From a number of mangoes, a man sells half the number of existing mangoes plus 1 to the first customer, then sells 1/3rd of the remaining number of mangoes plus 1 to the second customer, then 1/4th of the remaining number of mangoes plus 1 to the third customer and 1/5th of the remaining number of mangoes plus 1 to the fourth customer. He then finds that he does not have any mangoes left. How many mangoes did he have originally?
Question
From a number of mangoes, a man sells half the number of existing mangoes plus 1 to the first customer, then sells 1/3rd of the remaining number of mangoes plus 1 to the second customer, then 1/4th of the remaining number of mangoes plus 1 to the third customer and 1/5th of the remaining number of mangoes plus 1 to the fourth customer. He then finds that he does not have any mangoes left. How many mangoes did he have originally?
Solution
Let's solve this problem step by step:
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Let's denote the original number of mangoes as X.
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According to the problem, the man sells half of the mangoes plus 1 to the first customer. So, the remaining mangoes after the first customer are X - (X/2 + 1) = X/2 - 1.
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Then, he sells 1/3rd of the remaining mangoes plus 1 to the second customer. So, the remaining mangoes after the second customer are (X/2 - 1) - (X/2 - 1)/3 - 1 = 2X/3 - 4/3.
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Then, he sells 1/4th of the remaining mangoes plus 1 to the third customer. So, the remaining mangoes after the third customer are (2X/3 - 4/3) - (2X/3 - 4/3)/4 - 1 = 3X/4 - 16/4.
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Finally, he sells 1/5th of the remaining mangoes plus 1 to the fourth customer. So, the remaining mangoes after the fourth customer are (3X/4 - 16/4) - (3X/4 - 16/4)/5 - 1 = 4X/5 - 80/5.
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According to the problem, he does not have any mangoes left after selling to the fourth customer. So, we can set up the equation 4X/5 - 80/5 = 0.
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Solving this equation for X gives X = 80.
So, the man originally had 80 mangoes.
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