
A part of monthly expenses of a family is constant and the remaining part varies with the price of wheat. When the rate of wheat is Rs. 250 a quintal, the total monthly expenses of the family are Rs. 1000 and when it is Rs. 240 a quintal, the total monthly expenses are Rs 980. Find the total expenses of the family, when the cost of wheat is Rs 350 a quintal.
(a) Rs. 1500
(b) Rs. 2000
(c) Rs. 1200
(d) Rs. 1600
Answer
606.9k+ views
Hint:Assume that the constant part of monthly expenses of the family is Rs. x. Now, assume that the family consumes a total of ‘y’ quintals of wheat in one month. Find the total cost of consumption of wheat by multiplying the rate of wheat per quintal with the total whet consumed in quintals. Now, monthly expenses will be the sum of the constant part and the cost of consumption of wheat. Form two linear equations in two variables ‘x’ and ‘y’ according to the given information. Solve these equations to find the value of x and y and hence, find the value of total expenses of the family, when the cost of wheat is Rs 350 a quintal by substituting the values of x and y in the relation: $x+350y$.
Complete step-by-step answer:
Let us assume that the constant part of monthly expenses of the family is Rs. x and the family consumes a total of ‘y’ quintals of wheat in one month.
Now, total cost of wheat when the rate of wheat is Rs. 250 a quintal
$\begin{align}
& =\text{rate}\times \text{number of quintals} \\
& =250\times y \\
& =250y \\
\end{align}$
Since total monthly expenses of the family in this month is Rs. 1000, therefore,
$x+250y=1000.....................(i)$
Now, total cost of wheat when the rate of wheat is Rs. 240 a quintal
$\begin{align}
& =\text{rate}\times \text{number of quintals} \\
& =240\times y \\
& =240y \\
\end{align}$
Since total monthly expenses of the family in this month is Rs. 980, therefore,
$x+240y=980.....................(ii)$
Let us solve the two equations.
Subtracting equation (ii) from equation (i), we get,
$\begin{align}
& x+250y-\left( x+240y \right)=1000-980 \\
& \Rightarrow 10y=20 \\
& \Rightarrow y=2 \\
\end{align}$
Substituting y = 2 in equation (i), we get,
$\begin{align}
& x+250\times 2=1000 \\
& \Rightarrow x+500=1000 \\
& \Rightarrow x=1000-500 \\
& \Rightarrow x=500 \\
\end{align}$
Therefore, the constant part of the monthly expenses of the family is Rs. 500 and the family consumes a total of 2 quintals of wheat in a month.
Now, the total expenses of the family, when the cost of wheat is Rs 350 a quintal $=x+350y$.
Substituting the obtained values of x and y in the above relation, we get,
The total expenses of the family
$\begin{align}
& =x+350y \\
& =500+350\times 2 \\
& =500+700 \\
& =1200 \\
\end{align}$
Hence, option (c) is the correct answer.
Note: One may note that it is important to find the values of both x and y because we have to substitute these values in the equation: $x+350y$ to find the total expenses of the family, when the cost of wheat is Rs 350 a quintal. You may note that we have used elimination methods to solve the two linear equations. Do not use cross-multiplication methods to solve these equations because there can be calculation mistakes and you will get the wrong answer.
Complete step-by-step answer:
Let us assume that the constant part of monthly expenses of the family is Rs. x and the family consumes a total of ‘y’ quintals of wheat in one month.
Now, total cost of wheat when the rate of wheat is Rs. 250 a quintal
$\begin{align}
& =\text{rate}\times \text{number of quintals} \\
& =250\times y \\
& =250y \\
\end{align}$
Since total monthly expenses of the family in this month is Rs. 1000, therefore,
$x+250y=1000.....................(i)$
Now, total cost of wheat when the rate of wheat is Rs. 240 a quintal
$\begin{align}
& =\text{rate}\times \text{number of quintals} \\
& =240\times y \\
& =240y \\
\end{align}$
Since total monthly expenses of the family in this month is Rs. 980, therefore,
$x+240y=980.....................(ii)$
Let us solve the two equations.
Subtracting equation (ii) from equation (i), we get,
$\begin{align}
& x+250y-\left( x+240y \right)=1000-980 \\
& \Rightarrow 10y=20 \\
& \Rightarrow y=2 \\
\end{align}$
Substituting y = 2 in equation (i), we get,
$\begin{align}
& x+250\times 2=1000 \\
& \Rightarrow x+500=1000 \\
& \Rightarrow x=1000-500 \\
& \Rightarrow x=500 \\
\end{align}$
Therefore, the constant part of the monthly expenses of the family is Rs. 500 and the family consumes a total of 2 quintals of wheat in a month.
Now, the total expenses of the family, when the cost of wheat is Rs 350 a quintal $=x+350y$.
Substituting the obtained values of x and y in the above relation, we get,
The total expenses of the family
$\begin{align}
& =x+350y \\
& =500+350\times 2 \\
& =500+700 \\
& =1200 \\
\end{align}$
Hence, option (c) is the correct answer.
Note: One may note that it is important to find the values of both x and y because we have to substitute these values in the equation: $x+350y$ to find the total expenses of the family, when the cost of wheat is Rs 350 a quintal. You may note that we have used elimination methods to solve the two linear equations. Do not use cross-multiplication methods to solve these equations because there can be calculation mistakes and you will get the wrong answer.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

