The mean of 3 consecutive multiples of 3 is equal to A.The mean of 4 consecutive multiples of 4 is equal to A + 27.
The mean of the smallest and largest of these 7 numbers is 42. Find A
To solve this problem, we need to set up equations using the given information.
Let's start with the mean of 3 consecutive multiples of 3, which is equal to A. We can denote the smallest multiple as x, so the three multiples would be x, x + 3, and x + 6. The mean is the average of these three numbers:
(x + (x + 3) + (x + 6)) / 3 = A
Simplifying this equation gives us:
(3x + 9) / 3 = A
x + 3 = A
Next, let's consider the mean of 4 consecutive multiples of 4, which is equal to A + 27. Denoting the smallest multiple as y, the four multiples would be y, y + 4, y + 8, and y + 12. The mean of these four numbers is:
(y + (y + 4) + (y + 8) + (y + 12)) / 4 = A + 27
Simplifying this equation gives us:
(4y + 28) / 4 = A + 27
y + 7 = A + 27
Lastly, we're given that the mean of the smallest and largest of all seven numbers (x, x + 3, x + 6, y, y + 4, y + 8, y + 12) is 42. The smallest number is x, and the largest number is y + 12. Therefore, the mean is:
(x + y + 12) / 2 = 42
Simplifying this equation gives us:
x + y + 12 = 84
x + y = 72
We now have a system of two equations:
x + 3 = A
y + 7 = A + 27
x + y = 72
From the first equation, we can rewrite x in terms of A: x = A - 3.
Substituting this expression into the third equation:
(A - 3) + y = 72
A + y = 75
We can now solve this new equation for A:
A + A + 7 = 75
2A = 68
A = 34
Therefore, A is equal to 34.