Let's tackle the problems one by one.
First, we will solve the fraction problem. Let's use algebra to find the original fraction.
1. The fraction problem:
Let the denominator of the fraction be \(d\).
Since the numerator is 7 less than the denominator, the numerator will be \(d - 7\).
Now, the fraction is \(\frac{d - 7}{d}\).
According to the problem, if the denominator is increased by 9, the new denominator becomes \(d + 9\).
If the numerator is increased by 2, the new numerator becomes \(d - 7 + 2 = d - 5\).
It is given that even after increasing the numerator and denominator by 2 and 9 respectively, we get the same fraction, hence:
\[\frac{d - 5}{d + 9} = \frac{d - 7}{d}\]
Now, let's find the value of \(d\).
To find \(d\), we cross-multiply in the equation:
\((d - 5) \cdot d = (d - 7) \cdot (d + 9)\)
Expanding both sides, we get:
\[d^2 - 5d = d^2 + 9d - 7d - 63\]
Simplify the equation by subtracting \(d^2\) from both sides (it will be eliminated):
\[-5d = 2d - 63\]
Now, let's move 2d to the left side to combine like terms:
\[-5d - 2d = -63\]
This simplifies to:
\[-7d = -63\]
To find \(d\), we now divide both sides by -7:
\[d = \frac{-63}{-7}\]
So:
\[d = 9\]
Once we have \(d\), we substitute back in to find the numerator:
Numerator \(= d - 7 = 9 - 7 = 2\)
Therefore, the original fraction is \(\frac{2}{9}\).
Now, onto the second problem:
2. The sum of the ages of Ajeet and Ajay:
This problem seems incomplete because we don't have additional information or conditions in order to find out the individual ages of Ajeet and Ajay. We know their sum:
\( \text{Ajeet's age} + \text{Ajay's age} = 10 \)
Without more information, we can't find the specific ages of Ajeet and Ajay. Normally, we would need additional information, like the difference in their ages or a ratio of their ages, to solve for their individual ages. If there were a follow-up condition or equation provided, we could create a system of equations to solve for their individual ages. However, with the current information, there are infinitely many solutions to this problem.
