How many applicants need to apply in order to meet the hiring target?

Question

Cam needs to hire $30$ new employees. Ten percent $(10\%)$ of applicants do not meet the basic business requirements for the job, $12\%$ of the remaining applicants do not pass the pre-screening assessment, $23\%$ of those remaining applicants do not show up for the interview, and $5\%$ of those remaining applicants fail the background investigation. How many applicants need to apply in order to meet the hiring target?

$$A)\ 30\ \ \ \ \ \ B)\ 45\ \ \ \ \ \ C)\ 50\ \ \ \ \ \ D)\ 52\ \ \ \ \ \ E)\ 60$$

My answer:

I added $10+12+23+5=50$

That gave us $50\%$. I took a look at the answers and getting $50\%$ of $E)\ 60$ is $30.$

However, when I tried to solve it, I took $(0.50)(30)=15$ I then added $15$ to $30$ and it gave me $B)$ $45.$ Can someone please show me how to solve this?

Answer 1

You want $0.95 \cdot 0.77 \cdot 0.88 \cdot 0.9 \cdot N = 30$.

The first cut (not meeting basic requirements) only lets through $100 - 10 = 90$ percent of $N$, where $N$ is the total number of applicants.

The next cut (prescreening) only lets through $100 - 12 = 88$ percent of who's left.

And so forth.

So $N \approx 52$.

Answer 2

To elaborate on John's answer: You cannot simply add $10$, $12$, $23$, and $5$ percent, because the percentages are "in tandem". After you cut away the first $10$ percent, you only cut away $12$ percent of those who remain, not of those who started the selection process.

As a simpler example, consider the following selection process: You first eliminate those who are born in an odd-numbered month, and then you eliminate those who are born on an odd-numbered day. If we assume (for the sake of argument) that each step removes $50$ percent of the remaining candidates, it still is not true that at the end of the two steps, you have eliminated $50+50 = 100$ percent of the candidates.

Instead, what happens is that after the first step, $50$ percent have been eliminated, and $50$ percent remain. The second step then eliminates $50$ percent of the remaining $50$ percent, or $25$ percent. That leaves only the last $25$ percent.

In the same way, in your actual problem, the first step eliminates $10$ percent, leaving $90$ percent. The second step eliminates $12$ percent of that $90$ percent, leaving $88$ percent of $90$ percent, or $79.2$ percent. And so on.

Answer 3

Assuming $N$ as the number of all the applicants, if we consider four groups as the fail groups, named $A,\ B,\ C,$ and $D$ as you mentioned in the question, respectively. According to your given data, the percentages of these groups are as follows.

$$A=\frac{1}{10}\times N=0.1\times N$$ $$B=\frac{12}{100}\times(1-0.1)\times N=0.108\times N$$ $$C=\frac{23}{100}\times(1-(0.1+0.108))\times N\approx 0.205\times N$$ $$D\approx\frac{5}{100}\times(1-(0.1+0.108+0.205))\times N\approx.029\times N$$

So, the total number of failed applicants is $(0.1+0.108+0.205+0.029)N=0.442N$, and hence we have $(1-0.442)\times N\approx30$ which leads to $N\approx 53$, and so according to the options, it is concluded that $\color{red}{N=52}$.

Answer 4

If you have studied successive discounts,
the question is exactly equivalent to finding the whole dollar marked price of something
that is sold for $\$30$ at successive discounts of $10\%, 12\%, 23\%\; and\; 5\%$

$N(1-0.1)(1-0.12)(1-0.23)(1-0.05) = 30\; \Rightarrow N = 52$