[tex]\large\underline{\sf{Solution-}}[/tex]

Let assume that

Initially the price of petrol be Rs x

Given that, it increases by 10%

So, new price of petrol is

[tex]\sf \: = \: x + 10\% \: of \: x \\ \\ [/tex]

[tex]\sf \: = \: x + \frac{10}{100} \: \times \: x \\ \\ [/tex]

[tex]\sf \: = \: x + \frac{x}{10} \: \\ \\ [/tex]

[tex]\sf \: = \: \frac{10x+ x}{10} \: \\ \\ [/tex]

[tex]\sf \: = \: \frac{11x}{10} \: \\ \\ \\ [/tex]

Let further assume that

Initially the consumption of petrol be y litres.

Let assume that to keep the expenditure same, the consumption of petrol reduces by z%.

So, Consumption of petrol is

[tex]\sf \: = \: y - z\% \: of \: y \\ \\ [/tex]

[tex]\sf \: = \: y - \frac{z}{100} \: \times \: y \\ \\ [/tex]

[tex]\sf \: = \: y\bigg(1 - \dfrac{z}{100} \bigg) \\ \\ [/tex]

Now, As price of petrol increases, the consumption of petrol decreases.

So,

It means price of petrol and consumption are in inverse variation.

[tex]\begin{array}{|c|c|c|c} \hline \rm Price \: in \: Rs&\rm x \: \: \: \: \: \: &\rm \dfrac{11x}{10} \\ \\ \hline\rm Consumption \: (in \: litres)&\rm y \: \: \: \: \: \: &\rm y\bigg(1 - \dfrac{z}{100} \bigg) \\ \\ \hline \end{array} \\ \\ [/tex]

So, By using law of inverse variation, we get

[tex]\sf \: xy = \dfrac{11x}{10} \times y\bigg(1 - \dfrac{z}{100} \bigg) \\ \\ [/tex]

[tex]\sf \: 1 = \dfrac{11}{10} \bigg(1 - \dfrac{z}{100} \bigg) \\ \\ [/tex]

[tex]\sf \: \dfrac{10}{11} = \bigg(1 - \dfrac{z}{100} \bigg) \\ \\ [/tex]

[tex]\sf \: \dfrac{z}{100} = 1 - \frac{10}{11} \\ \\ [/tex]

[tex]\sf \: \dfrac{z}{100} = \frac{11 - 10}{11} \\ \\ [/tex]

[tex]\sf \: \dfrac{z}{100} = \frac{1}{11} \\ \\ [/tex]

[tex]\sf \:z = \frac{100}{11} \\ \\ [/tex]

[tex]\bf\implies \:z = 9.09 \: \% \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Step-by-step explanation:

Let the price be p,

When the price is increased by 10%

Increased price = (100% + 10%) p = 1.1p

So reduction percentage :-

1.1p - p/1.1p × 100

0.1p/1.1p × 100

9.090909091% = 9.09% (rounded to nearest

hundredth)