1) A tractor wheel is 88 inches in circumference. How many complete turns will the wheel make in rolling one mile on the ground? (1 mile = 5,280 feet)
2) In the addition problem below, each letter represents a digit and different letters represent different digits. What fourdigit number does D E E R represent? I N + R I D D E E R 3) Alice and Betty run a 50meter race and Alice wins by 10 meters. They then run a 60meter race, and each girl runs at the same speed she ran in the first race. By how many meters will Alice win? Solutions 1) When the wheel makes one complete turn, it has rolled a distance of 88 inches. The number of turns equals 1 mile divided by 88 inches (1 mile = 5280 feet = 5280 x 12 inches). Dividing yields 5280 x 12 inches = 720 88 inches 2) D in the sum DEER must be 1. The R in the second addend must be 9. It follows that E must be 0 and DEER represents 1009. 3) Method 1 Alice runs 50m in the same time that Betty runs 40m. Thus Alice runs 5m for every 4m that Betty runs. Therefore in the 60m race, Alice will run 12 x 5m or 60m, and Betty will run 12 x 4 = 48m. So Alice wins by 12m. Method 2 Since Alice wins the 50m race by 10m, Alice must gain 2m over Betty for every 10m that Alice runs. Therefore, in a 60m races, Alice will gain 6 x 2m or 12m.
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1) When certain numbers are placed in the empty boxes, the sum of the three numbers in each of the three rows, three columns, and two diagonals is the same. What number should be placed in the center box?
2) I am less than 6 feet tall but more than 2 feet tall. My height in inches is a multiple of 7 and is also 2 inches more than a multiple of 6. What is my height in inches? 3) The square below is divided into four congruent rectangles. The perimeter of each of the four congruent rectangles is 25 units. How many units are there in the perimeter of the square? Solutions 1) Since the sum of the three numbers in each diagonal is the same, and since they have the same middle number, the sum of each pair of numbers in opposite corners must be the same: 9 + 13 = 5 + ?. Clearly ? = 17. The bottom row now has a sum of 33. The middle number must be 11. 2) List multiples of 7 greater than 24 and less than 72. Also list the multiples of 6 which are less than and closest to each of the corresponding multiples of 7. 3) Each of the four congruent rectangles has a perimeter equivalent to 2 1/2 sides of the square. Since the length of 2 1/2 sides = 25 units, then, by doubling, we get the length of 5 sides = 50 units. Clearly, the length of one side is 10 units. Therefore the perimeter of the square is 40 units. 1) Suppose the counting numbers from 1 through 100 are written on paper. What is the total numbers of 3s and 8s that will appear on the paper?
2) In the "magicsquare" below, five more numbers can be placed in the boxes so that the sum of the three numbers in each row, in each column, and in each diagonal is always the same. What value should X have? 3) The Ushaped figure at the right contains 11 squares of the same size. The area of the Ushaped figure is 176 square inches. How many inches are in the perimeter of the Ushaped figure? Solutions 1) (a) 3 occurs in the units place once in each group of ten consecutive numbers. Therefore 3 will appear 10 times in the units place. (b) 3 occurs in the tens place ten times in 100 consecutive numbers: 30, 31, 32, 33, . . . , 38, 39. (c) Statements (a) and (b) are also true for occurrences of 8. Therefore there will be a total of 40 occurrences of 3s and 8s in the numbers from 1 through 100. 2) Method 1 From the first column, we see that the sum for each row, column, and diagonal should be 90. Then the missing number in the 3rd row is 40, and the missing number in the lead diagonal is 30. The sum for the second column is 40 + 30 + X or 70 + X. Then X must be 20. Method 2 Since the numbers in the two diagonals have equal sums and the same middle number, then the two numbers in the corners of each diagonal have equal sums; 25 + 35 = 15 + ? It follows that ?, the number in the lower right corner, is 45. Since the sums of the numbers in any row, column, or diagonal is 90, the bottom row sum 25 + X + 45 = 90. Thus X = 20. 3) The area of each square is 176/11 or 16 inches. Then the length of each side of a square is 4 inches. The perimeter of the Ushaped figure is equivalent to the total length of 24 sides or 96 inches. 1) Suppose 6 days after the day before yesterday is Thursday. What day of the week is the day after tomorrow?
2) There are 4 separate boxes, and inside each large box there are 3 separate small boxes, and inside each of these small boxes there are 2 separate smaller boxes. How many boxes, counting all sizes, are there altogether? 3) When asked how many gold coins he had, the collector said: If I arrange them in stacks of five, none are left over. If I arrange them in stacks of six, none are left over. If I arrange them in stacks of seven, one is left over. What is the least number of coins he could have? Solutions 1) Make a linediagram with the following shown on the diagram: N representing now or today, T for tomorrow, and T + 1 for the day after tomorrow, Y for yesterday, and Y  1 for the day before yesterday. Count 6 days to the right of Y  1 (the day before yesterday). Mark that day in the diagram as H for Thursday. Count back 2 days to T + 1 (the day after tomorrow). That day is Tuesday. 2) 3) The number of coins must be a multiple of 30. The multiples of 30 are 30, 60, 90, 120, 150, and so on. The smallest of these multiples that leaves a remainder of 1 when divided by 7 is 120. 1) A square is divided into three congruent rectangles as shown below. Each of the three rectangles has a perimeter of 16 meters. How many meters are in the perimeter of the square? 2) A person made a purchase for D dollars and C cents, and gave the cashier a $20 bill. The cashier incorrectly charged the person C dollars and D cents, and returned $4.88 in change. If the cashier had charged the correct price, what would the correct change have been? 3) In the addition problem below, different letters stand for different digits. What fivedigit number does SERVE represent? V C R + V C C T S E R V E Solutions 1) Let 3S represent the length of a side of the square as shown in the diagram below. Then the perimeter of one of the congruent rectangles is 8S or 16. It follows that S = 2, and 3S, the length of a side of the square, is 6. Therefore the perimeter of the square is 24 meters. 2) To find the incorrect charge, subtract $4.88 from $20. $20.00  $ 4.88 incorrect change $15.12 incorrect charge The incorrect charge, $15.12, represents C dollars and D cents. Therefore the correct charge is $12.15. To get the correct change, subtract $12.15 from $20. $20.00  $12.15 correct charge $ 7.85 correct change 3) Find the values of S, V, and E in that order: S = 1, V = 9, E = 0. Substitute those values in the cryptogram. V C R 9 C R + V C C T + 9 C C T S E R V E 1 0 R 9 0 Clearly R + T in the units column must equal 10 which will yield a carry of 1 in the tens column. So C + C + 1 must end in 9. C can be 9 or 4. But V = 9. So C must be 4. Substitute 4 for C. 9 4 R + 9 4 4 T 1 0 R 9 0 Since there is no carry from the tens column, R in the hundreds column must be 3 (and T in the units column must be 7) SERVE is the number 10390. 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 chime clock strikes 1 chime at one o'clock, 2 chimes at two o'clock, 3 chimes at three o'clock, and so forth. What is the total number of chimes the clock will strike in a twelvehour period?
2) A boy has the following seven coins in his pocket: 2 pennies, 2 nickels, 2 dimes, and 1 quarter. He takes out two coins, records the sum of their values, and then puts them back with the other coins. He continues to take out two coins, record the sum of their values, and put them back. How many different sums can he record at most? 3) The product of two numbers is 144 and their difference is 10. What is the sum of the two numbers? Solutions: 1) Method 1: T, the total number of chimes, equals 1 + 2 + 3 + 4 + ... + 10 + 11 + 12. The sum of this series is 78. Method 2: T = 1 + 2 + 3 + 4 + ... + 10 + 11 + 12 T = 12 + 11 + 10 + 9 + ... + 3 + 2 + 1 2T = 13 + 13 +13 +13 + ... + 13 + 13 + 13 Notice that the series on the second line is the reverse of the series on the first line. Each time we add a term and the corresponding term above, the sum is 13. then 2T is equal to 12 x 13 or 156. But 2T is twice the sum of the first series. Therefore, T = 78. 2) The following pairs of numbers represent the values of the two coins the boy could take from his pocket: (1,1) (1,5) (1,10) (1,25) (5,5) (5,10) (5,25) (10,10) (10,25) Each of the above pairs has a sum that is different from the sum of each of the other pairs. Therefore there are 9 different sums. 3) Examine the pairs of whole number whose product if 144. The only pair that has a difference of 10 is 18 and 8. Their sum is 26. 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. it. 1) Five brothers, each born in a different year, share a gift of $100 according to the following arrangement: each boy, except the youngest, gets $5 more than his next younger brother. How much does the youngest boy get?
2) When I open my math book, two pages face me and the sum of the two page numbers is 317. What is the number of the very next page? 3) 6, 14, and 15 are factors of the natural numbers N. What is the smallest value that N can have? Solutions: 1) Method 1: The average amount received by the 5 boys is $20. In order of age, the third boy receives the average amount of $20. The youngest receives $10 less than the third boy. Therefore, the youngest receives $10. Method 2: Let Y represent the amount received by the youngest boy. Then the amounts received are Y, Y + 5, Y + 10, Y + 15, and Y + 20. These amounts can be regrouped and totaled as Y + Y + Y +Y + Y + 5 + 10 + 15 + 20, or more simply as 5Y + 50. Then, 5Y + 50 = 100, 5Y = 50, and Y = 10. The youngest receives $10. 2) The average of the two consecutive page numbers is 317/2 = 158.5. Then the page numbers are 158 and 159, with 159 being the number of the righthand page. The next page number is 160. 3) Method 1: The LCM of a set of numbers is the smallest number N which each number of the set will divide exactly. LMC(6,14,15) = 210. Method 2: List the prime factors of each of the given factors of N: 6 = 2 x 3; 14 = 2 x 7, 15 = 3 x 5. N must have 2, 3, 5, and 7 as factors. Then N = 2 x 3 x 5 x 7 = 210. 1) A box contains over 100 marbles. The marbles can be divided into equal shares among 6, 7, or 8 children with 1 marble left over each time. What is the least number of marbles that the box can contain?
2) A fisherman sold some big fish at $4 each and twice as many small fish at $1 each. He received a total of $72 for the big and small fish. How many big fish did he sell? 3) In the addition example below, different letters represent different digits. What digit does A represent? A A + A A C A B Solutions: 1) Use a simpler problem. Instead of one marble being left over each time, assume that no marbles were left over each time. Then the least number of marbles is the LCM(6, 7, 8) which is equal to 3 x 7 x 8 = 168. However, since one marble should be left over each time, the least number of marbles in the box is 168 + 1 = 169. 2) For each big fish sold for $4, two small fish were sold for $1 each. One big fish and two small fish were sold for a total of $6. Since the total received for big and little fish was $72, there were 72/6 = 12 sets of 1 big fish and 2 little fish sold. Therefore 12 big fish were sold. 3) Method 1: A has to be 5 or more. Otherwise the sum will not be a threedigit number. In the second column, A + A + 1 ends in A. Then A is odd, so A = 5, 7, or 9. The only value that checks is 9. Therefore A is 9. Method 2: Use expanded notation: AA = 10A + A = 11A. Then AA + AA = 22A. Therefore C must be 1 and CAB = 100 + 10A + B. Then 22A = 100 + 10A + B. If we subtract 10A from both sides of the equality, the result is 12A = 100 + B. Since 12A has to be greater than 100, A has to be 9. 
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