Question

Seven times a two digit numbers equal four times the number with digits reversed. find the number, if the sum of its digit is 3​

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

Required Answer:-

GiveN:

• Seven times a two digit numbers equal four times the number with digits reversed.
• Sum of its digit is 3.

To FinD:

• The original number?

Step-by-step Explanation:

Let the ten's place digit be x and the ones place digit be y. Then the required number is 10x + y.

Now,

As given in the question. Seven times a two digit numbers equal four times the number with digits reversed. The number obtained reversing it = 10y + x. Then,

• 7(10x + y) = 4(10y + x)

⇒ 70x + 7y = 40y + 4x

⇒ 66x = 33y

⇒ 33(2x - y) = 0

⇒ 2x - y = 0 ----------(1)

Also it's given that, Sum of the digits is 3$.$

⇒ x + y = 3 -----------(2)

Adding (1) and (2),

⇒ 2x - y + x + y = 3

⇒ 3x = 3

⇒ x = 1

Then,

⇒ y = 3 - 1

⇒ y = 2

Therefore:

The required two digit number is:

$\huge{\underline{\boxed{\sf{\purple{12}}}}}$