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Basic Maths: Test 21

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Question 1
Find the value of
$ \displaystyle \frac{1}{1+\frac{1}{1-\frac{1}{2}}}\times \frac{1}{\frac{5}{6}\,\,of\,\frac{3}{2}\div \,1\frac{1}{4}\,}$
A
1/2
B
1/7
C
1/6
D
1/3
Question 1 Explanation: 
The equation can be simplified as
$ \displaystyle \begin{array}{l}\frac{1}{1+\frac{1}{\frac{1}{2}}}\times \frac{1}{\left( \frac{5}{6}\,\times \frac{3}{2} \right)\div \,\frac{5}{4}\,}\\=\frac{1}{1+2}\times \frac{1}{\frac{5}{4}\div \frac{5}{4}}\\=\frac{1}{3}\times \frac{1}{\frac{5}{4}\times \frac{4}{5}}\\=\frac{1}{3}\end{array}$
So the answer for the question is d
Question 2
$ \displaystyle \frac{1}{2+\frac{2}{3+\frac{2}{3+\frac{2}{3}}}\times 0.78}$ is simplified to
A
7/3
B
100/11
C
111/100
D
None of these
Question 2 Explanation: 
$ \displaystyle \begin{array}{l}\frac{1}{2+\frac{2}{3+\frac{2}{\frac{11}{3}}}\times 0.78}\\=\frac{1}{2+\frac{2}{3+\frac{6}{11}}\times 0.78}\\=\frac{1}{2+\frac{2}{\frac{33+6}{11}}\times 0.78}\\=\frac{1}{\begin{array}{l}2+\frac{11\times 2}{39}\times 0.78\\=\frac{1}{2+\frac{11\times 2}{39}\times \frac{78}{100}}\end{array}}\\=\frac{1}{2+\frac{11}{25}}\\=25/61\end{array}$
Question 3
$ \displaystyle 3+\frac{1}{1+\frac{1}{2}}$
is equal to
A
3/5
B
3/2
C
11/3
D
7/3
Question 3 Explanation: 
The Expression can be solved like as
$ \displaystyle 3+\frac{1}{1+\frac{1}{2}}$
$ \displaystyle \begin{array}{l}=3+\frac{1}{\frac{2+1}{2}}\\=3+\frac{2}{3}\\=\frac{9+2}{3}\\=\frac{11}{3}\end{array}$
Question 4
The value of
$ \displaystyle 5+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{2}{3}}}}}$ is
A
134/21
B
132/23
C
143/21
D
118/21
Question 4 Explanation: 
$ \displaystyle \begin{array}{l}5+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{3+2}{3}}}}}\\\end{array}$
$ \displaystyle \begin{array}{l}=5+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{3}{5}}}}\\=5+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{5+3}{5}}}}\\=5+\frac{1}{1+\frac{1}{1+\frac{5}{8}}}\\=5+\frac{1}{1+\frac{1}{\frac{8+5}{8}}}\\=5+\frac{1}{1+\frac{8}{13}}=5+\frac{1}{\frac{13+8}{13}}\\=5+\frac{13}{21}=\frac{105+13}{21}=\frac{118}{21}\end{array}$
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