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At a certain factory, twice as many men as women apply for work. If 5% of the people who apply are hired and 3% of the men who apply are hired, what percent of the women who apply are hired?

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    $\begingroup$ What have you tried? $\endgroup$ – David G. Stork May 16 at 18:53
  • $\begingroup$ I don't understand the question... I have been trying to solve the Word problems for days but only few i did, there are 35 WPs i need to solve. My word problem solving skill is not good. $\endgroup$ – Askani May 16 at 18:57
  • $\begingroup$ Why you need from us to solve this problem? $\endgroup$ – MarianD May 16 at 18:58
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    $\begingroup$ Try to write equations. For example, number of men applying is $m$, number of women applying is $w$, then $m=2w$. Continue with the other quantities $\endgroup$ – Andrei May 16 at 19:05
  • $\begingroup$ Start with this: if $x$ people applied for work, how many were hired? Can you write an expression for this number? $\endgroup$ – Vasya May 16 at 19:08
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Two thirds are men of which $3\%$ apply and hired, that makes $2\%$ of all. As $5\%$ are hired this leaves $3\%$ of all which are women and hired. Hence the percentage of women getting hired is $\frac{3\%}{1/3}=9\%$.

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  • $\begingroup$ U did it directly. I solved it through Equation: 0.03m + x(m/2) = 0.05(m+m/2) And x = 9%. $\endgroup$ – Askani May 17 at 18:32
  • $\begingroup$ Right: in this case Algebra isn't needed. $\endgroup$ – Michael Hoppe May 18 at 8:11
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$$300\%\cdot5\%=15\%\\15\%-6\%=9\%$$ 9% of women are hired. All I did was convert total population into women popultation equivalents. Then did the math in those.

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Work through these:

  • Total number applied: $x$
  • Number of men applied: $m(x)$
  • Number of women applied: $w(x)$
  • Total number hired: $h(x)$
  • Total number of women hired: $h_w(x)$
  • Total number of men hired: $h_m(x)$

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  • $\begingroup$ m = number of men apply for work. m/2 = number of women apply for work. 5℅ = people hired 3% = men are hired 2% = women are hired Equation: 0.03m + 0.02(m/2) = 0.05 ......Am i correct? $\endgroup$ – Askani May 16 at 21:01
  • $\begingroup$ So far yes... but you have a few more steps.... $\endgroup$ – David G. Stork May 16 at 21:21
  • $\begingroup$ I did it -->> Equation: 0.03m + x(m/2) = 0.05(m+m/2) And n = 9%. Thank you 🙂. It is hard for me to understand the Question. $\endgroup$ – Askani May 17 at 17:45
  • $\begingroup$ @Askani: You might find that drawing a picture helps. (See my amended solution.) $\endgroup$ – David G. Stork May 17 at 21:42

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