# 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.

Answer

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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 \[\dfrac{N}{D}\].

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.

Last updated date: 29th Sep 2023

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