Question

73. Three times the first of
e times the first of three consecutive odd integers
more than twice the third. The third integer is
(M.B.A., 1998)​

Answer 1

||✪✪ Correct QUESTION ✪✪||

Three times the first of three consecutive odd integers

(Here Number is Missing now, we can Take anything ) (Since You didnt Give This i m assuming it Y now,) more than twice the third. The third integer is

(M.B.A., 1998) ?

|| ✰✰ ANSWER ✰✰ ||

Let us assume That, our First odd Number is = x .

Since All Three Numbers are consecutive ,

So,

→ Second odd Number = (x + 2).

→ Third odd Number = (x + 4).

Now, Given That, Three times the first of three consecutive odd integers y more than twice the third.

So,

→ 3 * (First Odd integer) = 2 * (x+4) + y

→ 3*x = 2(x+4) + y

→ 3x = 2x + 8 + y

→ 3x - 2x = 8 + y

→ x = (8+y) .

So, The Third Number is = (x + 4) = (8+y) + 4 = (12+y). (Ans).

Now Here You can Put any value of Y and You will Get Your Answer..

Answer 2

Question should be :

If three times the first of three consecutive odd integers is three more than twice the third. The third integer is?

Solution :

Let the odd integers be x, (x+2) and (x+4)

According to the question,

3x = 2(x+4) + 3................(1)

=> 3x = 2x + 8 + 3

=> 3x = 2x + 11

=> 3x - 2x = 11

=> x = 11

_________________

First odd integer = x => 11

Second odd integer = x+2 => 11+2 => 13

Third odd integer = x+4 = 11 + 4 = 15

Therefore, third odd integer is 15.

_______________________

Verification :

Put the values of x in 1.

=> 3(11) = 2(11+4) + 3

=> 33 = 22 + 8 + 3

=> 33 = 33

Hence verified!