Question

Three consecutive even numbers have a sum where one half of that sum is between 90 and 105.

Answer 1

$n;\ n+2;\ n+4-three\ consecutive\ even\ numbers\\\\90 \ \textless \ \dfrac{1}{2}(n+n+2+n+4) \ \textless \ 105\\\\90 \ \textless \ \dfrac{1}{2}(3n+6) \ \textless \ 105\ \ \ |multiply\ both\ sides\ by\ 2\\\\180 \ \textless \ 3n+6 \ \textless \ 210\ \ \ |subtract\ 6\ from\ both\ sides\\\\174 \ \textless \ 3n \ \textless \ 204\ \ \ |divide\ both\ sides\ by\ 3\\\\58 \ \textless \ n \ \textless \ 68\\\\Answer:\\60;\ 62;\ 64\ or\ 62;\ 64;\ 66;\ or\ 64;\ 66;\ 68\ or\ 66;\ 68;\ 70.$

Answer 2

Answer: a) $90$

b)n can be 60,62,64; 62,64,66; 64,66,68; 66,68,70.

Step-by-step explanation:

Let the three consecutive even numbers be n, n+2, n+4.

Sum is given by

$n+n+2+n+4\\\\=3n+6$

So, half of that sum is given by

$\frac{1}{2}\times (3n+6)$

And it is between 90 and 105.

a)Write an inequality to find the three numbers. Let n represent the smallest even number.

so, it becomes,

$90$

b)Solve the inequality:

$90$

Hence, n is somewhere between 58 and 68.

Therefore, n can be 60,62,64; 62,64,66; 64,66,68; 66,68,70.