Question

6. Seven times a given two digit number is equal
to four times the number obtained by reversing the order of digits. the sum of digits of the number is 3 find the number ​

Answer 1

Answer:

Let numbers be x at onces place & y at tens place so,

10y+x is the digit.

Reversed digit =10x+y

According to the question ,

∴7(10y+x)=4(10x+y)

⇒x=2y⟶(i)

Now,

Given , x−y=3

from equation (i)2y=x substitute in above equation.

2y−y=3

y=3

So ,

x=2y=2⋅3=6

∴ Required original no: is =10y+x=10⋅3+6=36.

Answer 2

Answer:

The required two digit number is 12.

Step-by-step-explanation:

Let the digit at tens place of the two digit number be x.

And the digit at units place of the number be y.

∴ Two digit number = xy

i. e. Two digit number = 10x + y

Now,

The number obtained by reversing the digits = yx

∴ The number obtained by reversing the digits = 10y + x

From the first condition,

x + y = 3

⇒ x = 3 - y

⇒ x = - y + 3 - - - - - ( 1 )

From the second condition,

7 ( 10x + y ) = 4 ( 10y + x )

⇒ 70x + 7y = 40y + 4x

⇒ 70x - 4x = 40y - 7y

⇒ 66x = 33y

⇒ x = 33y ÷ 66

⇒ x = y / 2

⇒ - y + 3 = y / 2 - - - - [ From ( 1 ) ]

⇒ 2 ( - y + 3 ) = y

⇒ - 2y + 6 = y

⇒ 6 = y + 2y

⇒ 3y = 6

⇒ y = 6 ÷ 3

⇒ y = 2

By substituting y = 2 in equation ( 1 ), we get,

x = - y + 3 - - - - ( 1 )

⇒ x = - 2 + 3

⇒ x = 1

Now,

Two digit number = 10x + y

⇒ Two digit number = 10 * 1 + 2

⇒ Two digit number = 10 + 2

⇒ Two digit number = 12

∴ The required two digit number is 12.