Question

A sum of money was distributed equally in a class of boys. Had there been 10 boys more, each would have received a rupee less. If there had been 15 boys less, each would have received 3 rupees more. Find the sum of money and the number of boys.

Answer 1

Hint: Here, we need to find the sum of money and the number of boys. The sum of money received by each boy is equal to the quotient of the total sum of money and the total number of boys. First, we will find the sum of money received by each boy. Then, using the given information, we will form two equations. Finally, we will solve these equations to find the number of boys and the sum of money.

Complete step-by-step answer:
We will use two variables $$x$$ and $$y$$ to form equations in two variables using the given information.
Let the number of boys be $$x$$ and the total sum of money be $$y$$ rupees.
The sum of money received by each boy is equal to the quotient of the total sum of money and the total number of boys.
Therefore, we get
Sum of money received by each boy $$={\displaystyle \frac{y}{x}}$$
Now, it is given that if there were 10 boys more, each boy would have received a rupee less.
Here, the number of boys becomes $$x+10$$.
The total sum of money remains the same, that is $$y$$ rupees.
Therefore, we get
Sum of money received by each boy $$={\displaystyle \frac{y}{x+10}}$$
The sum of money received by each boy (where the number of boys is $$x+10$$) is Re. 1 less than the sum of money received by each boy (where the number of boys is $$x$$).
Therefore, we get the equation
$${\displaystyle \frac{y}{x+10}}={\displaystyle \frac{y}{x}}-1$$
We will simplify this equation.
Taking the L.C.M., we get
$$\Rightarrow {\displaystyle \frac{y}{x+10}}={\displaystyle \frac{y-x}{x}}$$
Simplifying by cross multiplication, we get
$$\Rightarrow xy=\left(y-x\right)\left(x+10\right)$$
Multiplying $$\left(y-x\right)$$ by $$\left(x+10\right)$$ using the distributive law of multiplication, we get
$$\Rightarrow xy=xy+10y-x^2-10x$$
Subtracting $$xy$$ from both sides, we get
$$\begin{array}{l}\Rightarrow xy-xy=xy+10y-x^2-10x-xy\\ \Rightarrow 0=10y-x^2-10x\end{array}$$
Rewriting the equation, we get
$$\Rightarrow x^2+10x=10y\dots \dots \dots \left(1\right)$$
Next, it is given that if there were 15 boys less, each boy would have received 3 rupees more.
Here, the number of boys becomes $$x-15$$.
The total sum of money remains the same, that is $$y$$ rupees.
Therefore, we get
Sum of money received by each boy $$={\displaystyle \frac{y}{x-15}}$$
The sum of money received by each boy (where the number of boys is $$x-15$$) is Rs. 3 more than the sum of money received by each boy (where the number of boys is $$x$$).
Therefore, we get the equation
$${\displaystyle \frac{y}{x-15}}={\displaystyle \frac{y}{x}}+3$$
Taking the L.C.M., we get
$$\Rightarrow {\displaystyle \frac{y}{x-15}}={\displaystyle \frac{y+3x}{x}}$$
Simplifying by cross multiplication, we get
$$\Rightarrow xy=\left(y+3x\right)\left(x-15\right)$$
Multiplying $$\left(y+3x\right)$$ by $$\left(x-15\right)$$ using the distributive law of multiplication, we get
$$\Rightarrow xy=xy-15y+3x^2-45x$$
Subtracting $$xy$$ from both sides, we get
$$\begin{array}{l}\Rightarrow xy-xy=xy-15y+3x^2-45x-xy\\ \Rightarrow 0=-15y+3x^2-45x\end{array}$$
Dividing both sides by 3 and rewriting the equation, we get
$$\begin{array}{l}\Rightarrow 0=-5y+x^2-15x\\ \Rightarrow x^2-15x=5y\dots \dots \dots \left(2\right)\end{array}$$
We can observe that the equations $$\left(1\right)$$ and $$\left(2\right)$$ are equations in two variables. We will solve these to get the values of $$x$$ and $$y$$.
Subtracting equation $$\left(2\right)$$ from equation $$\left(1\right)$$, we get
$$\Rightarrow x^2+10x-\left(x^2-15x\right)=10y-5y$$
Adding and subtracting the like terms, we get
$$\begin{array}{l}\Rightarrow x^2+10x-x^2+15x=5y\\ \Rightarrow 25x=5y\end{array}$$
Dividing both sides by 5, we get
$$\begin{array}{l}\Rightarrow {\displaystyle \frac{25x}{5}}={\displaystyle \frac{5y}{5}}\\ \Rightarrow 5x=y\end{array}$$
Substituting $$y=5x$$ in the equation $${\displaystyle \frac{y}{x+10}}={\displaystyle \frac{y}{x}}-1$$, we get
$$\Rightarrow {\displaystyle \frac{5x}{x+10}}={\displaystyle \frac{5x}{x}}-1$$
Thus, we get
$$\begin{array}{l}\Rightarrow {\displaystyle \frac{5x}{x+10}}=5-1\\ \Rightarrow {\displaystyle \frac{5x}{x+10}}=4\end{array}$$
Simplifying the equation by cross multiplying, we get
$$\begin{array}{l}\Rightarrow 5x=4\left(x+10\right)\\ \Rightarrow 5x=4x+40\end{array}$$
Subtracting $$4x$$ from both sides, we get
$$\begin{array}{l}\Rightarrow 5x-4x=4x+40-4x\\ \Rightarrow x=40\end{array}$$
Thus, the number of boys is 40.
Substituting $$x=40$$ in the equation $$y=5x$$, we get
$$y=5\times 40=\mathrm{R}\mathrm{s}.200$$
$$\therefore$$ The total sum of money is Rs. 200.

Note: We have used the distributive law of multiplication in the solution. The distributive law of multiplication states that $$\left(a+b\right)\left(c+d\right)=a\cdot c+a\cdot d+b\cdot c+b\cdot d$$.
We can verify our answer using the given information.
The number of boys is 40 and the sum of money is Rs. 200.
Therefore, the sum of money received by each boy is $${\displaystyle \frac{200}{40}}=\mathrm{R}\mathrm{s}.5$$.
If there were 10 more boys, the number of boys would be 50.
The sum of money received by each boy (where number of boys is 50) will be $${\displaystyle \frac{200}{50}}=\mathrm{R}\mathrm{s}.4$$, that is one rupee less than Rs. 5.
If there were 15 boys less, the number of boys would be 25.
The sum of money received by each boy (where number of boys is 25) will be $${\displaystyle \frac{200}{25}}=\mathrm{R}\mathrm{s}.8$$, that is 3 rupees more than Rs. 5.
Therefore, we have verified that our answer satisfies the given information.