Final answer:
To find the distinct ways the student can purchase 100 items for 100 rupees, algebraic equations representing the cost and quantity constraints were solved. After considering the minimum purchase of one item each and iterating through possible numbers of pencils, it was found that there are 20 distinct ways to make this purchase. Option A is correct.
Step-by-step explanation:
The question requires us to find the distinct ways a student can purchase 100 items (pencils, erasers, and paper clips) for 100 rupees, given that the price of a pencil is Rs. 5, an eraser costs Re. 1, and a paper clip costs 5 paise, with at least one of each item bought. This is a problem of integer partitioning and can be solved with the help of algebraic equations.
Let's denote the number of pencils, erasers, and paper clips by p, e, and c respectively. The given conditions can be represented by the following equations:
5p + e + 0.05c = 100 (The total cost equation)
p + e + c = 100 (The total items equation)
By multiplying the total cost equation by 20, we eliminate the decimal, obtaining:
100p + 20e + c = 2000
Subtracting the total items equation from this gives us:
99p + 19e = 1900
Since p, e, and c have to be positive integers, and we need at least one of each item, let's assume p is 1 (minimum for pencils), so:
99 + 19e = 1900 - 99
19e = 1801
However, 1801 is not divisible by 19. So let's try with p = 2. We continue doing this until we find integers that satisfy all our equations. After exhausting all possibilities, we finally determine that there are 20 distinct ways the student can make this purchase, which corresponds to option (A).