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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 51

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 Question 1
$\frac{{{\left( 6.679 \right)}^{3}}+{{\left( 3.321 \right)}^{3}}}{6.679\times 6.679-\left( 6.679\times 3.321 \right)+3.321\times 3.321}$
 A 10 B 1.248 C 20.44 D 1
Question 1 Explanation:
Let 6.679= a
And 3.321= b
Therefore given expression
\begin{align} & =\frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{2}}-ab+{{b}^{2}}} \\ & =\frac{\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)}{{{a}^{2}}-ab+{{b}^{2}}} \\ & =a+b=6.679+3.321 \\ & =10 \\ \end{align}
 Question 2
$\frac{127\times 127+127\times 123+123\times 123}{127\times 127\times 127-123\times 123\times 123}$ is equal to
 A 4 B 270 C $\frac{1}{4}$ D $\frac{1}{270}$
Question 2 Explanation:
Let 127 =a and 123= b
Given expression
\begin{align} & \frac{a\times a+a\times b+b\times b}{a\times a\times a-b\times b\times b} \\ & =\frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{3}}-{{b}^{3}}} \\ & =\frac{{{a}^{2}}+ab+{{b}^{2}}}{\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)} \\ & =\frac{1}{a-b}=\frac{1}{127-123}=\frac{1}{4} \\ \end{align}
 Question 3
The value of $\frac{0.512+0.343}{0.64-0.56+0.49}$ is
 A 0.2 B 0.25 C 0.3 D 0.8
Question 3 Explanation:
\begin{align} & \frac{{{\left( 0.8 \right)}^{3}}+{{\left( 0.7 \right)}^{3}}}{{{\left( 0.8 \right)}^{2}}-0.8\times 0.7+{{\left( 0.7 \right)}^{2}}} \\ & Let\,\,0.8=a\,,\,\,\,and\,\,\,0.7=b \\ & Therefore\,\,\,\exp ression \\ & =\frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{2}}-ab+{{b}^{2}}} \\ & =\frac{\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)}{{{a}^{2}}-ab+{{b}^{2}}} \\ & =a+b=0.8+0.7=1.5 \\ \end{align}
 Question 4
$3002\times 66+72716=?\times 128$
 A 2177 B 2167 C 2467 D 2116
Question 4 Explanation:
\begin{align} & 3002\times 66+72716=?\times 128 \\ & \Rightarrow 198132+72716=?\times 128 \\ & \Rightarrow 270848=?\times 128 \\ & \Rightarrow ?=270848\div 128=2116 \\ \end{align}
 Question 5
$\left[ \left( 3\sqrt{11}-\sqrt{11} \right)\times \left( 5\sqrt{11}+2\sqrt{11} \right) \right]-{{\left( 12 \right)}^{2}}=?$
 A 100 B $10\sqrt{11}$ C 10 D $\sqrt{11}$
Question 5 Explanation:
\begin{align} & ?=2\sqrt{11}\times 7\sqrt{11}-{{12}^{2}} \\ & =154-144=10 \\ \end{align}
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