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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:$
2 | |
4 | |
6 | |
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
144 | |
0.0144 | |
1.44 | |
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
$ \displaystyle \sqrt{{{\left( c-1 \right)}^{2}}}\,+\sqrt{{{\left( c-3 \right)}^{2}}}\,$ is
3 | |
2x−4 | |
1 | |
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}$
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
40 | |
42 | |
30 | |
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
1.1039 | |
2.1039 | |
0.1039 | |
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}$
$ \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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