Question

i. A two digit number and the number with digits interchanged add up to 143. In the given number the digit in
unit's place is 3 more than the digit in the ten's place. Find the original number?(full in detail)​

Answer 1

Answer:

58......

Step-by-step explanation:

Let the number be..(xy) y=x+3

Then, 10x +y +10y + x= 143

x+y= 13

x+x+3= 13

x=5

hence, no. is 58......

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Answer 2

Let the units digit of the number be 'x'.

Let the ten's place digit be 'y'.

The number would be ⇒ 10y + x

After we interchange the digit our number would be ⇒ 10x + y

According to the question, the sum of the number and the interchanged number would be 143.

The equation becomes ⇒ 10y + x + 10x + y = 143

(i)

⇒ 10y + x + 10x + y = 143

⇒ 10y + y + 10x + x = 143

⇒ 11y + 11x = 143

⇒ x + y = 3

(ii)

x - y = 3

After adding (i) and (ii) we get ⇒ 2x = 16

⇒ 2x = 16

⇒ x = 8

Now we'll put the value of 'x' and find 'y'.

⇒ 8 + y = 13

⇒ y = 5

∴ The required number is 58.