Given the word problem above, the maximum number of oranges she could have started with is x < 1,142. See the solution below.
<h3>
What is a word problem?</h3>
In math, sometimes, mathematical problems are stated using words rather than equations. When this occurs, we call it a word problem. The best way to solve this kind of problem is by putting it in form of an equation so that it is easy to solve.
<h3>What is the solution to the problem above?</h3>
Let the total number of Oranges that the vendor has be x.
⇒ She sells to the first customer (Half of all and half of one)
that is, (x/2) + 1/2
= (x+1)/2
stock balance = x - ((x+1)/2))
= (x-1)/2
⇒ She sells to the second customer (Half of the balance and half of one)
that is, 1/2 ( (x-1)/2) + 1/2
= (x-1)/4 + 1/2
= (x-1+2)/4
= (x+1)/4
Stock balance = ((x-1)/2) - ((x-1)/4)
= (2x−2−x−1)/4
= (x-3)/4
⇒ She sells to the third customer (Half of the balance and half of one)
That is 1/2 ((x-3)/4) + 1/2
= ((x-3)/8) + 1/2
= (x−3+4)/8
= (x+1)/8
Thus if x < 1000 the original number of oranges she could have started with is:
[ ((x+1)/2) + ((x+1)/4) + ((x+1)/8)] < 1,000
From here, we solve by first finding the common denominator
[(4(x+1)/8) + 2(x+1)/8) + ((x+1)/8) < 1000
Next - combine fractions with common denominators
[(4x+1) + 2(x+1) + x +1]/8 < 1000
Next, Simplify
[(4x + 7 + 2x + x)/8] <1000
⇒ [(7x + 7)/8] < 1000
⇒ 7x < 7993
⇒ 7x/x < 7993/7
Thus,
x < 7993/7
x < 1,141.8571428
x < 1,142
Thus the maximum number of oranges she could have started with is 1,142 oranges.
Learn more about word problems at:
brainly.com/question/21405634
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