The total population of a village is $5000$. The number of males and females increases by $10\%$ and $15\%$ respectively and consequently, the population of the village becomes $5600$. What was the number of males in the village?
Answer
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Hint: We start solving this question by assuming the number of males to be $x$ and the number of females is $y$. By using the given information that the number of males and females increases by $10\%$ and $15\%$ respectively, we form equation. Then, by solving the equations we get the values of $x\And y$.
Complete step-by-step solution:
We have been given that the total population of a village is $5000$.
We have to find the number of males in the village.
Let us assume the number of males be $x$ and the number of females be $y$.
So, we have $x+y=5000..............(i)$
Now, when the number of males and females increases by $10\%$ and $15\%$ respectively and consequently, the population of the village becomes $5600$.
So, we have
$\begin{align}
& x+10\%x+y+15\%y=5600 \\
& x+\dfrac{10}{100}x+y+\dfrac{15}{100}y=5600 \\
& \dfrac{100x+10x+100y+15y}{100}=5600 \\
& 110x+115y=560000.............(ii) \\
\end{align}$
Now, we have two equations
$x+y=5000..............(i)$
$110x+115y=560000.............(ii)$
To solve these equations we multiply the equation (i) with $115$ and subtract equation (ii) from equation (i), we get
$115x+115y=115\times 5000$
$\left( 115x+115y-575000 \right)-\left( 110x+115y-560000 \right)$
Now, simplifying the above equation we have
$\begin{align}
& 115x+115y-575000-110x+115y+560000=0 \\
& 5x-575000+560000=0 \\
& 5x-15000=0 \\
& 5x=15000 \\
& x=\dfrac{15000}{5} \\
& x=3000 \\
\end{align}$
So, the number of males in the village is $3000$.
Note: In this question the equations we get are linear equations. We use the elimination method to solve this equation. Alternatively, we can use the substitution method to solve the equations. The possibility of a mistake may be in forming the equation from the given equation. As given in the question the number of males and females increases by $10\%$ and $15\%$ respectively, the population of the village becomes $5600$. Some students may form the equation as $\dfrac{10}{100}x+\dfrac{15}{100}y=5600$ and get incorrect answers.
Complete step-by-step solution:
We have been given that the total population of a village is $5000$.
We have to find the number of males in the village.
Let us assume the number of males be $x$ and the number of females be $y$.
So, we have $x+y=5000..............(i)$
Now, when the number of males and females increases by $10\%$ and $15\%$ respectively and consequently, the population of the village becomes $5600$.
So, we have
$\begin{align}
& x+10\%x+y+15\%y=5600 \\
& x+\dfrac{10}{100}x+y+\dfrac{15}{100}y=5600 \\
& \dfrac{100x+10x+100y+15y}{100}=5600 \\
& 110x+115y=560000.............(ii) \\
\end{align}$
Now, we have two equations
$x+y=5000..............(i)$
$110x+115y=560000.............(ii)$
To solve these equations we multiply the equation (i) with $115$ and subtract equation (ii) from equation (i), we get
$115x+115y=115\times 5000$
$\left( 115x+115y-575000 \right)-\left( 110x+115y-560000 \right)$
Now, simplifying the above equation we have
$\begin{align}
& 115x+115y-575000-110x+115y+560000=0 \\
& 5x-575000+560000=0 \\
& 5x-15000=0 \\
& 5x=15000 \\
& x=\dfrac{15000}{5} \\
& x=3000 \\
\end{align}$
So, the number of males in the village is $3000$.
Note: In this question the equations we get are linear equations. We use the elimination method to solve this equation. Alternatively, we can use the substitution method to solve the equations. The possibility of a mistake may be in forming the equation from the given equation. As given in the question the number of males and females increases by $10\%$ and $15\%$ respectively, the population of the village becomes $5600$. Some students may form the equation as $\dfrac{10}{100}x+\dfrac{15}{100}y=5600$ and get incorrect answers.
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