The average of four possible numbers is 15. The second number is 2 times the first number. The third number is twice the second. What is the largest number?
Would these be consecutive integers?
And then you find the median??
n, 2n, 4n, x ... 7n + x = 60
if the 4th is twice the 3rd
... then ... 15n = 60 ... n = 4
but there is not enough information for a unique solution
X=22?
To find the largest number in this scenario, we can use the information given about the numbers and their relationships.
Let's assume the first number is x.
According to the problem, the second number is 2 times the first number, so the second number would be 2x.
Similarly, the third number is twice the second number, which means it would be 2 * 2x = 4x.
Now, we have the first number as x, the second number as 2x, the third number as 4x, and the fourth number, which we don't know yet.
The average of these four numbers is given as 15. To find the average, we sum up all the numbers and divide by 4:
(x + 2x + 4x + fourth number) / 4 = 15
Now, we can simplify the equation:
7x + fourth number = 60
Since we are looking for the largest number, we need to find the maximum value for the fourth number. For that, we can assume that the other numbers are minimized. So, we take the minimum value of x as 1 (since they are possible numbers).
Plugging in the values, we get:
7 * 1 + fourth number = 60
7 + fourth number = 60
To isolate the fourth number, we subtract 7 from both sides:
fourth number = 60 - 7
fourth number = 53
Therefore, the largest number in this scenario is 53.
To summarize:
1. Assume the first number as x.
2. The second number would be 2 * x.
3. The third number would be 4 * x.
4. Set up the equation using the average of the numbers.
5. Solve the equation by isolating the fourth number.
6. Find the maximum value for the fourth number by assuming the other numbers are minimized.
7. Calculate the value of the fourth number based on the equation.
8. The resulting value is the largest number.