average....?

Question

the average of 5 consecutive integers is 10. the sum of the largest and the smallest of these five integers is

Answer 1

(x + (x+1) + (x+2) + (x+3) + (x+4)) / 5 = 10
(5x + 10) / 5 = 10
5x +10 = 50
5x = 40
x = 8

Smallest = 8
Hightest = (x + 4) = 12

8 + 12 = 20 <-- ANSWER

Answer 2

If the average of 5 consecutive integers is x, it should be obvious that the largest and smallest of these integers is (x-2) and (x+2). Adding these would get you 2x. In this case 2*10 is 20.

Answer 3

Average equals mean

The numbers are

8 + 9 + 10 + 11 + 12 = 50

Divide 50 by the total number of integes. Which is 5

50 / 5 = 10

The average is 10

The sum of the largest and the smallest integers is

8 + 12 = 20

Click on the URL below for additional information concerning average and mean.

en.wikipedia.org/wiki/Mean

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Answer 4

20

Answer 5

it's easy.

x
x+1
x+2
x+3
x+5
add them n u get 5x+10

then this is the equation to slove the x

5x+10 over 5 = 10

slove it step by step n x = 8

the largest is x+4
plug in x
8+4 =12
now u no 12 is largest

smallest is x
x =8
smallest is 8

then add 8+12
u get 20!

20 is ur final answer!

Answer 6

let nos. be X, X+1, X+2, X+3, X+4
Average = (5X + 10)/5 = 10
5X + 10 = 50
X = 8
Smallest no. is 8 and largest no. is 12
Sum = 20

Answer 7

if the average is 10.. the middle number (the mode) must be 10 (if and only if they are consecutive).. so..
8, 9, 10, 11, 12.

Answer is 20.

Answer 8

20
The integers are 8, 9, 10, 11 and 12.

Answer 9

(x+x+1+x+2+x+3+x+4)/5=10
(5x+10)/5=10
x+2=10
x=8
Integers are:8,9,10,11,12
8+12=20
Solution is 20

Answer 10

[x+(x+1)+(x+2)+(x+3)+(x+4)]/5=10
[x+x+1+x+2+x+3+x+4]=10 *(times/multiply) 5
[5x + 10]=10*5
-10 = -10
5x=10*5-10
x=(10*5-10)/5
x=

x + (x+4) = ?
Plug the value you found for x into this equation