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The arithmetic mean (average) M of 4 terms is an integer. When M is di

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The arithmetic mean (average) M of 4 terms is an integer. When M is di  [#permalink]

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New post 11 May 2019, 00:56
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The arithmetic mean (average) M of 4 terms is an integer. When M is divided by 16, the remainder is 10. If each of the terms is increased by 100%, what is the remainder when the new mean is divided by 16?


A. 1
B. 4
C. 10
D. 12
E. 14
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The arithmetic mean (average) M of 4 terms is an integer. When M is di  [#permalink]

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New post Updated on: 02 Jun 2019, 10:41
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Given that- When M is divided by 16, the remainder is 10
Hence M= 16b+10
Now, when we increasing each number by 100 percent or doubling each number, their average mean will also be doubled.
New average mean, 2M= 32b+20= [16(2b+1)]+4
when we divide [16(2b+1)]+4, we will get remainder 4.

Originally posted by nick1816 on 11 May 2019, 01:12.
Last edited by nick1816 on 02 Jun 2019, 10:41, edited 2 times in total.
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Re: The arithmetic mean (average) M of 4 terms is an integer. When M is di  [#permalink]

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New post 11 May 2019, 13:25
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kiran120680 wrote:
The arithmetic mean (average) M of 4 terms is an integer. When M is divided by 16, the remainder is 10. If each of the terms is increased by 100%, what is the remainder when the new mean is divided by 16?


A. 1
B. 4
C. 10
D. 12
E. 14


let terms=7,9,11,13
mean=10
10/16 gives remainder of 10
if each term is doubled, mean will double also
20/16 gives a remainder of 4
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Re: The arithmetic mean (average) M of 4 terms is an integer. When M is di  [#permalink]

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New post 29 May 2020, 05:49
B.
M= (a+b+c+d)/4


Also, M= 16*k +10; k is constant

Now all numbers are doubles (100% increase = twice the original)
AM for new terms: (2a+2b+2c+2d)/4 = 2(a+b+c+d)/4 = 2M

Now 2M = 2*16*k + 2*10 = 2*16*k + 20 = 2*16*k + (16 +4)
= 2*16*(k+1) + 4

4 is the remainder
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Re: The arithmetic mean (average) M of 4 terms is an integer. When M is di   [#permalink] 29 May 2020, 05:49

The arithmetic mean (average) M of 4 terms is an integer. When M is di

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