有这样一题请教大家

grace1225

in a survey of parents, exactly 3/4 of the mothers and 9/10 of the farthers held full-time jobs. if 40%of the parents surveyed were women, what percent of the parents did not hold full-time jobs?

A27% B21%  C19%  D18%  E16%

tiger1234

parents = mothers + fathers, 3/4 * m and 9/10 * f held full-time jobs, if m = .4 * parents = .4 * (m + p), then what percent of the parents did not hld full-time jobs?

since we want to know what percent of parents did not hold full-time jobs, and we know that 3/4 * m and 9/10 * f held full-time jobs, then we know that 1/4 * m and 1/10 * f did not hold full-time jobs, since m = 40% of the parents = .4 * ( mothers + fathers) =  .4m + .4 f  

=> m = .4m + .4f  => .6m = .4f  => f = 3/2 m = 3/2 * .4 parents = .6*parents, thus, the answer is: 1/4 * .4 + 1/10 * .6 = .1 + .06 = .16 = 16%

[此贴子已经被作者于2004-9-13 11:00:37编辑过]

miaoyin_tx

我做这种题，用的是比较笨的办法，画个表格，然后假定一个总量，再根据条件填空就行了。

	Mother	Father	Sub-total
Full-time job	Step 3 ¾ x 40 = 30	Step 6 9/10 x 60 = 54
No F/T Job	Step 4 40-30=10	Step 7 60-54=6	? Final step 10+6 /100 = .16 Answer is 16%
Sub-total	Step 2: 40% x100=40	Step 5: 100-40=60	Step 1: Total 100

[此贴子已经被作者于2004-9-13 11:40:57编辑过]

grace1225

哇　！

好清楚。

谢谢￣￣￣```````