Simplification Questions in English

(D) To solve the expression $198.995 \times 12.005 + 16.25 \times 6.95 = ?$,we can use the method of approximation.
First,approximate the given values to the nearest whole numbers:
$198.995 \approx 199$
$12.005 \approx 12$
$16.25 \approx 16$
$6.95 \approx 7$
Now,substitute these values into the expression:
$? = (199 \times 12) + (16 \times 7)$
Calculate the products:
$199 \times 12 = 2388$
$16 \times 7 = 112$
Add the results:
$? = 2388 + 112 = 2500$
Therefore,the approximate value is $2500$.

(A) To solve the expression $1.2 \times 1.02 \times 1.002$,we perform the multiplication step by step.
First,multiply $1.2$ and $1.02$:
$1.2 \times 1.02 = 1.224$
Next,multiply the result by $1.002$:
$1.224 \times 1.002 = 1.226448$
Therefore,the final answer is $1.226448$.

(D) To solve the expression $6.4 - 1.6 \div 0.2$,we follow the order of operations ($BODMAS$/$PEMDAS$).
First,perform the division: $1.6 \div 0.2 = 1.6 \times \frac{1}{0.2} = 1.6 \times 5 = 8$.
Next,perform the subtraction: $6.4 - 8 = -1.6$.
Therefore,the correct answer is $-1.6$.

(D) To find the product of $8.88 \times 88.8 \times 88$,we perform the multiplication step-by-step.
First,multiply $8.88$ by $88.8$:
$8.88 \times 88.8 = 788.544$
Next,multiply the result by $88$:
$788.544 \times 88 = 69391.872$
Therefore,the correct answer is $69391.872$.

(C) To solve the expression $989.001 + 1.00982 \times 76.792$,we follow the order of operations ($BODMAS$/$PEMDAS$).
First,perform the multiplication: $1.00982 \times 76.792 \approx 1.01 \times 76.8 \approx 77.55$.
Next,perform the addition: $989.001 + 77.55 = 1066.551$.
Comparing this result with the given options,the closest value is $1065$.

(C) To solve the expression $3.75 + 2.832 - 1.001 + 1.803$,follow the order of operations (addition and subtraction from left to right):
Step $1$: Add $3.75$ and $2.832$: $3.750 + 2.832 = 6.582$.
Step $2$: Subtract $1.001$ from the result: $6.582 - 1.001 = 5.581$.
Step $3$: Add $1.803$ to the result: $5.581 + 1.803 = 7.384$.
Therefore,the final answer is $7.384$.

(B) To find the smallest fraction,we compare the given fractions:
$1$. Compare $\frac{6}{11}$ and $\frac{13}{17}$:
$6 \times 17 = 102$ and $11 \times 13 = 143$.
Since $102 < 143$,$\frac{6}{11} < \frac{13}{17}$.
$2$. Compare $\frac{6}{11}$ and $\frac{19}{27}$:
$6 \times 27 = 162$ and $11 \times 19 = 209$.
Since $162 < 209$,$\frac{6}{11} < \frac{19}{27}$.
$3$. Compare $\frac{6}{11}$ and $\frac{21}{23}$:
$6 \times 23 = 138$ and $11 \times 21 = 231$.
Since $138 < 231$,$\frac{6}{11} < \frac{21}{23}$.
$4$. Compare $\frac{6}{11}$ and $\frac{5}{7}$:
$6 \times 7 = 42$ and $11 \times 5 = 55$.
Since $42 < 55$,$\frac{6}{11} < \frac{5}{7}$.
Therefore,the smallest fraction is $\frac{6}{11}$.

(C) First,simplify the given expression: $\frac{2}{5} \times \frac{15}{21} \times \frac{7}{10} \times \frac{3}{8}$.
$= (\frac{2}{5} \times \frac{15}{21}) \times (\frac{7}{10} \times \frac{3}{8})$
$= (\frac{2 \times 3}{1 \times 21}) \times (\frac{21}{80})$ is incorrect,let's calculate step by step:
$= \frac{2 \times 15 \times 7 \times 3}{5 \times 21 \times 10 \times 8} = \frac{630}{8400} = \frac{63}{840} = \frac{3}{40}$.
Since $\frac{3}{40} < 1$,the smallest fraction to add to get the next whole number (which is $1$) is $1 - \frac{3}{40}$.
$= \frac{40 - 3}{40} = \frac{37}{40}$.

(A) The given expression is of the form $\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}}$.
We know the algebraic identity $a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2})$.
Therefore,the expression simplifies to $\frac{(a-b)(a^{2}+ab+b^{2})}{a^{2}+ab+b^{2}} = a-b$.
Here,$a = 0.96$ and $b = 0.1$.
Substituting these values,we get $a-b = 0.96 - 0.1 = 0.86$.

(C) The given expression is $\frac{(2.3)^{3}-0.027}{(2.3)^{2}+0.69+0.09}$.
We can rewrite $0.027$ as $(0.3)^{3}$ and $0.09$ as $(0.3)^{2}$. Also,$0.69 = (2.3)(0.3)$.
So,the expression becomes $\frac{(2.3)^{3}-(0.3)^{3}}{(2.3)^{2}+(2.3)(0.3)+(0.3)^{2}}$.
Using the algebraic identity $a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2})$,we have:
$\frac{(a-b)(a^{2}+ab+b^{2})}{a^{2}+ab+b^{2}} = a-b$.
Here,$a = 2.3$ and $b = 0.3$.
Therefore,$a-b = 2.3 - 0.3 = 2$.

(C) To simplify the expression $\frac{0.2 \times 0.2 + 0.2 \times 0.02}{0.044}$,follow these steps:
$1$. Calculate the products in the numerator:
$0.2 \times 0.2 = 0.04$
$0.2 \times 0.02 = 0.004$
$2$. Add the results in the numerator:
$0.04 + 0.004 = 0.044$
$3$. Divide the numerator by the denominator:
$\frac{0.044}{0.044} = 1$

(B) The given expression is in the form $\frac{a^{2}-b^{2}}{a-b}$.
Using the algebraic identity $a^{2}-b^{2} = (a-b)(a+b)$,we can simplify the expression as follows:
$\frac{(a-b)(a+b)}{a-b} = a+b$.
Here,$a = 2.39$ and $b = 1.61$.
Substituting these values,we get:
$a + b = 2.39 + 1.61 = 4.00$.
Therefore,the value of the expression is $4$.

(C) The expression is in the form $\frac{a^2 - b^2}{a - b}$,where $a = 2.644$ and $b = 2.356$.
Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$,we can rewrite the numerator as $(2.644 - 2.356)(2.644 + 2.356)$.
Substituting this into the expression: $\frac{(2.644 - 2.356)(2.644 + 2.356)}{0.288}$.
Since $2.644 - 2.356 = 0.288$,the expression becomes $\frac{0.288 \times (2.644 + 2.356)}{0.288}$.
Canceling $0.288$ from the numerator and denominator,we get $2.644 + 2.356 = 5$.

(B) Given expression: $\frac{1.49 \times 14.9 - 0.51 \times 5.1}{14.9 - 5.1}$
Rewrite the terms to simplify:
$14.9 = 1.49 \times 10$ and $5.1 = 0.51 \times 10$
Substitute these into the expression:
$= \frac{1.49 \times (1.49 \times 10) - 0.51 \times (0.51 \times 10)}{14.9 - 5.1}$
$= \frac{10 \times (1.49^2 - 0.51^2)}{14.9 - 5.1}$
Note that $14.9 - 5.1 = 9.8$ and $1.49 - 0.51 = 0.98$. Thus,$14.9 - 5.1 = 10 \times (1.49 - 0.51)$.
$= \frac{10 \times (1.49 - 0.51)(1.49 + 0.51)}{10 \times (1.49 - 0.51)}$
$= 1.49 + 0.51 = 2.00$

(B) The given expression is $\frac{4.2 \times 4.2 - 1.9 \times 1.9}{2.3 \times 6.1}$.
Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$,where $a = 4.2$ and $b = 1.9$:
Numerator $= (4.2)^2 - (1.9)^2 = (4.2 - 1.9)(4.2 + 1.9) = (2.3)(6.1)$.
Substituting this back into the expression:
$\frac{(2.3)(6.1)}{2.3 \times 6.1} = 1$.

(D) The expression is $(7.5 \times 7.5 + 37.5 + 2.5 \times 2.5)$.
We can rewrite $37.5$ as $2 \times 7.5 \times 2.5$ because $2 \times 7.5 = 15$ and $15 \times 2.5 = 37.5$.
Thus,the expression becomes $(7.5)^2 + 2(7.5)(2.5) + (2.5)^2$.
This is in the form of the algebraic identity $a^2 + 2ab + b^2 = (a + b)^2$,where $a = 7.5$ and $b = 2.5$.
Substituting the values,we get $(7.5 + 2.5)^2 = (10)^2 = 100$.

(C) The given expression is of the form $\frac{a^{3} + b^{3}}{a^{2} - ab + b^{2}}$,where $a = 0.051$ and $b = 0.041$.
We know the algebraic identity: $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$.
Substituting this into the expression,we get: $\frac{(a + b)(a^{2} - ab + b^{2})}{a^{2} - ab + b^{2}} = a + b$.
Now,substitute the values of $a$ and $b$: $0.051 + 0.041 = 0.092$.
Therefore,the value of the expression is $0.092$.

(C) The given expression is $\frac{0.125+0.027}{0.5 \times 0.5+0.09-0.15}$.
We can write the numerator as $(0.5)^3 + (0.3)^3$ because $0.5^3 = 0.125$ and $0.3^3 = 0.027$.
We can write the denominator as $(0.5)^2 + (0.3)^2 - (0.5)(0.3)$ because $0.5^2 = 0.25$,$0.3^2 = 0.09$,and $(0.5)(0.3) = 0.15$.
Using the algebraic identity $a^3 + b^3 = (a+b)(a^2 + b^2 - ab)$,where $a = 0.5$ and $b = 0.3$,the expression becomes:
$\frac{(0.5+0.3)(0.5^2 + 0.3^2 - 0.5 \times 0.3)}{0.5^2 + 0.3^2 - 0.5 \times 0.3}$.
Canceling the common term $(0.5^2 + 0.3^2 - 0.5 \times 0.3)$,we get $0.5 + 0.3 = 0.8$.

(C) The given expression is $\frac{8(3.75)^{3}+1}{(7.5)^{2}-6.5}$.
We can rewrite $8(3.75)^{3}$ as $(2 \times 3.75)^{3} = (7.5)^{3}$.
So,the expression becomes $\frac{(7.5)^{3} + 1^{3}}{(7.5)^{2} - 6.5}$.
Note that $6.5 = 7.5 - 1$,so the denominator is $(7.5)^{2} - (7.5 - 1) = (7.5)^{2} - 7.5 + 1$.
Actually,using the identity $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2})$,where $a = 7.5$ and $b = 1$:
Numerator $= (7.5 + 1)((7.5)^{2} - (7.5)(1) + 1^{2}) = (8.5)((7.5)^{2} - 7.5 + 1)$.
Denominator $= (7.5)^{2} - 6.5 = 56.25 - 6.5 = 49.75$.
Wait,let us re-evaluate the denominator: $(7.5)^{2} - 6.5 = 56.25 - 6.5 = 49.75$.
Using the identity: $\frac{(7.5)^{3} + 1^{3}}{(7.5)^{2} - 7.5 + 1} = \frac{(7.5+1)(7.5^{2} - 7.5 + 1)}{7.5^{2} - 7.5 + 1} = 7.5 + 1 = 8.5$.

(C) The given expression is in the form of $\frac{a^3 + b^3}{a^2 - ab + b^2}$,where $a = 10.3$ and $b = 1$.
Using the algebraic identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$,we can rewrite the expression as:
$\frac{(a + b)(a^2 - ab + b^2)}{(a^2 - ab + b^2)} = a + b$.
Substituting the values $a = 10.3$ and $b = 1$:
$10.3 + 1 = 11.3$.

(D) Given expression $= \frac{5 \times 1.6 - 2 \times 1.4}{1.3}$
First,perform the multiplications in the numerator:
$5 \times 1.6 = 8.0$
$2 \times 1.4 = 2.8$
Now,substitute these values back into the expression:
$= \frac{8.0 - 2.8}{1.3}$
$= \frac{5.2}{1.3}$
$= 4$

(C) To find the approximate value,we round the numbers to their nearest convenient values:
$3.157 \approx 3.2$
$3.198 \approx 3.2$
$63.972 \approx 64$
$2835.121 \approx 2835$
Substituting these values into the expression:
$\frac{3.2 \times 4126 \times 3.2}{64 \times 2835} = \frac{10.24 \times 4126}{181440} \approx \frac{42250}{181440} \approx 0.2328$
Rounding to the nearest option,the value is approximately $0.2$.

(A) To solve the expression $\frac{0.0203 \times 2.92}{0.0073 \times 14.5 \times 0.7}$,we first remove the decimals by multiplying the numerator and denominator by appropriate powers of $10$.
Numerator: $0.0203 \times 2.92 = \frac{203}{10000} \times \frac{292}{100} = \frac{203 \times 292}{10^6}$.
Denominator: $0.0073 \times 14.5 \times 0.7 = \frac{73}{10000} \times \frac{145}{10} \times \frac{7}{10} = \frac{73 \times 145 \times 7}{10^6}$.
Dividing the two,the $10^6$ terms cancel out:
$\frac{203 \times 292}{73 \times 145 \times 7}$.
Simplifying the fractions:
$203 / 7 = 29$.
$292 / 73 = 4$.
So,the expression becomes $\frac{29 \times 4}{145}$.
Since $145 = 29 \times 5$,we have $\frac{29 \times 4}{29 \times 5} = \frac{4}{5} = 0.8$.

(B) To solve the expression $\frac{3.6 \times 0.48 \times 2.50}{0.12 \times 0.09 \times 0.5}$,we first remove the decimals by multiplying the numerator and denominator by appropriate powers of $10$.
Numerator: $3.6 \times 0.48 \times 2.50 = \frac{36}{10} \times \frac{48}{100} \times \frac{250}{100} = \frac{36 \times 48 \times 250}{100000}$
Denominator: $0.12 \times 0.09 \times 0.5 = \frac{12}{100} \times \frac{9}{100} \times \frac{5}{10} = \frac{12 \times 9 \times 5}{100000}$
Dividing the two,the $100000$ terms cancel out:
$\frac{36 \times 48 \times 250}{12 \times 9 \times 5} = \frac{36}{12} \times \frac{48}{9} \times \frac{250}{5} = 3 \times 5.333... \times 50$ (Wait,let's simplify differently):
$\frac{36}{12} = 3$
$\frac{48}{9} = \frac{16}{3}$
$\frac{250}{5} = 50$
Result $= 3 \times \frac{16}{3} \times 50 = 16 \times 50 = 800$.

(D) Given expression $= \frac{(0.1667)(0.8333)(0.3333)}{(0.2222)(0.6667)(0.1250)}$
Converting the decimals to their fractional equivalents:
$0.1667 \approx \frac{1}{6}$
$0.8333 \approx \frac{5}{6}$
$0.3333 \approx \frac{1}{3}$
$0.2222 \approx \frac{2}{9}$
$0.6667 \approx \frac{2}{3}$
$0.1250 = \frac{1}{8}$
Substituting these values into the expression:
$= \left( \frac{1}{6} \times \frac{5}{6} \times \frac{1}{3} \right) \div \left( \frac{2}{9} \times \frac{2}{3} \times \frac{1}{8} \right)$
$= \left( \frac{5}{108} \right) \div \left( \frac{4}{216} \right)$
$= \frac{5}{108} \times \frac{216}{4}$
$= \frac{5}{108} \times 54 = 5 \times 0.5 = 2.5$

(C) Given that $1^{3}+2^{3}+\cdots+9^{3}=2025$.
We need to find the value of $(0.11)^{3}+(0.22)^{3}+\cdots+(0.99)^{3}$.
This can be written as $(0.11 \times 1)^{3} + (0.11 \times 2)^{3} + \cdots + (0.11 \times 9)^{3}$.
Factoring out $(0.11)^{3}$,we get $(0.11)^{3} \times (1^{3} + 2^{3} + \cdots + 9^{3})$.
Substituting the given sum,we have $(0.11)^{3} \times 2025$.
Since $(0.11)^{3} = 0.001331$,the expression becomes $0.001331 \times 2025$.
Calculating the product: $0.001331 \times 2025 = 2.695$.

(C) Given that $\frac{1}{6.198} = 0.16134.$
We need to find the value of $\frac{1}{0.0006198}.$
Observe that $0.0006198 = \frac{6.198}{10000}.$
Therefore,$\frac{1}{0.0006198} = \frac{1}{\frac{6.198}{10000}} = \frac{10000}{6.198}.$
This can be written as $10000 \times \frac{1}{6.198}.$
Substituting the given value,we get $10000 \times 0.16134 = 1613.4.$

(C) Given that $\frac{2994}{14.5} = 172$.
To find the value of $\frac{29.94}{1.45}$,we can rewrite the expression as:
$\frac{29.94}{1.45} = \frac{2994 \div 100}{14.5 \div 10} = \frac{2994}{14.5} \times \frac{10}{100}$.
Substituting the given value:
$= 172 \times \frac{1}{10} = 17.2$.

(A) To solve $(0 . \overline{09} \times 7 . \overline{3})$,we first convert the repeating decimals into fractions.
$1.$ Convert $0 . \overline{09}$ to a fraction:
Let $x = 0 . \overline{09} = 0.090909...$
$100x = 9.090909...$
$100x - x = 9.090909... - 0.090909... = 9$
$99x = 9 \implies x = \frac{9}{99} = \frac{1}{11}$.
$2.$ Convert $7 . \overline{3}$ to a fraction:
$7 . \overline{3} = 7 + 0 . \overline{3} = 7 + \frac{3}{9} = 7 + \frac{1}{3} = \frac{21+1}{3} = \frac{22}{3}$.
$3.$ Multiply the fractions:
$\frac{1}{11} \times \frac{22}{3} = \frac{1 \times 22}{11 \times 3} = \frac{22}{33} = \frac{2}{3}$.
$4.$ Convert $\frac{2}{3}$ back to decimal:
$\frac{2}{3} = 0.666... = 0 . \overline{6}$.

(D) To solve $3 . \overline{87} - 2 . \overline{59}$,we convert the recurring decimals into fractions.
$3 . \overline{87} = 3 + \frac{87}{99} = \frac{3 \times 99 + 87}{99} = \frac{297 + 87}{99} = \frac{384}{99}$.
$2 . \overline{59} = 2 + \frac{59}{99} = \frac{2 \times 99 + 59}{99} = \frac{198 + 59}{99} = \frac{257}{99}$.
Now,subtract the two fractions:
$\frac{384}{99} - \frac{257}{99} = \frac{384 - 257}{99} = \frac{127}{99}$.
Converting $\frac{127}{99}$ back to a recurring decimal:
$\frac{127}{99} = 1 + \frac{28}{99} = 1 . \overline{28}$.

(C) Let $x = 2.1\overline{36}$.
This can be written as $x = 2 + 0.1\overline{36}$.
Let $y = 0.1\overline{36} = 0.1363636...$
Multiplying by $10$,we get $10y = 1.363636...$
Multiplying by $1000$,we get $1000y = 136.363636...$
Subtracting the two equations: $1000y - 10y = 136.3636... - 1.3636...$
$990y = 135$
$y = \frac{135}{990} = \frac{27}{198} = \frac{3}{22}$.
Therefore,$x = 2 + \frac{3}{22} = 2\frac{3}{22}$.

(D) Let $x = 6.\overline{46}$.
This can be written as $x = 6.464646...$
Multiply by $100$ to shift the decimal point: $100x = 646.464646...$
Subtract the original equation $x = 6.464646...$ from $100x = 646.464646...$:
$100x - x = 646.464646... - 6.464646...$
$99x = 640$
$x = \frac{640}{99}$
Alternatively,$6.\overline{46} = 6 + 0.\overline{46} = 6 + \frac{46}{99} = \frac{6 \times 99 + 46}{99} = \frac{594 + 46}{99} = \frac{640}{99}$.

(C) Let $x = 0.125125 \ldots$ (Equation $1$).
Since there are $3$ repeating digits,multiply both sides by $1000$:
$1000x = 125.125125 \ldots$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$1000x - x = 125.125125 \ldots - 0.125125 \ldots$
$999x = 125$
$x = \frac{125}{999}$
Thus,the rational number is $\frac{125}{999}$.

(C) Let $x = 0.232323 \ldots$ (Equation $1$)
Since there are two repeating digits,multiply both sides by $100$:
$100x = 23.232323 \ldots$ (Equation $2$)
Subtract Equation $1$ from Equation $2$:
$100x - x = 23.232323 \ldots - 0.232323 \ldots$
$99x = 23$
$x = \frac{23}{99}$
Thus,$0.232323 \ldots = 0.\overline{23} = \frac{23}{99}$.

(A) Given the equation: $\frac{144}{0.144} = \frac{14.4}{x}$
First,simplify the left side: $\frac{144}{0.144} = \frac{144000}{144} = 1000$
Now,substitute this back into the equation: $1000 = \frac{14.4}{x}$
Rearrange to solve for $x$: $x = \frac{14.4}{1000}$
$x = 0.0144$

(C) Let the missing value be $x$.
Given the equation: $\frac{0.009}{x} = 0.01$.
Rearranging the equation to solve for $x$:
$x = \frac{0.009}{0.01}$.
To simplify,multiply the numerator and denominator by $100$:
$x = \frac{0.009 \times 100}{0.01 \times 100} = \frac{0.9}{1} = 0.9$.
Therefore,the missing value is $0.9$.

(D) Let $x = 5 . \overline{45} = 5.454545...$
Multiply by $100$ to shift the decimal point after the repeating part:
$100x = 545.454545...$
Subtract the original equation $x = 5.454545...$ from the new equation:
$100x - x = 545.454545... - 5.454545...$
$99x = 540$
Therefore,$x = \frac{540}{99}$.

(A) Step $1$: Calculate the value of the first quotient: $\frac{368.39}{17} = 21.67$.
Step $2$: Calculate the value of the second quotient: $\frac{170.50}{62} = 2.75$.
Step $3$: Calculate the value of the third quotient: $\frac{875.65}{83} = 10.55$.
Step $4$: Compare the values: $21.67 > 10.55 > 2.75$.
Therefore,the descending order is $1, 3, 2$. The correct option is $(A)$.

(B) To solve $0.04 \times 0.0162$,we can express the decimals as powers of $10$:
$0.04 = 4 \times 10^{-2}$
$0.0162 = 162 \times 10^{-4}$
Now,multiply these values:
$(4 \times 10^{-2}) \times (162 \times 10^{-4}) = (4 \times 162) \times (10^{-2} \times 10^{-4})$
$= 648 \times 10^{-6}$
To express this in scientific notation,shift the decimal point two places to the left:
$648 \times 10^{-6} = 6.48 \times 10^{2} \times 10^{-6} = 6.48 \times 10^{-4}$

(C) To solve the expression $40.83 \times 1.02 \times 1.2$,we first multiply the numbers as if they were integers:
$4083 \times 102 \times 12 = 4997592$
Next,we count the total number of decimal places in the original expression:
$40.83$ has $2$ decimal places.
$1.02$ has $2$ decimal places.
$1.2$ has $1$ decimal place.
Total decimal places = $2 + 2 + 1 = 5$.
Now,place the decimal point in the product $4997592$ such that there are $5$ digits to the right of the decimal point:
$49.97592$

(B) To multiply $16.02$ by $0.001$,we shift the decimal point in $16.02$ three places to the left.
Starting from $16.02$,moving the decimal point three places to the left results in $0.01602$.
Therefore,$16.02 \times 0.001 = 0.01602$.

(B) To multiply $0.002$ by $0.5$,we can ignore the decimal points initially and multiply the numbers as integers:
$2 \times 5 = 10$
Next,count the total number of decimal places in the original numbers:
$0.002$ has $3$ decimal places.
$0.5$ has $1$ decimal place.
Total decimal places $= 3 + 1 = 4$.
Now,place the decimal point in the result $(10)$ such that there are $4$ digits to the right of the decimal point:
$0.0010$,which is equivalent to $0.001$.

(A) To arrange the fractions in ascending order,convert them into decimal form:
$\frac{1}{3} \approx 0.333$
$\frac{2}{5} = 0.4$
$\frac{4}{7} \approx 0.571$
$\frac{3}{5} = 0.6$
$\frac{5}{6} \approx 0.833$
$\frac{6}{7} \approx 0.857$
Comparing the decimal values: $0.333 < 0.4 < 0.571 < 0.6 < 0.833 < 0.857$.
Thus,the ascending order is $\frac{1}{3}, \frac{2}{5}, \frac{4}{7}, \frac{3}{5}, \frac{5}{6}, \frac{6}{7}$.

(A) To find the biggest and smallest fractions,we convert them into decimal form:
$\frac{2}{3} \approx 0.666$
$\frac{3}{4} = 0.75$
$\frac{4}{5} = 0.8$
$\frac{5}{6} \approx 0.833$
The biggest fraction is $\frac{5}{6}$ and the smallest fraction is $\frac{2}{3}$.
The required difference is $\frac{5}{6} - \frac{2}{3}$.
Taking the least common multiple $(LCM)$ of $6$ and $3$,which is $6$:
$\frac{5}{6} - \frac{4}{6} = \frac{1}{6}$.

(D) To determine the descending order,convert each fraction into its decimal form:
$\frac{5}{9} \approx 0.555$
$\frac{7}{11} \approx 0.636$
$\frac{8}{15} \approx 0.533$
$\frac{11}{17} \approx 0.647$
Comparing the decimal values:
$0.647 > 0.636 > 0.555 > 0.533$
Therefore,the descending order is:
$\frac{11}{17} > \frac{7}{11} > \frac{5}{9} > \frac{8}{15}$

(D) Given equation: $832.58 - 242.31 = 779.84 - ?$
First,calculate the left-hand side: $832.58 - 242.31 = 590.27$
Now,the equation becomes: $590.27 = 779.84 - ?$
Rearranging to solve for $?$: $? = 779.84 - 590.27$
Performing the subtraction: $779.84 - 590.27 = 189.57$
Since $189.57$ is not among the options $A, B,$ or $C$,the correct answer is $D$.

(D) Given equation: $138.009 + 341.981 - 146.305 = 123.6 + ?$
First,calculate the sum of the first two terms: $138.009 + 341.981 = 479.990$
Next,subtract the third term: $479.990 - 146.305 = 333.685$
Now,the equation becomes: $333.685 = 123.6 + ?$
To find the value of $?$,subtract $123.6$ from $333.685$: $? = 333.685 - 123.6 = 210.085$
Since $210.085$ is not among the options $A, B,$ or $C$,the correct choice is $D$.

(C) To solve the expression $12.1212 + 17.0005 - 9.1102$,follow the order of operations (addition and subtraction).
First,add the positive numbers: $12.1212 + 17.0005 = 29.1217$.
Next,subtract the third number from the sum: $29.1217 - 9.1102 = 20.0115$.
Thus,the final result is $20.0115$.

(D) To subtract $32.006$ from $48.95$,we first align the decimal points:
$48.950 - 32.006$
Performing the subtraction:
$48.950 - 32.006 = 16.944$
Thus,the correct answer is $16.944$.

(C) To find the sum,align the decimal points of all the numbers:
$617.0000$
$6.0170$
$0.6170$
$6.0017$
----------
$629.6357$
Therefore,the sum is $629.6357$.