Question

Bhanu has a total of 40 coins of denominations 30 paise and 10 paise. The total amount with him is Rs. 9. Find the number of 10 paise coins with him.

Answer 1

Hint:
Here, we need to find the number of 10 paise coins that Bhanu has. We will solve this question by assuming the number of 30 paise coins and 10 paise coins to be \[x\] and \[y\] respectively. We will then add the terms and equate it with the total number of coins. We will again take the denominations in terms of \[x\] and \[y\] and equate it to the total amount to form another equation. Then we will solve both equations obtained to find the number of coins.

Complete step by step solution:
Let the number of 30 paise coins and 10 paise coins be \[x\] and \[y\] respectively.
We will solve the problem by forming two linear equations in two variables using the information given, and solving those equations.
Now, we know that the sum of the number of 30 paise coins and 10 paise coins is 40.
Thus, we get
\[x + y = 40 \ldots \ldots \ldots \left( 1 \right)\]
Next, the amount with Bhanu in the form of 30 paise coins is the product of the number of 30 paise coins and the value of each coin, that is 30 paise, or Rs. \[0.3\].
Thus, we get the amount with Bhanu in the form of 30 paise coins as Rs. \[0.3x\].
Similarly, the amount with Bhanu in the form of 10 paise coins is the product of the number of 10 paise coins and the value of each coin, that is 10 paise, or Rs. \[0.1\].
Thus, we get the amount with Bhanu in the form of 10 paise coins as Rs. \[0.1y\].
The sum of the amount in form of 30 paise and 10 paise coins is the total amount Bhanu has, that is Rs. 9.
Therefore, we get
\[0.3x + 0.1y = 9 \ldots \ldots \ldots \left( 2 \right)\]
Now, we can observe that the equations \[\left( 1 \right)\] and \[\left( 2 \right)\] are linear equations in two variables.
We will use the substitution method to solve the two linear equations.
Rewriting equation \[\left( 1 \right)\], we get
\[x = 40 - y \ldots \ldots \ldots \left( 3 \right)\]
Substituting \[x = 40 - y\] in equation \[\left( 2 \right)\], we get
\[ \Rightarrow 0.3\left( {40 - y} \right) + 0.1y = 9\]
Multiplying the terms in the equation, we get
\[ \Rightarrow 12 - 0.3y + 0.1y = 9\]
Simplifying the equation, we get
\[ \Rightarrow 12 - 0.2y = 9\]
Subtracting 9 from both sides and rewriting the equation, we get
\[\begin{array}{l} \Rightarrow 12 - 0.2y - 9 = 9 - 9\\ \Rightarrow 3 - 0.2y = 0\\ \Rightarrow 0.2y = 3\end{array}\]
Dividing both sides by \[0.2\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{0.2y}}{{0.2}} = \dfrac{3}{{0.2}}\\ \Rightarrow y = 15\end{array}\]

\[\therefore\] The number of 10 paise coins is 15 coins.

Note:
We can also solve the two equations using the elimination method.
Multiplying both sides of equation \[\left( 1 \right)\] by \[0.3\], we get
\[\begin{array}{l}0.3\left( {x + y} \right) = 0.3 \times 40\\ \Rightarrow 0.3x + 0.3y = 12 \ldots \ldots \ldots \left( 3 \right)\end{array}\]
Subtracting both sides of equation \[\left( 2 \right)\] from equation \[\left( 3 \right)\], we get
\[\begin{array}{l}0.3x + 0.3y = 12\\\underline {0.3x + 0.1y = {\text{ }}9} \\{\text{ }}0.2y = {\text{ 3}}\end{array}\]
Dividing both sides by \[0.2\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{0.2y}}{{0.2}} = \dfrac{3}{{0.2}}\\ \Rightarrow y = 15\end{array}\]
\[\therefore\] The number of 10 paise coins is 15 coins.