Question

The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.

Answer 1

Given : The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3.

Solution:

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

Number = 10y + x

The number obtained by reversing the order of the digits is = 10x + y

ATQ :

Condition : 1

x - y = ±3 ………….(1)

 

Condition : 2

(10x +  y) + (10y + x) = 99

10x + y + 10y + x = 99

11x + 11y = 99

11(x + y) = 99

x + y =99/11

x + y = 9  ………….(2)

Thus , we obtain two following systems of linear equations :  

(i) x - y = 3 ………….(3)

x + y = 9………….(4)

 

 

 (ii) x –  y = -3………….(5)

x + y = 9………….(6)

 

(i) First, we solve eq. (3) & (4) by adding :

x - y = 3

x + y = 9

----------------

2x  = 12

x = 12/2

x = 6

 

On putting x = 6in eq (3)  we obtain :  

x - y = 3

6 - y = 3

-y = 3 - 6

-y = - 3

y = 3

Number = 10y + x = 10 × 3 + 6 =  30 + 6 = 36

Hence, the number is 36.

(ii) Now, we solve eq. (5) & (6) by adding :

x - y = - 3

x + y = 9

---------------

2x = 6

x = 6/2

x = 3

On putting x = 3 in eq (5)  we obtain :  

x - y = - 3

3 - y = - 3

-y = -3 - 3

-y = - 6

y = 6

Number = 10y + x = 10 × 6 + 3 =  60 + 3 = 63

Hence, the number is 36 and 63.

Hope this answer will help you…

 

Some more questions from this chapter :  

The sum of two numbers is 1000 and the difference between their squares is 256000.Find the numbers.

https://brainly.in/question/17177185

 

The sum of a two-digit number and the number formed by reversing the order of digit is 66. If the two digits differ by 2, find the number. How many such numbers are there?

https://brainly.in/question/17176636

Answer 2

$\huge{ \mathrm{ \star{ \underline{ \underline{ \blue{Question...}}}}}}$

${ \: The \: sum \: of \: a \: two \: digit }$${ \: number \: and \: the \: number }$ ${ \: obtained \: by \:reversing \: the }$${\: order \: of \: its \: digits \: is \: 99. \: If }$${ \: the \: digits \: differ \: by \: 3\: find }$${\: the \: number }$

$\huge{ \mathrm{ \star{ \underline{ \underline{ \pink{Answer...}}}}}}$

$\large{ \mathrm{ \bullet{ \: = \: 63 }}}$

$\huge{ \star{ \underline{ \underline{ \pink{ \mathrm{Solution....}}}}}}$

${ \: Let \: the \: digits \: = \: x \: , \: y}$

${ \: So \: number \: = \: 10x \: + \: y}$

$\huge { A.T.Q }$

${ (10x + y) + (10y + x) = 99}$

${11x + 11y = 99}$

${ 11(x + y) = 99}$

${x + y = \frac{99}{11} }$

${x + y = 9\: \: \: .........(1) }$

${x - y = 3(given) \: \: \: .........(2)}$

${ \: After \: solving \: (1) \: and \: (2) \: we \: get }$

$\huge { \: x \: = \: 6 }$

$\huge { \: y \: = \: 3 }$

${ \mathrm{ \underline{ \pink{ \: So \: number \: is \: 10 \: x \: + \: y \: = \: 60 \: + \: 3 \: = \: 63}}}}$

$\therefore{ \orange{ \mathrm{ \large{ \underline{ \underline{ \: required \: answer \: is \: 63 }}}}}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\huge { \: @itzBrainlyTanuj}$