edited by
675 views
1 votes
1 votes

 

In an office, $30\%$ of the employees were women and $70\%$ of the employees were above the age of $40$ years, out of which $60\%$ are men. Find the percentage of women employees who are above $40$ years out of the total number of women employees.

  1. $96\%$
  2. $93.33\%$
  3. $70.44\%$
  4. $80.66\%$
edited by

2 Answers

0 votes
0 votes

30% female    70% male

above age 40   70% persons out of which 60% (42% of total)  are male

                          so rest 40% (28% of total) are female are above age 40

desired ratio 28%:30%

=28/30=93.33 option B

0 votes
0 votes
Given that, in an office, $30\%$of the employees were women, so the rest  $70\%$ of the employees in an office is men.

$70\%$ of the employees were above the age of $40$ years, out of which $60\%$ are men.

$\therefore$ The men above the age of $40$ years $ = \dfrac{70}{100} \times \dfrac{60}{100} = 42\%.$

So, the $70\%$ of the employees were above the age of $40$ years, out of which $40\%$ are women.

$\therefore$ The women above the age of $40$ years $ = \dfrac{70}{100} \times \dfrac{40}{100} = 28\%.$

Required percentage $ = \dfrac{28\%}{30\%} \times 100 \% = \dfrac{28}{30} \times 100\% = 93.33\%.$

So, the correct answer is $(B).$
Answer:

Related questions

3 votes
3 votes
4 answers
1
2 votes
2 votes
2 answers
2
gatecse asked Dec 9, 2020
730 views
Number of letter repeated in the given word $’MEASUREMENTS’$ are indicated in front of each alternative. Identify the correct alternative.$M_2E_2A_2S_2U_1R_1N_1T_1$$M...
2 votes
2 votes
1 answer
3
gatecse asked Dec 9, 2020
712 views
If $09/12/2001(DD/MM/YYYY)$ happens to be Sunday, then $09/12/1971$ would have been a:WednesdayTuesdaySaturdayThursday
2 votes
2 votes
2 answers
4
gatecse asked Dec 9, 2020
778 views
If a cube with length, height and width equal to $10\; cm$, is reduced to a smaller cube of height, length and width of $9\; cm$ then reduction in volume is :$172\;cm^3$$...