Answer
Verified
478.2k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE