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Algebra: Basics Test-3

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Question 1
$ \displaystyle if\,\,\,\frac{a}{3}=\frac{b}{4}=\frac{c}{7}\,\,then\,\,\frac{a+b+c}{c}\,\,is\,\,equal\,\,\,to:$
A
2
B
4
C
6
D
7
Question 1 Explanation: 
$ \displaystyle \begin{array}{l}\frac{a}{3}=\frac{b}{4}=\frac{c}{7}=1\,(let)\\a=3k,\,=4k\,,c=7k\\\therefore \frac{a+b+c}{c}=\frac{3k+4k+7k}{7k}\\=\frac{14k}{7k}=2\end{array}$
Question 2
$ \displaystyle f\,\,\frac{144}{0.144}=\frac{14.4}{p}$,then the value of p is
A
144
B
0.0144
C
1.44
D
14.4
Question 2 Explanation: 
$ \displaystyle \begin{array}{l}\frac{144}{0.144}=\frac{14.4}{p}\\\Rightarrow 144\times p=14.4\times 0.144\\\Rightarrow p=\frac{14.4\times 0.144}{144}\\=\frac{144\times 144}{144\times 10000}=0.0144\end{array}$
Question 3
If 1 < c < 2, then the value of
$ \displaystyle \sqrt{{{\left( c-1 \right)}^{2}}}\,+\sqrt{{{\left( c-3 \right)}^{2}}}\,$ is
A
3
B
2x−4
C
1
D
2
Question 3 Explanation: 
Since 1< c < 2, we have
c−1 > 0 and
c−3 < 0
or, 3 –c > 0
$ \displaystyle \therefore \sqrt{{{\left( c-1 \right)}^{2}}}+\sqrt{{{\left( 3-c \right)}^{2}}}\,$
$ \displaystyle \begin{array}{l}=\sqrt{{{\left( c-1 \right)}^{2}}}\,+\sqrt{{{\left( 3-c \right)}^{2}}}\\\left[ \because \,{{\left( c-3 \right)}^{2}}={{\left( 3-c \right)}^{2}}\, \right]\\=c-1+3-c=2\end{array}$
Question 4
p x q = (p × q) +q, then 5 x 7 equals to
A
40
B
42
C
30
D
32
Question 4 Explanation: 
$ \displaystyle \begin{array}{l}p\times r=\left( p\times q \right)+q\\\therefore 5\times 7=\left( 5\times 7 \right)+7=35+7=42\end{array}$
Question 5
If f = 0.1039, then the value of $ \displaystyle \sqrt{4{{f}^{2}}-4f+1\,\,}\,$ +3f is
A
1.1039
B
2.1039
C
0.1039
D
0.2078
Question 5 Explanation: 
f= 0.1039 (given)
$ \displaystyle \begin{array}{l}Now,\,\sqrt{4{{f}^{2}}-4f+1+3f}\,\\=\sqrt{{{\left( 1-2f \right)}^{2}}}\,+3f\\=1-2f+3f\\=1+f=1+0.1039\\=1.1039\end{array}$
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