MOEMS posts a problem of the month each month at http://www.moems.org/zinger.htm
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Here are some challenging word problems that are similar to what 4th and 5th graders experience at a Math Olympiad. Math team members can use these for extra practice and everyone else can use them to help extend your problem solving skills. Solutions are at the bottom of this post.
1) Suppose today is Tuesday. What day of the week will it be 100 days from now? 2) The fourdigit numeral 3AA1 is divisible by 9. What does A represent? 3) A purse contains 4 pennies, 2 nickels, 1 dime, and 1 quarter. Different values can be obtained by using one or more coins in the same purse. How many different values can be obtained? Solutions 1) Every 7 days from "today" will be Tuesday. Since 98 is a multiple of 7, the 98th day from today will be Tuesday. Then the 100th day from today will be Thursday. 2) If the number is divisible by 9, then the sum of its digits is divisible by 9. The digit sum is 3 + A + A + 1 = 4 + 2A. The digit sum cannot by 9, otherwise A = 2 1/2. So 4 + 2A = 18 which produces A = 7. 3) The largest amount that can be made is $0.49. Using the given set of coins, any amount from $0.01 to $0.49 can be made. Therefore there are 49 different amounts that can be made. 1) The weight of a glass bowl and the marbles it contains is 50 ounces. If the number of marbles in the bowl is doubled, the total weight of the bowl and marbles is 92 ounces. What is
the weight of the bowl? (Assume that each of the marbles has the same weight). 2) Square ABCD and rectangle AEFG each have an area of 36 square meters. E is the midpoint of AB. What is the perimeter of rectangle AEFG? 3) A slow clock loses 3 minutes every hour. Suppose the slow clock and a correct clock both show the correct time at 9 A.M. What time will the slow clock show when the correct clock shows 10 o'clock the evening of the same day? Solutions 1) Method 1: Since the weight of one bowl and the marbles it contains is 50 ounces, the weight of two bowls and twice as many marbles is 100 ounces. It is given that the weight of one bowl and twice as many marbles is 92 ounces. Then the difference of weights between 100 ounces and 92 ounces is the weight of the bowl, or 8 ounces. Method 2: The increase in weight from 50 ounces to 92 ounces occurs because the weight of the marbles is doubled. That increase therefore is the weight of the original set of marbles, 42 ounces. If follows that the bowl weighs 8 ounces. 2) The length of a side of the square is 6m. The dimensions of rectangle AEFG are 3m and 12m. The perimeter of the rectangle is 2 x (3m + 12m) which is equivalent to 30 m. 3) There are 13 hours between 9 AM and 10 PM. Thus the slow clock will lose 13 x 3 or 39 minutes and show 9:21 or 21:21 on a 24hour clock 1) Suppose the time is now 2 o'clock on a twelvehour which runs continuously. What time will it show 1,000 hours from now?
2) When a natural number is multiplied by itself, the result is a perfect square. For example, 1, 4, and 9 are perfect squares because 1 x 1 = 1, 2 x 2 = 4 and 3 x 3 = 9. How many perfect squares are less than 10,000? 3) A work team of four people completes half of a job in 15 days. How many days will it take a team of ten people to complete the remaining half of the job? (Assume that each person of both teams works at the same rate as each of the other people). Solutions 1) The time 2 o'clock is repeated every twelve hours. There are 83 twelves in 1,000 plus a remainder of 4. Therefore the clock will show a time of 6 o'clock 1,000 hours from now. 2) Since 10,000 = 100 x 100, each of the perfect squares 1 x 1, 2 x 2, 3 x 3, . . . , 99 x 99 is less than 100 x 100. There are 99 numbers in the above sequence. 3) Four people working 15 days is equivalent to one person working 60 days. To complete the other half of the job, ten people would have to work 6 days which is also equivalent to one person working 60 days 1) A motorist made a 60mile trip averaging 20 miles per hour. On the return trip, he averaged 30 miles per hour. What was the motorist's average speed for the entire trip?
2) 100 pounds of chocolate is packaged into boxes each containing 1 1/4 pounds of chocolate. Each box is then sold for $1.75. What is the total selling price for all of the boxes of chocolate? 3) In the multiplication problem below, A and B stand for different digits. Find A and B. A B x B A 1 1 4 3 0 4 3 1 5 4 Solutions: 1) The average speed for any trip is the total distance divided by the total time spent in traveling. The total distance was 120 miles and the total time was 5 hours. The average speed equals (120 miles)/(5 hours) or 24 miles/hour or 24 mph. 2) We need to know the number of boxes that were packaged in order to find the total selling price. The number of boxes is obtained by dividing 100 by 1 1/4: 100 divided by 5/4 = 100 x 4/5 = 80. 3) The first partial product 114 is equal to the product of AB and A. The second partial product 304 is equal to the product of AB and B. Then A must be less than B. Method 1: Since the product of AB and A is 114, A is a divisor of 114. Therefore A may be 2, 3, or 6. Since AB x A = 114, A cannot be 2 because AB x A would then be less 60. Similarly, A cannot be 6 since AB x A would then be greater than 360. Therefore A must be 3 and AB must be 114/3 or 38. Thus A = 3 and B = 8. Method 2: From the first partial product, observe that B x A must end in 4. Since A is less than B, A = 2 and B = 7, or A = 3 and B = 8, or A = 4 and B = 6. But 27 x 2 = 54, 46 x 4 = 184, and 38 x 3 = 114. Only the third equation satisfies the given conditions. So A = 3 and B = 8. Method 3: From the second partial product 304, we see that B x B ends in 4. Then B = 2 or 8. If B =2, then A must be 1 because A is less than B. But 12 x 21 = 252. If B = 8 and AB x B = 304, then AB = 304/8 or 38 and 38 x 83 = 3154. Therefore A = 3 and B = 8.
1) If a kindergarten teacher places her children 4 on a bench, there will be 3 children who will not have a place. However, if 5 children are placed on each bench, there will be 2 empty spaces.
What is the smallest number of children the class could have? 2) If the digits A, B, and C are added, the sum is the twodigit number AB as shown below. What is the value of C? A B + C A B 3) When I open my mathematics book, there are two pages which face me and the product of the two page numbers is 1806. What are the two page numbers? Solutions: 1) Since the number of children is 3 more than a multiple of 4, that number could be 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, . . . Since the number of children is 2 less than a multiple of 5, the number could be 3, 8, 13, 18, 23, 28, 33, 28, 43, . . . The numbers satisfying both conditions are 23, 43, 63, 83, and so forth. The smallest of these numbers is 23. Thus, there are 23 children in the class. 2) In the units column, notice that the sum of A, B, and C ends in B. Then A + C = 10. Since A is also the tens digit of the sum, A must be 1. Therefore C = 9. 3) If page numbers are in the 40s, then the product is greater than 1,600. If the page numbers are in the 50s, then the product is greater than 2,500. Clearly the page numbers must be in the 40s. Since the two page numbers are consecutive numbers, the units digits must be 2 and 3, or 7 and 8. Try 42 and 43. They work! (Page numbers 47 and 48 don't work). 
AuthorAmanda Lovett Archives
August 2017
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