Question

If a mixture contains honey and water both in an integral number of liters. When 91L of water is added to the mixture the ratio becomes square of the initial ratio. What was the initial ratio of honey and water?

Answer 1

very simple 36:78 will be answer if you simplify more if will be 12:26

Answer 2

Answer:

Step-by-step explanation:

Consider the ratio of honey and water is $1:x$

When 91L of water is added to the mixture the ratio becomes square of the initial ratio.

therefore the new ratio

$\text{honey}:\text{water}=1:(x+91)$

As per the problem,

$\text{honey}:\text{water}=1:(x+91)=1^2:x^2$

$1:(x+91)=1^2:x^2\\\\\Rightarrow \frac{1}{(x+91)}=\frac{1}{x^2}\\\\\Rightarrow {x^2}=x+91\\\\\Rightarrow {x^2}-x-91=0$

For Quadratic Equation $ax^{2}+bx+c=0 \text{ [where x is the variable and a, b and c are known values]}$

the value of $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$b^{2}-4ac \text{ is called Discriminant (D)}$

when Discriminant (D) is positive, we get two Real solutions  for x

when Discriminant (D) is zero we get just ONE real solution (both answers are the same)

when Discriminant (D) is negative we get a pair of Complex solutions

In the above equation  Discriminant $D=b^{2}-4ac=(-1)^2-4\times1\times(-91)=1+364=365$

As the Discriminant (D) is positive, we get two Real solutions  for x

therefore the value of x is either $x=\frac{-b+\sqrt{D}}{2a}\text{or}\text{ } x=\frac{-b-\sqrt{D}}{2a}$

$x=\frac{-b+\sqrt{D}}{2a}=\frac{-(-1)+\sqrt{365}}{2}=\frac{1+19.10}{2}=\frac{20.10}{2}=10.05$

$x=\frac{-b-\sqrt{D}}{2a}=\frac{-(-1)-\sqrt{365}}{2}=\frac{1-19.10}{2}=\frac{-18.10}{2}=-9.05$

As quantity of water is not consider as a negative value so we take the positive value of x=10.05

therefore the initial ration of honey and water is $1:10.05$(approx)