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Let s represent the son’s age, and f represent the father’s age. We know right now that s + f = 43. In n years, the son would be s + n and the father would be f + n. We also know that: 4(s+n) = f+n. Simplifying this gives 4s + 4n = f + n. Rearranging the first equation we can gain another equation: f = 43 - s. Now, we can substitute this into the simplified equation.
Isolate the s (because n isn’t what we are looking for). s = (43 - 3n)/5 For s, we are looking for values that would make 43 - 3n divisible by 5. Now it’s mainly just guess and check.
(10 minutes worth of guess and check that may or may not mainly have been me remembering the math test tomorrow and trying to find algebra rwas)
If n = 1, s would be 8. This would yield 35 as the father’s age. CHECK: In one year, they would be 9 and 36. 4 x 9 = 36. This is correct. If n = 6, s would be 5. This would yield 38 as the father’s age. CHECK: In 6 years, they would be 44 and 11. 4 x 11 = 44. This is correct. If n= 11, s would be 2. This would yield 41 as the father’s age. CHECK: In 11 years, the dad will be 52 and the son would be 13. 13 x 4 = 52. Wow, correct.
After a bit, I noticed that the son’s age drops by 3 and the father’s age rises by 3. After one year, they will be 8 and 35. Then it becomes 5 and 38. After 6 years, they will be 2 and 41. The son’s possible age drops by 3, the father’s rises by 3.
Therefore, they can be 8 and 35, 5 and 38, and 2 and 41.
Grade 7/8 POTW I will go straight into the math. Let f represent the father’s age and s represent the son’s age. The sum of both their ages is 43: f + s = 43. 43 - f = s. 43 - s = f. In n years, the father is four times the son’s age: 4(s + n) = f + n. 4(s + n) = 4s + 4n. We can now use substitution. 4s + 4n = 43 - s + n 4s + 4n + s - n = 43 - s + n + s - n 5s + 3n = 43 5s + 3n - 3n = 43 - 3n 5s = 43 - 3n 5s/5 = (43 - 3n)/5 s = (43 - 3n)/5 This means every time 43 - 3n = a multiple of 5, that is one solution. To do this, we first need to find one value of n. The easiest one to find is when n = 1, s = 8, and f = 35. From here, instead of just doing trial and error, we can simply just find every multiple of 5 (+ the additional 1) until the age can longer extend/reduce. 1. n = 1, s = 8, f = 35 s = (43 - 3n)/5 s = (43 - 3(1))/5 s = (43 - 3)/5 s = 40/5 s = 8 43 - 8 = 35 4(8 + 1) = 35 + 1 4(9) = 36 36 = 36 2. n = 6, s = 5, f = 38 s = (43 - 3n)/5 s = (43 - 3(6))/5 s = (43 - 18)/5 s = 25/5 s = 5 43 - 5 = 38 4(5 + 6) = 38 + 6 4(11) = 44 44 = 44 3. n = 11, s = 2, f = 41 s = (43 - 3n)/5 s = (43 - 3(11))/5 s = (43 - 33)/5 s = 10/5 s = 2 43 - 2 = 41 4(2 + 11) = 41 + 11 4(13) = 52 52 = 52 Therefore, there are 3 possible ages for the father and son including 35 and 8, 38 and 5, and 41 and 2 respectively.
Grade 7/8 POTW:
ReplyDeleteLet s represent the son’s age, and f represent the father’s age.
We know right now that s + f = 43. In n years, the son would be s + n and the father would be f + n.
We also know that: 4(s+n) = f+n. Simplifying this gives 4s + 4n = f + n.
Rearranging the first equation we can gain another equation: f = 43 - s. Now, we can substitute this into the simplified equation.
4s + 4n = (43 - s) + n
4s + 3n = 43 - s
5s + 3n = 43
Isolate the s (because n isn’t what we are looking for).
s = (43 - 3n)/5
For s, we are looking for values that would make 43 - 3n divisible by 5.
Now it’s mainly just guess and check.
(10 minutes worth of guess and check that may or may not mainly have been me remembering the math test tomorrow and trying to find algebra rwas)
If n = 1, s would be 8. This would yield 35 as the father’s age.
CHECK: In one year, they would be 9 and 36. 4 x 9 = 36. This is correct.
If n = 6, s would be 5. This would yield 38 as the father’s age.
CHECK: In 6 years, they would be 44 and 11. 4 x 11 = 44. This is correct.
If n= 11, s would be 2. This would yield 41 as the father’s age.
CHECK: In 11 years, the dad will be 52 and the son would be 13. 13 x 4 = 52. Wow, correct.
After a bit, I noticed that the son’s age drops by 3 and the father’s age rises by 3. After one year, they will be 8 and 35. Then it becomes 5 and 38. After 6 years, they will be 2 and 41. The son’s possible age drops by 3, the father’s rises by 3.
Therefore, they can be 8 and 35, 5 and 38, and 2 and 41.
Grade 7/8 POTW
ReplyDeleteI will go straight into the math. Let f represent the father’s age and s represent the son’s age. The sum of both their ages is 43: f + s = 43. 43 - f = s. 43 - s = f.
In n years, the father is four times the son’s age: 4(s + n) = f + n. 4(s + n) = 4s + 4n. We can now use substitution.
4s + 4n = 43 - s + n
4s + 4n + s - n = 43 - s + n + s - n
5s + 3n = 43
5s + 3n - 3n = 43 - 3n
5s = 43 - 3n
5s/5 = (43 - 3n)/5
s = (43 - 3n)/5
This means every time 43 - 3n = a multiple of 5, that is one solution. To do this, we first need to find one value of n. The easiest one to find is when n = 1, s = 8, and f = 35. From here, instead of just doing trial and error, we can simply just find every multiple of 5 (+ the additional 1) until the age can longer extend/reduce.
1. n = 1, s = 8, f = 35
s = (43 - 3n)/5
s = (43 - 3(1))/5
s = (43 - 3)/5
s = 40/5
s = 8
43 - 8 = 35
4(8 + 1) = 35 + 1
4(9) = 36
36 = 36
2. n = 6, s = 5, f = 38
s = (43 - 3n)/5
s = (43 - 3(6))/5
s = (43 - 18)/5
s = 25/5
s = 5
43 - 5 = 38
4(5 + 6) = 38 + 6
4(11) = 44
44 = 44
3. n = 11, s = 2, f = 41
s = (43 - 3n)/5
s = (43 - 3(11))/5
s = (43 - 33)/5
s = 10/5
s = 2
43 - 2 = 41
4(2 + 11) = 41 + 11
4(13) = 52
52 = 52
Therefore, there are 3 possible ages for the father and son including 35 and 8, 38 and 5, and 41 and 2 respectively.
POTW:
ReplyDeleteMy POTW was done on paper, possible ages include:
41 and 2
35 and 8
38 and 5