Two winners in the May Puzzler contest

Ten-year-old Clayton Vance is a repeat winner in the May Math Puzzler contest. And the other winner, 9-year-old Eric Hsu, is the brother of a past winner.

Clayton is from Mason Ridge Elementary School. Eric attends Wild Horse Elementary School.

These were the only two May entrants who answered all six of the questions correctly.

Congratulations to Clayton and Eric. They both will receive $10 Borders gift certificates as an extra bonus for their successful math figuring.

You'll remember the May competition had some unusual questions. Mr. Math Puzzler recommended the use of educated guessing to get the correct answers.

If you haven't participated in Young Saint Louis.com's Math Puzzler competition, why don't you try the June questions. The Math Puzzlers will continue right through the summer vacation period.

For those who haven't entered before, you might like to do some checking in our achives. YSL.com started the Math Puzzlers in September, 2001. We give the questions one month and then the answers the next.

That means, you can check past competitions to learn how Mr. Math Puzzler thinks.

Just go to the Past Stories tab on the top of the home page. Pick a month after September, 2001, and check past questions. Then, you can move forward to learn those answers.

After doing that for a few months, you'll have an idea on what kind of questions Mr. Math Puzzler likes. He is Wayne Hesse, an 8th grade math teacher at Green Park Lutheran School in south St. Louis County.

The answers to the May, 2003, questions are included below.

If you're ready for the June Puzzler competition, click here.

Print out the entry blank and questions. After noting your answers, mail the entry to YSL.com.

The Answers to April's Math Puzzlers

1. How many ways can you read ACE off the diagram below? You can move horizontally, vertically or any combination of horizontal or vertical as long as the letters are adjacent.

            A
          A C A
        A C E C A
          A C A
            A

Answer: 12

Explanation: This is one of the questions that needs an educated guess; there's no formula to achieve the answer. Remember, you can go backwards and even use right-angle turns to achieve the ACE word.

 

2. Timmy rents a car to drive to a city 100km away. He stops halfway and pick up a friend, who rides the last 50km with him. Returning in the evening with his friend, Timmy drops him where he picked him up, then drives on to his starting point, where he is charged $24 for car rental. Timmy and his friend share expenses equitably. How much should each pay?

Answer: Timmy, $16; Friend, $8

Explanation: Timmy went the whole 200km round-trip. His friend only went 100km. That made a total of 300 passenger-kilometers. Timmy's share was two-thirds of the total kilometers and therefore he needs to pay two-thirds of the car rental.

 

3. Tammy is preparing for a 42,000km trip in her car, a traditional four-wheel model. Buying tires which each last 24,000km, Timmy contends that 7 would be enough. Is she right? Prove it.

Answer: Yes

Explanation: A 42,000km trip amounts to 168,000 tire-kilometers. If each tire lasts 24,000km, seven tires should have a total of 168,000kms in them. Therefore, by rotating tires judiciously, Tammy could make the entire trip within the allotted kilometers.

 

4. Jenny is having dinner with a friend. She brought five dishes and her friend three dishes. At the last minute, another friend comes and eats with them. The second friend pays $4 as her share. If all dishes have the same value, how can the money be divided between Jenny and her first friend? (Be careful.)

Answer: Jenny gets $3.50; friend gets $.50

Explanation: Based on the $4 payment, the eight dishes are worth $12 ($4 times three people). The $12 divided into eight dishes, makes the average cost at $1.50 per dish. Therefore, Jenny's five dishes are worth $7,50. When Jenny's $4 is subtracted, that leaves $3.50. The first friend's $4 is subtracted from her dishes' $4.50 value, that leaves $.50.

 

5. Nine schoolchildren form a circle. To choose a leader, they decide to start from one of them, count up to 5 clockwise, ask the fifth player to leave the circle, and so on. The last player left in the circle is the leader. Andrew does the counting. He wants to take advantage of this to become the leader. Let's call him and his friends by the first letter of each child's first name using the letters A (for Andrew) through I, clockwise. From which spot should Andrew start his counting so he becomes the leader?

Answer: C

Explanation: The best way is to make a circle and identify each with a letter, from A through I. Then, by going around and around, taking out the fifth player on each round, you find that Andrew should have positioned himself in the C location to make sure he became the leader.

 

6. The locomotive, which is 24 feet long, plus a Pullman car equal the length of 3 coaches. The four Pullman cars equal the length of the locomotive plus the length of the 3 coaches. The diner car is 2 feet longer than a Pullman car. One of the 3 coaches is 1 foot longer than the other two. How long is each car?

Answer:Locomotive, 24 ft.;
2 coaches, 13 ft.each;
1 coach, 14 ft. Pullman, 16 ft, and
Diner, 18 ft.

Explanation: This answer does involve a collection of formulas. The locomotive is 24 ft (L = 24); the Pullman is 24 + P = 3C; the Diner is D = 2+ P. Combining those and then using the substitution method, you arrive at the Pullman being 16 ft and the Diner 18 ft. You then arrive at three coaches being a total of 40 ft. Using the last clue about one coach being a foot longer than the other two, you get 13 ft, 13 ft and 14 ft.