Question

A two digit number and the number with digit 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.​

Answer 1

Answer:

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Let the digit in unit place be x.

let the digit in the tens place is y.

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Therefore the number = 10 y + x

The number obtained by interchanging the digit is 10 x + y

According to the first condition two digit number + the number obtained by interchanging the digits = 143

Step-by-step explanation:

$therefore \: 10y + x + 10x + y = 143$

Therefore 11 x + 11 y = 143

=>X + Y = 13 ______(i)

From first condition ,

digit in unit place = digit in the tens place +3

Therefore X =y + 3

There fore X -y= 3_____(ii)

Adding equation (i) and (ii)

we get x= 8

Substitute values of x in equation (i)

x + y = 13

8 + 4 = 13

=>Therefore Y = 5

The original number is 10 y+ x

10 × 5 +8

=> 58 is your answer

Answer 2

Answer:58

Step-by-step explanation:

Given : A two digit number and the number with digits interchanged add up 143. In the given number the digit in unit's place is 3 more than digit in the ten's place.

Find the original number.

Solution:

Let the digit on ten's place be x

Let digit on unit's place be y

Number = xy

Since we are given that  the digit in unit's place is 3 more than digit in the ten's place.

y-x=3 ---a

Since we are given that  A two digit number and the number with digits interchanged add up 143.

⇒(10x+y)+(10y+x)=143

⇒11x+11y=143 ---b

Solving a and b using substitution method

Substitute value of y from a in b

⇒11x+11(x+3)=143

⇒11x+11x+33=143

⇒22x=110

⇒x=110/22

⇒x=5

So, ten's place digit is 5

Unit's place digit = x+3=5+3=8

So, the number is 58.....

I hope it's helpful

Step-by-step explanation: