Division J Sample Answers
A. - 300
METHOD 1: Make a simpler problem...
- Notice that each number being added is 100 more than one of the numbers being subtracted.
The value is 100 + 100 + 100 = 300
METHOD 2: Group by operation...
- Add the numbers 268 + 1375 + 6179 = 7822.
Then add the numbers 168 + 1275 + 6079 = 7522.
Finally, subtract the totals: 7822 - 7522 = 300
B. - 29 marbles
Draw a picture...
Draw 8 boxes. Then put the smallest possible number of marbles in each box. Put 1 marble in box 1. Then put 1 marble in box 2. You can't just put 1 marble in box 3, becuase that would make three boxes with the same number of marbles. So put 2 marbles in box 3, 3 marbles in box 4, and so on.
The smallest total number of marbles is 1+ 1 + 2 + 3 + 4 + 5 + 6 + 7 = 29 marbles.
C. - 45
Proceed one statement at a time. Eliminate those numbers which fail to satisfy all the conditions.
WHOLE NUMBERS THAT SATISFY ALL CONDITIONS
- Less than 100
1, 2, 3, ..., 99 - Multiple of 3
3, 6, 9, ..., 99 - Also multiple of 5
15, 30, 45, 75, 90 - Odd
15, 45, 75 - Sum of digits is odd
45
D. - 79
METHOD 1: Start with a simpler problem...
- (a) Lengths formed by 1 cm blocks: 1, 2, 3, 4.
(b) Lengths formed by remaining blocks: 5, 10, 15; 25, 30, 35, 40; 50, 55, 60, 65; 75, 80, 85, 90.
(c) Each of the fifteen (b) length bocks can be combined with the four (a) lengths, thus producing 15 x 4 = 60 different amounts.
TOTAL AMOUNTS:
(a) 4
(b) 15
(c) 60
GRAND TOTAL:
79 different amounts
METHOD 2
- Number of choices for 1cm lengths, including 0, is 5: (0, 1, 2, 3, 4).
Number of choices for 5cm lengths, including 0, is 4: (0, 1, 2, 3).
Number of chioces for 25cm lengths, including 0, is 4: (0, 1, 2, 3).
Total number of choices for all lengths is 5 x 4 x 4 = 80. However, 80 includes the choice of having none of the lengths as a choice. Since it is given that each length must be 1cm or longer, there are 80 - 1 = 79 amounts of at least 1cm.
METHOD 3: Establish a maximum and then eliminate all impossibilities.
- Find that largest possible length that can be made with the blocks, and then subtract the number of smaller values that cannot be made.
The maximum that can be made is 4 + 15 + 75 = 94cm. The 15 lengths less than 94cm that cannot be made are those that require four 5cm blocks. These are 20, 21, 22, 23, 24, 45, 46, 47, 48, 49, 70, 71, 72, 73 and 74cm lengths. The number of possible lengths is 94 - 15 = 79
E. - 180 cm²
Find the length of one side of the figure...
Because of the common dies, all the squares are congruent to each other. The perimeter consists of 12 equal sides. The length of a side is 72 / 12 = 6cm. The area of each square is 6 x 6 = 36cm².
The area of the figure is 5 x 36 = 180cm².
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