nine numbers are written in ascending order .the middle number is also the average of the nine numbers the?

Question

average of the 5 largest numbers is 68 and the average of the 5 smallest numbers is 44 the sum of all the numbers is
a)540
b)450
c)504
d)501

Answer 1

The average of all 9 will be the number halfway between the two other averages. (68 + 44) / 2 = 56.

To figure the total of all the numbers, multiply 9 times the average:
9 x 56 = 504

P.S. One sequence that exhibits this property is:
32, 38, 44, 50, 56, 62, 68, 74, 80

The answer is c) 504

Answer 2

let x be the average of all the numbers

total number of 5 heaviest = 4 heaviest +x =5*68 =340

total number of 5 lightest = 4 lightest +x =5*44 =220

as x is the average,10x = 340+220 = 560>>>>>>x=56

but, there are only 9 numbers

therefore, sum of all the numbers = 9*56 =504

Answer 3

Let the 9 numbers be x1,x2,....,x9 (in ascending order)

Let A=x1+x2+x3+x4
Let B=x5
Let C=x6+x7+x8+x9

Then (A+B)/5=44,
(B+C)/5=68, and
(A+B+C)/9=B.

Rearranging:
A-8B+C=0.
A+B=220,
B+C=340,

Gaussian elimination:
A-8B+C=0
9B-C=220
10C=2840

Therefore C=284, B=56, and A=164.

We conclude that the sum of the 9 numbers is A+B+C=504, choice (c).

Answer 4

Let S the unknown sum of the nine integers a1, a2, . . . a9. Then:

a5 = S/9. Also we have:

a1 + a2 + a3 + a4 + a5 = 5*44 = 220 (1)
a5 + a6 + a7 + a8 + a9 = 5*68 = 340 (2)

Add (1) and (2):

a1 + a2 + a3 + a4 + 2*a5 + a6 + a7 + a* + a9 = 560

S + S/9 = 560, 10S/9 = 560, S = 56*9, S = 504.

So the right answer is (c)

Answer 5

with out even seeing the question i would say C 504 -- but -- just so you know how to do it -- 68-44 / 2 = 12 -- 44 + 12 = 56(your middle number) 56 is your average so 56*9=504

Good Luck

Answer 6

68 * 5 = 340
and 44 * 5 = 220
so total of the no. is 340 + 220 =560
now as one number is repeated and average also 560 - 56 = 504
In this case it has to be less than 504 so answer is 450 i.e (b)

Answer 7

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Answer 8

let the 9 numbers be x1,x2,...,x5,...x8, x9 arranged in the ascending order

then we have, the average of (x5+..+x9) = 68 (5 largest)
so the sum of x5+...+x9 = 68*5 = 340

similarly, we have, the average of (x1+..+x5) = 44 (5 smallest)
so the sum of x1+...+x5 = 44*5 = 220

adding both, we get

x1+...+x5+x5+..+x9 = 560
or, (x1+...+x9) + x5 = 560 ......(A)
but the middle number, x5, is the average of x1....x9
so (x1+...+x9)/9 = x5 or x1+...+x9 = 9x5

so, 9x5+x5 = 560 (from A)
or, 10x5 = 560
or x5 = 56

so the sum of all the numbers = 9x5 = 9*56 = 504

note: a recent junior olympiad problem

Answer 9

c) 504