Question

The sum of the digits of a two-digit number is 7. If the digits are reversed, the new number increased by
3 equals 4 times the original number. Find the original number.​

Answer 1

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

$\sf \therefore \: The \: number \: = \: 10 \: x + y$

$\sf \: Sum \: of \: its \: digits = x + y$

$\sf \: On \: reversing \: the \: digits,$

$\sf \: The \: number \: becomes \: 10 \: y + x.$

$\sf \: A/Q, \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x + y = 7 \\ \sf \: and, \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 10y + x + 3 = 4(10x + y) \\ \\ \sf \red{x + y = 7 }\: \: \: \: \: \\ \sf \: 10y + x + 3 = 4(10x + y) \\ \sf \implies \: 10y + x + 3 = 40x + 4y \\ \sf \implies \: 10y - 4y + x - 40x = 3 \\ \sf \implies \red{6y - 39x = 3} \: \: \: \: \blue{} \\ \\ \\ \sf \: x + y = 7 \: \: \red{\times 6} \\ \sf \: 6x + 6y = 42 \: \: \: \: \blue{eq.(i)}\\ \\ \sf \: 6y - 39x = 3 \: \: \red{ \times 1} \\ \sf \: 6y - 39x = 3 \: \: \blue{...eq.(ii)} \\ \\ \sf \: on \: solving \: we \: get \: - \\ \sf \: x = 1 \: \: \: \: \: \: \: \: y = 6 \\ \\ \sf \therefore \: Required \: number = 10x + y \\ \sf \: = \: 10 \times 1 + 6 \\ \boxed{ \red{ \underline{ \sf \: = 16 \: \: ....(ans) }}}$

Answer 2

Answer:

Let’s take the digit at tens place = x

And let the digit at unit place = y

So, the number = 10 × x + 1 × y = 10x + y

Reversing the number = 10 × y + 1 × x = 10y + x

Now, according to the conditions given in the problem, we have

step by step explanation

x + y = 7… (i)

And,

10y + x = 4(10x + y) – 3

10y + x = 40x + 4y – 3

40x – x + 4y – 10y = 3

39x – 6y = 3

13x – 2y = 1 … (ii)

Performing 2 x (i) + (ii) to solve, we have

2x +2 y = 14

13x – 2y = 1

15x = 15

x = 15/15

x = 1

On substituting the value of x in equation (i), we have

1 + y = 7

y = 7 – 1

y = 6

Therefore, the number is 10x + y = 10(1) + 6 = 16

.