• This is an assessment test.
  • These tests focus on the basics of Maths and are meant to indicate your preparation level for the subject.
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Basic Maths: Test 2

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Your answers are highlighted below.
Question 1

$ \displaystyle -\frac{1}{\sqrt{16+6\sqrt{7}}}+\sqrt{43-12\sqrt{7}}=?$

A
$ \displaystyle 2\sqrt{7}-1$
B
$ \displaystyle -2\sqrt{7}+3$
C
$ \displaystyle -3$
D
$ \displaystyle \frac{9-\sqrt{7}}{2}$
Question 1 Explanation: 
$ \displaystyle \begin{array}{l}Expression\\=\sqrt{43-12\sqrt{7}}-\frac{1}{\sqrt{16+6\sqrt{7}}}\\=\sqrt{43-2\times 6\times \sqrt{7}}-\frac{1}{\sqrt{16+2+\times 3\times \sqrt{7}}}\\=\sqrt{36+7-2\times 6\sqrt{7}}-\frac{1}{\sqrt{9+7+2\times 3\times \sqrt{7}}}\\=\sqrt{{{\left( 6-\sqrt{7} \right)}^{2}}}-\frac{1}{\sqrt{{{\left( 3+\sqrt{7} \right)}^{2}}}}\\=6-\sqrt{7}-\frac{1}{3+\sqrt{7}}\\=6-\sqrt{7}-\frac{1\times \left( 3-\sqrt{7} \right)}{\left( 3+\sqrt{7} \right)\left( 3-\sqrt{7} \right)}\\=6-\sqrt{7}-\frac{\left( 3-\sqrt{7} \right)}{9-7}\\=\frac{12-2\sqrt{7}-3+\sqrt{7}}{2}\\=\frac{9-\sqrt{7}}{2}\end{array}$
Question 2
$ \displaystyle \frac{\sqrt[3]{729}}{\sqrt[3]{1728}}\times \frac{7}{15}\times \frac{4}{8}=?$
A
0.175
B
0.186
C
1.026
D
0.246
Question 2 Explanation: 
$ \displaystyle \begin{array}{l}\frac{\sqrt[3]{9\times 9\times 9}}{\sqrt[3]{12\times 12\times 12}}\times \frac{7}{15}\times \frac{4}{8}\\=\frac{9}{12}\times \frac{7}{15}\times \frac{4}{8}\\=\frac{7}{40}=0.175\end{array}$
Question 3
If
$ \displaystyle {{\left( \frac{729}{1000} \right)}^{\frac{2}{3}}}+\left[ \frac{{{\left( 12996 \right)}^{\frac{1}{2}}}}{\sqrt{625}} \right]\,=p\times {{10}^{-3}}$
then the value of 'p' is:
A
5730
B
5370
C
7530
D
7350
Question 3 Explanation: 
$ \displaystyle \begin{array}{l}{{\left( \frac{{{9}^{3}}}{{{10}^{3}}} \right)}^{\frac{2}{3}}}+\frac{\sqrt{12996}}{\sqrt{625}}=p\,\,\times {{10}^{-3}}\\\Rightarrow {{\left( \frac{9}{10} \right)}^{2}}+\frac{114}{25}=p\times {{10}^{-3}}\\\Rightarrow \frac{81}{100}+\frac{114}{25}=\frac{p}{1000}\\\Rightarrow \frac{81+456}{100}=\frac{p}{1000}\\\Rightarrow \frac{537}{100}=\frac{p}{1000}\\\Rightarrow p=5370\\\end{array}$
Question 4
128
find p
A
16421
B
16411
C
14641
D
16441
Question 4 Explanation: 
$ \displaystyle \begin{array}{l}\frac{5}{18}\times \frac{54}{100}\times 864=\sqrt{p}+\sqrt{73.96}\\\Rightarrow \frac{12960}{100}=\sqrt{p}+8.6\\\Rightarrow 129.6-8.6=\sqrt{p}\\\Rightarrow \sqrt{p}=121\\\Rightarrow p=121\times 121\\=14641\end{array}$
Question 5
764 –4263 ÷147 = p × 36
A
21
B
20.25
C
21.25
D
20.41
Question 5 Explanation: 
$ \displaystyle \begin{array}{l}764-4263\times \frac{1}{147}=p\times 36\\\Rightarrow 764-29=p\times 36\\\Rightarrow p\times 36=735\\\Rightarrow p=\frac{735}{36}=20.41\end{array}$
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