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
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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). 1) A bag contains 500 beads, each of the same size, but in 5 different colors. Suppose there are 100 beads of each color, and I am blindfolded. What is the fewest number of beads I must pick to
be absolutely sure there are 5 beads of the same color among the beads I have picked blindfolded? 2) A number has a remainder of 1 when divided by 4, a remainder of 2 when divided by 5, and a remainder of 3 when divided by 6. What is the smallest number that has the above properties? 3) A total of fifteen pennies are put into four piles so that each pile has a different number of pennies. What is the smallest possible number of pennies that could be in the largest pile? Solutions: 1) Suppose I select 5 beads blindfolded. The 5 beads could contain1 bead of each color. Suppose I select 5 beads three more times with the same result. I will then have a total 20 beads consisting of 4 beads of each color. The selection of the 21st bead guarantees that there are now at least 5 beads of the same color among the 21 beads. 2) Let N be the required number. Notice that when N is divided by 4, 5, or 6, the remainder is 3 less than the divisor in each case. If N is increased by 3, this new number N + 3 will be divisible by 4, 5, and 6. The smallest number that N + 3 can be is the least common multiple of 4, 5, and 6 which is 60. Therefore the required number is 57. 3) Method 1: The different ways 15 pennies can be distributed into 4 piles are (1, 2, 3, 9), (1, 2, 4, 8), (1, 2, 5, 7), (1, 3, 4, 7), (1, 3, 5, 6), (2, 3, 4, 6). Thus, the number of pennies in the largest pile of each distribution may be 6, 7, 8, or 9. The smallest of these is 6. Method 2: To find the smallest possible number of pennies in the largest pile, make the sum of the numbers of pennies in the other three piles as large as possible. This occurs when the first three piles contain 1, 3, and 5 pennies, or 2, 3, and 4 pennies. In either case the fourth pile will contain 6 pennies. 1) There are many numbers that divide 109 with a remainder of 4. List all twodigit numbers that have that property.
2) If a number ends in zeros, the zeros are called terminal zeros. For example, 520,000 has four terminal zeros, but 502,000 has just three terminal zeros. Let N equal the product of all natural numbers from 1 through 20: N = 1 x 2 x 3 x 4 x . . . x 20. How many terminal zeros will N have when it is written in standard form? 3) The XYZ club collected a total of $1.21 from its members with each member contributing the same amount. If each member paid for his or her share with 3 coins, how many nickels were contributed? Solutions 1) If 4 is subtracted from 109, the result is 105. Then each of the twodigit numbers that will divide 109 with a remainder of 4 will divide 105 exactly. Thus, the problem is equivalent to finding all twodigit divisors of 105. Since the prime factors of 105 are 3, 5, and 7, the divisors are 3 x 5, 3 x 7, and 5 x 7, or 15, 21, and 35. 2) Multiplication by 10 produces an additional terminal zero when a product is written in standard form. If 1 x 2 x 3 x . . . x 20 is written as a product of prime factors, it will contain four 5s and more than four 2s among many factors. Then part of the product can be written as: (5 x 2) (5 x 2) (5 x 2) (5 x 2) which can also be represented as 10 x 10 x 10 x 10. Therefore there are four terminal zeros in the product. 3) Since 121 = 11 x 11, the club has 11 members and each contributed $0.11. Each $0.11 share was paid in 3 coins which had to be 2 nickels and 1 penny. Then the 11 members contributed a total of 22 nickels. 
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