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• This is an assessment test.
• These tests focus on the basics of Maths and are meant to indicate your preparation level for the subject.
• Kindly take the tests in this series with a pre-defined schedule.

## Basic Maths: Test 13

Congratulations - you have completed Basic Maths: Test 13.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
 Question 1
$\displaystyle 999\frac{98}{49}\times 49$Â is equal to:
 A 49049 B 49349 C 49449 D 49249
Question 1 Explanation:
$\displaystyle \begin{array}{l}\left( 999+\frac{98}{49} \right)\times 49\\=(999+2)\times 49\\=\left( 1001 \right)49\\=\text{49049}\end{array}$
 Question 2
The value ofÂ $\displaystyle 49\frac{95}{49}\times 49$ is
 A 2296 B 2396 C 2196 D 2496
Question 2 Explanation:
Expression $\displaystyle \begin{array}{l}=\left( 49+\frac{95}{49} \right)\times 49\\=49\times 49+95=\text{2496}\end{array}$
 Question 3
The value of $\displaystyle 499\frac{995}{499}\times 499$Â is
 A 249796 B 249996 C 249986 D 249906
Question 3 Explanation:
$\displaystyle \begin{array}{l}=\left( 499+\frac{995}{499} \right)\times 499\\={{\left( 499 \right)}^{2}}+995\\={{\left( 500-1 \right)}^{2}}+995\\=250000+1-1000+995\\=\text{249996}\end{array}$
 Question 4
$\displaystyle \frac{{{\left( 998 \right)}^{2}}-{{\left( 997 \right)}^{2}}-240}{{{\left( 98 \right)}^{2}}-{{\left( 97 \right)}^{2}}}$ equals
 A 1995 B 9 C 95 D 10
Question 4 Explanation:
$\displaystyle \begin{array}{l}=\frac{\left[ {{\left( 998 \right)}^{2}}-{{\left( 997 \right)}^{2}} \right]-240}{{{\left( 98 \right)}^{2}}-{{\left( 97 \right)}^{2}}}\\=\frac{\left( 998+997 \right)\,\left( 998-997 \right)-240}{\left( 98+97 \right)\,\left( 98-97 \right)}\\=\frac{1995-240}{195}=\frac{1755}{195}=9\end{array}$
 Question 5
$\displaystyle 3+\left( 3+1 \right)\,({{3}^{2}}+1)\,\left( {{3}^{4}}+1 \right)\,\left( {{3}^{8}}+1 \right)\,\left( {{3}^{16}}+1 \right)\,\left( {{3}^{32}}+1 \right)$Â is equal to
 A $\displaystyle \frac{{{3}^{64}}-1}{2}$ B $\displaystyle \frac{{{3}^{64}}+5}{2}$ C $\displaystyle {{3}^{64}}-1$ D $\displaystyle {{3}^{64}}+1$
Question 5 Explanation:
$\displaystyle \begin{array}{l}3+\left( 3+1 \right)\,\left( {{3}^{2}}+1 \right)\,\left( {{3}^{4}}+1 \right)\left( {{3}^{8}}+1 \right)\left( {{3}^{16}}+1 \right)\left( {{3}^{32}}+1 \right)\\=3+\frac{\left( 3-1 \right)\,\left( 3+1 \right)}{3-1}\left( {{3}^{2}}+1 \right)\,\left( {{3}^{4}}+1 \right)......\left( {{3}^{32}}+1 \right)\\=3+\frac{\left( {{3}^{2}}-1 \right)\,\left( {{3}^{2}}+1 \right)\,\left( {{3}^{4}}+1 \right).....\left( {{3}^{32}}+1 \right)}{2}\\=3+\frac{\left( {{3}^{4}}-1 \right)\,\left( {{3}^{4}}+1 \right)\,\left( {{3}^{8}}+1 \right).....\left( {{3}^{32}}+1 \right)}{2}\\=3+\frac{\left( {{3}^{8}}-1 \right)\,\left( {{3}^{8}}+1 \right)\,\left( {{3}^{16}}+1 \right).....\left( {{3}^{32}}+1 \right)}{2}\\=3+\frac{\left( {{3}^{16}}-1 \right)\,\left( {{3}^{16}}+1 \right)\,\left( {{3}^{32}}+1 \right)}{2}\\=3+\frac{\left( {{3}^{32}}-1 \right)\,\left( {{3}^{32}}+1 \right)}{2}\\=3+\frac{{{3}^{64}}-1}{2}=\frac{{{3}^{64}}+5}{2}\end{array}$
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