Question

1000 children in a hostel had enough food for 28 days. After 4 days, some students were shifted to another hostel. As a result the food now lasted for 32 days. How many students were shifted?

Answer 1

Let the number of students shifted be $x$.

Number of students(x)-:  1000         1000-x

Number of days-:             28             32

As the number of students increase food remains for less days (inverse variation)

1000 × 28  = (1000 - x) × 32

$1000-x = \frac{1000* 28}{32} = 875$

x = 1000 - 875 = 125

∴ 125 students were shifted.

Answer 2

$\huge{\underline{Solution}}$

→ The amount of provisions (call this p) required is directly proportional to the number of students (call this n) and the number of days (call this t) in the hostel.

⇝ Mathematically, p ∝ nt or

⇝ p=knt (k being a constant for provisions per student-day)

⇝ Then p1 = kn1t1 (for 1st situation with 100 st for 20 d) and

⇝ p2 = kn2t2 (for 2nd situation with 125 st for an unknown number of days, t2)

★ Note that the k is the same for both situations because it is assumed that both groups need the same amount of provisions/student-day and

★ p1 = p2 because the group of 100 students (n1) for 20 days (t1) will have the same amount of provisions as the group of 100 students (n2) for t2 days (the question you are asking for a solution). Therefore,

$\sf{kn1t1 = kn2t2\: or}$

$\sf{n1t1 = n2t2}$(constants k cancel each other out)

Isolating the variable t2 algebraically,

$⟹\sf{t2 = n1t1/n2\: or}$

$⟹\sf{t2 = \dfrac{(100 students)(20 days)}{125 students}}$

$⟹\sf{t2 = 16 days (Final answer)}$

Note that the 1’s and 2’s should appear as subscripts.