The digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. What can be the original number?
Answers
Answer:
Let the number be xy=10x+y
x-y=3 (equation 1)
Now xy becomes yx=10y+x
xy+yx=143
10x+y+10y+x=143
11x+11y=143
11(x+y)=143
x+y=13 (equation 2)
x-y=3
By substitution method
x=3+y
substitute it in equation 1
3+y+y=13
3+2y=13
2y=10
y=5
x=3+y
x=3+5
x=8
Number is 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.
Hope it helps you ❤️