
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.
Answer
581.7k+ views
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.
Complete step-by-step answer:
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%.
Complete step-by-step answer:
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%.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

