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

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Bunuel
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Three times the first of three consecutive odd integers is 3 more than [#permalink] New post   05 Mar 2018, 21:25
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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
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E
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Re: Three times the first of three consecutive odd integers is 3 more than [#permalink] New post   05 Mar 2018, 23:14
Use x, y, z for the integers.
x<y<z
3x=2z+3
Plug in the answer choices as Z. X is 2 odd integers before Z.
A) Incorrect -- If Z=7, then X=3. 3(3) /= 2(7)+3 >>> 9 /= 17
B) Incorrect -- If Z=9, then X=5. 3(5) /= 2(9)+3 >>> 15/=21
C) Incorrect -- If Z=11, then X=7. 3(7) /= 2(11)+3 >>> 21/=25
D) Incorrect -- If Z=13, then X=9. 3(9) /= 2(13)+3 >>> 27/=29
E) Correct -- If Z=15, then X=11. 3(11) = 2(15)+3 >>> 33=33
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Re: Three times the first of three consecutive odd integers is 3 more than [#permalink] New post   05 Mar 2018, 23:15
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Bunuel wrote:
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



Let the three consecutive odd integers be x, x+2, and x+4

From the question stem, we get
3x = 2(x+4) + 3 -> 3x = 2x + 8 + 3 -> x = 11

Therefore, the third consective odd integer is x+4 = 15(Option E)
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Re: Three times the first of three consecutive odd integers is 3 more than [#permalink] New post   05 Mar 2018, 23:24
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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Re: Three times the first of three consecutive odd integers is 3 more than [#permalink] New post   07 Mar 2018, 00:18
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Bunuel wrote:
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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Three times the first of three consecutive odd integers is 3 more than [#permalink] New post   07 Mar 2018, 14:35
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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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Re: Three times the first of three consecutive odd integers is 3 more than [#permalink] New post   08 Mar 2018, 17:58
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Bunuel wrote:
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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Three times the first of three consecutive odd integers is 3 more than [#permalink] New post   01 Sep 2020, 02:14
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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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Three times the first of three consecutive odd integers is 3 more than [#permalink]
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