Past winners have Mr. Math Puzzler figured out

Four past winners of Math Puzzler competitions got all of the August Puzzlers correct. They ought to be ready for whatever math classes they have for the 2003-2004 school year.

All the August Puzzlers could be answered with numbers.

Unlike July, all August Puzzlers had only one answer. The previous month, Mr. Math Puzzler included some questions which had more than one right answer. Not this time.

But, for Drew Fendler, brothers Eric and Phillip Hsu and Thomas Van Horn, they only needed one chance per question.

Then, we put the four winners in a hat and drew for the three $10 Borders book certificates. Thomas Van Horn's name was the one still left in the hat. But, he had been the only winner in the July contest.

Young Saint Louis.com is now starting its third year of Math Puzzlers. Math teacher Wayne Hesse of Green Park Lutheran School agreed to be Mr. Math Puzzler in September, 2001.

YSL.com always is looking for more entrants for the Math Puzzlers. If this is your first time with the Puzzlers, you might like to review past questions and answers before answering. A review would help you figure out how Mr. Math Puzzler thinks.

For review, use the Past Stories tab on the home page to bring up the archives of past issues. Then, using any month since September, 2001, you can check both questions and answers.

For the August Puzzler questions and answers, see below:

The August, 2003, Math Puzzlers

1. Each of three towns was menaced by a dragon living in a cave above each town. The wizards Malefano, Sagareth and Thaumater created these three dragons. Sagareth's dragon will menace its town for the same number of years as the square root of the number of years Thaumater's dragon curse is on its town. Sagareth's dragon curse also will last the number of years equal to half the square root of the number of years Malefano's dragon curse is on the third town. Thaumater's dragon curse will last the number of years equal to twice the square root of the number of years of Malefano's curse. How long will each curse last? (Hint: All curses will be expressed in whole years, no fractions. Also, you need to set up formulas and a table of values for all.)

Answer: M=64 yrs; S= 4, and T=16

The explanation: First you set up a formula for each age.

Then, a table of values:

        4  1   1
       16  2   4
       36  3   9
     * 64  4  16

 

2. Five glubs that tried to enter the town of Galvinchy were slain by 10 Knights of the Golden Sword. The knights laid the five glubs' bodies tail to head with five feet between one tail and the next head. Stretched out like that, the glubs covered a distance of 200 feet. The first, third and fifth glubs were all the same length, as were the second and fourth glubs. Each glub was either 10 feet longer or 10 feet shorter than its neighboring glub. Furthermore, each glub's length was a multiple of 10. What was the length of each glub? (Hint: Make a line drawing featuring the glubs and the gaps between them.)

Answer: 1,3,5=40; 2,4=30

The explanation: First lay out the glubs (marking 3 as X and 2 as Y) in a line with 5 feet between each, and total length at 200 feet (X+5+Y+5+X+5+Y+5+X=200). Then, you subtract 20 feet which represents the four 5-foot spaces between the five glubs. That leaves 180 feet in length left for the five glubs. You need to figure out whether the X glubs or the Y glubs are longer. Assuming the X glubs are longer, we'll subtract 30 feet (equal the 10-feet per glub we need to add to the longer glubs.) That's 150 divided by 5 glubs or 30 feet per glub. If you add 10 feet to three X glubs and put 5-foot spaces between glubs, you get (40+5+30+5+40+5+30+5+40=200.) (If you had figured the Y glubs were longer, you'd have subtracted 20 feet from 180 and divide by 5 to get 32 feet for a shorter glub and 42 feet for the longer. Those lengths aren't multiples of 10.)

 

3. Four cave dwarves, each of whom works at the same rate, were to complete a mining job according to a schedule. However, because of an argument, two of the four quit after working only one day. The remaining two dwarves finished the job, but it required two more days than originally had been scheduled. How many days were originally scheduled for completion of the job? (Hint: Create a table.)

Answer: 3 days

The explanation:

You can figure out the work schedule with a chart. If the 3rd and 4th workers effort was transferred to the first two workers and the job went two more days, you can see by the transferred work that the original job would have taken three days.

 

4. A certain gardener had a number of skilled workers, each of whom was getting $28 a day. He also had a number of semi-skilled workers, each earning $12 a day. This amounted to a combined daily payroll of $264 per day. Since many of the jobs didn't require much supervision, he figured that by reducing the number of skilled workers by one-half and doubling the number of semi-skilled employees, he could actually get much more work done at a cost of only $12 a day more. After making these changes, how many total employees does he now have? (Hint: Set up a system of linear equations and then find the answers by using the addition method.)

Answer: 19 employees

The explanation: This is a problem that can be solved by an addition method formula. Use X for the skilled workers and Y for the unskilled.

-2 (28X + 12Y = 264) ——> -56X - 24Y = -528
    14X + 24Y = 276  ——>  14X + 24Y =  276
                          ————————————————
                         -42X       = -252
                         —————                ————
                            X       =  6/2

                            X = 3 skilled

Using a similar formula for the unskilled, you get 8 unskilled, which when doubled comes to 16. Then, 3 skilled and 16 unskilled makes a total of 19 employees.

 

5. Art, Boyd and Carl were sitting at a bar and, to pass time, Art suggested that they flip coins. Whoever tossed the only head or tail of the three coins thrown wins 1/2 of the money that each of the others then has. They all start out with the same amount of money. Art won the first flip; Boyd won the second flip, and Carl won the third. Carl then counted his money and had exactly $13. How much did each have when they started? (Hint: Use a table and think in terms of common denominators of fractions.)

Answer: $8

The explanation: The best way to figure this is use a table and figure the various steps using decimals.

              A      B       C
            ———————————————————————
Start         x      x       x
First flip   2x    .5x     .5x
Second flip   x  1.75x    .25x
Third flip  .5x  .875x  1.625x

 

1.625X    $13
———————— = ————————   or X = $8
1.625    1.625

 

6. After playing poker for a few hours, George realized that he had lost 3/4ths of his money, so he stopped playing and went to the cafe for a bite to eat. He spent $3 for lunch and then returned to play a few more hours. During this time, he won back 4/5ths of the money he had lost. He stopped playing for the night and discovered that he now had $21 less than when he had started. How much money did he have when he started?

Answer: $120

The explanation: This is solved by using a collecting of like items formula. George started with X amount of money. Before he stopped he was left with 1/4X. Then, he spent $3 for lunch. He came back to win 4/5th of what he lost. The formula looks like this:

   1/4 X  - 3 + 3/5 X = X - 21

 

 17/20 X  - 3         = X - 21
          +21              +21
          ————                                   ————
 17/20 X + 18         = X
-17/20 X                -17/20 X
—————————————————           ——————————
           18         = 3/20 X

 

20/3 (18) = 20/3 (3/20 X)

      120 = X