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A set contains 5 distinct positive integers with a sum S
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28 Nov 2021, 22:56
28 Nov 2021, 22:56
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A set contains 5 distinct positive integers with a sum S. If each entry of the set is increased by 5 and then multiplied by 6, then
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
The difference between the average of the new and the old set |
S+10 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
A set contains 5 distinct positive integers with a sum S
[#permalink]
29 Nov 2021, 07:30
29 Nov 2021, 07:30
1
Let \(a, b, c, d, \) and \(e\) be the five distinct positive integers.
\(a + b + c + d + e = S\)
Since \(a, b, c, d, \) and \(e\) are positive, \(S\) will also be positive.
each integer is increased by 5 and then multiplied by 6. So,
\(a_{new} = 6(a+5)\)
\(b_{new} = 6(b+5)\)
\(c_{new} = 6(c+5)\)
\(d_{new} = 6(d+5)\)
\(e_{new} = 6(e+5)\)
The difference between the average of the new and the old set = \(\frac{a_{new} + b_{new} + c_{new} + d_{new} + e_{new}}{5} - \frac{a + b + c + d + e}{5}\)
\(= \frac{6(a+5) + 6(b+5) + 6(c+5) + 6(d+5) + 6(e+5)}{5} - \frac{S}{5}\)
\(= \frac{6(a + b + c + d + e + 5*5)}{5} - \frac{S}{5}\)
\(= \frac{6(S + 25)}{5} - \frac{S}{5}\)
\(= \frac{6(S + 25) - S}{5}\)
\(= \frac{6S + 150 - S}{5} = \frac{5S + 150}{5} = \textbf{S + 30} > S + 10\)
Hence, Answer is A
\(a + b + c + d + e = S\)
Since \(a, b, c, d, \) and \(e\) are positive, \(S\) will also be positive.
each integer is increased by 5 and then multiplied by 6. So,
\(a_{new} = 6(a+5)\)
\(b_{new} = 6(b+5)\)
\(c_{new} = 6(c+5)\)
\(d_{new} = 6(d+5)\)
\(e_{new} = 6(e+5)\)
The difference between the average of the new and the old set = \(\frac{a_{new} + b_{new} + c_{new} + d_{new} + e_{new}}{5} - \frac{a + b + c + d + e}{5}\)
\(= \frac{6(a+5) + 6(b+5) + 6(c+5) + 6(d+5) + 6(e+5)}{5} - \frac{S}{5}\)
\(= \frac{6(a + b + c + d + e + 5*5)}{5} - \frac{S}{5}\)
\(= \frac{6(S + 25)}{5} - \frac{S}{5}\)
\(= \frac{6(S + 25) - S}{5}\)
\(= \frac{6S + 150 - S}{5} = \frac{5S + 150}{5} = \textbf{S + 30} > S + 10\)
Hence, Answer is A
gmatclubot
A set contains 5 distinct positive integers with a sum S [#permalink]
29 Nov 2021, 07:30
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