How to solve Algebra age problems?
We are having a quiz on them and I cant solve them, I am horrible! Here is an example (on old HW) I need to pass this quiz if i want to pass the course!
Please go step by step!
For six consecutive years, a man's age was a multiple of his grand-daughters age. How old was each during the sixth year?
AND
Al is 3/4 as old as Ann. Six years ago Al was 1/2 as old as Ann. How old is each now?
Thanks for solving these and putting them out step by step!
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The man's age during the 6 years was
n, n+1, n+2, n+3, n+4, n+5
and the girls age was
m, m+1, m+2, m+3, m+4, m+5
with each n being a multiple of the corresponding m.
We want to avoid prime numbers:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101,
And grandpa is probably at least 40, so we trim the list to
41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101
Gaps of at least 5 are
61 to 67, 73 to 79, 83 to 89, 89 to 97.
None of these have more than 5 consecutive composite numbers,
so the girl's initial age is 1, and we need to find consecutive multiples of 2,3, 4, 5, 6
61, 62, 63, 64, 65, 66 works as those are multiples of
1, 2, 3, 4, 5, 6 respectively.
Let A = Al's age now,
let N = Ann's age now
A = 3/4 * N (now)
(A - 6) = (1/2) (N - 6) (six years ago)
Solve by substitution:
3N/4 - 6 = N/2 - 3
1/4 N = 3
N = 12
A = 9
Six years ago they were 6 and 3.