St. Gabriel kids clean up in January

There were six kids who got all of January's Math Puzzlers correct. Five were from St. Gabriel Catholic School in south St. Louis.

And several kids who entered also found out there was more than one way to answer Question 3. That's the one that asked you to rearrange numbers around a triangle so the three numbers on each side all equaled the same total.

Depending on how you arranged the numbers, the legs would be equal at 9, 10, 11 or 12.

We are getting some repeat winners in the Math Puzzlers. That means you're starting to think like Mr. Math Puzzler. He's math teacher Wayne Hesse from Green Park Lutheran School in south St. Louis County.

The five St. Gabriel winners in January included one girl who didn't want her name listed. The other four were Dominic DeVasto, Amy Lang, Leslie Hlavaty and Angela Mazzuce.

The only non-Gabriel winner was Tim Hakenewerth of Immaculate Conception School in Old Monroe, Mo.

We put the six winning entries into a hat and drew out three to get the $10 Borders gift certificates. The three were Leslie Hlavaty and Angela Mazzuce and the unnamed winner.

Young Saint Louis.com likes it when Math Puzzler entrants get all answers correct. We also like to award the Borders gift certificates as an extra bonus.

After you read the answers to January questions, you'll want to enter the February competition. (To see the February questions, click here.)

When you enter the February contest, remember to get your entries in the mail before the 15th of the month. That's the deadline for entries. Again in January, we had entries come in after the deadline. They couldn't be considered.

One of those who entered late in January was a past winner.

Math Puzzler answers for January, 2003

1. While you are raiding your refrigerator, you look behind the stove and discover a slice of bread that you misplaced several weeks ago. Needless to say, it is covered with mold. Since the mold started growing, the area it has covered has doubled each day. By the end of the eighth day, the entire surface of the bread was covered. When was the bread half-covered with mold?

Answer: End of 7th day

The explanation: This answer comes up almost before you start figuring. If the mold doubles each day, the way to get the answer is to start backwards from Day 8. By dividing 100% by 2, you find the bread was half covered on Day 7. Of course, that's the answer.

 

2. Suppose you have a three-wheeled car with one spare tire. You rotate tires regularly enough that each tire gets equal usage over a 50,000-mile span. How many miles will there be on each tire at the end of that distance?

Answer: 37,500 miles

The explanation: You have four tires and use only three at any one time. Therefore, each tire will be on three-fourths of the time. Three-fourths of 50,000 is 37,500 miles.

 

3. Place the numbers 1,2,3,4,5 and 6 in the circles below so that the sums of the three numbers on each side are equal?

Answer: Several answers

The explanation: This is a puzzler where you need to do some educated guesses and try different combinations. As it turned out there are four correct answers. And, depending on how you arrange the numbers, you can have side totals that equal 9, 10, 11 or 12.

 

4. Once a week, a wagon driver leaves his hut and drives his oxen to the river dock to pick up supplies for his town. At 4:05 p.m., one-fifth of the way to the dock, he passes the smithy. At 4:15 p.m., one-third of the way, he passes the miller. At what time does he leave his home?

Answer: Left home at 3:50 p.m.

The explanation: You need to find the relationship between one-fifth of the way and one-third of the way in minutes. The leg of the trip you can measure in minutes is the 10 minutes it takes to go from one-fifth of the distance to one-third of the distance. That 10 minutes is 2/15th of the total distance, therefore 1/15th is five minutes. The first leg was 3/15ths of the distance and therefore took 15 minutes. Subtracting 15 minutes from the arrival time of the first leg (4:05 p.m.) means the wagon driver left his hut at 3:50 p.m.

 

5. The dwarfs Dobbit and Mobbit are building a bridge over a narrow stream. Dobbit can do the job alone in 30 hours; Mobbit can do the job alone in 45 hours. How long would it take them if they work together?

Answer: 18 hours

The explanation: If Dobbit could do the whole job in 30 hours, that means he did 1/30th of the job each hour. If Mobbit did the job in 45 hours, that meant he did 1/45th of the job each hour. Finding a common denominator of 90, Dobbit does 3/90ths of the job in one hour and Mobbit does 2/90th. Together, they do 5/90th of the job in one hour. Therefore, they could do the job together in 18 hours.

 

6. Suppose Dobbit (from problem No. 5 above) worked on the project alone for 5 hours. How long would it take Dobbit and Mobbit to complete the rest of the job working together?

Answer: 15 hours

The explanation: Using the Dobbit's rate of work from Question 5, you know he could do 1/6th of the job in five hours. That leaves 5/6th of the job left. Five/sixth of 18 hours is 15 hours.