Answer
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Hint: Assume the cost of one pen and one pencil as some variable. Form simultaneous equations based on the data given and solve them.
Let the cost of one pen and one pencil is $Rs.x$ and $Rs.y$ respectively.
Then, according to the question, the cost of one pen and one pencil is Rs. 12. So, we have:
$
\Rightarrow x + y = 12, \\
\Rightarrow x = 12 - y .....(i) \\
$
Next, it is given that three pencils and five pens cost Rs.56. So, we have:
$ \Rightarrow 5x + 3y = 56 .....(ii)$
Putting the value of $x$ from equation $(i)$ into equation $(ii)$. We’ll get:
$
\Rightarrow 5\left( {12 - y} \right) + 3y = 56, \\
\Rightarrow 60 - 5y + 3y = 56, \\
\Rightarrow 2y = 4, \\
\Rightarrow y = 2 \\
$
Putting the value of $y$ in equation $(i)$ we’ll get:
$
\Rightarrow x = 12 - 2, \\
\Rightarrow x = 10 \\
$
Thus, the cost of one pen is Rs. 10 and that of one pencil is Rs. 2.
Note: We can solve simultaneous equations using the addition method also. In this, we try to make the coefficient of one of the variables same in both the equations by multiplying them with suitable constants. And then simply subtract any one equation from another to get a linear equation in one variable.
Let the cost of one pen and one pencil is $Rs.x$ and $Rs.y$ respectively.
Then, according to the question, the cost of one pen and one pencil is Rs. 12. So, we have:
$
\Rightarrow x + y = 12, \\
\Rightarrow x = 12 - y .....(i) \\
$
Next, it is given that three pencils and five pens cost Rs.56. So, we have:
$ \Rightarrow 5x + 3y = 56 .....(ii)$
Putting the value of $x$ from equation $(i)$ into equation $(ii)$. We’ll get:
$
\Rightarrow 5\left( {12 - y} \right) + 3y = 56, \\
\Rightarrow 60 - 5y + 3y = 56, \\
\Rightarrow 2y = 4, \\
\Rightarrow y = 2 \\
$
Putting the value of $y$ in equation $(i)$ we’ll get:
$
\Rightarrow x = 12 - 2, \\
\Rightarrow x = 10 \\
$
Thus, the cost of one pen is Rs. 10 and that of one pencil is Rs. 2.
Note: We can solve simultaneous equations using the addition method also. In this, we try to make the coefficient of one of the variables same in both the equations by multiplying them with suitable constants. And then simply subtract any one equation from another to get a linear equation in one variable.
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