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Solved the problem by substituting the answer choices and back solving.

7 - if 7 is third integer, first needs to be 3, which does not match question stem - 3 more than twice the third. Same applies to all 4 options.

Lets see option E) 15 - first odd integer is 11 and matches our condition ' 3 more than twice the third'

Ans: E
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Bunuel
Three times the first of three consecutive odd integers is 3 more than twice the third. Find the third integer.

(A) 7

(B) 9

(C) 11

(D) 13

(E) 15
IMO E
3 consecutive odd integer
if 1st is A
then second is A+2
and third is A+4
now according to question 3A=2(A+4)+3
3A=2A+8+3
A=11 then 3rd number is A+4=11+4=15.
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Taking the three odd integers as 2n+1, 2n+3 and 2n+5 (where n>=0)

3(2n+1) = 3 + 2(2n+5)
On solving we get n=5
Putting n=5 in 2n+5 (third number), we get 15 (option E)
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Bunuel
Three times the first of three consecutive odd integers is 3 more than twice the third. Find the third integer.

(A) 7

(B) 9

(C) 11

(D) 13

(E) 15

We can let the first integer = x and the third integer = x + 4; thus:

3x = 3 + 2(x + 4)

3x = 3 + 2x + 8

x = 11

Thus, the third integer is 11 + 4 = 15.

Answer: E
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Lets consider the three consecutive odd integer to be (x-2), x,(x+2).

Three times the first of three consecutive odd integers is 3 more than twice the third :

=> 3(x-2) = 2 (x+2) + 3

=> 3x-6 = 2x+4 + 3

=> x = 13

Third integer : x+ 2 = 13 + 2 = 15

Answer E
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