Simplification Questions in English

(C) To find the sum,align the decimal points and add the numbers:
$34.950$
$240.016$
$23.980$
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$298.946$
Thus,$34.95 + 240.016 + 23.98 = 298.946$.

(D) Given the equation: $3889 + 12.952 - ? = 3854.002$
Rearranging the terms to solve for $?$,we get:
$? = (3889 + 12.952) - 3854.002$
First,calculate the sum:
$3889 + 12.952 = 3901.952$
Next,subtract $3854.002$ from the sum:
$3901.952 - 3854.002 = 47.95$
Therefore,the value of $?$ is $47.95$.

(C) To solve the expression $337.62 + 8.591 + 34.4$,we align the decimal points and add the numbers:
$337.620$
$+\quad 8.591$
$+\quad 34.400$
. . . . . .
$380.611$
Thus,the sum is $380.611$.

(A) The $n$-th term of the series is given by $T_n = \frac{1}{(n+4)(n+5)}$.
We can write this as $T_n = \frac{1}{n+4} - \frac{1}{n+5}$.
For the first $20$ terms,the sum $S_{20} = \sum_{n=1}^{20} \left( \frac{1}{n+4} - \frac{1}{n+5} \right)$.
Expanding the sum: $S_{20} = \left( \frac{1}{5} - \frac{1}{6} \right) + \left( \frac{1}{6} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{8} \right) + \cdots + \left( \frac{1}{24} - \frac{1}{25} \right)$.
This is a telescoping series where intermediate terms cancel out.
$S_{20} = \frac{1}{5} - \frac{1}{25}$.
$S_{20} = \frac{5 - 1}{25} = \frac{4}{25}$.
$S_{20} = \frac{16}{100} = 0.16$.

(C) Given that $\frac{52416}{312} = 168.$
We need to find the value of $x = \frac{52.416}{0.0168}.$
To simplify,we can write the decimals as fractions:
$x = \frac{52416 / 1000}{168 / 10000}$
$x = \frac{52416}{1000} \times \frac{10000}{168}$
$x = \frac{52416}{168} \times \frac{10000}{1000}$
Since $\frac{52416}{312} = 168,$ it follows that $\frac{52416}{168} = 312.$
Substituting this value:
$x = 312 \times 10 = 3120.$

(C) To arrange the fractions in ascending order,convert each fraction into its decimal form:
$\frac{2}{3} = 0.666...$
$\frac{3}{5} = 0.6$
$\frac{7}{9} = 0.777...$
$\frac{9}{11} = 0.8181...$
$\frac{8}{9} = 0.888...$
Comparing these values: $0.6 < 0.666... < 0.777... < 0.8181... < 0.888...$
Therefore,the ascending order is: $\frac{3}{5} < \frac{2}{3} < \frac{7}{9} < \frac{9}{11} < \frac{8}{9}$.

(C) The given expression is: $1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{64} + \frac{1}{1024}$
Step $1$: Calculate each term individually:
$1 = 1.0000$
$\frac{1}{2} = 0.5000$
$\frac{1}{8} = 0.1250$
$\frac{1}{64} = 0.015625$
$\frac{1}{1024} \approx 0.000976$
Step $2$: Sum these values:
$1 + 0.5 + 0.125 + 0.015625 + 0.000976 = 1.641601$
Rounding to four decimal places,we get $1.6416$.

(A) Let the numerator be $x$ and the denominator be $x+3$.
The fraction is represented as $\frac{x}{x+3}$.
According to the problem,if both the numerator and the denominator are increased by $4$,the new fraction becomes $\frac{4}{5}$.
So,$\frac{x+4}{(x+3)+4} = \frac{4}{5}$.
$\frac{x+4}{x+7} = \frac{4}{5}$.
Cross-multiplying gives: $5(x+4) = 4(x+7)$.
$5x + 20 = 4x + 28$.
$5x - 4x = 28 - 20$.
$x = 8$.
Therefore,the numerator is $8$ and the denominator is $8+3 = 11$.
The original fraction is $\frac{8}{11}$.

(B) Let the two fractions be $x$ and $y$,where $x > y$.
Given:
$x y = \frac{14}{15}$ ............$(1)$
$\frac{x}{y} = \frac{35}{24}$ ............$(2)$
To find $x$ and $y$,multiply $(1)$ and $(2)$:
$(x y) \times (\frac{x}{y}) = \frac{14}{15} \times \frac{35}{24}$
$x^2 = \frac{14 \times 35}{15 \times 24} = \frac{2 \times 7 \times 5 \times 7}{3 \times 5 \times 3 \times 8} = \frac{49}{36}$
$x = \sqrt{\frac{49}{36}} = \frac{7}{6}$
Now,substitute $x = \frac{7}{6}$ into $(1)$:
$(\frac{7}{6}) y = \frac{14}{15}$
$y = \frac{14}{15} \times \frac{6}{7} = \frac{2 \times 2}{5} = \frac{4}{5}$
Comparing $x = \frac{7}{6} \approx 1.166$ and $y = \frac{4}{5} = 0.8$,the greater fraction is $\frac{7}{6}$.

(A) Let the fraction be $\frac{x}{y}$.
According to the first condition,$\frac{x+2}{y+3} = \frac{7}{9}$.
Cross-multiplying gives $9(x+2) = 7(y+3)$,which simplifies to $9x + 18 = 7y + 21$,or $9x - 7y = 3$ ..........$(1)$
According to the second condition,$\frac{x-1}{y-1} = \frac{4}{5}$.
Cross-multiplying gives $5(x-1) = 4(y-1)$,which simplifies to $5x - 5 = 4y - 4$,or $5x - 4y = 1$ ..........$(2)$
To solve the system of equations,multiply $(1)$ by $4$ and $(2)$ by $7$:
$36x - 28y = 12$ ..........$(3)$
$35x - 28y = 7$ ..........$(4)$
Subtracting $(4)$ from $(3)$ gives $x = 5$.
Substituting $x = 5$ into $(2)$:
$5(5) - 4y = 1 \Rightarrow 25 - 4y = 1 \Rightarrow 4y = 24 \Rightarrow y = 6$.
Thus,the original fraction is $\frac{5}{6}$.

(C) Let the fraction $B$ be $x$ and the fraction $A$ be $2x.$
Given that the product of the two fractions is $\frac{2}{25}.$
Therefore,$x \times 2x = \frac{2}{25}.$
This simplifies to $2x^2 = \frac{2}{25}.$
Dividing both sides by $2$,we get $x^2 = \frac{1}{25}.$
Taking the square root,$x = \frac{1}{5}$ (considering positive fractions).
Thus,the value of fraction $A = 2x = 2 \times \frac{1}{5} = \frac{2}{5}.$

(B) Let the fraction be $\frac{x}{y}$.
According to the problem,the difference between the numerator and denominator is $5$,so $x - y = 5$,which implies $x = y + 5$.
When $5$ is added to the denominator,the new fraction becomes $\frac{x}{y+5}$.
Given that the fraction decreases by $1 \frac{1}{4} = \frac{5}{4}$,we have:
$\frac{x}{y} - \frac{x}{y+5} = \frac{5}{4}$.
Substitute $x = y + 5$ into the equation:
$\frac{y+5}{y} - \frac{y+5}{y+5} = \frac{5}{4}$
$\frac{y+5}{y} - 1 = \frac{5}{4}$
$\frac{y+5}{y} = 1 + \frac{5}{4} = \frac{9}{4}$
$4(y+5) = 9y$
$4y + 20 = 9y$
$5y = 20 \Rightarrow y = 4$.
Since $x = y + 5$,we have $x = 4 + 5 = 9$.
The fraction is $\frac{x}{y} = \frac{9}{4} = 2 \frac{1}{4}$.

(B) Let the fraction be $\frac{x}{y}$.
According to the first condition,adding $1$ to the denominator gives $\frac{x}{y+1} = \frac{1}{2}$,which implies $2x = y + 1$,or $2x - y = 1$ ........$(1)$.
According to the second condition,adding $1$ to the numerator gives $\frac{x+1}{y} = 1$,which implies $x + 1 = y$,or $y - x = 1$ ........$(2)$.
Adding equations $(1)$ and $(2)$:
$(2x - y) + (y - x) = 1 + 1$
$x = 2$.
Substituting $x = 2$ into equation $(2)$:
$y - 2 = 1$
$y = 3$.
Therefore,the required fraction is $\frac{x}{y} = \frac{2}{3}$.