Find two consecutive even numbers such that the difference of one-half the larger and two-fifths the smaller is equal to five. . . Which equation could be used to determine the number?. 1/2(n - 2) + 2/5 n = 5. 1/2(n + 2) - 2/5 n = 5. 1/2(n + 2) + 2/5 n = 5

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SJ2006

Then, it would be: 1/2 (n+2) - 2/5 n = 5

So, OPTION B is your answer.....

Answer:  The correct option is (B) [tex]\dfrac{1}{2}(n+2)-\dfrac{2}{5}n=5.[/tex]

Step-by-step explanation:  We are given to select the correct equation to determine two consecutive even numbers such that the difference of one-half the larger and two-fifths the smaller is equal to five.

Let, 'n' and '(n + 2)' be the two consecutive even numbers.

Then, according to the given information, the equation can be written as

[tex]\dfrac{1}{2}\times (n+2)-\dfrac{2}{5}\times n=5\\\\\\\Rightarrow \dfrac{1}{2}(n+2)-\dfrac{2}{5}n=5\\\\\\\Rightarrow \dfrac{5(n+2)-4n}{10}=5\\\\\Rightarrow 5n+10-4n=50\\\\\Rightarrow n=50-10\\\\\Rightarrow n=40.[/tex]

So, n = 40  and  n + 2 = 40 + 2 = 42.

Thus, the two even numbers are 40 and 42.

And, the required equation is [tex]\dfrac{1}{2}(n+2)-\dfrac{2}{5}n=5.[/tex]

Option (B) is correct.