We need to find a fraction based on two conditions.
Let the fraction be represented as $\frac{x}{y}$, where $x$ is the numerator and $y$ is the denominator.
We can solve this system of equations. First, express $y$ from the first equation:
$y = 11 - x$Now, substitute this expression for $y$ into the second equation:
$\frac{x-1}{11-x} = \frac{1}{4}$Cross-multiply to solve for $x$:
$4(x-1) = 1(11-x)$ $4x - 4 = 11 - x$Combine like terms:
$4x + x = 11 + 4$ $5x = 15$ $x = \frac{15}{5}$ $x = 3$Now, find the value of $y$ using $y = 11 - x$:
$y = 11 - 3$ $y = 8$The numerator ($x$) is 3 and the denominator ($y$) is 8. Therefore, the fraction is:
$\frac{3}{8}$Both conditions are satisfied.
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: