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Washington State Math Olympiad
Hints and Solutions
2011 Grade 5 Algebra
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| Problem |
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1) On January 1st Linda has $14 in savings and John has $62. Linda saves $4.50 each month. John notices he spends $3.50 out of savings every month to build his tournament game card collection. On the first of which month will they have the same amount if this pattern continues?
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| Solution(s) |
There are 3 ways to solve this problem:
Method 1:Write an equation:
1. Let N = number of the month past January that they will have the same amount.
2. Linda's side of the equation must equal Johns, so:
3. Linda's expression:
$14 + $4.50 x N
4. John's expression:
$62 - $3.50 x N
5. Total equation:
14 + 4.50 x N = 62 - 3.50 x N
8 N = 62 - 14
8 N = 48
N = 6 Month = July
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Method 2:By deductive logic:
1. Every month the difference between Linda's and John's savings decreases by
($4.50 + $3.50) = $8 .
2. The difference between their January savings is
$62 - $14 = $48
3. Divide this difference by the decrease and you have the number of months until they are equal =
$48/8 = 6 (months past January)
4. That month is July.
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Method 3:Make a table:
Complete this table to find the month their savings accounts equal:
| Month | Linda's
acct. | John's
acct |
| January | $14 | $62 |
| February | $18.50 | $58.50 |
| March | $23.00 | $55.00 |
| April | $27.50 | $52.50 |
| May | $32.00 | $49.00 |
| June | $36.50 | $45.50 |
| July | $41.00 | $41.00 |
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| Problem |
Solution |
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2) Jim wants to swim 36 yards out into the ocean from the shore. He swims out 6 yards in 4 seconds but then in one second a wave pushes him back 3 yards. If this cycle continues, how long will it take Jim to get 36 yards out into the ocean from the shore for the first time?
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Method 1: Make a table:
| Cycle | time
(sec) | Start
(yards) | Swim
out
(yards) | Push
back
(yards) |
| 1 | 0 | 0 | 6 | 3 |
| 2 | 5 | 3 | 9 | 6 |
| 3 | 10 | 6 | 12 | 9 |
| 4 | 15 | 9 | 15 | 12 |
| 5 | 20 | 12 | 18 | 15 |
| 6 | 25 | 15 | 21 | 18 |
| 7 | 30 | 18 | 24 | 21 |
| 8 | 35 | 21 | 27 | 24 |
| 9 | 40 | 24 | 30 | 27 |
| 10 | 45 | 27 | 33 | 30 |
| 11 | 50 | 30 | 36 | 33 |
| 12 | 55 | 33 | 39 | 36 |
On the 11th cycle, Jim reaches 36 yards on the swim out. That cycle starts at 50 seconds and the swim out to 36 yards takes 4 seconds, so Jim reaches 36 yards at 54 seconds.
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Method 2: Use analysis:
1. Compute how far Jim progresses after a swim out and then a push back = 3 yards.
2. Divide the total distance, 36 yards, by this amount = 12 cycles
3. This means that after 12 full cycles he has progressed 36 yards.
4. On the previous cycle, number 11, he went out 36 before he got pushed back 3, so he achieved 36 yards after 11 cycles.
5. It takes 11 x 5 = 55 seconds for cycle 11, but he didn't get pushed back, so the total time is 55 - 1 = 54 seconds
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| Problem |
Solution |
3) In the pattern of figures shown below
how many unit squares are needed to build the 8th figure?
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OK, this is tricky because there are actually 2 sequences that are added together:
1. The top horizontal top row (black squares)
2. The blocks below the top row (red squares)
There are 2 ways to analyze this:
Method 1:Make a table
Here they are in a table
(N is the figure number):
| N | Top row | bottom
blocks | total |
| 1 | 4 | 2 = 1x2 | 6 |
| 2 | 7 | 6 = 2x3 | 13 |
| 3 | 10 | 12= 3x4 | 22 |
| 4 | 13 | 20= 4x5 | 33 |
| 5 | 16 | 30= 5x6 | 46 |
| 6 | 19 | 42= 6x7 | 61 |
| 7 | 22 | 56= 7x8 | 78 |
| 8 | 25 | 72= 8x9 | 97 |
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Method 2: Find the equations:
1. Top row (A): Every figure adds 3 blocks so the top row equation is:
An = A1 + 3 (n - 1)
A8 = 4 + 3 x 7 = 25
2. Bottom blocks (B): Every figure contains
N x (N - 1) blocks, so :
B8 = 8 x 9 = 72 blocks.
3. Add these 2 together:
A8 + B8 =
25 + 72 = 97 blocks
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| Problem |
Solution |
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4) Each week, Roxy mows Mrs. Smith's lawn and does some extra yard work. Mrs. Smith pays Roxy a fixed amount for mowing her lawn. He earns an additional hourly amount for extra work. Based on the table below, how much is the hourly amount for the extra work?
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You only need the first 2 columns to solve this problem.
1. The increase from the first week to the second week is $45 - $27 = $18
2. The number of additional hours for extra work in the second week over the first week was 3 hours.
3. That increase divided by the number of additional hours gives you the extra hourly rate = $18/3 = $6/hour for extra work.
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5) In the letter puzzle given, each letter stands for a different number from the list 1, 2, 3, 4, 5, 6. The value of O is less than the value of W. What number is WOW?
NET
+TEN
WOW
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- Only 5 letters are in the sum (E,T,N,O,W), so one of the digits 1-6 is not there.
- The second column is E + E so E must be 3 or less or their sum would add to more than 6. So E is 1, 2, or 3.
- If E = 3, then O is 6 which cannot be because it is less than W, so E is either 1 or 2.
- Columns 1 and 3 both add T + N , so they, too, must add to 6 or less.
- Start with the possibilities for E and see if values for O, T and N are possible:
- E = 1; If E is 1 then O is 2 and there are no possibilities for T or N because they would add to more than 6.
- E = 2; If E is 2 then O is 4 and the only possibility for T and N are: 1 & 5
so, W = 6 and O = 4, making the problem:
125
+521
646 or
521
+125
646,
WOW = 646
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