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Basic Maths: Test 28

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Question 1
$ \frac{4\frac{3}{8}-\frac{1}{2}}{3\frac{2}{3}+2\frac{1}{7}}\div \frac{1}{3+\frac{3}{3+\frac{1}{3-\frac{1}{3}}}}$
is equal to
A
$ \frac{7699}{2928}$
B
$ \frac{7595}{2928}$
C
$ \frac{75}{2928}$
D
$ \frac{759}{2928}$
Question 1 Explanation:
image-27
Question 2
$ If\,\,{{\left[ 4-\frac{5}{3-\frac{1}{\frac{5}{9}+\frac{1}{2+\frac{1}{4}}}} \right]}^{th}}$
part of a journey takes 10 minutes, then to compete 6/5th of that journey, it will take
A
3 minutes
B
13 minutes
C
24 minutes
D
8 minutes
Question 2 Explanation:
We are given by a fraction which is a part of any Distance
First we will solve that fraction part then we will proceed for the further question
$ \displaystyle \begin{array}{l}4-\frac{5}{3-\frac{1}{\frac{5}{9}+\frac{1}{\frac{9}{4}}}}=4-\frac{5}{3-\frac{1}{\frac{5}{9}+\frac{4}{9}}}\\=4-\frac{5}{3-1}=4-\frac{5}{2}\\=\frac{8-5}{2}\\=\frac{3}{2}\\Time\,\,\,taken\,\,in\,\,\,completing\,\,\,=10\,\,\min utes\\Therefore\,\,\,time\,\,taken\,\,\,in\,\,\,completing\,\,\frac{6}{5}\,\,part\\=10\times \frac{2}{3}\times \frac{6}{5}\\=8\,\,Minutes\end{array}$
Question 3
$ \left( 1001\frac{1001}{1000}\times 7 \right)$
is equals to
A
$ 6998\frac{1}{1000}$
B
$ 7008\frac{7}{1000}$
C
$ 7003\frac{1}{1000}$
D
$ 7014\frac{7}{1000}$
Question 3 Explanation:
$ \begin{array}{l}1001\frac{1001}{1000}\times 7\\=\left( 1001+\frac{1001}{1000} \right)\times 7\\=7007+\frac{7007}{1000}\\=7007+7\frac{7}{1000}\\=7014\frac{7}{1000}\end{array}$
Question 4
$ \sqrt{\frac{15\frac{1}{8}-\frac{38}{16}}{5+\frac{1}{4}+1\frac{1}{8}}\div \frac{179}{3+\frac{1}{3+\frac{1}{6-\frac{1}{3}}}}}$
is equal to
A
$ \frac{1}{3\sqrt{3}}$
B
$ 3\sqrt{3}$
C
$ 3$
D
$ \sqrt{3}$
Question 4 Explanation:
image-1
Question 5
The sum ‘9+…..+19’ is equal to:
A
150
B
160
C
154
D
164
Question 5 Explanation:
$ \begin{array}{l}1+2+3+.....+n=\frac{n\left( n+1 \right)}{2}\\Therefore\,9+10+11+....+19\\=\left( 1+2+3+.....+19 \right)-\left( 1+2+3....+8 \right)\\=\frac{19\left( 19+1 \right)}{2}-\frac{8\left( 8+1 \right)}{2}=190-36=154\end{array}$
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