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 7 of 8.

Namrata said:   1 decade ago
Zero is neither odd nor even number.

Utkarsh said:   1 decade ago
If we put x=1,then three consecutive no's x,x+2,x+4 gets an odd no.

Virendra said:   1 decade ago
Forget everything just go option wise then answer should be 9 or 15.

Shivaraj said:   1 decade ago
Explain this problm please.

Srinu said:   1 decade ago
I don't understand this problem. Can any one explain this problm please.

Vamshidhar reddy said:   1 decade ago
Three consecutive integers x, x+1, x+2

3x=2(x+2)+3
x=7
x+2=9

Swetha said:   1 decade ago
In this problem, they had given three consecutive odd integers.

If we take that 3 odd integers as x , x+2 and x+4,

if we put x = 2 => the numbers will become 2 , 4, 6

these nos are even numbers..then how we assume X,X+2,X+4 as three consecutive odd integers?

Can anyone Clear my doubt??

Santhosh said:   1 decade ago
Rahman is right.

Because the given numbers should be odd numbers.

So we have to take...the three consecutive odd numbers as 2x+1, 2x+3, 2x+5

Md. Saifur Rahman said:   1 decade ago
2x+1, 2x+3, 2x+5
3(2x+1)=2(2x+5)+3
... x=5
(2x+5)=15

FRANCO said:   1 decade ago
The first part is 3 more than the second part.

So to equate it we have to add 3 to the second part

I part = II part + 3

The ans 15 is correct


Post your comments here:

Your comments will be displayed after verification.