Question

the sum of a two digit number and the number obtained by reversing the order of it's digit is 121 and the two digits differ by 3. Find the number

Answer 1

Let no at ones= x 

Let no. at tens=10(x+3)

                       =10x+ 30

New no.

Let no. at ones place x+3

let no. ta tens place 10x

10x +30+x + 10x + x +3= 121

22x=88

x=88/22

x=4

the original no. is 74

the new no. is 47

Hope it helps :) 

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Answer 2

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

$:\implies\sf Number = 10y + x$

⠀⠀⠀☯ The number obtained by reversing the order of the digits is 10x ,+ y.

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\qquad\qquad\boxed{\bf{\mid{\overline{\underline{\bigstar\: According\: to \: the \: Question :}}}}}\mid$

$:\implies\sf (10y + x) + (10x + y) = 121$

$\qquad \: \: \: :\implies\sf 11x + 11y = 121$

$\qquad \: \: \::\implies\sf 11(x + y) = 121$

$\qquad \: \: \: \: \: \: :\implies\sf x + y = \cancel{ \dfrac{121}{11}}$

$\sf And,\;\; x - y = \pm 3\qquad\qquad\bigg\lgroup\sf \because\; Difference\;of\;digits\;is\;3.\bigg\rgroup$

★ Thus, We have the following sets of simultaneous equations

x + y = 11⠀⠀⠀...eq [1]

x - y = 3⠀⠀⠀ ...eq [2]

or

x + y = 11⠀⠀⠀...eq [3]

x - y = 3⠀⠀ ⠀...eq [4]

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

☯ On solving eq. [1] and eq. [2], we get,

$\dashrightarrow\sf x = 7, y = 4$

☯ On solving eq. [3] and eq. [4], we get,

$\dashrightarrow\sf x = 4, y = 7$

✇ When x = 7, y = 4, we have

$\qquad\quad\sf Number = 10y + x = 10 \times 4 + 7 = \red{47}$

✇ When x = 4, y = 7, we have

$\qquad\quad\sf Number = 10y + x = 10 \times 7 + 4 = \purple{74}$

$\therefore\;{\underline{\sf{Hence,\;the\; required\;number\;is\;either\;47\;or,\;74.}}}$