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Emil is running a lemonade stand. He sells half of his lemonade in the morning, a quarter of what was left at lunchtime, and a third of what ws left in the afternoon. When he closes for the day, he has 4 liters left. How many liters of lemonade did he start the day with? math

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Anuksha0456idn

Emil starts the day with 16 liters of lemonade if he is left with 4 liters by the end of the day. Computed using the fractional values given.

Let the initial quantity of lemonade with Emil be x liters.

Quantity sold by Emil in the morning = Half of his lemonade = (1/2)x liters, that is half fraction of x.

Quantity left with Emil = x - x/2 = x/2 liters, that is half fraction of x.

Quantity sold by Emil at lunchtime = Quarter of what was left = (1/4)(x/2) liters = x/8 liters, that is the one-eight fraction of x.

Quantity left with Emil = x/2 - x/8 = 3x/8 liters, that is the three-eight fraction of x.

Quantity sold by Emil in the afternoon = One-third of what was left = (1/3)(3x/8) liters = x/8 liters, that is the one-eight fraction of x.

Quantity left with Emil = 3x/8 - x/8 = x/4 liters, that is the quarter fraction of x.

Now, we are said that Emil closes for the day with 4 liters remaining.

Therefore, x/4 = 4, or, x = 4*4 = 16 liters.

Therefore, Emil started the day with 16 liters of lemonade.

Learn more about fractions at

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