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The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is

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Bunuel
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The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is [#permalink] New post   31 Mar 2019, 21:01
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The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is x?

A. 49/101
B. 50/101
C. 1/2
D. 51/101
E. 50/99
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B
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Re: The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is [#permalink] New post   31 Mar 2019, 22:25
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Bunuel wrote:
The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is x?

A. 49/101
B. 50/101
C. 1/2
D. 51/101
E. 50/99


Sum of first n natural numbers = (n*(n+1))/2

Hence for first 99 numbers = (99 * 100)/2

From the question stem, we know:

(1+2+3+4....98+99+x)/100 = 100x
=> 1+2+3....99+x = 100*100x
=> (99*100)/2 + x = 100*100x
=> (99*100)/2 = 100*100x - x = 10000x - x = 9999x = 9(1111)x
=> 99*100 = 2*9*1111x
=> x = (99*100)/(2*9*1111) = 50/101

Hence B is the correct answer.
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Re: The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is [#permalink] New post   01 Apr 2019, 03:09
Bunuel wrote:
The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is x?

A. 49/101
B. 50/101
C. 1/2
D. 51/101
E. 50/99


sum of integers 1 to 99 ; 99 *100/2 = 4950
4950+x/100 = 100x
so
solve for x we get 50/101
IMO B
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Re: The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is [#permalink] New post   01 Apr 2019, 03:32
Sum of numbers 1 to 99 = 4950 ( find the sum of AP 1, 2 ,3 ... 99 where a = 1 and d =1 , n = 99)

So sum of all numbers including x = 4950+x

and total numbers = 99+1 = 100 ( 99 are the original numbers and one extra is x ,so total numbers = 100)

average = total sum /total numbers = (4950+x)/100 = 100x ( given in question)

So we get 4950+x = 10000x
=> 4950 = 10000x-x
=> 4950 = 9999x

so x = 4950/9999 = 50/101

IMO B


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Re: The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is [#permalink] New post   01 Apr 2019, 12:16
Bunuel wrote:
The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is x?

A. 49/101
B. 50/101
C. 1/2
D. 51/101
E. 50/99


There are a total of 100 terms i.e. 1 to 99 and x and given average of these 100 terms is 100x

\(\frac{4950 +x}{100}={100x}\) where 4950 sum of terms from 1 to 99 using \(\frac{n(n+1)}{2}\) \(\Rightarrow\)\(\frac{99(99+1)}{2}\) \(\Rightarrow\) \(\frac{99(100)}{2}\)\(\Rightarrow\) 4950

on solving we get x=\(\frac{4950}{9999}\) on further reduction we get x =\(\frac{50}{101}\)
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Re: The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is [#permalink] New post   11 Dec 2019, 11:24
Bunuel wrote:
The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is x?

A. 49/101
B. 50/101
C. 1/2
D. 51/101
E. 50/99


Sum first N natural numbers: n(n+1)/2, N = 1 to 99 = 99-1+1 = 99
Sum first 99: 99(100)/2 = 99(50)
terms: 99 and x = 100 terms
sum = avg * terms
sum = 100x*100 = 100^2*x
sum = 99(50)+x
100^2*x = 99(50)+x, x(100^2-1)=99(50), x=99(50)/(100+1)(100-1), x=50/101

Ans (B)
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Re: The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is [#permalink] New post   12 Dec 2019, 19:52
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Bunuel wrote:
The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is x?

A. 49/101
B. 50/101
C. 1/2
D. 51/101
E. 50/99


The sum of the numbers from 1 to 99, inclusive, is:

(99 + 1)/2 x 99 = 50 x 99 = 4,950

So, now we have:

(4,950 + x)/100 = 100x

4,950 + x = 10,000x

4,950 = 9,999x

x = 4,950/9,999 = 550/1111 = 50/101

Answer: B
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Re: The average of the numbers 1, 2, 3,..., 98, 99, and x is 100x. What is [#permalink]
12 Dec 2019, 19:52
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