1)FING THREE CONSECUTIVE ODD NUMBERS WHOSE SUM IS:45.
2)ONE NUMBER IS THREE TIMES ANOTHER NUMBER.IF 15 IS ADDED TO BOTH THE NUMBERS,THEN ONE OF THE NUMBER BECOMES TWICE THAT OF THE NEW NUMBER.FIND THE NUMBERS.
3)SUM OF THE DIGITS OF A TWO DIGIT NUMBER IS 9,THE NUMBER OBTAINED BY INTER CHANGING THE DIGITS EXCEEDS THE GIVEN NUMBER BY 27.FIND THE NUMBER..


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2014-09-02T19:16:11+05:30
1) a + (a + 2) + (a + 4) = 45 
3a + 6 = 45 
3a = 39 
a = 13 
Ans: 13, 15, 17
2) Let one number be x.
Then the other number be = 3x
If 15 is added to both numbers, then one of the new numbers becomes twice that of the other new number.
When 15 is added to the numbers, the new numbers are x + 15 and 3x + 15
⇒ 2(x + 15) = 3x + 15
⇒ 2x + 30 = 3x + 15
⇒ 2x - 3x = 15 - 30 (Transposing 3x to the LHS and 30 to the RHS)
⇒ - x = - 15
⇒ x = 15
Therefore one of the number = 15
⇒ The other number = 3x
= 3 x 15
= 45
The two numbers are 15 and 45.
3) First Number at unit place is is x and tens place is y.
So original number is 10y + x 
After interchanging digits number will be 10x + y
Number obtained by interchanging the digits exceed the given number by 27.
10y + x + 27 = 10x + y 
10y - y + x - 10x + 27 = 0 
9y - 9x + 27 = 0
9     (3y - 3x + 9 = 0)
3y - 3x + 9 = 0     ---------(1) 
The sum of digit of a 2 digit number is 9 
x + y = 9             ---------(2) 
By elimination method:
Multiplying eq 1 by 1 3y - 3x = 9
Multiplying eq 2 by 3 3y + 3x = 27 
- 6x = -18 
x= - 18/ -6
x = 3
Substituting value in eq 2
x + y = 9
3 + y = 9
y = 6 
The number is 36.
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2014-09-02T19:26:37+05:30

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1 ) N + (N+2) + (N+4) = 45        3N+ 6 = 45    N = 39/3 = 13

2) M = 3 N         
   M+15 = 2 (N+15)
   M = 2 N + 15
   3 N = 2 N + 15                  N = 15      M = 45

3) AB          A+B = 9  -- eq 1

   BA - AB = 27         
  10*B+ A  - (10*A+B) = 27
   9 B - 9 A = 27
   B- A = 3        --- eq 2
   Solving eq 1 and eq 2                 B = 6  A = 3

  given number  36

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