Simplification Questions in English

(D) Step $1$: Simplify the numerator of the first fraction: $2 \frac{1}{7} - 2 \frac{1}{2} = \frac{15}{7} - \frac{5}{2} = \frac{30-35}{14} = \frac{-5}{14}$.
Step $2$: Simplify the denominator of the first fraction: $2 \frac{1}{4} + 1 \frac{1}{7} = \frac{9}{4} + \frac{8}{7} = \frac{63+32}{28} = \frac{95}{28}$.
Step $3$: Divide the results: $\frac{-5}{14} \div \frac{95}{28} = \frac{-5}{14} \times \frac{28}{95} = \frac{-1}{1} \times \frac{2}{19} = \frac{-2}{19}$.
Step $4$: Simplify the second part: $2 - \frac{1}{2} = \frac{3}{2}$.
Then,$2 + \frac{1}{3/2} = 2 + \frac{2}{3} = \frac{8}{3}$.
Then,$2 + \frac{1}{8/3} = 2 + \frac{3}{8} = \frac{19}{8}$.
So,the second part is $\frac{1}{19/8} = \frac{8}{19}$.
Step $5$: Perform the final division: $\frac{-2}{19} \div \frac{8}{19} = \frac{-2}{19} \times \frac{19}{8} = \frac{-2}{8} = \frac{-1}{4}$.

(B) Let $a = 2.75$ and $b = 2.25$.
The given expression is in the form $\frac{a^3 - b^3}{a^2 + ab + b^2}$.
We know the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.
Substituting this into the expression,we get $\frac{(a - b)(a^2 + ab + b^2)}{a^2 + ab + b^2}$.
Canceling the common term $(a^2 + ab + b^2)$,the expression simplifies to $a - b$.
Substituting the values back: $2.75 - 2.25 = 0.50$.

(C) Given expression is $\frac{\frac{1}{2} \div 4 + 20}{\frac{1}{2} \times 4 + 20}$.
First,solve the numerator: $\frac{1}{2} \div 4 + 20 = \frac{1}{2} \times \frac{1}{4} + 20 = \frac{1}{8} + 20 = \frac{1 + 160}{8} = \frac{161}{8}$.
Next,solve the denominator: $\frac{1}{2} \times 4 + 20 = 2 + 20 = 22$.
Now,divide the numerator by the denominator: $\frac{161/8}{22} = \frac{161}{8} \times \frac{1}{22} = \frac{161}{176}$.

(C) The given expression is in the form $\frac{a^2 - 2ab + b^2}{a - b}$,where $a = 0.53$ and $b = 0.41$.
Using the algebraic identity $a^2 - 2ab + b^2 = (a - b)^2$,the numerator becomes $(0.53 - 0.41)^2$.
Substituting the values,we get $\frac{(0.53 - 0.41)^2}{0.53 - 0.41}$.
This simplifies to $(0.53 - 0.41) = 0.12$.
Therefore,the correct answer is $0.12$.

(C) Given expression is $\frac{9^{2} \times 18^{4}}{3^{16}}$.
Expressing the bases in terms of prime factors:
$9 = 3^{2}$ and $18 = 2 \times 3^{2}$.
Substituting these into the expression:
$= \frac{(3^{2})^{2} \times (2 \times 3^{2})^{4}}{3^{16}}$
$= \frac{3^{4} \times 2^{4} \times (3^{2})^{4}}{3^{16}}$
$= \frac{3^{4} \times 2^{4} \times 3^{8}}{3^{16}}$
$= \frac{3^{4+8} \times 2^{4}}{3^{16}}$
$= \frac{3^{12} \times 2^{4}}{3^{16}}$
$= \frac{2^{4}}{3^{16-12}}$
$= \frac{16}{3^{4}}$
$= \frac{16}{81}$.

(C) To simplify the expression $1+\frac{1}{2+\frac{1}{1-\frac{1}{3}}}$,we solve from the bottom up.
First,simplify the denominator of the lowest fraction: $1-\frac{1}{3} = \frac{3-1}{3} = \frac{2}{3}$.
Now,the expression becomes $1+\frac{1}{2+\frac{1}{2/3}}$.
Next,simplify the term $\frac{1}{2/3}$,which is $\frac{3}{2}$.
Now,the expression becomes $1+\frac{1}{2+\frac{3}{2}}$.
Simplify the denominator $2+\frac{3}{2} = \frac{4+3}{2} = \frac{7}{2}$.
Finally,the expression becomes $1+\frac{1}{7/2} = 1+\frac{2}{7} = \frac{7+2}{7} = \frac{9}{7}$.

(D) Let the missing number be $x$.
The given equation is: $108 \div (x \text{ of } \frac{1}{3}) + \frac{2}{5} \times 3 \frac{3}{4} = 10 \frac{1}{2}$.
First,convert the mixed fractions to improper fractions: $3 \frac{3}{4} = \frac{15}{4}$ and $10 \frac{1}{2} = \frac{21}{2}$.
The equation becomes: $108 \div (x \times \frac{1}{3}) + \frac{2}{5} \times \frac{15}{4} = \frac{21}{2}$.
Simplify the multiplication term: $\frac{2}{5} \times \frac{15}{4} = \frac{1}{1} \times \frac{3}{2} = \frac{3}{2}$.
Now the equation is: $108 \div \frac{x}{3} + \frac{3}{2} = \frac{21}{2}$.
Convert division to multiplication by the reciprocal: $108 \times \frac{3}{x} + \frac{3}{2} = \frac{21}{2}$.
Subtract $\frac{3}{2}$ from both sides: $\frac{324}{x} = \frac{21}{2} - \frac{3}{2} = \frac{18}{2} = 9$.
Solve for $x$: $9x = 324$,so $x = \frac{324}{9} = 36$.

(D) The given expression is $1 + 1 + \frac{1}{12} + \frac{1}{4 \times 9} + \frac{1}{4 \times 27}$.
This simplifies to $2 + \frac{1}{12} + \frac{1}{36} + \frac{1}{108}$.
To add these fractions,find the least common multiple of $12, 36,$ and $108$,which is $108$.
The expression becomes $2 + \frac{9}{108} + \frac{3}{108} + \frac{1}{108}$.
Summing the fractions gives $2 + \frac{9 + 3 + 1}{108} = 2 + \frac{13}{108}$.
Calculating the decimal value: $2 + 0.12037... = 2.12037...$.
Rounding to four decimal places,we get $2.1204$.

(D) Let the required part be $x$.
According to the problem,$x$ of $\frac{1}{12} = \frac{3}{48}$.
First,simplify $\frac{3}{48}$ by dividing both numerator and denominator by $3$,which gives $\frac{1}{16}$.
So,$x \times \frac{1}{12} = \frac{1}{16}$.
Solving for $x$,we get $x = \frac{1}{16} \times 12$.
$x = \frac{12}{16}$.
Dividing both by $4$,we get $x = \frac{3}{4}$.

(C) To simplify the expression,we solve from the bottom up:
$1$. First,simplify the innermost fraction: $1 - \frac{2}{3} = \frac{3-2}{3} = \frac{1}{3}$.
$2$. Substitute this back into the expression: $1 + \frac{1}{1 + \frac{1}{1/3}} = 1 + \frac{1}{1 + 3}$.
$3$. Simplify the denominator: $1 + 3 = 4$.
$4$. The expression becomes: $1 + \frac{1}{4}$.
$5$. Finally,add the terms: $1 + \frac{1}{4} = \frac{4+1}{4} = \frac{5}{4}$.

(D) First,convert the given fractions into decimal form:
$\frac{1}{3} \approx 0.333$
$\frac{7}{8} = 0.875$
We need a fraction $x$ such that $0.333 < x < 0.875$.
Check the options:
$A) \frac{1}{4} = 0.25$. Since $0.25 < 0.333$,this is incorrect.
$B) \frac{23}{24} \approx 0.958$. Since $0.958 > 0.875$,this is incorrect.
$C) \frac{11}{12} \approx 0.916$. Since $0.916 > 0.875$,this is incorrect.
$D) \frac{17}{24} \approx 0.708$. Since $0.333 < 0.708 < 0.875$,this is the correct fraction.

(A) To find how many $\frac{1}{8}$'s are in $37 \frac{1}{2}$,we need to divide $37 \frac{1}{2}$ by $\frac{1}{8}$.
First,convert the mixed fraction $37 \frac{1}{2}$ into an improper fraction: $37 \frac{1}{2} = \frac{37 \times 2 + 1}{2} = \frac{75}{2}$.
Now,perform the division: $\frac{75}{2} \div \frac{1}{8} = \frac{75}{2} \times 8$.
Calculating the result: $\frac{75 \times 8}{2} = 75 \times 4 = 300$.

(D) Let the number of girls be $G$ and the number of boys be $B$.
Number of girls who took part in the camp = $\frac{1}{5}G$.
Number of boys who took part in the camp = $\frac{1}{8}B$.
Total number of students who took part in the camp = $\frac{1}{5}G + \frac{1}{8}B$.
Total number of students in the college = $G + B$.
The fraction of students who took part in the camp is $\frac{\frac{1}{5}G + \frac{1}{8}B}{G + B}$.
Since the ratio of girls to boys is not given,the exact fraction cannot be determined.
Therefore,the data is inadequate.

(C) Using the $BODMAS$ rule,we solve the expression step by step:
First,simplify the terms inside the parentheses: $(1 \frac{2}{5} - 1 \frac{1}{3}) = (\frac{7}{5} - \frac{4}{3}) = \frac{21-20}{15} = \frac{1}{15}$.
Next,evaluate the 'of' operations:
$\frac{1}{2} \text{ of } \frac{1}{4} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.
$1 \frac{7}{8} \text{ of } \frac{1}{15} = \frac{15}{8} \times \frac{1}{15} = \frac{1}{8}$.
Now substitute these back into the expression:
$= \{\frac{15}{2} + \frac{1}{2} \div \frac{1}{8} - \frac{2}{5} \times \frac{7}{3} \div \frac{1}{8}\}$.
Perform the division operations:
$= \{\frac{15}{2} + (\frac{1}{2} \times 8) - (\frac{2}{5} \times \frac{7}{3} \times 8)\}$.
$= \{\frac{15}{2} + 4 - \frac{112}{15}\}$.
$= \{\frac{23}{2} - \frac{112}{15}\}$.
$= \frac{345 - 224}{30} = \frac{121}{30} = 4 \frac{1}{30}$.

(C) The given expression is $\left(2-\frac{1}{3}\right)\left(2-\frac{3}{5}\right)\left(2-\frac{5}{7}\right) \ldots \left(2-\frac{999}{1001}\right)$.
Simplifying each term:
$\left(\frac{6-1}{3}\right) \times \left(\frac{10-3}{5}\right) \times \left(\frac{14-5}{7}\right) \times \ldots \times \left(\frac{2002-999}{1001}\right)$
$= \frac{5}{3} \times \frac{7}{5} \times \frac{9}{7} \times \ldots \times \frac{1003}{1001}$.
Observing the pattern,the numerator of each fraction cancels with the denominator of the subsequent fraction:
$= \frac{{5}}{3} \times \frac{{7}}{{5}} \times \frac{{9}}{{7}} \times \ldots \times \frac{1003}{{1001}}$
$= \frac{1003}{3}$.

(A) The given expression is $S = \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \frac{1}{4 \cdot 5 \cdot 6}$.
Calculating each term individually:
$\frac{1}{6} + \frac{1}{24} + \frac{1}{60} + \frac{1}{120}$.
To add these fractions,find the least common multiple $(LCM)$ of $6, 24, 60, 120$,which is $120$.
$S = \frac{20}{120} + \frac{5}{120} + \frac{2}{120} + \frac{1}{120}$.
$S = \frac{20 + 5 + 2 + 1}{120} = \frac{28}{120}$.
Simplifying the fraction by dividing both numerator and denominator by $4$,we get $\frac{7}{30}$.

(D) Convert the mixed fractions into improper fractions:
$15 \frac{2}{3} = \frac{47}{3}$,$3 \frac{1}{6} = \frac{19}{6}$,$6 \frac{1}{3} = \frac{19}{3}$,$11 \frac{7}{18} = \frac{205}{18}$.
Let the missing value be $x$.
The equation becomes: $\frac{47}{3} \times \frac{19}{6} + \frac{19}{3} = \frac{205}{18} + x$.
Calculate the product: $\frac{47 \times 19}{3 \times 6} = \frac{893}{18}$.
Now,$\frac{893}{18} + \frac{19}{3} = \frac{205}{18} + x$.
Convert $\frac{19}{3}$ to a fraction with denominator $18$: $\frac{19 \times 6}{3 \times 6} = \frac{114}{18}$.
So,$\frac{893}{18} + \frac{114}{18} = \frac{205}{18} + x$.
$\frac{1007}{18} = \frac{205}{18} + x$.
$x = \frac{1007}{18} - \frac{205}{18} = \frac{802}{18}$.
Simplifying $\frac{802}{18}$ by dividing by $2$: $\frac{401}{9}$.
Converting $\frac{401}{9}$ to a mixed fraction: $44 \frac{5}{9}$.

(A) Follow the order of operations ($BODMAS$/$PEMDAS$):
First,solve the innermost parenthesis: $(2 + \frac{8}{13}) = \frac{26+8}{13} = \frac{34}{13}$.
Next,solve the curly brackets: $(4 - 2) \div \frac{34}{13} = 2 \div \frac{34}{13} = 2 \times \frac{13}{34} = \frac{13}{17}$.
Then,solve the square brackets: $(8 - 5) \div \frac{13}{17} = 3 \div \frac{13}{17} = 3 \times \frac{17}{13} = \frac{51}{13}$.
Finally,solve the whole expression: $3 \div \frac{51}{13} = 3 \times \frac{13}{51} = \frac{13}{17}$.

(C) To compare the fractions,convert them into decimal form:
$\frac{3}{5} = 0.60$
$\frac{2}{3} \approx 0.666...$
$\frac{3}{4} = 0.75$
Comparing these values: $0.75 > 0.666... > 0.60$.
Therefore,$\frac{3}{4} > \frac{2}{3} > \frac{3}{5}$.

(C) Given expression is $\frac{(272-32)(124+176)}{17 \times 15-15}$.
First,solve the terms inside the parentheses:
$272 - 32 = 240$
$124 + 176 = 300$
Next,solve the denominator:
$17 \times 15 - 15 = 15(17 - 1) = 15 \times 16 = 240$.
Now,substitute these values back into the expression:
$\frac{240 \times 300}{240} = 300$.

(NONE) Given that $\frac{a}{b} = \frac{1}{2}.$
To find the value of $\frac{3a + 2b}{3a - 2b},$ divide both the numerator and the denominator by $b$:
$\frac{3a + 2b}{3a - 2b} = \frac{3(\frac{a}{b}) + 2}{3(\frac{a}{b}) - 2}.$
Substitute $\frac{a}{b} = \frac{1}{2}$ into the expression:
$= \frac{3(\frac{1}{2}) + 2}{3(\frac{1}{2}) - 2} = \frac{\frac{3}{2} + 2}{\frac{3}{2} - 2}.$
$= \frac{\frac{3 + 4}{2}}{\frac{3 - 4}{2}} = \frac{\frac{7}{2}}{-\frac{1}{2}}.$
$= \frac{7}{2} \times (-\frac{2}{1}) = -7.$

(A) To solve the expression $(20 \div 5) \div 2 + (16 \div 8) \times 2 + (10 \div 5) \times (3 \div 2)$,we follow the $BODMAS$ rule:
Step $1$: Solve the operations inside the parentheses:
$(20 \div 5) = 4$
$(16 \div 8) = 2$
$(10 \div 5) = 2$
$(3 \div 2) = 1.5$
Step $2$: Substitute these values back into the expression:
$4 \div 2 + 2 \times 2 + 2 \times 1.5$
Step $3$: Perform division and multiplication:
$2 + 4 + 3$
Step $4$: Perform addition:
$2 + 4 + 3 = 9$
Therefore,the correct answer is $9$.

(B) Let the missing value be $x$. The equation is: $\frac{5}{6} \div \frac{6}{7} \times x - \frac{8}{9} \div \frac{8}{5} + \frac{3}{4} \times \frac{10}{3} = \frac{25}{9}$.
Applying the $BODMAS$ rule,we first solve the divisions and multiplications:
$\frac{5}{6} \times \frac{7}{6} \times x - \frac{8}{9} \times \frac{5}{8} + \frac{3}{4} \times \frac{10}{3} = \frac{25}{9}$.
Simplifying each term:
$\frac{35}{36} x - \frac{5}{9} + \frac{5}{2} = \frac{25}{9}$.
Rearranging to solve for $x$:
$\frac{35}{36} x = \frac{25}{9} + \frac{5}{9} - \frac{5}{2}$.
$\frac{35}{36} x = \frac{30}{9} - \frac{5}{2} = \frac{10}{3} - \frac{5}{2}$.
$\frac{35}{36} x = \frac{20 - 15}{6} = \frac{5}{6}$.
$x = \frac{5}{6} \times \frac{36}{35} = \frac{6}{7}$.

(NONE) Convert the mixed fractions into improper fractions:
$4 \frac{1}{2} = \frac{9}{2}$,$3 \frac{1}{6} = \frac{19}{6}$,$2 \frac{1}{3} = \frac{7}{3}$,and $13 \frac{2}{3} = \frac{41}{3}$.
The equation becomes: $\frac{9}{2} + \frac{19}{6} + x + \frac{7}{3} = \frac{41}{3}$.
Find a common denominator for the fractions on the left side,which is $6$:
$\frac{27}{6} + \frac{19}{6} + \frac{14}{6} + x = \frac{41}{3}$.
Sum the fractions: $\frac{27 + 19 + 14}{6} + x = \frac{41}{3}$.
$\frac{60}{6} + x = \frac{41}{3}$.
$10 + x = \frac{41}{3}$.
$x = \frac{41}{3} - 10$.
$x = \frac{41 - 30}{3} = \frac{11}{3} = 3 \frac{2}{3}$.

(A) Let $a = 0.67$ and $b = 0.1$.
Then $a^3 = 0.67 \times 0.67 \times 0.67$ and $b^3 = 0.1 \times 0.1 \times 0.1 = 0.001$.
Also,$ab = 0.67 \times 0.1 = 0.067$ and $b^2 = 0.1 \times 0.1 = 0.01$.
The expression is of the form $\frac{a^3 - b^3}{a^2 + ab + b^2}$.
Using the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,the expression simplifies to:
$\frac{(a - b)(a^2 + ab + b^2)}{a^2 + ab + b^2} = a - b$.
Substituting the values,we get $0.67 - 0.1 = 0.57$.

(A) Let the total number of students be $x$.
Since the number of girls is twice the number of boys,let the number of boys be $b$ and girls be $2b$.
Total students $x = b + 2b = 3b$,so $b = \frac{x}{3}$ and girls $= \frac{2x}{3}$.
Number of girls who took part in the camp $= \frac{1}{5} \times \frac{2x}{3} = \frac{2x}{15}$.
Number of boys who took part in the camp $= \frac{1}{8} \times \frac{x}{3} = \frac{x}{24}$.
Total students who took part in the camp $= \frac{2x}{15} + \frac{x}{24}$.
To add these,find the least common multiple of $15$ and $24$,which is $120$.
Total $= \frac{16x + 5x}{120} = \frac{21x}{120} = \frac{7x}{40}$.
Thus,the fraction of total students who took part is $\frac{7}{40}$.

(B) Let the fraction be $x$.
According to the problem,we multiply the fraction by itself and divide by its reciprocal:
$x^2 \div (1/x) = 18 \frac{26}{27}$
$x^2 \times x = \frac{18 \times 27 + 26}{27}$
$x^3 = \frac{486 + 26}{27} = \frac{512}{27}$
$x^3 = (\frac{8}{3})^3$
Taking the cube root on both sides,we get:
$x = \frac{8}{3} = 2 \frac{2}{3}$

(B) To determine the greatest number,let us calculate the value of each option:
$A) (0.3)^{2} = 0.3 \times 0.3 = 0.09$
$B) 1 \div 0.3 = 10 \div 3 \approx 3.33$
$C) 1/8 = 0.125$
$D) \sqrt{0.49} = 0.7$
Comparing the values: $0.09, 3.33, 0.125, 0.7$.
The greatest value is $3.33$,which corresponds to option $B$.

(D) Let the fraction to be subtracted be $x$.
According to the problem,the average of the three fractions is $\frac{1}{12}$.
The three fractions are $\frac{1}{4}$,$\frac{1}{6}$,and $-x$.
The sum of these three fractions is $\frac{1}{4} + \frac{1}{6} - x$.
The average is given by $\frac{\text{Sum}}{3} = \frac{1}{12}$.
Therefore,$\frac{\frac{1}{4} + \frac{1}{6} - x}{3} = \frac{1}{12}$.
Multiplying both sides by $3$,we get $\frac{1}{4} + \frac{1}{6} - x = \frac{3}{12} = \frac{1}{4}$.
Subtracting $\frac{1}{4}$ from both sides,we get $\frac{1}{6} - x = 0$.
Thus,$x = \frac{1}{6}$.

(B) Let the fraction be $x$.
According to the problem,the person divided by $\frac{6}{7}$ instead of multiplying by $\frac{6}{7}$.
This means the calculated value is $x \div \frac{6}{7} = \frac{7}{6}x$.
The correct answer should have been $\frac{6}{7}x$.
Given that the calculated value exceeds the correct answer by $\frac{1}{7}$,we have:
$\frac{7}{6}x - \frac{6}{7}x = \frac{1}{7}$
To solve for $x$,find a common denominator for the fractions on the left side,which is $42$:
$\frac{49x - 36x}{42} = \frac{1}{7}$
$\frac{13x}{42} = \frac{1}{7}$
$x = \frac{1}{7} \times \frac{42}{13} = \frac{6}{13}$
Now,calculate the correct answer:
Correct answer $= \frac{6}{7}x = \frac{6}{7} \times \frac{6}{13} = \frac{36}{91}$.

(A) Given expression: $2+\sqrt{2}+\frac{1}{2+\sqrt{2}}+\frac{1}{\sqrt{2}-2}$
Combine the last two terms by finding a common denominator:
$\frac{1}{2+\sqrt{2}}+\frac{1}{\sqrt{2}-2} = \frac{(\sqrt{2}-2) + (2+\sqrt{2})}{(2+\sqrt{2})(\sqrt{2}-2)}$
Simplify the numerator and denominator:
Numerator: $\sqrt{2}-2+2+\sqrt{2} = 2\sqrt{2}$
Denominator: $(\sqrt{2})^2 - (2)^2 = 2 - 4 = -2$
So,the sum of the last two terms is $\frac{2\sqrt{2}}{-2} = -\sqrt{2}$
Now,substitute this back into the original expression:
$2 + \sqrt{2} + (-\sqrt{2}) = 2$
Therefore,the correct option is $A$.

(C) We calculate each term individually:
$\frac{1}{2} = 0.50000$
$\frac{1}{2 \cdot 3} = \frac{1}{6} \approx 0.16667$
$\frac{1}{2 \cdot 3 \cdot 4} = \frac{1}{24} \approx 0.04167$
$\frac{1}{2 \cdot 3 \cdot 4 \cdot 5} = \frac{1}{120} \approx 0.00833$
Summing these values: $0.50000 + 0.16667 + 0.04167 + 0.00833 = 0.71667$
Rounding to three decimal places,we get $0.717$.

(C) Let the missing value be $x$.
The given equation is $\frac{x \div 12}{0.2 \times 3.6} = 2$.
Multiplying both sides by the denominator $(0.2 \times 3.6)$:
$x \div 12 = 2 \times 0.2 \times 3.6$
$x \div 12 = 0.4 \times 3.6$
$x \div 12 = 1.44$
Now,multiplying both sides by $12$ to solve for $x$:
$x = 1.44 \times 12$
$x = 17.28$

(A) Let the missing value be $x$.
$\sqrt{x \times 7} \times 18 = 84$
Divide both sides by $18$:
$\sqrt{x \times 7} = \frac{84}{18} = \frac{14}{3}$
Square both sides to remove the square root:
$x \times 7 = \left(\frac{14}{3}\right)^2$
$x \times 7 = \frac{196}{9}$
Solve for $x$:
$x = \frac{196}{9 \times 7}$
$x = \frac{196}{63}$
$x = \frac{28}{9} \approx 3.111...$
Rounding to two decimal places,we get $x = 3.11$.

(A) First,convert the mixed fractions into improper fractions:
$1 \frac{3}{4} = \frac{7}{4}$,$2 \frac{1}{3} = \frac{7}{3}$,$3 \frac{5}{12} = \frac{41}{12}$,$5 \frac{1}{5} = \frac{26}{5}$,$2 \frac{1}{6} = \frac{13}{6}$.
Sum $= \frac{7}{4} + \frac{7}{3} + \frac{41}{12} + \frac{26}{5} + \frac{13}{6}$.
The least common multiple $(LCM)$ of $4, 3, 12, 5, 6$ is $60$.
Sum $= \frac{105 + 140 + 205 + 312 + 130}{60} = \frac{892}{60}$.
Simplifying $\frac{892}{60} = \frac{223}{15} = 14 \frac{13}{15}$.
The nearest whole number to $14 \frac{13}{15}$ is $15$.
The difference is $15 - 14 \frac{13}{15} = \frac{2}{15}$.

(C) Given the equation: $(2 + \frac{3}{x}) \times (y + \frac{1}{2}) = \frac{31}{4}$.
Expanding the expression: $(\frac{2x+3}{x}) \times (\frac{2y+1}{2}) = \frac{31}{4}$.
This simplifies to: $\frac{(2x+3)(2y+1)}{2x} = \frac{31}{4}$.
Cross-multiplying: $4(2x+3)(2y+1) = 62x$,which reduces to $2(2x+3)(2y+1) = 31x$.
For $x=14$: $2(2(14)+3)(2y+1) = 31(14) \Rightarrow 2(31)(2y+1) = 31(14)$.
Dividing both sides by $62$: $2y+1 = 7 \Rightarrow 2y = 6 \Rightarrow y = 3$.
Thus,the values are $x=14$ and $y=3$.

(C) Let the fraction be $x$.
According to the problem,we multiply the fraction by itself and divide by the square of its reciprocal:
$x \times x \div \left(\frac{1}{x}\right)^2 = 3 \frac{13}{81}$
Convert the mixed fraction to an improper fraction:
$3 \frac{13}{81} = \frac{3 \times 81 + 13}{81} = \frac{243 + 13}{81} = \frac{256}{81}$
Now,simplify the equation:
$x^2 \div \frac{1}{x^2} = \frac{256}{81}$
$x^2 \times x^2 = \frac{256}{81}$
$x^4 = \frac{256}{81}$
Express the right side as a power of $4$:
$x^4 = \left(\frac{4}{3}\right)^4$
Therefore,$x = \frac{4}{3}$.

(B) Given the operation $x \times y = (x + 2)^2 (y - 2)$.
Substituting $x = 7$ and $y = 5$ into the expression:
$7 \times 5 = (7 + 2)^2 (5 - 2)$
$= (9)^2 \times (3)$
$= 81 \times 3$
$= 243$

(C) Given that $m^{n} = 121$.
We know that $121 = 11^{2}$.
Comparing $m^{n} = 11^{2}$,we get $m = 11$ and $n = 2$.
Now,we need to find the value of $(m-1)^{n+1}$.
Substituting the values of $m$ and $n$:
$(11-1)^{2+1} = 10^{3}$.
Thus,the value is $10^{3}$.

(D) The given fractions are $\frac{1}{8} = 0.125$ and $\frac{1}{2} = 0.5$.
In mathematics,between any two distinct rational numbers,there exist infinitely many rational numbers.
Since any rational number can be expressed as a fraction,there are infinitely many fractions between $\frac{1}{8}$ and $\frac{1}{2}$.
For example,we can find fractions like $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}$ and many more.

(D) Let the given number be $x$.
According to the problem,the boy divided the number by $\frac{8}{17}$ instead of multiplying it by $\frac{8}{17}$.
Dividing by $\frac{8}{17}$ is equivalent to multiplying by $\frac{17}{8}$.
The difference between the result obtained and the expected result is $225$.
So,the equation is: $x \times \frac{17}{8} - x \times \frac{8}{17} = 225$.
Taking $x$ as a common factor: $x \times (\frac{17}{8} - \frac{8}{17}) = 225$.
Calculating the difference inside the bracket: $\frac{17^2 - 8^2}{8 \times 17} = \frac{289 - 64}{136} = \frac{225}{136}$.
Thus,$x \times \frac{225}{136} = 225$.
Dividing both sides by $225$,we get: $\frac{x}{136} = 1$.
Therefore,$x = 136$.

(C) The given expression is $\sum_{n=1}^{9} \frac{1}{n(n+1)}$.
We know that $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
Substituting this into the series,we get:
$\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots + \left(\frac{1}{9} - \frac{1}{10}\right)$.
This is a telescoping series where all intermediate terms cancel out.
$= 1 - \frac{1}{10} = \frac{10 - 1}{10} = \frac{9}{10}$.

(C) Let the missing value be $x$.
The given equation is $\frac{\sqrt{1296}}{x} = \frac{x}{2.25}$.
Cross-multiplying the terms,we get $x^2 = \sqrt{1296} \times 2.25$.
Since $\sqrt{1296} = 36$,the equation becomes $x^2 = 36 \times 2.25$.
Calculating the product,$x^2 = 81$.
Taking the square root on both sides,$x = \sqrt{81} = 9$.
Therefore,the missing value is $9$.

(B) Let the fraction be $x$.
According to the problem,we multiply the fraction by itself $(x \times x = x^2)$ and divide the product by its reciprocal $(\frac{1}{x})$.
So,the equation is: $x^2 \div \frac{1}{x} = 18 \frac{26}{27}$.
$x^2 \times x = \frac{18 \times 27 + 26}{27}$.
$x^3 = \frac{486 + 26}{27} = \frac{512}{27}$.
Taking the cube root on both sides: $x = \sqrt[3]{\frac{512}{27}} = \frac{8}{3}$.
Converting to a mixed fraction: $x = 2 \frac{2}{3}$.

(C) Given that $\frac{a}{a+b} = \frac{17}{23}$.
If we assume $a = 17$,then $a + b = 23$.
From $a + b = 23$,we can find $b$ by substituting $a = 17$:
$17 + b = 23 \implies b = 23 - 17 = 6$.
Now,calculate $a - b$:
$a - b = 17 - 6 = 11$.
Therefore,the value of $\frac{a+b}{a-b}$ is:
$\frac{a+b}{a-b} = \frac{23}{11}$.

(D) Let the total capacity of the tin be $x$ bottles.
Initially,the tin was $\frac{4}{5}$ full,so the amount of oil was $\frac{4}{5}x$.
When $6$ bottles were removed,the amount became $\frac{4}{5}x - 6$.
Then,$4$ bottles were added,making the amount $\frac{4}{5}x - 6 + 4 = \frac{4}{5}x - 2$.
According to the problem,this final amount is $\frac{3}{4}$ of the total capacity,so $\frac{4}{5}x - 2 = \frac{3}{4}x$.
Rearranging the equation: $\frac{4}{5}x - \frac{3}{4}x = 2$.
Finding a common denominator $(20)$: $\frac{16x - 15x}{20} = 2$.
$\frac{x}{20} = 2$.
$x = 40$.
Therefore,the tin can contain $40$ bottles of oil.

(C) Let the number be $x$.
According to the problem,the student calculated $\frac{3}{4}x$ instead of $\frac{3}{14}x$.
Given that the incorrect answer is $150$ more than the correct answer,we have the equation:
$\frac{3}{4}x - \frac{3}{14}x = 150$
To solve this,find the least common multiple of $4$ and $14$,which is $28$.
$\frac{3 \times 7}{28}x - \frac{3 \times 2}{28}x = 150$
$\frac{21}{28}x - \frac{6}{28}x = 150$
$\frac{15}{28}x = 150$
$x = \frac{150 \times 28}{15}$
$x = 10 \times 28 = 280$
Thus,the number is $280$.

(A) The given expression is $\left(2-\frac{1}{3}\right)\left(2-\frac{3}{5}\right)\left(2-\frac{5}{7}\right) \ldots \left(2-\frac{999}{1001}\right)$.
Simplifying each term:
$\left(\frac{6-1}{3}\right)\left(\frac{10-3}{5}\right)\left(\frac{14-5}{7}\right) \ldots \left(\frac{2002-999}{1001}\right)$
$= \frac{5}{3} \times \frac{7}{5} \times \frac{9}{7} \times \ldots \times \frac{1003}{1001}$.
Observing the pattern,the numerator of each fraction cancels out the denominator of the subsequent fraction:
$= \frac{{5}}{3} \times \frac{{7}}{{5}} \times \frac{{9}}{{7}} \times \ldots \times \frac{1003}{{1001}}$.
After cancellation,we are left with the numerator of the last term and the denominator of the first term:
$= \frac{1003}{3}$.

(C) Given the equation $\sqrt{2^{n}} = 64$.
We know that $64$ can be written as $2^{6}$.
Also,$\sqrt{2^{n}}$ can be written as $(2^{n})^{1/2} = 2^{n/2}$.
Equating the powers of the same base $2$,we get $\frac{n}{2} = 6$.
Multiplying both sides by $2$,we get $n = 12$.

(B) Given the equation: $10^{2y} = 25$.
To find the value of $10^{y}$,we take the square root of both sides of the equation.
$\sqrt{10^{2y}} = \sqrt{25}$.
Using the exponent rule $\sqrt{a^{2n}} = a^n$,we get:
$10^{y} = 5$.