Question

31. The digits of a two digit number differ by 3. If the digits are interchanged and the two numbers are
added, their sum is 77. Find the original number,
32. .
... in linear equations in one variable and according to class 8​

Answer 1

Given

The digits of a two digit number differ by 3. If the digits are interchanged and the two numbers are

added, their sum is 77.

Find out

Find the original number.

Solution

★Let the tens digit be x and ones digit be y

• Original number = 10x + y

According to the given condition

✰Two digit number differ by 3

• x - y = 3 ---(i)

✰If the digits are interchanged and the two numbers are added, their sum is 77.

• Interchanged number = 10y + x

➞ 10x + y + 10y + x = 77

➞ 11x + 11y = 77

➞ 11(x + y) = 77

➞ x + y = 7 ----(ii)

Add both the equations

➞ (x + y) + (x - y) = 7 + 3

➞ x + y + x - y = 10

➞ 2x = 10

➞ x = 10/2 = 5

Putting the value of x in equation (ii)

➞ x + y = 7

➞ 5 + y = 7

➞ y = 7 - 5 = 2

Hence,

• Original number = 10x + y = 52
• Interchanged number =10y + x= 25

$\rule{200}3$

Answer 2

Answer

The original number is 25

Given

• The digits of a two number differ by 3
• When the digits are interchanged and the two numbers are added , their sum becomes s 77

To Find

• The original number

Solution

Let us consider the digits be x and y respectively

Therefore , the number is

$\tt \longrightarrow 10x + y$

$\sf \underline{\underline{ \dag \: \: According \: \: to \: \: question}}$

$\tt \implies y-x = 3 \\\\ \tt\implies y = x+3\dashrightarrow(1)$

$\sf \underline{ \underline{ \dag \: \: Again \: by \: question}}$

$\tt \implies 10x + y + 10y + x = 77 \\\\ \tt\implies 11x +11y = 77 \\\\ \tt\implies 11(x+y)= 77 \\\\ \tt\implies x+y= 7 \\\\ \tt\implies x+x+3 = 7 \\\\ \tt\implies 2x=7-3\\\\ \tt\implies 2x=4\\\\ \tt\implies x=2$

$\sf\underline{\underline{ \dag \: \: Putting \: the \: value \: of \: \bf{x} \sf{ \: in \: (1)} }}$

$\tt\implies y = 2+3 \\\\ \tt\implies y = 5$

Thus , the original number is ,

➝10×2 + 5

➝ 20 + 5

➝ 25