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Algebra: Basics Test-4

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Question 1
If p = 0.25, q = −0.05, r = 0.5, then the value of
$ \displaystyle \frac{{{p}^{2}}-{{q}^{2}}-{{r}^{2}}-2pq}{{{p}^{2}}+{{q}^{2}}-2qr-{{r}^{2}}}$
A
$ \displaystyle \frac{7}{8}$
B
$ \displaystyle \frac{8}{9}$
C
$ \displaystyle \frac{1}{8}$
D
$ \displaystyle \frac{5}{8}$
Question 1 Explanation: 
$ \displaystyle \begin{array}{l}\frac{{{p}^{2}}-{{q}^{2}}-{{r}^{2}}-2qr}{{{p}^{2}}+{{q}^{2}}-2pq-{{r}^{2}}}\\=\frac{{{p}^{2}}-\left( {{q}^{2}}+{{r}^{2}}+2qr \right)}{\left( {{p}^{2}}+{{q}^{2}}-2pq \right)-{{r}^{2}}}\\=\frac{{{p}^{2}}-{{\left( q+r \right)}^{2}}}{{{\left( p-q \right)}^{2}}-{{r}^{2}}}\\=\frac{\left( p+q+r \right)\left( p-q-r \right)}{\left( p-q+r \right)\left( p-q-r \right)}\\=\frac{p+q+r}{p-q+r}=\frac{0.25-0.05+0.5}{0.25+0.05+0.5}\\=\frac{0.7}{0.8}=\frac{7}{8}\end{array}$
Question 2
If a x b = a2+ b2–ab, then the value of 9 × 11 is
A
103
B
113
C
119
D
129
Question 2 Explanation: 
$ \displaystyle \begin{array}{l}a\times b={{a}^{2}}+{{b}^{2}}-ab\,\left( Given \right)\\\Rightarrow 9\times 11={{9}^{2}}+{{11}^{2}}-3\times 11\\=81+121-99\\=202-99=103\end{array}$
Question 3
If p = 999, then the value of
$ \displaystyle 3\sqrt{p\left( {{p}^{2}}+3p+3 \right)+1}$
A
1000
B
999
C
998
D
1002
Question 3 Explanation: 
p= 999 (Given)
$ \displaystyle \begin{array}{l}Now,\,3\sqrt{p\left( {{p}^{2}}+3p+3 \right)+1}\,\\3\sqrt{{{p}^{3}}+3{{p}^{2}}+3p+1}\,\\=3\sqrt{{{\left( p+1 \right)}^{3}}}\,=p+1\\=333+1=1000\end{array}$
Question 4
If 55a + 5=1, then a equals
A
-1
B
-2
C
3/5
D
-4/5
Question 4 Explanation: 
$ \displaystyle \begin{array}{l}{{5}^{5a+5}}=1\\\Rightarrow {{5}^{5a}}\times {{5}^{5}}\Rightarrow {{5}^{5a}}=\frac{1}{{{5}^{5}}}\\\Rightarrow {{5}^{5a}}={{5}^{-5}}\Rightarrow 5a-5\\\Rightarrow a=-1\end{array}$
Question 5
If 3r+3+7=250, then r is equal to
A
4
B
6
C
2
D
8
Question 5 Explanation: 
$ \displaystyle \begin{array}{l}{{3}^{r+3}}+7=250\\\Rightarrow {{3}^{r+3}}=243\,\,\,\,\,\,\,\,\Rightarrow {{3}^{r+3}}={{3}^{5}}\\\Rightarrow r+3=5\Rightarrow r=2\end{array}$
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