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 Rs50, whereas 7 pencils and 5 pens together cost Rs46. Find the cost of one pencil and that of one pen.

Updated On: Oct 31, 2023
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Solution and Explanation

(i) Let the number of girls be \(x\) and the number of boys be \(y.\)

According to the question, the algebraic representation is \(x + y = 10 x − y = 4\) For \(x + y = 10, x = 10 − y;\)

\(x\)\(5\)\(4\)\(6\)
\(y\)\(5\)\(6\)\(4\)


For \(x − y = 4, x = 4 + y\)

\(x\)\(5\)\(4\)\(3\)
\(y\)\(1\)\(0\)\(-1\)


Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at points \((7, 3)\).

Therefore, the number of girls and boys in the class is \(7\) and \(3\) respectively.


(ii) Let the cost of \(1\) pencil be Rs \(x\) and the cost of \(1\) pen be Rs \(y\). According to the question, the algebraic representation is 

\(5x + 7y = 50\)
\(7x + 5y = 46\)

For \(5x + 7y = 50,\)
\(x= 50-\dfrac{7y}{5}\)

\(x\)\(3\)\(10\)\(-4\)
\(y\)\(5\)\(0\)\(10\)


\(7x + 5y = 46\)
\(x= 46-\dfrac{5y}{7}\)

\(x\)\(8\)\(3\)\(-2\)
\(y\)\(-2\)\(5\)\(12\)

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point \((3, 5).\)

Therefore, the cost of a pencil and a pen are Rs \(3\) and Rs \(5\) respectively.

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