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Explanation:

Let the number of students in section B is x and that in A is (x – 10).
(x-10)×32+x×602x-10 = integer

92x-3202x-10 = integer

46x-160x-5 = integer

46(x-5)+70x-5 = integer

⇒ (x - 5) + 70x-5 = integer

70x-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.

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