The average of 5 consecutive integers starting with x is a. What is the average of 9 consecutive integers that start with x+3 ? A - x+a B - a+5 C - 2x D - 2a E - 2x+a

Let x=1. 5 consec int - (1,2,3,4,5) Avg = a =3. Now if first number is x+3 = 4 We have (4,5,6,7,8,9,10,11,12) And Avg = 8 (or) a+5 B

x, x+1, x+2, x+3, x+4 5x+10 = 5a 5x = 5a-10 x = a-2 Sum of 9 consecutive integers = 9x+3+4+5+6+7+8+9+10+11 =9x+9(3+11)/2 [sum of AP] =>9x + 63 average = (9x+63)/9 = x+7 = a-2 +7 [because x=a-2] avg = a+5 B is the correct option

silverflamein wrote:The average of 5 consecutive integers starting with x is a. What is the average of 9 consecutive integers that start with x+3 ? A - x+a B - a+5 C - 2x D - 2a E - 2x+a With evenly spaced integers, the average = the median. Let x=1. The integers are 1,2,3,4,5. a = 3. Given x+3 = 1+3 = 4: The integers are 4,5,6,7,8,9,10,11,12. Average = 8. This is our target. Now we plug x=1 and a=3 into the answers to see which yields our target of 8. Only answer choice B works: a+5 = 3+5 = 8. The correct answer is B.

silverflamein wrote:The average of 5 consecutive integers starting with x is a. What is the average of 9 consecutive integers that start with x+3 ? Average of 5 consecutive integers starting with x will be the 3rd integer = (x + 2) Hence, (x + 2) = a Average of 9 consecutive integers starting with (x + 3) will be the 5th integer = (x + 3) + 4 = (x + 7) = (a + 5) The correct answer is B

used mitch's method as shown above.. but while finding the avg did - 4+12/2 = 8 (but i wasted a lil time at first coz i did 4+13/2 then i was stuck so be careful )