Find all pairs of consecutive even positive i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $x$ be the smaller of the two consecutive even positive integers.Then,the other integer is $x + 2$.Since both integers are larger than $5$,we

Vedclass · Accepted Answer

$(6, 8), (8, 10), (10, 12)$

Vedclass · Answer

(A) Let $x$ be the smaller of the two consecutive even positive integers.Then,the other integer is $x + 2$.Since both integers are larger than $5$,we have $x > 5$ $(1)$.Also,the sum of the two integers is less than $23$,so $x + (x + 2) $2x + 2 $2x $x From $(1)$ and $(2)$,we obtain $5 Since $x$ is an even number,$x$ can take the values $6, 8,$ and $10$.Thus,the required possible pairs are $(6, 8), (8, 10),$ and $(10, 12)$.

Find all pairs of consecutive even positive integers,both of which are larger than $5$,such that their sum is less than $23$.

Similar Questions

For a positive integer $n$,if $10^{n-2} > 91n$,then the complete set of values of $n$ is:

When $R$ is the set of all real numbers,$\{x \in R: \frac{\sqrt{12-x-x^2}}{x+10} \leq \frac{\sqrt{12-x-x^2}}{2x+9}\} = $

The number of integral values of $x$ satisfying $5x - 1 < (x + 1)^2 < 7x - 3$ is

The inequality $\frac{x-1}{3x+4} < \frac{x-3}{3x-2}$ holds for all $x$ in the interval:

The set of values of $x$ which satisfy $5x + 2 < 3x + 8$ and $\frac{x + 2}{x - 1} < 4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module