Question

seven times a two-digits number equals four times the number with digits reversed. find the number, if the sum of its digits is 3.​

Answer 1

Let the given number be x at tenths place and y at ones place

Therefore the number will be = (10x+y)

According to the question

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

70x +7y = 40y+ 4x

66x = 33y

2x=y

x = y/2

It is given that the difference between them is 3. Since x is less then

y - x = 3

Putting the value of x

y - y/2 = 3

y/2 = 3

y = 6

Therefore x= 3

Hence the original number is 36 and the reversed number be 63.

Answer 2

AnSwer:

• Let x numbers be at ones place.
• Let y numbers be at tens place.

So, accordingly 10y+x is the digit.

Number obtained by reversing digits is 10x + y.

★ According to the question,

$\sf \implies \: 7(10y+x)=4(10x+y) \\ \\ \sf \implies \: x=2y - - - (i)$

Now,

$\sf \implies \: x - y=3(given)$

Substitute the value of x= 2y in above eqn,

$\sf \implies \: x - y = 3 \\ \\ \sf \implies \:2y - y = 3 \\ \\ \sf \implies \:y = 3$

Now, let us find the value of x.

$\sf \implies \:x = 2y \\ \\ \sf \implies \:x = 2(3) \\ \\ \sf \implies \:x = 6$

★ Original Number :-

$\sf \implies \: 10y+x \\ \\ \sf \implies \:10 \times 3+6 \\ \\ \sf \implies \: 36.$

So, the original number is 36.