Question

If $ \left( a-b \right):\left( a+b \right)=1:11 $ , find the ratio $ \left( 5a+4b+15 \right):\left( 5a-4b+3 \right) $ .

Answer 1

Hint: We first try to find the relation between the numbers $ a $ and $ b $ from the given equation of $ \left( a-b \right):\left( a+b \right)=1:11 $ . The solution gives \[5a=6b\]. We put the value in $ \dfrac{5a+4b+15}{5a-4b+3} $ to find the simplified form and find the final solution.

Complete step-by-step answer:
It is given that $ \left( a-b \right):\left( a+b \right)=1:11 $ .
We convert the proportionality to a division form.
 $ \left( a-b \right):\left( a+b \right)=1:11 $ becomes \[\dfrac{\left( a-b \right)}{\left( a+b \right)}=\dfrac{1}{11}\].
Now we try to solve the equation with the help of cross-multiplication.
\[\begin{align}
  & \dfrac{\left( a-b \right)}{\left( a+b \right)}=\dfrac{1}{11} \\
 & \Rightarrow 11a-11b=a+b \\
 & \Rightarrow 10a=12b \\
 & \Rightarrow 5a=6b \\
\end{align}\]
We have expressed the relation between the numbers $ a $ and $ b $ .
We have \[5a=6b\] and we use that in the expression of $ \left( 5a+4b+15 \right):\left( 5a-4b+3 \right) $ to find the solution.
 $ \left( 5a+4b+15 \right):\left( 5a-4b+3 \right) $ can be converted to $ \dfrac{5a+4b+15}{5a-4b+3} $ .
So, $ \dfrac{5a+4b+15}{5a-4b+3}=\dfrac{6b+4b+15}{6b-4b+3} $ .
We complete the binary operations both in the numerator and denominator and get
 $ \dfrac{6b+4b+15}{6b-4b+3}=\dfrac{10b+15}{2b+3} $ .
We now take 5 common from the numerator to factorise it.
So, \[\dfrac{10b+15}{2b+3}=\dfrac{5\left( 2b+3 \right)}{2b+3}\].
We can now divide the $ 2b+3 $ part from both numerator and denominator.
We get $ \dfrac{5a+4b+15}{5a-4b+3}=5 $ . The ratio form of 5 will be $ 5:1 $ .
Therefore, the ratio $ \left( 5a+4b+15 \right):\left( 5a-4b+3 \right) $ will be $ 5:1 $ .
So, the correct answer is “ $ 5:1 $ ”.

Note: We need to remember that the ratio of $ \left( a-b \right):\left( a+b \right)=1:11 $ can also be expressed as \[\dfrac{\left( a-b \right)}{\left( a+b \right)}=\dfrac{1}{11}=k\] to find the value of the numbers $ a $ and $ b $ with respect to $ k $ . We get the same ratio of $ 5:1 $ for the final solution after replacing the variables.