In a town, 7/9 th of the men and 3/5th of the women are married. IN that town, what fraction of the adults are married?Assume that all married adults are the residents of the town.
the answer in the back of the book says it's 21/31, why is that?I couldn't get it. Can anyone help me out with this?
2007-02-05 21:32:08 · 4 answers · asked by quz_us 1 in Science & Mathematics ➔ Mathematics
Suppose you have 42 married adults. Then you have 21 men and 21 women to which correspond 6 unmarried men and 14 unmarried women. So for 42 married adulurs you have 20 un married adults, total 62. 42/62 = 21 / 31.
2007-02-05 21:39:28 · answer #1 · answered by gianlino 7 · 0⤊ 0⤋
You know the number of married men equals the number of married women (at least, in most states). So 7/9 the number of men = 3/5 the number of women. In this case you want to find a common numerator. 21/27 men and 21/35 women. So if the town has 27x men and 35x women, the population is 62x (we can drop the x's, they just cancel out). So 42/62 of the people are married, = 21/31
2007-02-06 05:44:11 · answer #2 · answered by sofarsogood 5 · 3⤊ 0⤋
7x/9x = married men to total men
3y/5y = married women to total women
9x + 5y = Total number of adults
7x + 3y = Total number of married adults
so the ratio we want is:
(7x + 3y)/(9x + 5y)
BUT we also know that it must be the case that 7x = 3y
14x/(9x + 5(7/3)x)
14/(62/3)
21/31
2007-02-06 06:07:11 · answer #3 · answered by cp_exit_105 4 · 0⤊ 0⤋
My guess would be they included those who have registered their marriage and the records are out dated.
2007-02-06 05:42:48 · answer #4 · answered by Conway 4 · 0⤊ 2⤋