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• 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

Congratulations - you have completed Basic Maths: Test 30.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
 Question 1
$\frac{0.08\times 0.08\times 0.08-0.03\times 0.03\times 0.03}{0.08\times 0.08+0.08\times 0.03+0.03\times 0.03}$
 A 0.05 B 0.001 C 0.01 D 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
$\begin{array}{l}\frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{2}}+ab+{{b}^{2}}}=a-b\\\operatorname{Re}quired\,\,\,answer\\=0.08-0.03=0.05\end{array}$
 Question 2
$1+\frac{1}{3}+\frac{1}{5}+\frac{1}{15}+\frac{1}{45}$
is equal to
 A 2 B 1.2 C 1.6 D 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
$\frac{0.02\times 0.02-0.01\times 0.01}{0.02\times 0.02+0.01\times 0.01-2\times 0.02\times 0.01}$
 A 3 B 0.3 C 0.03 D 0.003
Question 3 Explanation:
Using the formula
$\begin{array}{l}{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\\{{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}\\?=\frac{0.02\times 0.02-0.01\times 0.01}{0.02\times 0.02+0.01\times 0.01-2\times 0.02\times 0.01}\\=\frac{{{\left( 0.02 \right)}^{2}}-{{\left( 0.01 \right)}^{2}}}{{{\left( 0.02 \right)}^{2}}+{{\left( 0.01 \right)}^{2}}-2\times 0.02\times 0.01}\\=\frac{0.02+0.01}{0.02-0.01}\\=\frac{0.03}{0.01}=3\end{array}$
 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:
 A $\displaystyle 5\frac{1}{10}$ B $\displaystyle 1$ C $\displaystyle 3\frac{1}{6}$ D $\displaystyle 3\frac{3}{10}$
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 $\sqrt{\frac{{{\left( 0.5 \right)}^{2}}+{{\left( 0.06 \right)}^{2}}+{{\left( 0.007 \right)}^{2}}}{{{\left( 0.05 \right)}^{2}}+{{\left( 0.006 \right)}^{2}}+{{\left( 0.0007 \right)}^{2}}}}$
 A 102 B 10 C 0.1 D 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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