# math homework?

A set of 10 positive numbers is combined with another set of 12 positive numbers. If the mean of the first set is 16, what is the third smallest possible mean of the second set so that the average of the combined set is an integer?

### 3 Answers

- Ian HLv 74 weeks ago
If the mean of 10 positive numbers is 16 their total is 160.

If the mean of 12 positive numbers is m their total is 12m

The combined total of the 22 positive numbers is 12m + 160

The average of the combined 22 positive numbers, A = (12m + 160)/22

A = 7 + (12m + 6)/22 = 7 + (6m + 3)/11

Let the second term in that be another integer, i

m = (11i – 3)/6, so i = 3, gives us m = 5 as the lowest value of the mean m

But we needed the THIRD smallest possible mean of the second set.

To find that, we need to increase i by two steps of 6.

i = 3 + 12 = 15 which produces m = (165 – 3)/6 = 27

The third smallest possible mean of the second set was 27

- Login to reply the answers

- stanschimLv 74 weeks ago
The sum of the first set of 10 positive numbers must be 10*16 = 160.

The average of the combined set would then be (160+12m) / 22, where m is the mean of the second set of 12 numbers.

For the average of the combined set to be an integer, m must be 5, 16, 27, 38, 49...

The third smallest possible mean of the second set of numbers would be 27.

- Login to reply the answers

- 4 weeks ago
The first set sums to 160. Let m be the mean of the 2nd set

(160 + 12m) / (10 + 12) = k

k is an integer

(160 + 12m) / 22 = k

160 + 12m = 22k

80 + 6m = 11k

80 + 6m is a multiple of 11

77 + 3 + 6m = 11k

7 + 3 * (1 + 2m) / 11 = k

(3/11) * (1 + 2m)

m = 5 works

m = 16 works

m = 27 works

(3/11) * (1 + 2 * (5 + 11j))

(3/11) * (1 + 10 + 22j) =>

(3/11) * (11 + 22j)

(3/11) * 11 * (1 + 2j)

3 * (1 + 2j)

3 + 6j

7 + 3 + 6j = k

10 + 6j = k

j is an integer

j = 0 , 1 , 2 , 3 , 4 , 5 , ......

10 + 6 * 0 , 10 + 6 * 1 , 10 + 6 * 2 , ....

10 , 16 , 22 , 28 , 34 , 40 , 46 , 52 , 58 , 64 , 70 , ...

- Login to reply the answers