- This is an assessment test.
- These tests focus on the basics of Maths and are meant to indicate your preparation level for the subject.
- Kindly take the tests in this series with a pre-defined schedule.
Basic Maths: Test 18
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Question 1 |
Simplify:$ \displaystyle \frac{1+\frac{1}{2}}{1-\frac{1}{2}}\div \frac{3}{14}\left( \frac{2}{5}+\frac{3}{10} \right)\,of\,\frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{2}-\frac{1}{3}}$
$ \displaystyle 4$ | |
$ \displaystyle 37\frac{1}{2}$ | |
$ \displaystyle \frac{3}{2}$ | |
$ \displaystyle 18\frac{3}{8}$ |
Question 1 Explanation:
Question 2 |
Simplify $ \displaystyle \left[ 0.9\div \left\{ 2.3-3.2-\left( 7.1-5.4-3.5 \right) \right\} \right]$
0.18 | |
1.8 | |
1 | |
2.6 |
Question 2 Explanation:
$ \displaystyle \begin{array}{l}\left[ 0.9\div \left\{ 2.3-3.2-\left( 7.1-8.9 \right) \right\} \right]\\=\left[ 0.9\div \left\{ 2.3-3.2+1.8 \right\} \right]\\=\left[ 0.9\div 0.9 \right]\\=1\end{array}$
Question 3 |
The value of $ \displaystyle \frac{{{\left( 75.8 \right)}^{2}}-{{\left( 55.8 \right)}^{2}}}{40}$ is
20 | |
40 | |
65.8 | |
131.6 |
Question 3 Explanation:
$ \displaystyle \begin{array}{l}\frac{{{(75.8)}^{2}}-{{\left( 55.8 \right)}^{2}}}{40}\\=\frac{\left( 75.8-55.8 \right)\left( 75.8+55.8 \right)}{40}\\=\frac{20\times 131.6}{40}=65.8\end{array}$
Question 4 |
$ \displaystyle \frac{1}{4}+\left\{ 4\frac{3}{4}-\left( 3\frac{1}{6}-2\frac{1}{3} \right) \right\}$Â is equal to
$ \displaystyle 3\frac{2}{3}$ | |
$ \displaystyle 1\frac{1}{4}$ | |
$ \displaystyle 4\frac{1}{6}$ | |
$ \displaystyle 1\frac{2}{3}$ |
Question 4 Explanation:
Expression
$ \displaystyle \begin{array}{l}=\frac{1}{4}+\left\{ \frac{19}{4}-\left( \frac{19}{6}-\frac{7}{3} \right)\, \right\}\\=\frac{1}{4}+\left\{ \frac{19}{4}-\left( \frac{19-14}{6} \right)\, \right\}\\=\frac{1}{4}+\left\{ \frac{19}{4}-\frac{5}{6} \right\}\\=\frac{1}{4}+\frac{19}{4}-\frac{5}{6}\\=\frac{50}{12}\\=4\frac{1}{6}\end{array}$
$ \displaystyle \begin{array}{l}=\frac{1}{4}+\left\{ \frac{19}{4}-\left( \frac{19}{6}-\frac{7}{3} \right)\, \right\}\\=\frac{1}{4}+\left\{ \frac{19}{4}-\left( \frac{19-14}{6} \right)\, \right\}\\=\frac{1}{4}+\left\{ \frac{19}{4}-\frac{5}{6} \right\}\\=\frac{1}{4}+\frac{19}{4}-\frac{5}{6}\\=\frac{50}{12}\\=4\frac{1}{6}\end{array}$
Question 5 |
If p represents a number, then the value of p in  $ \displaystyle 5\frac{3}{p}\times 3\frac{1}{2}=19$ is:
7 | |
4 | |
6 | |
2 |
Question 5 Explanation:
$ \displaystyle \begin{array}{l}5\frac{3}{p}\times \frac{7}{2}=19\\\Rightarrow 5\frac{3}{p}=\frac{19\times 2}{7}\\\Rightarrow 5\frac{3}{p}=\frac{38}{7}=5\frac{3}{7}\\\Rightarrow p=7\end{array}$
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