Discussion

Explanation:

Let the number of students in section B is x and that in A is (x – 10).
⇒ $\frac{(\mathrm{x}-10)\times 32+\mathrm{x}\times 60}{2\mathrm{x}-10}$ = integer

⇒ $\frac{92\mathrm{x}-320}{2\mathrm{x}-10}$ = integer

⇒ $\frac{46\mathrm{x}-160}{\mathrm{x}-5}$ = integer

⇒ $\frac{46(\mathrm{x}-5)+70}{\mathrm{x}-5}$ = integer

⇒ (x - 5) + $\frac{70}{\mathrm{x}-5}$ = integer

⇒ $\frac{70}{\mathrm{x}-5}$ should be an integer

∴ (x – 5) should be a factor of 70 while x > 10

⇒ Highest possible value of x = 75 while lowest possible value is 12

∴ Difference between highest and lowest values of x = 75 – 12 = 63.

Hence, 63.

» Your doubt will be displayed only after approval.