The digits of a two - digits number differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. What is the original number?
Answers
Answer:
let tens digit no as a
and ones digit as b
original no 10a + b
no by reversing digit is 10 b+ a
a- b=3
than a= b +3
according to statement
(10 a+ b) + ( 10 b+ a) = 143
{ 10( b+3) +b} +{10b + 3+ b}=143
(10 b+30+b) +(10 b +b+3) =143
(11 b+30) +(11 b +3) =143=
11 b+30+11 b+3=143
22 b+33=143
22 b=143-33
22 b=110
b=110/22
b=5
now a=b+3
=5+3
=8
hence original no is 10 a+b
=10(8) +5=85
Solve:-
According to question ❓:-
Take , for example , a 2 digit number , say, 56.
Take , for example , a 2 digit number , say, 56. It can be written as 56 = (10 × 5) + 6.
~If the digits in 56 are interchanged , We get 65 , which can be written as (10×6) + 5.
• Let us take the two digit number such that the digit in the unit place is b.
• The digit in the tens place is different from b by 3.
• Let us take it as b + 3.
• So the two digit number is 10 (b+3) + b
= 10b + 30 + b
= 11b + 30
With interchange the digits , the resulting two number will be
= 10b + (b+3) = 11b + 3
If we add these two two digit numbers , their sum is
(11b + 30) + (11b + 3) = 11b + 11b + 30 + 3
= 22b + 33
It is given that the sum is 143.
Therefore , 22b + 33 = 143
• 22b = 143 - 33
• 22b = 110
• b = 110/22
• b = 5
Now,
Unit place = b
The value of b is 5 .
So, The unit place is 5.
Unit place = 5
Tens place = b + 3
So , We have to sum the both numbers.
Value of b = 5
So ,
b + 3
= 5 + 3
= 8
hence,
the number is 85.
Answer is verified.
Answering check ✅ => On interchange of digits the number we get is 58.
The sum of 85 and 58 is 143 are given.