put the turkey into the oven is 6:50 a.m. However, many people would put the turkey in the oven by at least 6:30 a.m.,
if not sooner, just to be sure that it is cooked in time. (Keep in mind that there might be other food items that need to
be cooked in the oven, as well.)
Problem
Miles was peeling potatoes for the big
Thanksgiving dinner. He can peel 2 every minute. After he had worked for 25 minutes, his wife, Prudence, realized that he
would never finish all the potatoes in time for dinner, so she decided to help him. She is faster than Miles and can peel
3 potatoes every minute. They finished in time for dinner, and the two of them peeled exactly 400 potatoes. How much time
did each person spend peeling potatoes?
Solution
Miles 95 minutes; Prudence 70 minutes.
Miles peeled 50 potatoes in the first 25 minutes. During the time they worked together, Miles and Prudence peeled 350 potatoes.
Together they peeled 2 + 3, or 5, potatoes a minute, so dividing 350 by 5, we find that they were working together for 70
minutes. Miles worked 70 + 25 = 95 minutes. Prudence worked 70 minutes.
Problem
Coach Crowe has a certain number of students
in her PE class. If there had been as many students again as there actually were in the class, plus 1/2 as many more, plus
1/3 as many more, plus 1/6 as many more, plus 8 more students, there would have been 170 students. How many students were
actually in the class?
Solution
54 students. Let x equal the number
of students in the class. Then write the following equation based on the conditions of the problem: x + x + (1/2)x + (1/3)x
+ (1/6)x + 8 = 170. Solving for x, you get 54 students who were actually in the class. (Essentially, you are dividing the
difference between 170 and 8 by the sum 1 + 1 + 1/2 + 1/3 + 1/6, or 3.)
Problem
On a line, B is 1 unit to the right of
point A, point C is 2 units to the left of point A, point D is 3 units to the right of point B, and point E is 12 units to
the left of point D. How many units of distance is it from A to E?
Solution
8 units (to the left). One way to solve
this problem is to draw a line and use the conditions of the problem to insert points. You might want to use graph paper to
have a readymade scale.
Another way to solve this problem
without a drawing is to reason that point D (from which point E is measured) is four units to the right of point A (one unit
to point B and then three more units to point D). Since point E is 12 units to the left of point D, it must be eight units
to the left of point A.
Problem
At a Halloween party, your brother won
7 packages of colored candies with 24 pieces each. According to the manufacturer, for each 1 blue piece produced, the company
makes 3 brown, 2 yellow, 2 red, 1 orange, and 1 green. How many of each color do you expect to find in your brother’s
candy?
Solution
50 brown; 34 each of yellow and
red: 17 each of orange, blue, and green. The most direct method is to convert the ratios to percents and then use the percents
to find the number of each out of the 168 total. Note: this solution method results in rounding errors.
Problem
The United States acquired Alaska from
Russia on October 18, 1867, for about 2¢ an acre. Currently, Alaska has an area of 570,374 square miles. Use this current
area as an estimate for the 1867 area. About how much did Alaska cost? Rhode Island, the smallest state, is 1045 square miles.
At 2¢ an acre, how much would Rhode Island cost? How much would your state cost? Note: 1 square mile = 640 acres.
Solution
About $7,300,000 (the actual price
was $7,200,000). Rhode Island would cost about $13,000. Answers will vary for other states. Multiply the area in square miles
times the number of acres in a square mile times the cost per acre. For Alaska, you get (570,374 mi.2) × (640 acre/mi.2) ×
(2¢/acre) = $7,300,787, which should be reported as $7,300,000, with only two significant digits (at best), because the cost
per acre is only given to the nearest whole cent with one significant digit.
Problem
Prudence has been trying to decide which
pumpkin she should purchase to decorate for the Fall Festival. She has narrowed her choice to two, but she needs some help
from you to make the final decision. Each pumpkin is approximately spherical. Pumpkin A has a diameter that is 1/4 bigger
than pumpkin B, but it costs 1 1/2 times as much. Prudence wants to make the more practical purchase, the lower cost per amount
of pumpkin. Which pumpkin should she buy?
Solution
Pumpkin A. Say that pumpkin B has
a diameter of 1. Pumpkin A has a diameter of 1 1/4. Since the volume varies as the cube of the diameter (or the cube of the
radius depending on the formula you use), we have 5/4 × 5/4 × 5/4 = 125/64 = 1 61/64 for pumpkin A and 1 × 1 × 1 = 1 for pumpkin
B. Pumpkin A is almost twice as big and the price is only 1 1/2 times as much, so the cost per amount of pumpkin is less for
pumpkin A. Many students might not know about "varies as the cube of the diameter or as the radius," so they might substitute
realistic values for the two diameters and use the equation for the volume of a sphere to calculate the volumes.
Problem
Several years ago, Roger Maris broke Babe
Ruth’s 1927 home-run record of 60 home runs in one season by hittinghis 61st home run on October 1, the last day of
the season. In 1998, Mark McGwire broke Maris’s record with 70 home runs. In 2001, Barry Bonds broke McGwire’s
record with 73 home runs. The number of years between Maris’s and McGwire’s records is the same as the sum of
the number of years between Ruth’s and Maris’s records and the number of years between McGwire’s and Bonds’s
records. In what year did Maris break Ruth’s record?
Solution
1961. There are 3 years between
McGwire’s and Bonds’s records. There are 71 years between Ruth’s and McGwire’s records. These 71 years
represent the years between Ruth and Maris plus the years between Maris and McGwire. But we know that the number of years
between Ruth and Maris is 3 years less than the time between Maris and McGwire, so there must be 34 years between Ruth and
Maris and 37 years between Maris and McGwire. Adding 34 years to Ruth’s record date of 1927 gives us 1961. Algebraically,
let x = the number of years between Ruth and Maris. Then, x + 3 will be the number of years between Maris and McGuire. Solve
the equation x + (x + 3) = 71 years to get x = 34 years.
Problem
Lydia works at a dog kennel. She uses 5
pounds of dog food to feed 3 dogs for 4 days. At the same feeding rate, how many pounds of dog food will Lydia need to feed
12 dogs for 1 week?
Solution
35 pounds. Since 5 pounds feeds
3 dogs for 4 days, then 5 pounds will feed 12 dogs for 1 day. Then the dogs require 35 pounds for 7 days.
Problem
A total of 3 holstein cows and 3 guernsey
cows give as much milk in 3 days as 2 holstein cows and 4 guernsey cows in 4 days. Which breed of cow is the better milk producer,holstein
or guernsey?
Solution
Holstein.
Let H be the rate of milk production
(gal./day) for a holstein, and G the rate of milk production for a guernsey. Then we have (3 day)[(3H gal./day) + (3G gal./day)]
= (4 day)[(2H gal./day) + (4G gal./day)], or 9H + 9G = 8H + 16G. Solving, we find H gal./day = 7G gal./day, which shows that
a holstein cow produces milk 7 times faster than a guernsey cow. Therefore, holsteins are the better milk producers.
Problem
When 5 new girls joined a class, the percent
of girls increased from 40% to 50%. What is the number of boys in the class?
Solution
15 boys.
The original class was made up
of 2 girls for every 3 boys, that is, .40:.60 = 2:3. Adding 5 girls to the class results in the same number of girls as boys.
Looking at the table of possible number of boys and girls, we see that adding 5 girls to 10 girls gives 15, which is the same
as the number of boys. Girls 2 4 6 8 10 12 Boys 3 6 9 12 15 18
Problem
The gauge of an oil tank indicated that
the tank was 1/7 full. After 240 gallons of oil were added to the tank, the gauge indicated that the tank was 4/7 full. How
many gallons of oil does the tank hold, assuming that the gauge is accurate?
Solution
560 gallons.
Since 3/7 of the tank, or 240 gallons,
was added, then 1/7 of a tank is 80 gallons. The whole tank holds 7 times 80, or 560, gallons.
Problem
Kwame likes to think of puzzles for his
younger sister, Omarossa. His most recent one is this: I have an apple, a tangerine, and a peach. I weighed them two at a
time, getting weights of 14 ounces, 18 ounces, and 20 ounces for the three possible pairs of fruit. How much will each of
the three pieces of fruit weigh?
Solution
6 ounces, 8 ounces, and 12 ounces.
We do not need to know which fruit weighs
how much, since we can just assign variables A, B, and C for the weights.
A + B = 14, A + C = 18, and B + C = 20.
Adding the members of the first two equations
gives 2A + B + C = 32.
Substituting 20 for B + C in this equation
gives 2A + 20 = 32.
Solving gives A = 6.
Since A + B = 14, then B = 8. Since
A + C = 18, C = 12.
Problem
Hiking boots that cost $150 a pair are
reduced by 40% for a special weekend sale. By what percent must the special sales price be increased to bring the price back
to $150?
Solution
66 2/3%.
Since 40% of $150 is $60, we could subtract
to get the sale price of $90.
The increase needed to get from $90 to
$150 is $60; 60 is 2/3 of 90, so the increase is 66 2/3%.
(Students might ask why the percents
of decrease and increase are not the same. It is because they are percents of different numbers. The decrease is 40% of 150,
and the increase is 66 2/3% of 90.)
Problem
A square and a triangle each have an area
of 64 square centimeters. If the length of one side of the triangle is the same as the length of a side of the square, find
the length of the altitude of the triangle.
Solution
16 cm.
Since the area of the square is 64 cm2,
the side of the square is 8 cm.
The area of the triangle is 1/2bh = 1/2(8)h
= 4h = 64.
Solving gives us h = 16 cm.
Problem
What is the largest number less than 10,000
that is divisible by both 36 and 60?
Solution
9,900.
The least common multiple of 36 and 60
is 180. Dividing 10,000 by 180 gives us 55+ (we do not need the decimal).
To check: 180 × 55 = 9,900.
Problem
Mr. Brown, Mr. Green, and Mr. Black met
for lunch. One man wore a brown tie, one a green tie, and one a black tie. "Have you noticed," said the man with a green tie,
"that although our ties have colors that match our names, not one of us is wearing a tie that matches his own name?" "By golly,
you’re right!" exclaimed Mr. Brown. What color tie was each man wearing?
Solution
Mr. Brown, black tie; Mr. Green, brown
tie; Mr. Black, green tie.
Mr. Brown is not the man with the green
tie, so he must be wearing black.
So Mr. Green must be wearing brown,
and Mr. Black must be wearing green.
Problem
Mrs. Hernandez bought several items, all
the same price. The number of items was equal to the cost of each item in cents. The change that Mrs. Hernandez received from
$10 was $1 and 7 coins totaling less than $1. How much did each item cost?
Solution
The items cost 29 cents each.
From the amount of change received, she
must have spent between $8.00 and 9.00. We are, therefore, looking for a perfect square between 800 and 900.
Since 900 = 302, try 29; 29 × 29
= 841 cents, or $8.41. Her change of $1.59 could have been $1.00, 2 quarters, 1 nickel, and 4 pennies.
Problem
The rectangle shown below with unequal
sides is placed on a square so that each vertex lies on a side of the square at a trisection point of that side. What portion
of the area of the square is covered by the rectangle?
Solution
4/9 of the squares.
Divide the square into 9 equal squares
(remember that the vertices of the rectangle trisect the sides of the square). Count the small squares that are "covered"
by the rectangle.
There are 4 small squares (or 4/9) of the
big square "covered." Check by counting the parts not "covered." You will find 5 squares (or 5/9 of the big square) not "covered."
By adding, we find that the whole square
has been accounted for.
Problem
Suppose your salary could be raised 10%
and then a month later reduced by 10%. Or suppose that you may choose to have the cut first, followed by the raise one month
later. Which option is better? Why?
Solution
Both options will give you the same salary
for the second month. However, you will earn more during the first month under the first plan.
For example, assume that you are making
$100 a month. Under the first option, your salary will become $110 for one month. Reducing it by 10% at the end of a month
will give you a final salary of $99.
Under the second option, you will receive
$90 at first, then $99 when the 10% is added. Your final salary is the same under each plan, but you will earn more the first
month if you choose the first plan.
In general, if x is your beginning salary,
under the first plan your salary will be x, 1.10x, 0.99x, 0.99x, 0.99x, . . . .
Under the second plan, your salary
will be x, 0.90x, 0.99x, 0.99x, 0.99x. . . .
Problem
You have 7 full cartons, 7 half-full cartons,
and 7 empty cartons. The cartons are to be divided among three bins so that each bin contains the equivalent of 3 1/2 cartons.
How would you divide the cartons?
Solution
The total is 10 1/2 cartons, so each bin
must contain 10 1/2 ÷3 = 3 1/2 cartons.
Here are two solution possibilities:
Solution 1
A: 3 full, 1 half full, 3 empty
B: 3 full, 1 half full, 3 empty
C: 1 full, 5 half full, 1 empty
Solution 2
A: 3 full, 1 half full, 3 empty
B: 2 full, 3 half full, 2 empty
C: 2 full, 3 half full, 2 empty
Problem
At school, 9 students always take walks
in groups of three. How can the groups be arranged so that each person walks in a group with every other person exactly once
in four days?
Solution
Letting letters represent students, they
can be arranged in these twelve ways: ABC; DEF; GHI; ADG; BEH; CFI; AEI; BFG; CDH; AFH; BDI; and CEG.
One way to approach this problem is to
make a systematic table. Beginning with the three groups of ABC, DEF, and GHI, you could then begin to pair A with each possible
pair, B with each possible pair, and so on.
