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Hint: Let the number of students be x. Then, the number of apples received by each student is $\dfrac{300}{x}$. If there are 10 more students, i.e. (x + 10) students, then, the number of apples received by each student is $\dfrac{300}{x+10}$. According to the question, $\dfrac{300}{x}-\dfrac{300}{x+10}=1$. Solve this to find the value of x.
Complete step-by-step solution -
In this question, we are given that 300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less.
We need to find the number of students.
Let the number of students be x.
Then, the number of apples received by each student is $\dfrac{300}{x}$.
If there are 10 more students, i.e. (x + 10) students,
Then, the number of apples received by each student is $\dfrac{300}{x+10}$.
We are given that this number is one less than $\dfrac{300}{x}$.
So, $\dfrac{300}{x}-\dfrac{300}{x+10}=1$
$\dfrac{300x+3000-300x}{x\left( x+10 \right)}=1$
\[3000={{x}^{2}}+10x\]
\[{{x}^{2}}+10x-3000=0\]
Now, we will solve this equation by splitting the middle term. Doing this, we will get the following:
\[{{x}^{2}}+60x-50x-3000=0\]
Now, we will take x common from the first two terms and 50 common from the last two terms. Doing this, we will get the following:
\[x\left( x+60 \right)-50\left( x+60 \right)=0\]
This resultant has two terms.
Now, we will take (x + 50) common from these two terms. Doing this, we will get the following:
\[\left( x+60 \right)\left( x-50 \right)=0\]
\[\left( x+60 \right)=0\text{ or }\left( x-50 \right)=0\]
$x=-60,50$
But, the number of students cannot be negative.
So, the number of students = 50
Hence, option (c) is correct.
Note: In this question, it is very important to realize and understand that if we have 300 apples in total to divide equally among x students, then each student gets $\dfrac{300}{x}$ apples. This is the first step in the solution and if there is any error in this step, you will get a wrong answer. Also note that, after solving the equation, it is important to eliminate the negative value as the number of students cannot be negative. Some students may forget to do so.
Complete step-by-step solution -
In this question, we are given that 300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less.
We need to find the number of students.
Let the number of students be x.
Then, the number of apples received by each student is $\dfrac{300}{x}$.
If there are 10 more students, i.e. (x + 10) students,
Then, the number of apples received by each student is $\dfrac{300}{x+10}$.
We are given that this number is one less than $\dfrac{300}{x}$.
So, $\dfrac{300}{x}-\dfrac{300}{x+10}=1$
$\dfrac{300x+3000-300x}{x\left( x+10 \right)}=1$
\[3000={{x}^{2}}+10x\]
\[{{x}^{2}}+10x-3000=0\]
Now, we will solve this equation by splitting the middle term. Doing this, we will get the following:
\[{{x}^{2}}+60x-50x-3000=0\]
Now, we will take x common from the first two terms and 50 common from the last two terms. Doing this, we will get the following:
\[x\left( x+60 \right)-50\left( x+60 \right)=0\]
This resultant has two terms.
Now, we will take (x + 50) common from these two terms. Doing this, we will get the following:
\[\left( x+60 \right)\left( x-50 \right)=0\]
\[\left( x+60 \right)=0\text{ or }\left( x-50 \right)=0\]
$x=-60,50$
But, the number of students cannot be negative.
So, the number of students = 50
Hence, option (c) is correct.
Note: In this question, it is very important to realize and understand that if we have 300 apples in total to divide equally among x students, then each student gets $\dfrac{300}{x}$ apples. This is the first step in the solution and if there is any error in this step, you will get a wrong answer. Also note that, after solving the equation, it is important to eliminate the negative value as the number of students cannot be negative. Some students may forget to do so.
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