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A fraction becomes $\dfrac{9}{11}$, if 1 is added to both the numerator and denominator. If 3 is added to both the numerator and the denominator it becomes $\dfrac{5}{6}$. Find the fraction.

Last updated date: 13th Jul 2024
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Hint: Here, first of all, assume a fraction $\dfrac{N}{D}$. Now, add 1 to both N and D and equate it to $\dfrac{9}{11}$. Similarly, add 3 to both N and D and equate it to $\dfrac{5}{6}$. Solve these two equations to get the value of $\dfrac{N}{D}$.

We are given that a fraction becomes $\dfrac{9}{11}$ if 1 is added to both numerator and denominator whereas if 3 is added to both numerator and denominator, it becomes $\dfrac{5}{6}$. We have to find the value of the fraction.
Let us consider our original fraction as $\dfrac{N}{D}$, where N is the numerator, and D is the denominator of the original fraction.
Now, we are given that if we add 1 to both numerator and denominator, it becomes $\dfrac{9}{11}$. So, by adding 1 to both numerator and denominator of the original fraction $\dfrac{N}{D}$, we get,
$\dfrac{N+1}{D+1}=\dfrac{9}{11}$
By cross multiplying the above the equation, we get,
$11N+11=9D+9$
By simplifying the above equation, we get,
$9D11N=2\ldots ..\left( i \right)$
Now, we are also given that, if we add 3 to both numerator and denominator, it becomes $\dfrac{5}{6}$. So, by adding 3 to both the sides of the numerator and denominator of the original fraction $\dfrac{N}{D}$, we get,
$\dfrac{N+3}{D+3}=\dfrac{5}{6}$
By cross multiplying the above equation, we get,
$6N+18=5D+15$
By simplifying the above equation, we get,
$5D-6N=3.....\left( ii \right)$
By multiplying 5 on both the sides of the equation (i), we get,
$45D-55N=10....\left( iii \right)$
Also, by multiplying 9 on both the sides of equation (ii), we get,
$45D-54N=27....\left( iv \right)$

Now, by subtracting equation (iv) from (iii), we get,
$\left( 45D-55N \right)-\left( 45D-54N \right)=10-27$
By simplifying the above equation, we get,
$-55N+54N=-17$
$\Rightarrow -N=-17$
So, we get N = 17.
By substituting N = 17 in equation (i), we get,
$9D-11\left( 17 \right)=2$
$\Rightarrow 9D=2+187$
Or, $9D=189$
By dividing 9 on both the sides, we get,
$D=\dfrac{189}{9}=21$
So, we get D = 21.
We know that our fraction is $\dfrac{N}{D}$, so by substituting the values of N and D in the original fraction, we get,
Original fraction $=\dfrac{N}{D}=\dfrac{17}{21}$.
So, we get the original fraction as $\dfrac{17}{21}$.

Note: Students can cross-check their answer as follows:
We know that $\dfrac{N+1}{D+1}=\dfrac{9}{11}$
By substituting N = 17 and D = 21, we get,
$\dfrac{17+1}{21+1}=\dfrac{9}{11}$
$\Rightarrow \dfrac{18}{22}=\dfrac{9}{11}$
By simplifying LHS of the above equation, we get,
$\dfrac{9}{11}=\dfrac{9}{11}$
LHS = RHS
Since, LHS = RHS, therefore our answer is correct. Similarly, students can also check by substituting N and D in the other equations.