- 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 30
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Question 1 |
$ \frac{0.08\times 0.08\times 0.08-0.03\times 0.03\times 0.03}{0.08\times 0.08+0.08\times 0.03+0.03\times 0.03}$
0.05 | |
0.001 | |
0.01 | |
0.02 |
Question 1 Explanation:
$ \frac{0.08\times 0.08\times 0.08-0.03\times 0.03\times 0.03}{0.08\times 0.08+0.08\times 0.03+0.03\times 0.03}$
So we know that
$ \begin{array}{l}\frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{2}}+ab+{{b}^{2}}}=a-b\\\operatorname{Re}quired\,\,\,answer\\=0.08-0.03=0.05\end{array}$
So we know that
$ \begin{array}{l}\frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{2}}+ab+{{b}^{2}}}=a-b\\\operatorname{Re}quired\,\,\,answer\\=0.08-0.03=0.05\end{array}$
Question 2 |
$ 1+\frac{1}{3}+\frac{1}{5}+\frac{1}{15}+\frac{1}{45}$
is equal to
is equal to
2 | |
1.2 | |
1.6 | |
3 |
Question 2 Explanation:
$ \begin{array}{l}?=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{15}+\frac{1}{45}\\=\frac{45+15+9+3+1}{45}\\=\frac{73}{45}=1.6\end{array}$
Question 3 |
Simplify
$ \frac{0.02\times 0.02-0.01\times 0.01}{0.02\times 0.02+0.01\times 0.01-2\times 0.02\times 0.01}$
$ \frac{0.02\times 0.02-0.01\times 0.01}{0.02\times 0.02+0.01\times 0.01-2\times 0.02\times 0.01}$
3 | |
0.3 | |
0.03 | |
0.003 |
Question 3 Explanation:
Using the formula
$ \begin{array}{l}{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\\{{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}\\?=\frac{0.02\times 0.02-0.01\times 0.01}{0.02\times 0.02+0.01\times 0.01-2\times 0.02\times 0.01}\\=\frac{{{\left( 0.02 \right)}^{2}}-{{\left( 0.01 \right)}^{2}}}{{{\left( 0.02 \right)}^{2}}+{{\left( 0.01 \right)}^{2}}-2\times 0.02\times 0.01}\\=\frac{0.02+0.01}{0.02-0.01}\\=\frac{0.03}{0.01}=3\end{array}$
$ \begin{array}{l}{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\\{{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}\\?=\frac{0.02\times 0.02-0.01\times 0.01}{0.02\times 0.02+0.01\times 0.01-2\times 0.02\times 0.01}\\=\frac{{{\left( 0.02 \right)}^{2}}-{{\left( 0.01 \right)}^{2}}}{{{\left( 0.02 \right)}^{2}}+{{\left( 0.01 \right)}^{2}}-2\times 0.02\times 0.01}\\=\frac{0.02+0.01}{0.02-0.01}\\=\frac{0.03}{0.01}=3\end{array}$
Question 4 |
When
$ \left( \frac{1}{3}-\frac{1}{5}+\frac{1}{6}-\frac{1}{10} \right)$
is divided by
$ \left( \frac{2}{6}-\frac{7}{15}+\frac{3}{4}-\frac{5}{12} \right)$
the result is:
$ \left( \frac{1}{3}-\frac{1}{5}+\frac{1}{6}-\frac{1}{10} \right)$
is divided by
$ \left( \frac{2}{6}-\frac{7}{15}+\frac{3}{4}-\frac{5}{12} \right)$
the result is:
$ \displaystyle 5\frac{1}{10}$ | |
$ \displaystyle 1$ | |
$ \displaystyle 3\frac{1}{6}$ | |
$ \displaystyle 3\frac{3}{10}$ |
Question 4 Explanation:
$ \begin{array}{l}\left( \frac{1}{3}-\frac{1}{5}+\frac{1}{6}-\frac{1}{10} \right)\div \left( \frac{2}{6}-\frac{7}{15}+\frac{3}{4}-\frac{5}{12} \right)\\=\left( \frac{10-6+5-3}{30} \right)\div \left( \frac{20-28+45-25}{60} \right)\\=\left( \frac{1}{5} \right)\div \left( \frac{1}{5} \right)=1\end{array}$
Question 5 |
The value of $ \sqrt{\frac{{{\left( 0.5 \right)}^{2}}+{{\left( 0.06 \right)}^{2}}+{{\left( 0.007 \right)}^{2}}}{{{\left( 0.05 \right)}^{2}}+{{\left( 0.006 \right)}^{2}}+{{\left( 0.0007 \right)}^{2}}}}$
102 | |
10 | |
0.1 | |
0.01 |
Question 5 Explanation:
$ \begin{array}{l}\sqrt{\frac{{{\left( 0.5 \right)}^{2}}+{{\left( 0.06 \right)}^{2}}+{{\left( 0.007 \right)}^{2}}}{{{\left( 0.05 \right)}^{2}}+{{\left( 0.006 \right)}^{2}}+{{\left( 0.0007 \right)}^{2}}}}\\=\sqrt{\frac{{{\left( 0.5 \right)}^{2}}+{{\left( 0.06 \right)}^{2}}+{{\left( 0.007 \right)}^{2}}}{{{\left( 0.1 \right)}^{2}}[{{\left( 0.5 \right)}^{2}}+{{\left( 0.06 \right)}^{2}}+{{\left( 0.007 \right)}^{2}}]}}\\=\sqrt{\frac{1}{0.01}}=\sqrt{100}=10\end{array}$
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