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
x-y=3 =>x=3+y _1
when the digits are interchanged
10y+x+10x+y=143
11x+11y=143
11(x+y)=143
x+y=143/11
x+y=13 _2
putting the value of x from eq1 in eq 2
we get
3+y+y=13
2y=10
y=5
putting this value of y in eq 1 we recive that
x=8
substituting value of x and y in a we get
10×8+5=85
this is the original no.
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 ❤️