SOLUTION: June, an owner of a coffee stand, marked down the price of a latte between 7:00 A.M. and 8:00 A.M. from $2.00 a cup. If she grossed $98.69 from the latte sale and we know that she

Question 762291: June, an owner of a coffee stand, marked down the price of a latte between 7:00 A.M. and 8:00 A.M. from $2.00 a cup. If she grossed $98.69 from the latte sale and we know that she never sells a latte for less than a dollar, how many lattes did she sell between 7:00 A.M. and 8:00 A.M.? Explain your reasoning
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Originally price was $2.00 per cup but was then reduced to 2-d per cup using d as the amount discounted.

Unknown is, how many cups of this latte were sold. Call the number, n.

, and as expected condition. This condition is the same as and , or .

The answer is between two inclusive extremes.
(2-0.01)n=98.69 and 1*n=98.69

Need the whole number n between 1.99n=98.69 and 1*n=98.69.
Anything even near the 1$/latte value will not do, selling one or two or so few latte's should not be expected to achieve revenue over 90 dollars.
Looking at the other end, from the (2-0.01)n limit gives easily more meaningful possibilities.

Not being able to find an efficient algebraic way to solve this, I resorted to a simple BASIC program and this program shows 71 cups of latte at $1.39 per latte to be the best answer.

Program used was this code:
for i=1 to 99
n=98.69/((99+i)/100)
print n;" ";(99+i)/100
next i

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