3. Mr. White does not wear white, so he is wearing the blue shirt. 4. Then Mr. Red wears a white shirt.

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5A METHOD 1: Strategy: Use reasoning. 1. Mr. Red and Mr. White are older than the man in gray. Neither Mr. Red nor Mr. White wears gray. Mr. Gray does not wear gray. So Mr. Blue wears the gray shirt.

1 5A METHOD 1: Strategy: Use reasoning. 1. Mr. Red and Mr. White are older than the man in gray. Neither Mr. Red nor Mr. White wears gray. Mr. Gray does not wear gray. So Mr. Blue wears the gray shirt. 2. The man in red is next to Mr. White. Mr. White does not wear red, Mr. Red does not wear red, and Mr. Blue is known to be wearing gray. So Mr. Gray wears the red shirt. 3. Mr. White does not wear white, so he is wearing the blue shirt. 4. Then Mr. Red wears a white shirt. Some students may find it convenient to organize their thoughts by using a table, entering X for a combination that is ruled out by the clues and O for a match. One way to develop the table is shown below. The darkened boxes show new entries. 5B Strategy: Minimize the numerator; maximize the denominator. A fraction has minimum value if the numerator is as small as possible and the denominator is as large as possible. The least possible value of the fraction is 5C Strategy: Find the total area of the missing pieces. 3 = The original area of the piece of paper is = 1200 sq cm. The total area of the regions cut out is then = 220 sq cm. Then:

METHOD 1: Strategy: Change the figure to create a simpler problem. Slide the two shaded cutout rectangles together, as shown. None of the areas will change.

2 METHOD 1: Strategy: Change the figure to create a simpler problem. Slide the two shaded cutout rectangles together, as shown. None of the areas will change. The total area of the cutouts is the same as the area of a single rectangle with base = 20 cm. The area of this single rectangle is 220 so the height of each cut is = 11 cm. METHOD 2: Strategy: Use algebra. Let h = the height of each cut. The areas of the cut-out rectangles are 12h and 8h. 12h + 8h = 220 Add 12h and 8h: 20h = 220 Divide each side of the equation by 20: h = 11 The height of each cut is 11 cm. 5D For ease, call a pizza cut into large slices a large pie and call one cut into small slices a small pie. METHOD 1: Find a relationship between the numbers of large and small pies. Each small pie is cut into 8 slices, so the number of small slices is a multiple of 8. With 5 small slices for each 3 large slices, the number of small slices is also a multiple of 5. Then the number of small slices is a multiple of 40, the LCM of 8 and 5. Now, group the small slices by 40s. For each (8 5) = 40 small slices there are (8 3) = 24 large slices. But 40 small slices form 5 small pies and 24 large slices form 4 large pies. That is, of every 9 pies, 5 are small and 4 are large. Alexis has 10 groups of 9 pies, so there are 10 x 5 = 50 small pies (and 40 large pies). METHOD 2: Guess and check. With 5 small slices for every 3 large slices, then 1 of the number of small slices equals 1 of the 5 3 number of large slices. In the table at the right we try first 45 pies of each size and then adjust by fives. There are 50 small pies.

3 5E Strategy: Find three different counting numbers with a sum of 8. The numbers of each type of marble can be 1, 3, and 4, or they can be 1, 2 and 5. To get the least possible total weight, assign the greatest weight to the least number of marbles and the second greatest weight to the second least possible number of marbles. Then the least total weight occurs when Lin has one 50 gram marble, two 40 g marbles, and five 20 g marbles. The smallest possible total weight of Lin's marbles is (1 50) + (2 40) + (5 20) = 230 g. 5A METHOD 1: Strategy: Start with a simpler case and look for a pattern. 2 posts result in a fence 1 6 = 6 m long, 3 posts result in a fence 2 6 = 12 m long, and so on. Then 10 posts result in a fence 9 6 or 54 m long. METHOD 2: Strategy: Draw a picture. There are ten posts and nine sections offence between them. The fence is 6 9 or 54 m long. 5B METHOD 1: Strategy: Examine the prime factors. The prime factors of 15,600 are Of three consecutive numbers, no more than one can be a multiple of 5. Since there are two 5s in the list of prime factors, one number is a multiple of 25. One of the numbers is a multiple of 13, which suggests 26. Rearrange the prime factors as (2 13) (5 5) ( ) = x 24 = 15,600. The least of the three numbers is 24. METHOD 2: Strategy: Use the fact that the three factors are nearly equal. Since 10 3 = 1000, 20 3 = 8000, and 30 3 = 27,000, the numbers are between 20 and 30. One of the numbers must be a multiple of 5, so try 25 3, which is 15,625. Therefore, one possible answer could be Since that gives us the correct product of 15,600, the least of the three numbers is 24. 5C METHOD 1: Strategy: Find the percent represented by the given amount. Since 60% of the students are female, then 40% are male. The number of females exceeds the number of males by (60 40) = 20% of the total number of students. Then 72 is 20%, or onefifth, of the total. The total number of students is 5 72 =360. Thus 40% of 360 = 144 male students are enrolled.

METHOD 2: Strategy: Try simpler related cases and look for a pattern.

Notice that in each case the number of males (column 3) is twice the difference (column 4). Since the actual difference is 72, the number of male students is 144. METHOD 3: Strategy: Use algebra.

Set up a proportion: x+72 = 60 2x+72 100 Now cross-multiply: 60(2x +72) = 100 (x +72) Multiply out on each side of the equation: 120x + 4320 = 100x + 7200 Subtract 100x + 4320 from each side of the

4 METHOD 2: Strategy: Try simpler related cases and look for a pattern. Assuming a total of 100, 200, and 300 students, this table shows in each case the subtraction of the number of males from the number of females. Notice that in each case the number of males (column 3) is twice the difference (column 4). Since the actual difference is 72, the number of male students is 144. METHOD 3: Strategy: Use algebra. Let x = the number of males. Then x + 72 = the number of females and 2x + 72 = the total number of students. Set up a proportion: x+72 = 60 2x Now cross-multiply: 60(2x +72) = 100 (x +72) Multiply out on each side of the equation: 120x = 100x Subtract 100x from each side of the equation: 20x = 2880 Divide each side of the equation by 20: x = 144 Thus, 144 male students are enrolled. 5D Strategy: Split the region into more familiar shapes. Draw two lines as shown to divide the square into four congruent smaller squares. METHOD 1: Strategy: Find the area of each shaded region separately. Square A, fully shaded, has an area of 16 sq m. In squares B and C, each shaded region is found by removing the quarter circle from the square. Each of those areas is then about 16 ( ), which then simplifies to 3.44 sq m. 4 Adding 16, 3.44, and 3.44 and then rounding off, the area of the shaded region is then 22.9 sq m. METHOD 2: Strategy: Find the area of the unshaded region.

METHOD 2A: The unshaded area is the sum of the areas of regions D, C, and B. The area of square D is 16 sq m. The area of one quarter-circle is 1 4 3.14 42 12.56 sq m.

5 METHOD 2A: The unshaded area is the sum of the areas of regions D, C, and B. The area of square D is 16 sq m. The area of one quarter-circle is sq m. The area of the unshaded region is about = sq m. The area of the shaded region is then about = 22.88, which rounds to 22.9 sq m. METHOD 2B: Think of the complete unshaded region as the sum of two semicircles of radius 4 meters minus the overlap. The overlap, divided into 2 parts by the dotted line in the diagram, consists of two regions congruent to the region considered in problem 4D. Its area was computed as approximately 4.56, so the area of the unshaded region is about less than 2 ( ). As above, the area of the shaded region is = about 22.9 sq m. 5E METHOD 1: Strategy: List all possible values for the tens digit. Consider the second condition first. The possible two-digit numbers are 13, 25, 37, and 49. The table compares the results of the first condition against those of the second condition. The twodigit number is 49. METHOD 2: Strategy: Use algebra. Let t = the tens digit and u = the units digit. Then 10t + u = the two-digit number and t + u = the sum of the digits. English: The number is equal to 10 more than 3 times the sum of the digits. Equation 1: 10t + u = 3(t + u) + 10 English: The units digit is 1 more than twice the tens digit. Equation 2: u = 2t + 1 Substituting 2t + 1 for u in both places in equation 1, we get: 10t +2t + 1 = 3(t +2t + 1) + 10 Combine like terms on each side of the equation: 12t + 1 = 3 (3t + 1) + 10 Multiply 3t + 1 by 3: 12t + 1 = 9t Add 3 and 10: 12t + 1 = 9t + 13 Subtract 1 from each side of the equation: 12t = 9t + 12 Subtract 9t from each side of the equation: 3t =12 Divide each side of the equation by 3: t = 4 Because u = 2t+1, u = 2(4) + 1, so that u = 9. The two-digit number is 49.

Need more help with decimal subtraction? See T23. Note: The inequality sign is reversed only when multiplying or dividing by a negative number.

Need more help with decimal subtraction? See T23. Note: The inequality sign is reversed only when multiplying or dividing by a negative number. . (D) According to the histogram, junior boys sleep an average of.5 hours on a daily basis and junior girls sleep an average of. hours. To find how many more hours the average junior boy sleeps than the