List \(S\) consists of \(10\) consecutive odd integers, and list \(T\) consists of \(5\) consecutive even integers. If the least integer in \(S\) is \(7\) more than the least integer in \(T,\) how much greater is the average (arithmetic mean) of the integers in \(S\) than the average of the integers in \(T?\)

(A) 2

(B) 7

(C) 8

(D) 12

(E) 22

Answer: D

Source: Official Guide

## List \(S\) consists of \(10\) consecutive odd integers, and list \(T\) consists of \(5\) consecutive even integers. If

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Picked up numbers:VJesus12 wrote: ↑Sun Apr 11, 2021 12:58 pmList \(S\) consists of \(10\) consecutive odd integers, and list \(T\) consists of \(5\) consecutive even integers. If the least integer in \(S\) is \(7\) more than the least integer in \(T,\) how much greater is the average (arithmetic mean) of the integers in \(S\) than the average of the integers in \(T?\)

(A) 2

(B) 7

(C) 8

(D) 12

(E) 22

Answer: D

Source: Official Guide

\(T = 2 , 4 , {\color{red}6}, 8 , 10\) (Mean \(= 6\))

\(S = 9, 11, 13, 15, {\color{red}17, 19}, 21, 23, 25, 27\) \(\bigg(\)Mean \(=\dfrac{17+19}{2}=18\bigg)\)

Difference \(= 18-6 = 12\)

Therefore, D

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VJesus12 wrote: ↑Sun Apr 11, 2021 12:58 pmList \(S\) consists of \(10\) consecutive odd integers, and list \(T\) consists of \(5\) consecutive even integers. If the least integer in \(S\) is \(7\) more than the least integer in \(T,\) how much greater is the average (arithmetic mean) of the integers in \(S\) than the average of the integers in \(T?\)

(A) 2

(B) 7

(C) 8

(D) 12

(E) 22

Answer: D

Source: Official Guide

**Solution:**

We can let x = the least integer in T. Thus, T contains the following integers: x, x + 2, x + 4, x + 6, and x + 8.

Since the least integer in S is 7 more than the least integer in T, x + 7 = the least integer in S, and so S has the following integers: x + 7, x + 9, x + 11, x + 13, x + 15, x + 17, x + 19, x + 21, x + 23, and x + 25.

Since each list is an evenly spaced set, the average of each list is the respective median. Since the median of the integers in T is x + 4 and the median of integers in S is [(x +15) + (x + 17)]/2 = (2x + 32)/2 = x + 16, the averages of the integers in T and S are x + 4 and x +16, respectively.

Therefore, the average of list S is (x + 16) - (x + 4) = 12 more than the average of list T.

**Answer: D**

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