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Only A fits the bills. Others are too big.

Sent from my Moto G (5) Plus using GMAT Club Forum mobile app
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MathRevolution
If \(\frac{x}{100}+\frac{x}{1,000}+\frac{x}{10,000}+\frac{x}{100,000}=111\), what is the approximation of x?

A. 11,100
B. 111,000
C. 1,111,000
D. 11,111,000
E. 111,111,000

\(\frac{x}{100}+\frac{x}{1,000}+\frac{x}{10,000}+\frac{x}{100,000}=111\)
\(x\frac{(1000+100+10+10)}{100000} = 111\)
\(x\frac{(1111)}{100000} = 111\)
\(x\frac{(10.009)}{100000} = 1\)
\(x\frac{(1)}{10000} = 1\)
Hence Answer A
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==> Since it is approximation, you get _. Then, only x/100=111 is left, and thus you get x=11,100.

The answer is A.
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\(\frac{x}{100}+\frac{x}{1,000}+\frac{x}{10,000}+\frac{x}{100,000}=111\)

\((\frac{x}{100}) ( 1+ \frac{1}{10} + \frac{1}{100} + \frac{1}{1000}) = 111\)

\((\frac{x}{100}) ( 1 + 0.1 + 0.01 + 0.001) = 111\)

\((\frac{x}{100}) (1.11) = 111\) OR

\((\frac{x}{100}) (1) = 111\)

\(x = 111 * 100\)

\(x = 111,100\)

Hence, Answer is A
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MathRevolution
If \(\frac{x}{100}+\frac{x}{1,000}+\frac{x}{10,000}+\frac{x}{100,000}=111\), what is the approximation of x?

A. 11,100
B. 111,000
C. 1,111,000
D. 11,111,000
E. 111,111,000

10^3 x + 10^2 x + 10 x + x/10^5 = 111
or, x (1000 + 100 + 10 + 1) / 10^5 = 111
or, 1111 x = 111 * 10^5
or, x ~ 1/10 * 100,000
so, x ~ 10,000

A is the answer.
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