Basic Maths: Test 32

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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:

A	15
B	5
C	25
D	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:

A	$\sqrt{6}$
B	$\sqrt{3}/2$
C	$\sqrt{2}/4$
D	$\sqrt{6}/2$

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:

A	$\frac{3}{4}$
B	$\frac{2}{3}$
C	$1\frac{1}{2}$
D	$1\frac{2}{3}$

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:

A	6⅔
B	3
C	$3\frac{1}{2}$
D	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, $x\ne -3$

Question 5

$\left( 1.5\times 1.5+16.5+5.5\times 5.5 \right)$ is equal to:

A	99
B	49
C	100
D	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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