Aptitude - Problems on Numbers - Discussion

Discussion Forum : Problems on Numbers - General Questions (Q.No. 2)
2.
Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:
9
11
13
15
Answer: Option
Explanation:

Let the three integers be x, x + 2 and x + 4.

Then, 3x = 2(x + 4) + 3      x = 11.

Third integer = x + 4 = 15.

Video Explanation: https://youtu.be/_77C9YE321Y

Discussion:
75 comments Page 8 of 8.

Prashants.s.wagge said:   1 decade ago
Just solve in simp way.

1st 3 odd int are 1, 3, 5.

Add those 1+3+5 = 9.

Now take 3 odd num after 9 i.e. = 15.

Nani said:   1 decade ago
If we take three consecutive numbers are x, x+2, x+4....

If we substitute x as 1 (because 1 is the 1st odd num).

We get odd numbers.....1, 3, 5..

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

Twice the third is 2(X+4).

3x = 3+2(X+4).

x = 11.

Third integer = x + 4 = 15.

Kamal said:   1 decade ago
How can we take x, x+2, x+4 as odd positive integers. Kindly clear my doubt?

Nusohel said:   1 decade ago
Three conservatives odd integers are X, X+2, X+4

Three times the first of three consecutive odd integers is 3X.
3 more than is 3+
Twice the third is 2(X+4)

3x = 3+2(X+4).

x = 11.

Third integer = x + 4 = 15.

Santhosh said:   1 decade ago
Three consecutive odd integers are x+1, x+3, x+5

So,

3(x+1) = 3+2(x+5).

3x+3 = 3 + 2x + 10.

x = 10. So the third number is x+5.

10+5 = 15.

Answer 15.


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