Find two consecutive even integers such that the sum of the larger and twice the smaller is 62?

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Answer 1 · 2018-06-15T04:16:46

#20 and 22#

Let the two consecutive even numbers be: #2n and 2(n+1)# for some #n in NN#

Since #n in NN -> 2(n+1) > 2n#

We are told that the sum of the larger plus twice the smaller equals 62.

Hence, #2(n+1) + 2xx 2n =62#

#2n+2+4n =62#

#6n = 60 -> n=10#

Replacing #n=10# into #2n and 2(n+1)#

#-># Smaller number #=20# and larger number #=22#

Check:

#20 and 22# are consecutive even numbers.

#22+2xx20 = 62#