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12 Nov 2024, 03:49
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p: 25
q: 20
At the beginning of 2007, a company had \(x\) employees. To handle an increasing workload, it recruited \(p\%\) additional employees during the year. However, in 2008, due to the financial crisis, the company was compelled to lay off \(q\%\) of its workforce. As a result, the total number of employees returned to \(x\).
Select for \(p\) and for \(q\) values that are jointly consistent with the information provided. Make only two selections, one in each column.
Select for \(p\) and for \(q\) values that are jointly consistent with the information provided. Make only two selections, one in each column.
| p | q | |
| 10 | ||
| 15 | ||
| 20 | ||
| 25 | ||
| 50 | ||
| 75 |
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Official Answer
p: 25
q: 20
12 Nov 2024, 03:49
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Official Solution:
The number of employees after \(p\%\) increase and \(q\%\) decrease would be \(x(1 + \frac{p}{100})(1 - \frac{q}{100})\). Since at the end the company had the same number of employees, we get:
\(x(1 + \frac{p}{100})(1 - \frac{q}{100})=x\)
\((1 + \frac{p}{100})(1 - \frac{q}{100})=1\)
\((\frac{100+p}{100})(\frac{100-q}{100})=1\)
\((100+p)(100-q)=100*100\)
Now, we need to find the values of \(p\) and \(q\) that satisfy this equation. Notice that the right-hand side, \(100 * 100 = 10^4\), has only 2 and 5 as its prime factors. Thus, the left-hand side must also have only these primes. Substituting different values of \(p\) from the options for \(100 + p\), we get: 110, 115, 120, 125, 150, and 175. Only 125 does not have primes other than 2 and 5. Hence, \(p\) must be 25. Increasing something by \(25\%\) needs to be decreased by \(20\%\) to bring it back to the original value, making \(q\) equal to 20.
Correct answer:
p "25"
q "20"
Bunuel
The number of employees after \(p\%\) increase and \(q\%\) decrease would be \(x(1 + \frac{p}{100})(1 - \frac{q}{100})\). Since at the end the company had the same number of employees, we get:
\(x(1 + \frac{p}{100})(1 - \frac{q}{100})=x\)
\((1 + \frac{p}{100})(1 - \frac{q}{100})=1\)
\((\frac{100+p}{100})(\frac{100-q}{100})=1\)
\((100+p)(100-q)=100*100\)
Now, we need to find the values of \(p\) and \(q\) that satisfy this equation. Notice that the right-hand side, \(100 * 100 = 10^4\), has only 2 and 5 as its prime factors. Thus, the left-hand side must also have only these primes. Substituting different values of \(p\) from the options for \(100 + p\), we get: 110, 115, 120, 125, 150, and 175. Only 125 does not have primes other than 2 and 5. Hence, \(p\) must be 25. Increasing something by \(25\%\) needs to be decreased by \(20\%\) to bring it back to the original value, making \(q\) equal to 20.
Correct answer:
p "25"
q "20"
21 Jan 2025, 01:18
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I have revised the question, solution, and formatting by adding more details to enhance clarity.
