Question

Average of 6 consecutive even numbers is 221. Find the value of highest even number?

Answer 1

$\sf{\huge{Heya \: mate.\: Solution \\below.}}$
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$<font color = "red"><u><b>Answer -</b></u></font color>$




♠ $\tt{Even \: numbers\: -}$

Numbers which are divisible by 2 are called even numbers. Also, even numbers occur at an interval of two.



♠ $\tt{Consecutive\: numbers \: -}$

Numbers which follow each other in an order are called Consecutive numbers. They usually have a fixed interval between them.

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♠ So, let the consecutive even number as given in the question be,

$|♠| \: x; x + 2;x + 4;x + 6;x + 8;x + 10$



♠ Then, their mean (average) will be,

$mean \: = \: \frac{sum \: of \: observations}{total \: observations}$

$= > 221 = \frac{x +( x + 2) + (x + 4 )+ (x + 6) + (x + 8) + (x + 10)}{6}$

$= > 221 = \frac{6x + 30}{6}$

$= > 221(6) = 6x + 30$

$= > 1326 = 6x + 30$

$= > 1326 - 30 = 6x$

$= > 1296 = 6x$

$= > x = \frac{1296}{6}$

$= > x = 216$



♠ So, the numbers are,

$= x; x + 2;x + 4;x + 6;x + 8;x + 10$

$= (216);(216 + 2);(216 + 4);(216 + 6);(216 + 8);(216 + 10)$

$= 216;218;220;222;224;226$



♥ •°• $\boxed{\mathcal{Highest\: of \:them\: is\: 226. }}$♥
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Thank you... (^_-)

Answer 2

Answer:

${\large{\pink{\underline{\underline{\bf{Given : -}}}}}}$

• » Average of 6 consecutive even numbers is 221.

$\begin{gathered}\end{gathered}$

${\large{\purple{\underline{\underline{\bf{To \: Find : -}}}}}}$

• » The value of highest even number.

$\begin{gathered}\end{gathered}$

${\large{\pink{\underline{\underline{\bf{Using \: Formula : -}}}}}}$

${\bigstar{\underline{\boxed{\sf{\red{Average = \dfrac{Sum \: of \: Observations }{Total \: Numbers \: of \: Observations}}}}}}}$

$\begin{gathered}\end{gathered}$

${\large{\purple{\underline{\underline{\bf{Solution : -}}}}}}$

$\green\bigstar$ Let the consecutive even numbers be,

• ↠ 1 consecutive even number = x
• ↠ 2 consecutive even number = x+2
• ↠ 3 consecutive even number = x+4
• ↠ 4 consecutive even number = x+6
• ↠ 5 consecutive even number = x+8
• ↠ 6 consecutive even number = x+10

$\begin{gathered}\end{gathered}$

$\green\bigstar$ Now, According to the question :-

${\small{\dashrightarrow{\sf{Average = \dfrac{Sum \: of \: Observations }{Total \: Numbers \: of\: Observations}}}}}$

• Substuting the values

${\small{\dashrightarrow{\sf{221= \dfrac{(x) + (x + 2) + (x + 4) + (x + 6) + (x + 8) + (x + 10)}{6}}}}}$

${\small{\dashrightarrow{\sf{221= \dfrac{(x + x + x + x + x + x) + (2 + 4 + 6 + 8+ 10)}{6}}}}}$

${\small{\dashrightarrow{\sf{221= \dfrac{6x+ 30}{6}}}}}$

${\small{\dashrightarrow{\sf{221 \times 6= {6x+ 30}}}}}$

${\small{\dashrightarrow{\sf{1326= {6x+ 30}}}}}$

${\small{\dashrightarrow{\sf{1326 - 30= {6x}}}}}$

${\small{\dashrightarrow{\sf{1296= {6x}}}}}$

${\small{\dashrightarrow{\sf{x = \dfrac{1296}{6}}}}}$

${\small{\dashrightarrow{\sf{x = \cancel{\dfrac{1296}{6}}}}}}$

${\small{\dashrightarrow{\sf{x = 216}}}}$

$\bigstar{\underline{\boxed{\bf{\red{x = 216}}}}}$

∴ The value of x is 216.

$\begin{gathered}\end{gathered}$

$\green\bigstar$ Hence, the consecutive even numbers are :-

• ➯ 1 number = 216
• ➯ 2 number = (216+2) = 218
• ➯ 3 number = (216+4) = 220
• ➯ 4 number = (216+6) = 222
• ➯ 5 number = (216+8) = 224
• ➯ 6 number = (216+10) = 226

$\bigstar{\underline{\boxed{\bf{\red{Highest \: even \: number = 226}}}}}$

∴ The highest even number is 226.

$\begin{gathered}\end{gathered}$

${\large{\pink{\underline{\underline{\bf{Learn \: More : -}}}}}}$

$\green\bigstar$ Mean :

➠ The "mean" is the "average" you're used to, where you add up all the numbers and then divide by the number of numbers. 

The mean formula is given as the average of all the observations

${:\implies{\bf{\red{Mean = \dfrac{Sum \: of \: Observation}{Total \: numbers \: of \: Observations}}}}}$

$\begin{gathered}\end{gathered}$

$\green\bigstar$ Median :

➠ The "median" is the "middle" value in the list of numbers. 

If the total number of observation given is odd, then the formula to calculate the median is:

${: \implies{\bf{\red{Median ={\bigg(\dfrac{n + 1}{2}}\bigg)^{th}}}}}$

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