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Bunuel
The sum \(\frac{3}{10}+\frac{43}{100}+\frac{17}{1000}\) is equivalent to which of the following sums?

A. \(\frac{7}{1000}+\frac{4}{10}+\frac{7}{100}\)

B. \(\frac{6}{10}+\frac{12}{100}+\frac{37}{1000}\)

C. \(\frac{7}{100}+\frac{4}{10}+\frac{7}{1000}\)

D. \(\frac{7}{100}+\frac{4}{1000}+\frac{7}{10}\)

E. \(\frac{32}{100}+\frac{4}{10}+\frac{27}{1000}\)
The sum \(\frac{3}{10}+\frac{43}{100}+\frac{17}{1000}\) is equivalent to which of the following sums?

Express as decimals. (I did so for answer choices, too, below. Takes about 10 seconds.)

\(0.3 + 0.43 + 0.017 = .747\)

1) 7 must be in the tenths place.

Eliminate A and C. They have \(\frac{4}{10}=.4\). Both of their other digits are single (7s) such that no 3 will "carry" to the 0.4 and change it to 0.7

2) 7 must be in the thousandths place.

Eliminate D. It has \(\frac{4}{1000}= .004\) in the thousandths place

We're down to B and E.

(B) \(= .6 + .12 + .037= .757\) REJECT

(E) \(.32 + .4+ .027 =.747\) CORRECT

Answer E

Alternatively, express all as decimals and sum each.
Only E = .747

A. \(\frac{7}{1000}+\frac{4}{10}+\frac{7}{100}=.007 + .4 +.07\)

B. \(\frac{6}{10}+\frac{12}{100}+\frac{37}{1000}\)= \(.6 + .12+ .037\)

C. \(\frac{7}{100}+\frac{4}{10}+\frac{7}{1000}\)\(=.07+.4+.007\)

D. \(\frac{7}{100}+\frac{4}{1000}+\frac{7}{10}\)\(=.07+.004 +.7\)

E. \(\frac{32}{100}+\frac{4}{10}+\frac{27}{1000}\)\(=.32+.4+.027\)
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