Problem of the Week

2/27/17

House Prices

 The average house price in Boomtown rose 30 percent each year for the last five years. If the average house price is currently $250,000, what was the average house price five years ago?


2/6/17

Upright Integers 

An integer is defined as upright if the sum of its first two digits equals its third digit. For example, 145 is an upright integer since 1 + 4 = 5. How many positive three-digit integers are upright?

Solution

45. From the definition, the first and second digits of an upright integer automatically determine the third digit, which is the sum of the first two digits. Consider those upright integers beginning with 1: 101, 112, 123, 134, 145, 156, 167, 178, and 189; there is a total of 9 such numbers. (Note that the second digit may not be 9; otherwise, the last “digit” would be 1 + 9 = 10.) Beginning with 2, the upright integers are 202, 213, 224, 235, 246, 257, 268, and 279; there is a total of 8 such numbers. We may continue this pattern of analysis to show that the numbers of upright integers beginning with a digit of 3, 4, 5, 6, 7, 8, or 9 are 7, 6, 5, 4, 3, 2, and 1, respectively. Therefore, there is a total of 9+8+7+6+5+4+3+2+1 = 45 three-digit upright integers.

 

1/30/17

e-traders 

Three students open e-trade accounts and become day traders. Although they all work hard, they achieve the following steady rates of losing money: The first student loses $1000 in one hour, the second student loses $1000 in two hours, and the third student loses $1000 in three hours. Find the number of minutes it takes for the three students together to lose a total of $2000.

Solution

About 65 minutes (to the nearest minute). The losing rates are $1000/hr., $500/hr., and $333.33/hr. The combined losing rate is $1833.33/hr. Thus,


1833/1 = 2000/x  so x = 1 and 1/11 hours or 65 minutes



1/23/17

Investments

A man has $10,000 to invest. He invests $4,000 at 5 percent and $3,500 at 4 percent. To have a yearly income of $500 from the investment, at what rate must he invest the remainder of the money? 

Solution

At 6.4 percent or higher.

The man wants to earn $500 in interest each year. Investing $4000 at 5 percent yields yearly interest of 0.05 x $4000, or $200; whereas $3500 invested at 4 percent yields 0.04 x $3500, or $140; so the remaining $2500 needs to be invested at a rate that yields interest of $160 per year (to have a total interest of $500 each year). If r is the annual percentage rate for the $2500, then

 

so the man needs to invest the remaining $2500 at 6.4 percent or higher to guarantee an interest income of at least $500 a year.

 

1/16/17

What Is the Price?

Matthew and Matilda want to buy a set of DVDs. Matthew has $47 less than the purchase price, and Matilda has $2 less. If they pool their money, they still do not have enough to buy the DVDs. If the set costs a whole number of dollars, what is its price?

Solution

$48.

If P is the price of the DVDs, Matthew has P - 47 dollars and Matilda has P - 2 dollars.

We know that (P - 47) + (P - 2) < P, since the amount of their combined savings is still less than the price of the DVDs.

Solving for P, we have P < 49; but P > 47, since Matthew had some money. Therefore, P is a whole number such that 47 < P < 49, so the price of the DVDs is $48.

1/9/17

Tom and Bill

Tom is standing in a hole that is 4 feet deep. Bill asks him how much deeper he is going to dig the hole. Tom replies that he will dig 4 feet 2 inches deeper and that the top of his head will then be the same distance below ground level that it is now above ground level. How tall is Tom?

Solution

6 feet 1 inch. The top of his head will go down 4 feet 2 inches with the additional digging. Half that distance was above the hole before the additional digging, so he is 4 feet + 2 feet 1 inch, or 6 feet 1 inch, tall.

 

1/2/17

Wire Length

In the diagram, the rectangular wire grid contains 15 identical squares. The length of the grid is 10. What is the length of wire needed to construct the grid?

 

Solution

76. Since the length of the rectangular grid is 10, the side of each square in the grid is 10 ÷ 5 = 2. The height of the grid is therefore 6 (3 squares). There are four horizontal wires, each of length 10, and six vertical wires, each of length 6, for a total length of wire of 4 • 10 + 6 • 6 = 40 + 36 = 76.

12/19/16

How Tall Was the Tree?

Lightning hit a tree one-fourth of the distance up the trunk from the ground and broke the tree so that its top landed 60 feet from its base, thus creating a triangle. How tall was the tree before it broke?

Solution
Approximately 84.85 feet tall.

Let h be the height, in feet, of the tree before it fell. Then (h/4)2 + 602 = (3h/4)2, so h2/16 + 3600 = 9h2/16.

Solving for h reveals that h2/2 = 3600, so h = tree4, and the original height of the tree equals tree3, or approximately 84.85 feet.



12/12/16

Facetious

In how many possible arrangements of the letters in the word FACETIOUS are the vowels in alphabetical order?


Solution

3024 ways.

A total of 9! (9x8x7x6x5x4x3x2x1) ways exist to arrange the letters in the word FACETIOUS (nine ways of picking the first letter exist, eight ways to pick the second exist, and so on).

For each arrangement of the consonants in the word, the vowels can be arranged in 5! (5x4x3x2x1) ways. In only one of them are the vowels in alphabetical order.

Hence, 1/5! of the arrangements have the vowels in alphabetical order, for a total of (9!/5!), or 3024.


12/5/16

Townhouses

A contractor is asked to build a new set of townhouses in attached clusters of different sizes. He created plans for one-, two-, and three-house clusters, as shown in the diagram below. The builder used computer software to draw the line segments used to represent the houses. How many line segments are needed to draw 4 houses? 10 houses? 47 houses? n houses? Explain your strategy.

 

Solution

21; 51; 236; 5n + 1. One house takes 6 segments. Each additional house only requires 5 additional segments. Therefore, the sequence for the number of segments is 6, 11, 16, 21, 26, 31, 36, . . . . If n is the number of houses, then the nth term for this sequence is 5n + 1, or 5 times the number of houses plus 1. The 5th house would require 5(5) + 1 = 26 segments; the 10th house would require 5(10) + 1 = 51 segments; and the 47th house would require 5(47) + 1 = 236 segments. Another method is to begin with 6 segments for the first house and notice that each additional house adds 5 segments to the total. Therefore, the total number of segments is 6 + 5(n – 1) = 5n + 1.

11/28/16

Classmates' Birthdays 

Your mathematics teacher displays a monthly calendar, which highlights all the birthdays of your classmates. Your mathematics class has 25 students. What is the probability that 3 or more students were born in the same month?

Solution

1, or 100 percent. In a class of 24 students, it is possible for exactly 2 students to have been born in each of the 12 months. For 25 or more students, it is certain that there would be at least 1 month containing 3 or more birthdays. Therefore, the probability is 1.


11/14/16

Cards in a Hat 

Cards numbered 1 to 50 are put in a hat. What is the probability that the first two cards chosen at random have prime numbers on them?


Solution

3/35, or about .086. There are 15 prime numbers in the range of 1 to 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. Remember that 1 is not a prime number. The probability of selecting a prime on the first pick is 15 out of 50; for the second pick, the probability is 14/49. The probabilities are combined by multiplying them, so 15/50 × 14/49 = 210/2450, or 3/35, or about .086.

 

10/31/16

How Many? 

How many two-digit numbers exist such that when the product of its digits is added to the sum of its digits, the result is equal to the original two-digit number?

Solution

Nine two-digit numbers: 19, 29, 39, 49, 59, 69, 79, 89, and 99. Any two-digit number can be represented by 10a + b. The product of the digits is ab, and the sum is a + b. When does ab + (a + b) = 10a + b? Solving, when ab = 9a, or when b = 9.

 

10/24/16

Old Books 

A mathematics teacher collects old books. One day, a student asked him how many old math books he has. He replied, “If I divide the books into two unequal whole numbers, then 64 times the difference between the two numbers equals the difference between the squares of the two numbers.” How many old math books does the teacher have?

Solution

64. Let a and b represent the two unequal whole numbers. 
Then 64(a – b) = a2 – b2. 
Factoring the right side yields 64(a – b) = (a – b)(a + b). 
Since a and b are not equal, (a – b) does not equal zero. 
Therefore, you can divide both sides by (a – b), 
which yields 64 = a + b. 
To solve the problem, 
you do not have to know the values of a and b, 
just the sum of a and b, 
which is 64, the total number of books.