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| − | ==Problem==
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| − | Rounded to <math>2</math> decimal places, <math>\frac79</math> is
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| − | <math> \text{ (A) }\ 0.70 \qquad\text{ (B) }\ 0.77 \qquad\text{ (C) }\ 0.78 \qquad\text{ (D) }\ 0.79 \qquad\text{ (E) }\ 0.80 </math>
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| − | ==Solution==
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| − | We can use the fact that <math>\frac79 = 0.\overline{777}</math>.
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| − | Rounding this number to the nearest hundredth (to <math>2</math> decimal places), we get <math>\boxed {\textbf {(C) } 0.78}</math> because the thousandths place is greater than or equal to <math>5</math>, so it is closer to <math>0.78</math> than <math>0.77</math>.
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| − | ~anabel.disher
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| − | ==Solution 2 (answer choices)==
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| − | We can go backwards by multiplying answer choices by <math>9</math> and seeing how they compare to <math>7</math>
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| − | Multiplying <math>0.78</math> by <math>9</math>, we get:
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| − | <math>0.78 \times 9 = 7.02</math>
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| − | This makes <math>0.79</math> and <math>0.80</math> too large and <math>0.70</math> too small, eliminating them as answers
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| − | <math>0.77 \times 9 = 6.93</math>
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| − | We can now look at which answer <math>7</math> is closest to, using the distance to <math>7</math>:
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| − | <math>7 - 6.93 = 0.07</math>
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| − | <math>7.02 - 7 = 0.02</math>
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| − | <math>7.02</math> is closer to <math>7</math>, so the answer is <math>\boxed {\textbf {(C) } 0.78}</math>
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| − | ~anabel.disher
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