- 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 32
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Question 1 |
$\left( 7.5\times 7.5-37.5+2.5\times 2.5 \right)$ is equal to:
15 | |
5 | |
25 | |
35 |
Question 1 Explanation:
$\begin{align}
& \left( 7.5\times 7.5-37.5+2.5\times 2.5 \right) \\
& =\left[ {{\left( 7.5 \right)}^{2}}-2\times 7.5\times 2.5+{{\left( 2.5 \right)}^{2}} \right] \\
& ={{\left( 7.5-2.5 \right)}^{2}}={{\left( 5 \right)}^{2}}=25 \\
\end{align}$
Question 2 |
$\left( \sqrt{192}-\sqrt{147} \right)\div \sqrt{24}$ is equal to:
$\sqrt{6}$ | |
$\sqrt{3}/2$ | |
$\sqrt{2}/4$ | |
$\sqrt{6}/2$ |
Question 2 Explanation:
$\begin{align}
& \left( \sqrt{192}-\sqrt{147} \right)\div \sqrt{24} \\
& =\frac{\sqrt{192}-\sqrt{147}}{\sqrt{24}} \\
& =\frac{8\sqrt{3}-7\sqrt{3}}{2\sqrt{6}}=\frac{\sqrt{3}}{2\sqrt{2}\sqrt{3}}=\frac{1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\
& =\frac{\sqrt{2}}{4} \\
\end{align}$
Question 3 |
The value of
\[\frac{\sqrt{63}-\sqrt{99}}{\sqrt{28}-\sqrt{44}}\]is:
\[\frac{\sqrt{63}-\sqrt{99}}{\sqrt{28}-\sqrt{44}}\]is:
$\frac{3}{4}$ | |
$\frac{2}{3}$ | |
$1\frac{1}{2}$ | |
$1\frac{2}{3}$ |
Question 3 Explanation:
$\begin{align}
& \frac{\sqrt{63}-\sqrt{99}}{\sqrt{28}-\sqrt{44}} \\
& =\frac{\sqrt{9\times 7}-\sqrt{9\times 11}}{\sqrt{4\times 7}-\sqrt{4\times 11}} \\
& =\frac{3\sqrt{7}-3\sqrt{11}}{2\sqrt{7}-2\sqrt{11}}=\frac{3\left( \sqrt{7}-\sqrt{11} \right)}{2\left( \sqrt{7}-\sqrt{11} \right)} \\
& =\frac{3}{2}=1\frac{1}{2} \\
\end{align}$
Question 4 |
$\sqrt{12+\sqrt{12+\sqrt{12+.......}}}$is equal to:
6â…” | |
3 | |
$3\frac{1}{2}$ | |
4 |
Question 4 Explanation:
$\begin{align}
& Let\,\,x\,=\,\sqrt{12+\sqrt{12+\sqrt{12+.......}}} \\
& or,\,x=\sqrt{12+x} \\
& or,\,{{x}^{2}}=12+x \\
& or,\,{{x}^{2}}-x-12=0 \\
& or,\,{{x}^{2}}-4x+3x-12=0 \\
& or\,x\left( x-4 \right)+3\left( x-4 \right)=0 \\
& or,\,\left( x+3 \right)\,\left( x-4 \right)=0 \\
& Therefore\,\,x=-3\,\,or\,x=4 \\
\end{align}$
But the given expression is positive Hence, $x\ne -3$
But the given expression is positive Hence, $x\ne -3$
Question 5 |
$\left( 1.5\times 1.5+16.5+5.5\times 5.5 \right)$ is equal to:
99 | |
49 | |
100 | |
36 |
Question 5 Explanation:
$\begin{align}
& ?=\left( 1.5\times 1.5+16.5+5.5\times 5.5 \right) \\
& =\left[ {{\left( 1.5 \right)}^{2}}+2\times 1.5\times 5.5+{{\left( 5.5 \right)}^{2}} \right] \\
& ={{\left[ 1.5+5.5 \right]}^{2}}={{\left( 7 \right)}^{2}}=49 \\
\end{align}$
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