The digits of a two digit number differ by 3. If the digits are interchanged
and the two numbers are added, their sum is 77. Answer the following
questions:
(i) If the digit at one's place is r then, the digit at ten's place will be
(ii) The original number formed is
(iii) As per the question, the equation formed is
(iv) The two digit number will be
(a) (x+3); (11x + 30); (22 x + 3 = 71). 52
(b) (x-3); (11x + 30): (11x +30 285
(c) (x + 3): (11x + 30): (22 x - 33-269
(d) (3 + x): (11x + 30). (22 + 33 77 74
answer plzzzzz
Answers
Answer:
9568865365528565555566
Answer:
Let the digit at units place be x
Let the digit at tens place be x + 3
10(x + 3) + x
When digits interchanged, the number becomes 10x + x + 3
According to the question,
10(x + 3) + x + 10x + x + 3 = 77
10x + 30 + 12x + 3 = 77
22x + 33 = 77
22x = 77 - 33
22x = 44
x =
x = 2
∴ The number = 10(x + 3) + x = 10 * 5 + 3 + 5 = 50 + 3 + 5 = 53 + 5 = 58
∴ The number is 58
Step-by-step explanation:
Let the digit in units place be x.
Then, digit in ten's place =3x
So, original number =10(3x)+x=31x
On interchanging the digits, new number =10x+3x=13x
According to the given condition,
31x+13x=88
⇒44x=88
⇒x=2
So, the original number is 31x or 31×2=62.
On the other hand, if we consider the digit in ten's place as x, then the digit in unit's place will be 3x.
So, the resulting number we get is 26.
Here, both answers are correct, as 26+62=88.