Ex.3.2 Q1 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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Question

Form the pair of linear equations in the following problems and find their Solutions graphically.

(i) \(10\) students of Class X took part in a Mathematics quiz. If the number of girls is \(4\) more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) \(5\) pencils and \(7\) pens together cost ₹ \(50, \) whereas \(7\) pencils and \(5\) pens together cost ₹ \(46.\) Find the cost of one pencil and that of one pen.

 Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.2 | Question 1

Text Solution

(i)

What is Known?

(i) Number of students took part in Quiz \(= 10\)

(ii) Number of girls is \(4\) more than number of boys

What is Unknown?

Finding Solutions graphically for the given situation.

Reasoning:

Assuming the number of boys as \(x\) and the number of girls as \(y,\) two linear equations can be formed for the above situation.

Steps:

Total number of boys and girls is:

\[x + y = 10\]

Number of girls is \(4\) more than the number of boys,

Mathematically:

\[\begin{align}y &= x + 4\\ - x + y &= 4\end{align}\]

Algebraic representation where \(x\) and \(y\) are the number of boys and girls respectively.

\[\begin{align}x + y &= 10 \qquad(1)\\- x + y &= 4\qquad \;\;(2) \end{align}\] 

Therefore, the algebraic representation for equation 1 is:

\[\begin{align}x + y &= 10\\y &= 10-x\end{align}\]

And, the algebraic representation is for equation 2 is:

\[\begin{align} - x + y& = 4\\y &= x + 4\end{align}\]

Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in table shown below.

\(x\) \(2\) \(7\)
\(y = 10 - x\) \(8\) \(3\)

 

\(x\) \(1\) \(4\)
\(y = x + 4\) \(5\) \(8\)

The graphical representation is as follows.

Answer:

From graph solution \(\left( {x,y} \right) = \left( {3,7} \right)\)

Number of boys \(= 3\)

Number of girls\( = 7\)

(ii)

What is Known?

(i) \(5\) pencils and \(7\) pens cost ₹ \(50\) \(\left( {x,y} \right) = \left( {3,7} \right)\)

(ii) \(7\) pencils and \(5\) pens cost ₹ \(46\)

What is Unknown?

Finding Solutions graphically for the given situation.

Reasoning:

Assuming the cost of \(1\) pencil as ₹ \(x\) and the cost of \(1\) pen as ₹ \(y,\) two linear equations are to be formed for the above situation.

Steps:

Let us assume cost of \(1\) pencil be \(x\) and cost of \(1\) pen be \(y.\)

The cost of \(5\) pencils and \(7\) pens is ₹ \(50.\)

Mathematically,

\[5x + 7y = 50\]

And, the cost of \(7\) pencils and 5 pens is ₹ \(50.\)

Mathematically,

\[7x + 5y = 46\]

Algebraic representation where \(x\) and \(y\) are the cost of \(1\) pencil and \(1\) pen respectively.

\[\begin{align}5x + 7y &= 50 \qquad (1) \\7x + 5y &= 46 \qquad(2) \end{align}\]

Therefore, the algebraic representation for equation \(1\) is:

\[\begin{align}5x+ 7y &= 50\\7y &= 50-5x\\y &= \frac{{50 - 5x}}{7}\end{align}\]

And, the algebraic representation for equation \(2\) is:

\[\begin{align}7x + 5y &= 46\\5y &= 46-7x\\y &= \frac{{46 - 7x}}{5}\end{align}\]

Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in table shown below.

\(x\) \(3\) \( - 4\)
\(y = \frac{{50 - 5x}}{7}\) \(5\) \(10\)

 

\(x\) \(3\) \(8\)
 \(y = \frac{{46 - 7x}}{5}\) \(5\) \( - 2\)

From graph Solution \(\left( {x,y} \right) = \left( {3,5} \right)\)

Cost of one pencil \(=\)\(3\)

Cost of one pen \(=\)\(5\)