A machine makes (150) packets in the first hour and (10) fewer packets each next hour. In which hour will the production be (60) packets?

Correct answer: (10)

The governing concept is the nth-term formula for an arithmetic progression. Hourly production forms the sequence 150, 140, 130, and so on, so the first term is a = 150 and the common difference is d = -10 because production decreases by 10 each hour. For the nth hour, a_n = a + (n - 1)d. Setting the production equal to 60 gives 60 = 150 + (n - 1)(-10). Thus 60 = 150 - 10n + 10 = 160 - 10n, so 10n = 100 and n = 10. Therefore option D is correct. Options A, B and C represent earlier hours and give 90, 80 and 70 packets respectively, not 60.