Question

# Due to an increase in the price of sugar by 25%, by how much percent must a householder decrease the consumption of sugar as so that there is no increase in the expenditure of the sugar.

Hint: Assume the household consumption to be x and let it be decreased by y% so that net expenditure remains constant. Assume initial prices to be p. Now use the fact that net expenditure is constant to get a value for y.

Let the initial prices of sugar be p, the initial household consumption be x and let it be decreased by y% so that the net expenditure remains the same.
Initial expenditure $=prices\times consumption=px$
New prices = p + 25% of p = $p+\dfrac{25}{100}\times p=\text{ }p\text{ }+\text{ }0.25p\text{ }=\text{ }1.25p$
New consumption = x – y% of x $=x-\dfrac{y}{100}x=x-\dfrac{xy}{100}$
New expenditure = prices $\times$ consumption = $1.25p\times \left[ x-\dfrac{xy}{100} \right]=1.25px\left[ 1-\dfrac{y}{100} \right]$
Since new expenditure = initial expenditure we have $1.25px\left[ 1-\dfrac{y}{100} \right]=px$
Dividing both sides by px we get
$1.25\left[ 1-\dfrac{y}{100} \right]=1$
Dividing both sides by 1.25 we get
$1-\dfrac{y}{100}=\dfrac{1}{1.25}=\dfrac{100}{125}=\dfrac{4}{5}$
Subtracting 1 on both sides we get
$1-\dfrac{y}{100}-1=\dfrac{4}{5}-1$
$\Rightarrow -\dfrac{y}{100}=\dfrac{-1}{5}$
Multiplying both sides by -1 we get
$\dfrac{y}{100}=\dfrac{1}{5}$
Cross multiplying, we get
$5y=100$
Dividing both sides by 5, we get
$\dfrac{5y}{5}=\dfrac{100}{5}$
i.e. y = 20
Hence the consumption should be decreased by 20% so that there is no net change in the expenditure on sugar.

Note:This question can be solved directly using ratio and proportion.
We know that net expenditure $=prices\times consumption$
So that $prices\propto \dfrac{1}{consumption}$
$prices=\dfrac{k}{consumption}$
So we have $\dfrac{price{{s}_{f}}-price{{s}_{i}}}{price{{s}_{i}}}=0.25$ where $price{{s}_{i}}$ are the initial prices and $price{{s}_{f}}$ are the final prices.
Hence, we have
\begin{align} & \dfrac{\dfrac{k}{consumptio{{n}_{f}}}-\dfrac{k}{consumptio{{n}_{i}}}}{\dfrac{k}{consumptio{{n}_{i}}}}=\dfrac{25}{100} \\ & \Rightarrow \dfrac{consumptio{{n}_{i}}-consumptio{{n}_{f}}}{consumptio{{n}_{f}}}=\dfrac{25}{100} \\ \end{align}

Now we know that if $\dfrac{a}{b}=\dfrac{c}{d}$ then $\dfrac{a}{b+a}=\dfrac{c}{d+c}$
Using the above property, we get

\begin{align} & \dfrac{consumptio{{n}_{i}}-consumptio{{n}_{f}}}{consumptio{{n}_{f}}+\left( consumptio{{n}_{i}}-consumptio{{n}_{f}} \right)}=\dfrac{25}{100+25} \\ & \Rightarrow \dfrac{consumptio{{n}_{i}}-consumptio{{n}_{f}}}{consumptio{{n}_{i}}}=\dfrac{25}{125}=\dfrac{1}{5}=\dfrac{100}{5}\%=20\% \\ \end{align}
Hence the consumption should be decreased by 20%.