Let the original fraction be represented algebraically as $\frac{x}{y}$.
The numerator $x$ is increased by 30%. The new numerator is calculated as:
$ x + (30\% \times x) = x + (0.30 \times x) = 1.30x $
The denominator $y$ is decreased by 35%. The new denominator is calculated as:
$ y - (35\% \times y) = y - (0.35 \times y) = 0.65y $
The resulting fraction after these modifications is $\frac{1.30x}{0.65y}$.
We are given that this new fraction equals $\frac{3}{15}$, which simplifies to $\frac{1}{5}$.
$ \frac{1.30x}{0.65y} = \frac{1}{5} $
First, simplify the numerical coefficients in the equation:
$ \frac{1.30}{0.65} = \frac{130}{65} = 2 $
Substitute this back into the equation:
$ 2 \times \frac{x}{y} = \frac{1}{5} $
To find the original fraction $\frac{x}{y}$, divide both sides by 2:
$ \frac{x}{y} = \frac{1}{5 \times 2} $
$ \frac{x}{y} = \frac{1}{10} $
Thus, the original fraction is $\frac{1}{10}$.
Which fraction among the following is the least ?
\(\frac{5}{11}, \frac{7}{12}, \frac{8}{13}, \frac{9}{17}\)
Find the value of the following expression:
\(\frac{{3 \div 1 \times 2 + 5 - 2}}{{3 \times 3 - 2}}\)
Simplify the expression 441 ÷ \(\left[270 \div \frac{3}{7}+\left(17\div \frac{1}{3}\right)-\left(8\frac{1}{2}-\frac{5}{2}\right)\right]\)
If the sum of two positive numbers is 65 and the square root of their product is 26, then the sum of their reciprocals is:
The value of \(9 \div [\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}\div(\frac{3}{4}-\frac{1}{3})\;of\;\frac{2}{9}]\) is: