the sum of two digits and the number formed by interchanging it's digits in 110.if 10 is subtracted from the first number the new number is 4more than five times the sum of the digits in first number. find the number
Answers
Answer:
Step-by-step explanation:
Solution :-
Let the unit's place digit be x.
And the tens place digit be y.
Number = 10y + x.
Interchanged Number = 10x + y
According to the Question,
1st Part,
⇒ (10y + x) + (10x + y) = 110
⇒ 11x + 11y = 110
⇒ x + y = 10
⇒ y = 10 - x ..... (i)
2nd Part,
⇒ 10y + x - 10 = 5(x + y) + 4
⇒ 4x - 5y = - 14
⇒ 4x - 5(10 - x) = - 14 (From (i))
⇒ 5x - 50 - 5x = - 14
⇒ 9x = - 14 + 50
⇒ 9x = 36
⇒ x = 36/9
⇒ x = 4
Putting x's value in Eq (i), we get
⇒ y = 10 - x
⇒ y = 10 - 4
⇒ y = 6
Number = 10y + x = 10(6) + 4 = 60 + 4 = 64
Hence, the required number is 64.
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☞ Correct Question :-
The sum of two digit number and the number formed by interchanging it's digits is 110. If 10 is subtracted from the first number ,the new number is 4 more than five times the sum of the digits in first number.
☞ To Find :-
- Orginal Number
- Reversed or Interchanged number
☞ Solution
Let tens place digit be 'x'
And,
Ones place digit be 'y'
Now ,
➦ Original number = (10x + y)
➦ Reversed number = (10y + x)
Then,
- ✞ According to 1st condition :-
↦ ( 10x + y) + ( 10y + x) = 110
↦ 11x + 11y = 110
↦ 11 ( x + y) = 110
↦ x + y = 110/11 = 10
↦ x + y = 10
↦ x = 10 - y. -------(1)
✞ According to 2nd Condition :-
↦ (10x + y) - 10 = 4 + 5(x + y)
↦ 10x + y - 10 = 4 + 5x + 5y
↦ 10x - 5x + y - 5y = 4 + 10
↦ 5x - 4y = 14. ---(2)
On putting the value of x from (1) in (2) , we get,
↦ 5(10 - y) - 4y = 14
↦ 50 - 5y - 4y = 14
↦ -9y = 14 - 50
↦ -9y = -36
↦ y = -36/-9 = 4
Now,
Put this value of y in (1) , We get,
↦ x = 10 - y
↦ x = 10 - 4 = 6
Hence,
- ➾ Original number = 10x + y = 10×6 + 4 = 60 + 4 = 64
- ➾ Reversed Number = 10y + x = 10×4 + 6 = 46
So,
- Original number = 64
- Reversed number = 46