- Home
- Standard 13
- Quantitative Aptitude
4.Average
medium
There are $6$ consecutive odd numbers in increasing order. The difference between the average of the squares of the first $4$ numbers and the last four numbers is $64 .$ If the sum of the squares of the first and the last element (i.e., odd numbers) is $178 ,$ then the average of all the six numbers is
A
$7$
B
$8$
C
$9$
D
$10$
Solution
let the number be $(a-5),(a-3),(a-1),(a+1)(a+3)(a+5)$
Then the average of all six consecutive odd no.
$=\frac{(a-5)+(a-3)+(a-1)+(a+1)+(a+3)+(a+5)}{6}=a$
The value of "a' can be found by using the last statement $(a-5)^{2}+(a+5)^{2}=178 \Rightarrow a^{2}=64 a=8$
Average of all six number is $8.$
Standard 13
Quantitative Aptitude
