Mathew had some pens. On the first day, he sold 1/4 of them. On the second day, he sold 1/3 more than the number of pens sold on the first day. He sold the remaining pens on the third day. If he sold 15 more pens ont he second day than on the third day, how many pens did he have at first?
Question
Mathew had some pens. On the first day, he sold 1/4 of them. On the second day, he sold 1/3 more than the number of pens sold on the first day. He sold the remaining pens on the third day. If he sold 15 more pens ont he second day than on the third day, how many pens did he have at first?
Solution
Let's denote the initial number of pens Mathew had as P.
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On the first day, Mathew sold 1/4 of his pens, which is P/4.
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On the second day, he sold 1/3 more pens than he did on the first day. This means he sold (1 + 1/3) * (P/4) = P/3 pens on the second day.
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According to the problem, Mathew sold 15 more pens on the second day than on the third day. This means he sold P/3 - 15 pens on the third day.
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The total number of pens Mathew sold is the sum of the pens he sold on each day. So, P = P/4 (first day) + P/3 (second day) + (P/3 - 15) (third day).
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Simplifying the equation gives P = 4P/12 + 4P/12 + 4P/12 - 60.
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This simplifies to P = P - 60.
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Solving for P gives P = 60.
So, Mathew initially had 60 pens.
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