Question

300 apples are distributed equally among certain number of students. Had there been 10 more students each would have received one apple less . Find the number of students.

Answer 1

Let the no. of students be x.

When 300 apples are distributed equally=300/x

ATQ->

(300/x)-1= 300/(x+10)

=> (300/x)-300/(x+10)=1

=> 300[(1/x)-1/(x+10) =1

=> 300 [(x+10-x)/(x² + 10x)]=1

=> 300 {10/(x²+10x)]=1

=> 3000= x²+10x

x²+10x-3000=0

x= [-b±√(b²-4ac)]/2a

x=[-10 ± √{10²-4(1)(-3000)}]/2

x= [-10 ±√(100+12000)]/2

x=[-10±√12100]/2

x=(-10±110)/2

Case 1:

x=(-10+110)/2

x=100/2

=50

Case 2:

x=(-10-110)/2

x=-120/2

x=-60

∴x=50 and x=-60

As number of students can not be negative.

∴There were 50 students

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Answer 2

Let the original number of students be x

Number of apples received by each student = $\sf{\frac{300}{x}}$

Increased number of students = x + 10

Number of apples received by each student = $\sf{\frac{300}{x+10}}$

A/C,

$\sf\red{\frac{300}{x} - \frac{300}{x + 10} = }1$

$\sf\red{300x + 3000 - 300x = x(x + 10)}$

$\sf\red{ {x}^{2} + 10x - 3000 = 0}$

$\sf\red{(x - 50)(x + 60) = 0}$

$\sf\red{x = 50 \: or \: x = - 60}$

As the number of students cannot be negative, x ≠ -60 ,so x = 50

Hence, the original number of students = 50.