Question

the digit of a two digit no. differ by 3 if the digits are interchanged and resulting number is added to the original number we get 143 what can be the original number

Answer 1

Here is your Answer

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Let the unit digit be x

ten digit = X+3

so

10(X+3)+ X

digit interchange number

10x +X+3

=>
10(X+3) +X + 10x + x+ 3 = 143
=>
10x+ 30 +12x +3 = 143
22x + 33 = 143
22x = 143-33
X = 110/22
X = 5

unit digit number 5
10 digit number 5+3 = 8

so the original number 85


Hope it's helpful ✌✌

Answer 2

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 ❤️

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