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08 Jul 2011, 14:44
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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. \(a*\frac{k}{5} - 19a\)
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07 Oct 2023, 13:47
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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. \(a*\frac{k}{5} - 19a\)
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. \(a(1 + \frac{k}{5})\)
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, 23:09
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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?
Testing values can also help. A little longer way to get to the answer, however, it needs to be verified by checking conditions also.
Let's say a = 5 and k = 5
Initial sum = 19a = 85
New sum = 85 + 5(1+5%)
A. \(a(1 + \frac{k}{5})\)
\(5(1 + \frac{5}{5}) = 10\)
Can be the answer. Keep it
B. \(a*\frac{k}{100} - 20a\)
\(5*\frac{5}{100} - 20*5 = -99.75\)
Negative value, eliminate
C. \(20a(1 + \frac{k}{100})\)
\(20*5(1 + \frac{5}{100}) = 105\)
Can keep it but this needs to be verified since value seems to be too high considering the initial value of sum is only 85.
Two ways we can verify:
1. New sum we already know from above so this is not the right answer.
2. Also, if 105 is the value of new element then
total sum becomes 190 which gives us average of 9.5.
9.5 is more than 5% the jump we are looking for. The jump now is 90%.
So, this can't be answer.
D. \(20(1 + \frac{k}{100}) - 19a\)
\(20(1 + \frac{5}{100}) - 19*5 = -64 \)
Negative eliminate.
E. \(a*\frac{k}{5} - 19a\)
\(5*\frac{5}{5} - 19*5 = -80\)
Negative, eliminate.
Answer A.
Testing values can also help. A little longer way to get to the answer, however, it needs to be verified by checking conditions also.
Let's say a = 5 and k = 5
Initial sum = 19a = 85
New sum = 85 + 5(1+5%)
A. \(a(1 + \frac{k}{5})\)
\(5(1 + \frac{5}{5}) = 10\)
Can be the answer. Keep it
B. \(a*\frac{k}{100} - 20a\)
\(5*\frac{5}{100} - 20*5 = -99.75\)
Negative value, eliminate
C. \(20a(1 + \frac{k}{100})\)
\(20*5(1 + \frac{5}{100}) = 105\)
Can keep it but this needs to be verified since value seems to be too high considering the initial value of sum is only 85.
Two ways we can verify:
1. New sum we already know from above so this is not the right answer.
2. Also, if 105 is the value of new element then
total sum becomes 190 which gives us average of 9.5.
9.5 is more than 5% the jump we are looking for. The jump now is 90%.
So, this can't be answer.
D. \(20(1 + \frac{k}{100}) - 19a\)
\(20(1 + \frac{5}{100}) - 19*5 = -64 \)
Negative eliminate.
E. \(a*\frac{k}{5} - 19a\)
\(5*\frac{5}{5} - 19*5 = -80\)
Negative, eliminate.
Answer A.
