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

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

by VJesus12 » Sun Apr 11, 2021 12:58 pm

00:00

A

B

C

D

E

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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

Source: Official Guide

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### Re: List $$S$$ consists of $$10$$ consecutive odd integers, and list $$T$$ consists of $$5$$ consecutive even integers.

by swerve » Tue Apr 13, 2021 11:23 am
VJesus12 wrote:
Sun Apr 11, 2021 12:58 pm
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

Source: Official Guide
Picked up numbers:

$$T = 2 , 4 , {\color{red}6}, 8 , 10$$ (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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### Re: List $$S$$ consists of $$10$$ consecutive odd integers, and list $$T$$ consists of $$5$$ consecutive even integers.

by [email protected] » Mon Apr 26, 2021 6:43 am
VJesus12 wrote:
Sun Apr 11, 2021 12:58 pm
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

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.