Find three consecutive odd numbers such that eight more than the smallest integer is 23 less than the sum of the median and largest integer.

Question

Type 3 word problem and is hard for me

Answer 1

Let A = the first integer

Let B = the second integer

Let C = the third integer

 

Since the numbers are consecutive and odd, we know that the numbers must all be 2 apart.

 

Therefore we know

 

B = A + 2

C = B + 2 = A + 4

 

We also know that the median of a set of numbers is the middle value. Therefore, the median = B

 

Finally, from the problem, we know that

 

A + 8 = (B + C) - 23

 

Since we have B and C in terms of A, we can just use substitution:

 

A + 8 = ((A + 2) + (A + 4)) - 23

A + 8 = A + 2 + A + 4 - 23

8 = A + 2 + A + 4 - 23 - A

8 = A + 2 + 4 - 23

8 = A + 6 - 23

8 = A - 17

8 + 17 = A

25 = A

 

Therefore A = 25, B = 27, C = 29.

 

Answer 2

You can really break this down by writing it into the form of an equation.  First, we'll come up with variables for the three consecutive odd numbers

 

x1 - first (smallest) odd number

x2 - second (median) odd number

x3 - third (largest) odd number

 

Now, the problem tells us 8 more than the smallest is 23 less than the sum of the median and the largest.

 

So, x1+8=-23+(x2+x3)

 

Simplify this equation and you'll be well on your way to a solution!