Question

the mean of 6 numbers is 11. the first 4 numbers are 6,8,10,and 12 .find the last two numbers, if one of them is 2 more than the other..​

Answer 1

Answer:

14,28

Step-by-step explanation:

given the numbers are 6,8,10,12,

and their mean is equal to 11

let the numbers be 2+x,x

so,6+8+10+12+2+x+x/6=11

38+x+x/6=11

38+2x/6=11

38+2x=66

2x=66-38=28

2x=28

x=14

2x28

Answer 2

Given :-

• The mean of 6 numbers is 11
• The first 4 numbers are 6,8,10 and 12 respectively.
• In the last 2 numbers, one is greater than the other by 2.

Aim :-

• To find the remaining 2 numbers

Average :-

The average of any number of observations is calculated by using the formula :-

$\dfrac{sum \: of \: total \: number \: of \: observations}{total \: number \: of \: observations}$

Let the 6 numbers be x,y,z,a,b and c.

According to the question,

• x = 6
• y = 8
• z = 10
• a = 12

The remaining two numbers b and c can be expressed as :-

• b
• b + 2

(As one of the 2 numbers is 2 more than the other)

Substituting the values,

$\dfrac{6 + 8 + 10 + 12 + b + b + 2}{6} = 11$

By adding all the constants and variables in the numerator,

$\dfrac{2b + 38}{6} = 11$

By transposing 6 to the RHS (Right Hand Side),

$2b + 38 = 6 \times 11$

$2b + 38 = 66$

Transposing 38 to the other side, we get :-

$2b = 66 - 38$

$2b = 28$

By transposing 2,

$b = \dfrac{28}{2}$

By reducing to the lowest terms,

$b = 14$

Hence the 2 numbers are :-

• b = 14
• c = 16 (b+2 =≥ 14 + 2)

Verification :-

By substituting the values of b and b+2 as 14 and 16 respectively, let us verify the Equation.

$\dfrac{6 + 8 + 10 + 12 + 14 + 16}{6} = 11$

Adding all the numbers in the numerator,

$\dfrac{66}{6} = 11$

By cancelling, we get

LHS = RHS

Therefore verified