# 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:

Answer: Option

Explanation:

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

Then, 3*x* = 2(*x* + 4) + 3 *x* = 11.

Third integer = *x* + 4 = 15.

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

Discussion:

75 comments Page 1 of 8.
Aryan said:
3 years ago

Here

Three odd consecutive number.

3, 5, 7 = 3 + 5 + 7 then ans 15.

Three odd consecutive number.

3, 5, 7 = 3 + 5 + 7 then ans 15.

(25)

Lavanya said:
4 years ago

Let three consecutive odd integers are: x, x+2,x+4.

Three times the first of three integers, so--> 3x; is 3 more than twice the third so--> 3+ 2(x+4).

Then, we get an equation like 3x=3+2(x+4).

=3+2x+8.

3x-2x=11

X=11.

So, the third integer is x+4=15.

Three times the first of three integers, so--> 3x; is 3 more than twice the third so--> 3+ 2(x+4).

Then, we get an equation like 3x=3+2(x+4).

=3+2x+8.

3x-2x=11

X=11.

So, the third integer is x+4=15.

(7)

Mahendrakumar said:
3 months ago

The first three odd numbers,

X, X+2, X+4.

in the question they've mentioned 3 times the 1st odd number(which is X)

So, 3X.

Then they've mentioned 3 more than twice of the third number 3+2(X+4).

then;

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

3x = 3+ 2x + 8.

3x - 2x = 3 + 8.

x = 11.

So, the ans is 11 +4 => 15.

X, X+2, X+4.

in the question they've mentioned 3 times the 1st odd number(which is X)

So, 3X.

Then they've mentioned 3 more than twice of the third number 3+2(X+4).

then;

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

3x = 3+ 2x + 8.

3x - 2x = 3 + 8.

x = 11.

So, the ans is 11 +4 => 15.

(5)

Tashi Tshering said:
3 years ago

Here,

1,3,5 three odd consecutive no;

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

3x when the third inter is x=5.

3*5=15.

1,3,5 three odd consecutive no;

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

3x when the third inter is x=5.

3*5=15.

(5)

Megersa Tadale said:
2 years ago

Suppose we take x,x+2,x+4 for odd consecutive numbers;

Then, according to the question, the first among the consecutive number is 3 more than twice the third number.

It means 3x+3=2(x+4) then;

we will get 3x+3=2x+8(now collecting like terms);

then 3x-2x=8-3.

x=5

So the third number is x+4----->>>> 9.

Then, according to the question, the first among the consecutive number is 3 more than twice the third number.

It means 3x+3=2(x+4) then;

we will get 3x+3=2x+8(now collecting like terms);

then 3x-2x=8-3.

x=5

So the third number is x+4----->>>> 9.

(4)

Mahin said:
5 years ago

The Consecutive odd integers are (2x+1, 2x+3, 2x+5). Sub any no.for 'x', you'll get an odd integer.

Now, try solving the question with these three, you'll get X = 5;

That means the 3rd integer, 2X + 5 = 15.

Now, try solving the question with these three, you'll get X = 5;

That means the 3rd integer, 2X + 5 = 15.

(3)

Shanu kumar said:
5 years ago

Odd numbers are never be in form of x, x+2 etc.

Odd no Is in the form of 2x+1.

Odd no Is in the form of 2x+1.

(3)

Pinky said:
4 years ago

My answer is 15.

Let 2n+1 be an odd number for any value of n from 0 up to infinity.

Take a look: If n=0, 2n+1 is 1. If n=1, 2n+1 is 3. See, the resulting number is always odd whether n is odd or even.

Let (2n+1), (2n+3), (2n+5) be the three consecutive numbers.

Just to see if these three are really consecutive odd, let us try to substitute n=0. If n=0, the resulting three consecutive odd are 1,2,3. If n=1, the resulting three consecutive odd are 3,5,7. This is just to show you that these three will really result to three consecutive odd for any value of n from 0 up to infinity.

Since we do not know exactly what are these three consecutive odd, so let us take these (2n+1), (2n+3), (2n+5) as the three consecutive odd numbers.

Solution:

3(2n+1) = 2(2n+5)+3

6n+3 = 4n+10+3

6n-4n = 10+3-3

2n = 10

n = 5.

Since the third odd number is represented by 2n+5, just substitute 5 to n. So the third odd number is 15.

We can also say that the three consecutive odds are 11, 13, 15. The third is 15.

Let 2n+1 be an odd number for any value of n from 0 up to infinity.

Take a look: If n=0, 2n+1 is 1. If n=1, 2n+1 is 3. See, the resulting number is always odd whether n is odd or even.

Let (2n+1), (2n+3), (2n+5) be the three consecutive numbers.

Just to see if these three are really consecutive odd, let us try to substitute n=0. If n=0, the resulting three consecutive odd are 1,2,3. If n=1, the resulting three consecutive odd are 3,5,7. This is just to show you that these three will really result to three consecutive odd for any value of n from 0 up to infinity.

Since we do not know exactly what are these three consecutive odd, so let us take these (2n+1), (2n+3), (2n+5) as the three consecutive odd numbers.

Solution:

3(2n+1) = 2(2n+5)+3

6n+3 = 4n+10+3

6n-4n = 10+3-3

2n = 10

n = 5.

Since the third odd number is represented by 2n+5, just substitute 5 to n. So the third odd number is 15.

We can also say that the three consecutive odds are 11, 13, 15. The third is 15.

(2)

Ramya said:
6 years ago

Anyone can explain this problem? Please anyone help me to solve this problem.

(2)

Pratap said:
5 years ago

X+1, X+3, X+5.

Atq:

3(X+1) = 3+2(X+5),

X = 10.

Atq:

3(X+1) = 3+2(X+5),

X = 10.

(2)

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