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

Answer:

Let the ones digit be x.

then the tens digit=7-x

original no.=10(7-x)+x

=70-9x

no. formed after reversing the digits=10x+7-x

=9x+7

according to the question,

9x+7+3=4(70-9x)

9x+10=280-36x

9x+36x=280-10

45x=270

x=270/45

x=6

original no.=10(7-6)+6

=16

Answer 2

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

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

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

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

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

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