Find the fraction which is equal to 1/2 when both its numerator and de...
**Solution:**
Let the required fraction be x/y.
**Step 1:**
According to the problem, when both the numerator and denominator are increased by 2, the fraction becomes 1/2.
So, (x+2)/(y+2) = 1/2
Multiplying both sides by 2(y+2), we get:
2x + 4 = y + 2
2x - y = -2 ------Equation 1
**Step 2:**
According to the problem, when both the numerator and denominator are increased by 12, the fraction becomes 3/4.
So, (x+12)/(y+12) = 3/4
Multiplying both sides by 4(y+12), we get:
4x + 48 = 3y + 36
4x - 3y = -12 ------Equation 2
**Step 3:**
Now, we have two equations:
2x - y = -2 ------Equation 1
4x - 3y = -12 ------Equation 2
We can solve these equations simultaneously to find the values of x and y.
Multiplying Equation 1 by 3 and Equation 2 by 2, we get:
6x - 3y = -6 ------Equation 3
8x - 6y = -24 ------Equation 4
Subtracting Equation 3 from Equation 4, we get:
2x - 3y = -18
2x = 3y - 18
x = (3y - 18)/2
**Step 4:**
Substituting this value of x in Equation 1, we get:
2((3y - 18)/2) - y = -2
3y - 18 - y = -2
2y = 16
y = 8
**Step 5:**
Substituting the values of x and y in the original fraction x/y, we get:
x/y = (3y - 18)/2y = (3(8) - 18)/2(8) = -3/8
**Step 6:**
Therefore, the required fraction is -3/8.
**Explanation:**
We can solve the problem using the following steps:
- First, we need to set up two equations using the given information.
- The first equation comes from the fact that when both the numerator and denominator are increased by 2, the fraction becomes 1/2.
- The second equation comes from the fact that when both the numerator and denominator are increased by 12, the fraction becomes 3/4.
- Next, we can solve these equations simultaneously to find the values of x and y.
- Finally, we can substitute these values in the original fraction x/y to get the required answer.
- It is important to pay attention to the signs of the fractions while solving the equations, as they can affect the final answer.
Find the fraction which is equal to 1/2 when both its numerator and de...
To make sure you are not studying endlessly, EduRev has designed CA Foundation study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in CA Foundation.