Q. 23.8( 6 Votes )

Let’s write down three consecutive integers such that 5 less than the sum of the number is equal to 11 more than twice the second number. Let’s find the three consecutive integers.

Answer :

Let the three consecutive numbers be x, (x+1), (x+2).

Now according to the given criteria,


5 less than the sum of the number = x+(x+1)+(x+2)-5…..(i)


11 more than twice the second number = 2(x+1)+11…….(ii)


So from question equation (i) and equation (ii) are equal,


Hence


x+(x+1)+(x+2)-5 = 2(x+1)+11


3x+3-5=2x+2+11


3x-2=2x+13


3x-2x=13+2


x=15


x+1=15+1=16


x+2=15+2=17


Hence the three consecutive numbers are


15, 16 and 17


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