 |
|
|
|
|
Solutions for Session 6, Part C
See solutions for Problems: C2 | C3 | C4 | C5 | C6 | C7 | C8 | C9
 |
Problem C2 | |
a. | Let n = the number of blocks in each bag. Then the equation is: 3n + 2 = 2n + 7 because the two sides of the scale are equivalent in weight. |
b. | Remove two bags from each side. Then, remove 2 blocks from each side. There will now be 1 bag on the left and 5 blocks on the right, so it must be that there are 5 blocks in each bag. |
c. | Subtract 2n from each side (remove 2 bags from each side) to yield n + 2 = 7. Then, subtract 2 from each side (remove 2 blocks from each side) to yield n = 5. This means that each bag contains 5 blocks. |
<< back to Problem C2
|
|
| |
|
Problem C3 | |
|
One such problem has 3 bags and 2 blocks on the left, and 1 bag and 7 blocks on the right. Solving yields that each bag should have 2 1/2 blocks. One way to modify the model is to have the solution represent the relative weight of the bag rather than the number of blocks inside it.
<< back to Problem C3
|
|
|
| |
|
Problem C4 | |
|
The solution is h = 5. In the balance puzzle, this means that there would be 5 blocks in each bag, since h was the variable assigned to the unknown quantity.
<< back to Problem C4
|
|
|
| |
|
Problem C5 | |
|
There is no solution! One explanation is that no matter what value b represents, the left side of the equation will be larger than the right. Looking at the balance puzzle, there are always 5 more blocks on the left side than on the right. This is most similar to equation (f) of Problem A1, which had no solution.
<< back to Problem C5
|
|
|
| |
|
Problem C6 | |
|
Every number is a solution! No matter what value b represents, the left side and right side are equal. In the balance puzzle, no matter how many blocks are in each bag, the balance is maintained. Note that this means there is not enough information to find the value of b. This is most similar to equation (e) of Problem A1, which had an infinite number of solutions.
<< back to Problem C6
|
|
|
| |
|
Problem C7 | |
|
It can be done, but it requires some changes. One way is to represent "subtracting 2" by adding 2 blocks to the opposite side. A better way is to use items that would reduce the weight of a side, such as "negative blocks" or helium balloons. Each balloon cancels the weight of a block, so adding a block to a side would be identical to removing a balloon, and vice versa. Note that in this problem, "x" represents the number of blocks in each bag.
<< back to Problem C7
|
|
|
| |
|
Problem C8 | |
|
Add 3 to each side to yield 4x + 1 = 5x. Then, subtract 4x from each side to yield
1 = x, the solution.
<< back to Problem C8
|
|
|
| |
|
Problem C9 | |
|
Consider an equation like 3x + 5 = 23. Solving this by moving 5 to the right and switching the sign gives 3x = 23 + (-5), or 3x = 18. Solving by doing the same thing to both sides would require us to subtract 5 from both sides:
3x + 5 = 23
- 5 - 5
------------
3x = 18
Note that this is identical in result to the solution by moving, but it also explains the origin of the -5.
<< back to Problem C9
|
|
|
