Problem Solving

Sequence A consists of 10 consecutive odd numbers and B consists of 5 consecutive even numbers. If the
least number in A is 7 greater than the least number in B, the average of the numbers in A is how much
greater than average of numbers in B?

OA given is 14 but I think the answer should be 12. Thanks

the least number in A is a, and the least number in B is b; a=b+7
set A contains a+2*9 and set B contains b+4*2
the mean in set A is (a+a+18)/2 and the mean in set B is (b+b+8)/2

(b+7+b+7+18)/2 - (b+b+8)/2=24/2=12

i made mistake before

sampath wrote:Sequence A consists of 10 consecutive odd numbers and B consists of 5 consecutive even numbers. If the
least number in A is 7 greater than the least number in B, the average of the numbers in A is how much
greater than average of numbers in B?

OA given is 14 but I think the answer should be 12. Thanks

consider least number in set A = x
so set A = {x, x+2, ..., x+18}
So set B = {x-7, x-5, x-3, x-1, x+1}

Avg of set A = (first term + last term) / 2 = (2x + 18)/2 = x +9
Avg of set B = (first term + last term) / 2 = (x-7+x+1)/2 = x -3

Diff. = x+9 -(x-3) = 12

Please check OA. It should be 12.

Let's use numbers and see if this is true..

let's say x = 7
set A = {7,9,11,13,15,17,19,21,23,25}; Avg = (25+7)/2 = (15+17)/2 = 16
set B = {0,2,4,6,8} Avg = 4 (middle term)

Difference between avg = 16 - 4 = 12... hurraayyyyyyyy

algebra works...