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