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Basic Maths: Test 18

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 Question 1
Simplify:$\displaystyle \frac{1+\frac{1}{2}}{1-\frac{1}{2}}\div \frac{3}{14}\left( \frac{2}{5}+\frac{3}{10} \right)\,of\,\frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{2}-\frac{1}{3}}$
 A $\displaystyle 4$ B $\displaystyle 37\frac{1}{2}$ C $\displaystyle \frac{3}{2}$ D $\displaystyle 18\frac{3}{8}$
Question 1 Explanation:
 Question 2
SimplifyÂ $\displaystyle \left[ 0.9\div \left\{ 2.3-3.2-\left( 7.1-5.4-3.5 \right) \right\} \right]$
 A 0.18 B 1.8 C 1 D 2.6
Question 2 Explanation:
$\displaystyle \begin{array}{l}\left[ 0.9\div \left\{ 2.3-3.2-\left( 7.1-8.9 \right) \right\} \right]\\=\left[ 0.9\div \left\{ 2.3-3.2+1.8 \right\} \right]\\=\left[ 0.9\div 0.9 \right]\\=1\end{array}$
 Question 3
The value ofÂ $\displaystyle \frac{{{\left( 75.8 \right)}^{2}}-{{\left( 55.8 \right)}^{2}}}{40}$ is
 A 20 B 40 C 65.8 D 131.6
Question 3 Explanation:
$\displaystyle \begin{array}{l}\frac{{{(75.8)}^{2}}-{{\left( 55.8 \right)}^{2}}}{40}\\=\frac{\left( 75.8-55.8 \right)\left( 75.8+55.8 \right)}{40}\\=\frac{20\times 131.6}{40}=65.8\end{array}$
 Question 4
$\displaystyle \frac{1}{4}+\left\{ 4\frac{3}{4}-\left( 3\frac{1}{6}-2\frac{1}{3} \right) \right\}$Â is equal to
 A $\displaystyle 3\frac{2}{3}$ B $\displaystyle 1\frac{1}{4}$ C $\displaystyle 4\frac{1}{6}$ D $\displaystyle 1\frac{2}{3}$
Question 4 Explanation:
Expression
$\displaystyle \begin{array}{l}=\frac{1}{4}+\left\{ \frac{19}{4}-\left( \frac{19}{6}-\frac{7}{3} \right)\, \right\}\\=\frac{1}{4}+\left\{ \frac{19}{4}-\left( \frac{19-14}{6} \right)\, \right\}\\=\frac{1}{4}+\left\{ \frac{19}{4}-\frac{5}{6} \right\}\\=\frac{1}{4}+\frac{19}{4}-\frac{5}{6}\\=\frac{50}{12}\\=4\frac{1}{6}\end{array}$
 Question 5
If p represents a number, then the value of p in Â $\displaystyle 5\frac{3}{p}\times 3\frac{1}{2}=19$ is:
 A 7 B 4 C 6 D 2
Question 5 Explanation:
$\displaystyle \begin{array}{l}5\frac{3}{p}\times \frac{7}{2}=19\\\Rightarrow 5\frac{3}{p}=\frac{19\times 2}{7}\\\Rightarrow 5\frac{3}{p}=\frac{38}{7}=5\frac{3}{7}\\\Rightarrow p=7\end{array}$
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