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A recent survey examined the working arrangements of households of mar
[#permalink]
19 Jan 2022, 07:15
19 Jan 2022, 07:15
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A recent survey examined the working arrangements of households of married couples with no children. It was found that 82% of the households had at least one working member. In 30% of the household with the women not working, the man also does not work. In 60% of the households in which the man doesn't work, the woman also does not work. What is the probability that both the man and the woman work?
A. 13%
B. 22%
C. 28%
D. 32%
E. 34%
A. 13%
B. 22%
C. 28%
D. 32%
E. 34%
Re: A recent survey examined the working arrangements of households of mar
[#permalink]
19 Jan 2022, 23:30
19 Jan 2022, 23:30
1
Kudos
Let the total number of households be 100, the number of households in which women don't work be 'x' and the number of households in which the men don't work be 'y'. Also, let the number of households where only women work be 'w', where only men work be 'm' and where both women and men work be 'mw'.
(1) The number of households where neither the man nor the woman work is 100 - 82 = 18.
(2) "In 30% of the households with the woman not working the man also doesn't work". So:
3x/10=18...> x=60. T/4, the number of households in which women work is 40.
(3) "In 60% of the households in which the man doesn't work, the woman also doesn't work". So:
3y/5=18...> y=30. T/4, the number of households in which men work is 60.
(m+mw)+(w+mw)=70+40=110
m+w+mw=82
mw=110-82=28
ANS: C
(1) The number of households where neither the man nor the woman work is 100 - 82 = 18.
(2) "In 30% of the households with the woman not working the man also doesn't work". So:
3x/10=18...> x=60. T/4, the number of households in which women work is 40.
(3) "In 60% of the households in which the man doesn't work, the woman also doesn't work". So:
3y/5=18...> y=30. T/4, the number of households in which men work is 60.
(m+mw)+(w+mw)=70+40=110
m+w+mw=82
mw=110-82=28
ANS: C
Re: A recent survey examined the working arrangements of households of mar
[#permalink]
19 Jan 2022, 23:30
19 Jan 2022, 23:30
1
Kudos
Let the total number of households be 100, the number of households in which women don't work be 'x' and the number of households in which the men don't work be 'y'. Also, let the number of households where only women work be 'w', where only men work be 'm' and where both women and men work be 'mw'.
(1) The number of households where neither the man nor the woman work is 100 - 82 = 18.
(2) "In 30% of the households with the woman not working the man also doesn't work". So:
3x/10=18...> x=60. T/4, the number of households in which women work is 40.
(3) "In 60% of the households in which the man doesn't work, the woman also doesn't work". So:
3y/5=18...> y=30. T/4, the number of households in which men work is 60.
(m+mw)+(w+mw)=70+40=110
m+w+mw=82
mw=110-82=28
ANS: C
(1) The number of households where neither the man nor the woman work is 100 - 82 = 18.
(2) "In 30% of the households with the woman not working the man also doesn't work". So:
3x/10=18...> x=60. T/4, the number of households in which women work is 40.
(3) "In 60% of the households in which the man doesn't work, the woman also doesn't work". So:
3y/5=18...> y=30. T/4, the number of households in which men work is 60.
(m+mw)+(w+mw)=70+40=110
m+w+mw=82
mw=110-82=28
ANS: C
