Question

A vendor has a certain number of oranges. One customer comes along and buys half of the vendor's oranges plus half an orange. Another customer comes along and buys half of the vendor's remaining oranges plus half an orange. A third customer comes along and buys half of the vendor's remaining oranges plus half an orange. If the vendor originally had less than 1000 oranges, what is the maximum number of oranges she could have started with

Answer 1

Given the word problem above, the maximum number of oranges she could have started with is x < 1,142. See the solution below.

What is a word problem?

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.

What is the solution to the problem above?

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:
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