Basic Maths: Test 13

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Basic Maths: Test 13

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Question 1

$\displaystyle 999\frac{98}{49}\times 49$ 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 of

$\displaystyle 49\frac{95}{49}\times 49$ 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

$\displaystyle 499\frac{995}{499}\times 499$ 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

$\displaystyle 3+\left( 3+1 \right)\,({{3}^{2}}+1)\,\left( {{3}^{4}}+1 \right)\,\left( {{3}^{8}}+1 \right)\,\left( {{3}^{16}}+1 \right)\,\left( {{3}^{32}}+1 \right)$ 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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