bellaboo44
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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.
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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.
How would write general expression for a number divisible by 3 (multiple of 3)? → 3 * n (where n = 1,2,3,................)
How would write general expression for consecutive numbers divisible 3? 3n -3, 3n, 3n+3
What is their mean?
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Hello, bellaboo44!
Three consecutive multiples of 3: \(\displaystyle 3m,\:3m+3,\:3m+6\)
. . Their mean is \(\displaystyle A\!:\;\;\dfrac{3m + (3m+3) + (3m+6)}{3} \:=\:A \quad\Rightarrow\quad 3m - A \:=\:-3\)
Four consecutive multiples of 4: \(\displaystyle 4n,\:4n+4,\:4n+8,\:4n+12\)
. . Their mean is \(\displaystyle A+27\!:\;\;\dfrac{4n + (4n+4) + (4n+8) + (4n+12)}{4} \:=\:A+27 \quad\Rightarrow\quad 4n - A \:=\:21\)
The smallest is \(\displaystyle 3m\); the largest is \(\displaystyle 4n+12.\)
. . Their mean is 42: .\(\displaystyle \dfrac{3m + (4n+12)}{2} \:=\:42 \quad\Rightarrow\quad 3m + 4n \:=\:72\)
We have a system of equations: .\(\displaystyle \begin{Bmatrix}3m \qquad\;\; - A &=& \text{-}3 & [1] \\ \qquad\quad 4n - A &=& 21 & [2] \\ 3m + 4n \quad &=& 72 & [3] \end{Bmatrix}\)
\(\displaystyle \begin{array}{cccccc}\text{Subtract [1] - [2]:} & 3m - 4n &=& \text{-}24 \\ \text{Add [3]: } & 3m + 4n &=& 72 \end{array}\)
We have: .\(\displaystyle 6m \:=\:48 \quad\Rightarrow\quad m \:=\:8\)
Substitute into [1]: .\(\displaystyle 3(8) - A \:=\:\text{-}3 \quad\Rightarrow\quad \boxed{A \:=\:27} \)
The mean of 3 consecutive multiple of 3 is \(\displaystyle A.\)
The mean of 4 consecutive mutliples of 4 is \(\displaystyle A+27.\)
The mean of the smallest and largest of these 7 numbers is 42.
Find \(\displaystyle A.\)
Three consecutive multiples of 3: \(\displaystyle 3m,\:3m+3,\:3m+6\)
. . Their mean is \(\displaystyle A\!:\;\;\dfrac{3m + (3m+3) + (3m+6)}{3} \:=\:A \quad\Rightarrow\quad 3m - A \:=\:-3\)
Four consecutive multiples of 4: \(\displaystyle 4n,\:4n+4,\:4n+8,\:4n+12\)
. . Their mean is \(\displaystyle A+27\!:\;\;\dfrac{4n + (4n+4) + (4n+8) + (4n+12)}{4} \:=\:A+27 \quad\Rightarrow\quad 4n - A \:=\:21\)
The smallest is \(\displaystyle 3m\); the largest is \(\displaystyle 4n+12.\)
. . Their mean is 42: .\(\displaystyle \dfrac{3m + (4n+12)}{2} \:=\:42 \quad\Rightarrow\quad 3m + 4n \:=\:72\)
We have a system of equations: .\(\displaystyle \begin{Bmatrix}3m \qquad\;\; - A &=& \text{-}3 & [1] \\ \qquad\quad 4n - A &=& 21 & [2] \\ 3m + 4n \quad &=& 72 & [3] \end{Bmatrix}\)
\(\displaystyle \begin{array}{cccccc}\text{Subtract [1] - [2]:} & 3m - 4n &=& \text{-}24 \\ \text{Add [3]: } & 3m + 4n &=& 72 \end{array}\)
We have: .\(\displaystyle 6m \:=\:48 \quad\Rightarrow\quad m \:=\:8\)
Substitute into [1]: .\(\displaystyle 3(8) - A \:=\:\text{-}3 \quad\Rightarrow\quad \boxed{A \:=\:27} \)
Do we know that 3m is smaller than 4n? Do we know that the result is unique? Perhaps two other cases should be considered:The smallest is \(\displaystyle 3m\); the largest is \(\displaystyle 4n+12.\)
. . Their mean is 42: .\(\displaystyle \dfrac{3m + (4n+12)}{2} \:=\:42 \quad\Rightarrow\quad 3m + 4n \:=\:72\)
a) smallest is 4n, largest is 3m+6, --> 3m + 4n = . .
b) smallest is 4n, largest is 4n+12, --> n = . .
I would like to see bellaboo44 work out these two cases.