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(1 1/6)*(1 11/21) =

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Math Expert
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(1 1/6)*(1 11/21) =  [#permalink]

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New post 23 Oct 2018, 21:34
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

89% (01:09) correct 11% (02:13) wrong based on 61 sessions

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Re: (1 1/6)*(1 11/21) =  [#permalink]

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New post 23 Oct 2018, 23:50
Bunuel wrote:
\((1 \frac{1}{6})*(1 \frac{11}{21}) =\)


A. \(1\frac{7}{9}\)

B. \(1\frac{11}{14}\)

C. \(1\frac{29}{42}\)

D. \(1\frac{11}{126}\)

E. \(2\frac{29}{42}\)


\((1 \frac{1}{6})*(1 \frac{11}{21})\)

\(= \frac{7}{6}*\frac{32}{21}\)

\(= \frac{16}{9}\)

\(= 1\frac{7}{9}\) ; Answer must be (A)
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Re: (1 1/6)*(1 11/21) =  [#permalink]

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New post 24 Oct 2018, 00:11
Bunuel wrote:
\((1 \frac{1}{6})*(1 \frac{11}{21}) =\)


A. \(1\frac{7}{9}\)

B. \(1\frac{11}{14}\)

C. \(1\frac{29}{42}\)

D. \(1\frac{11}{126}\)

E. \(2\frac{29}{42}\)


We can write \(a \frac{b}{c}=a +\frac{b}{c}\)

Using the same logic, \((1 \frac{1}{6})*(1 \frac{11}{21})\)=\((1+\frac{1}{6})*(1+\frac{11}{21})\)=\(\frac{7}{6}*\frac{32}{21}=\frac{16}{9}=1+\frac{7}{9}=1\frac{7}{9}\)


Ans. (A)
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Re: (1 1/6)*(1 11/21) =  [#permalink]

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New post 28 Oct 2018, 18:22
Bunuel wrote:
\((1 \frac{1}{6})*(1 \frac{11}{21}) =\)


A. \(1\frac{7}{9}\)

B. \(1\frac{11}{14}\)

C. \(1\frac{29}{42}\)

D. \(1\frac{11}{126}\)

E. \(2\frac{29}{42}\)


We first convert the two mixed numbers to improper fractions and then simplify. We have:

7/6 x 32/21 = (7 x 32)/(6 x 21) = 32/(6 x 3) = 32/18 = 16/9 = 1 7/9

Answer: A
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Re: (1 1/6)*(1 11/21) =   [#permalink] 28 Oct 2018, 18:22
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