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16 Sep 2014, 00:59
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A set consists of 19 elements with an average of \(a\). If addition of a new element increases the average by \(k\%\), what is the value of the new element?
A. \(a(1 + \frac{k}{5})\)
B. \(a*\frac{k}{100} - 20a\)
C. \(20a(1 + \frac{k}{100})\)
D. \(20(1 + \frac{k}{100}) - 19a\)
E. \(a*\frac{k}{5} - 19a\)
A. \(a(1 + \frac{k}{5})\)
B. \(a*\frac{k}{100} - 20a\)
C. \(20a(1 + \frac{k}{100})\)
D. \(20(1 + \frac{k}{100}) - 19a\)
E. \(a*\frac{k}{5} - 19a\)
16 Sep 2014, 00:59
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Official Solution:
A set consists of 19 elements with an average of \(a\). If addition of a new element increases the average by \(k\%\), what is the value of the new element?
A. \(a(1 + \frac{k}{5})\)
B. \(a*\frac{k}{100} - 20a\)
C. \(20a(1 + \frac{k}{100})\)
D. \(20(1 + \frac{k}{100}) - 19a\)
E. \(a*\frac{k}{5} - 19a\)
Let \(x\) represent the value of the new element.
The new average is given by \(\frac{\text{new sum} }{20} = \frac{\text{old sum} + x}{20} = \frac{19a + x}{20}\).
Given that the average increased by \(k\%\), we'd get \(\frac{19a + x}{20} = (1 + \frac{k}{100})*a\).
\(19a + x = a*(20 + \frac{k}{5})\).
\(x = a*(20 + \frac{k}{5})-19a\).
\(x = a * (1 + \frac{k}{5})\).
Answer: A
A set consists of 19 elements with an average of \(a\). If addition of a new element increases the average by \(k\%\), what is the value of the new element?
A. \(a(1 + \frac{k}{5})\)
B. \(a*\frac{k}{100} - 20a\)
C. \(20a(1 + \frac{k}{100})\)
D. \(20(1 + \frac{k}{100}) - 19a\)
E. \(a*\frac{k}{5} - 19a\)
Let \(x\) represent the value of the new element.
The new average is given by \(\frac{\text{new sum} }{20} = \frac{\text{old sum} + x}{20} = \frac{19a + x}{20}\).
Given that the average increased by \(k\%\), we'd get \(\frac{19a + x}{20} = (1 + \frac{k}{100})*a\).
\(19a + x = a*(20 + \frac{k}{5})\).
\(x = a*(20 + \frac{k}{5})-19a\).
\(x = a * (1 + \frac{k}{5})\).
Answer: A
07 Oct 2023, 13:45
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
