Question

Sum of the digits of a two digit number is 7.When we interchange the digits,it is found that the resulting new number is greater than the original number by 45.​

Answer 1

Answer:

16

Step-by-step explanation:

Let the ten's place be x and unit's place be y.

According to the question,

x + y = 7        ....(1)

Original number: x(10) + y(1) = 10x + y

Reversed number: y(10) + x(1) = 10y + x

According to the second part of the question,

Reversed number = Original number + 45

10y + x = 10x + y + 45

9y - 9x = 45

9[y - x] = 45

y - x = 5        ....(2)

Now if we solve the two equations (1) and (2), we would get,

y = 6 and x = 1

Therefore, the original number is = 1(10) + 6(1) = 10 + 6 = 16[answer]

Hope this helps you!!!

Answer 2

Answer :

The original number is 16

Given :

• The sum of the digits of a two digit number is 7
• When we interchange the digits , it is found that resulting new number is greater than the original number by 45

To Find :

• The original number

Solution :

Let us consider the digits as x and y respectively

Therefore , the number becomes

$\sf \dashrightarrow 10x + y$

According to question :

$\sf \implies x + y = 7.........(1)$

Again by question :

$\sf \implies 10y + x = 10x + y + 45 \\\\ \sf \implies 10y - y + x - 10x = 45 \\\\ \sf \implies 9y - 9x = 45 \\\\ \sf \implies 9(y-x) = 45 \\\\ \sf \implies y - x = 5 ........(2)$

Adding (1) and (2) we have ,

$\sf \implies x + y + y - x = 7 + 5 \\\\ \sf \implies 2y = 12 \\\\ \sf \implies y = 6$

Using the value of y in (1) we have ,

$\sf \implies x + 6 = 7 \\\\ \sf \implies x = 7-6 \\\\ \sf \implies x = 1$

Thus , the required number is

$\sf \dashrightarrow 10\times 1 + 6 \\\\ \sf \dashrightarrow 16$