Question

Find two consecutive even integers such that two fifth of the smaller exceeds two eleventh of the larger by 4

Answer 1

Let on the numbers be x and x+2
2x/5 - 2(x+2)/11=4
(22x-10x-20)/55 =4
12x-10=55*4
12x=240
So, x=20
so numbers are 20 and 22.

Answer 2

Answer: The two consecutive even numbers are 20 and 22.

Step-by-step explanation:

Let the first even number be x.

Let the consecutive even number be x+2.

Two fifth of the smaller is $\dfrac{2}{5}x$

Two eleventh of the larger is $\dfrac{2}{11}(x+2)$

According to question, it becomes,

$\dfrac{2}{5}x-\dfrac{2}{11}(x+2)=4\\\\\dfrac{2x}{5}-\dfrac{2x}{11}-\dfrac{4}{11}=4\\\\\dfrac{22x-10x}{55}=4+\dfrac{4}{11}\\\\\dfrac{12x}{55}=\dfrac{44+4}{11}\\\\\dfrac{12x}{55}=\dfrac{48}{11}\\\\x=\dfrac{48}{11}\times \dfrac{55}{12}\\\\x=4\times 5\\\\x=20$

Hence, the two consecutive even numbers are 20 and 22.