Question

the sum of the digits of a two digit number is 13 the number formed by inter changing the digits is 45 more than the original number find the originalnumber​

Answer 1

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Q) the sum of the digits of a two digit number is 13 the number formed by inter changing the digits is 45 more than the original number find the original number

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Answer:-

• let the tens place of the two digit number be denoted by x and one place be denoted by y

• From the question,

x + y = 13 ------(1)

and 10y + x = 10x + y +45 ------(2) (interchanging digits gives a number 45 more than original number)

from (1)

x = 13 - y ------(3)

• substituting this in (2)

10y + 13 - y = 130 - 10y + y +45

18y = 162

hence, value of y = 162 / 18 = 9

y = 9

• substituting value of y in (3)

x = 13 - 9

x = 4

• The original two digit number is 49

• the number made by interchanging digits is 94 which is 45 more than 49.

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Answer 2

Let the digit at tens place be x and digit at ones place be y.

$\star \: \sf Original \: Number =10x + y$

$\star \: \sf Interchanged \: Number =10y + x$

Given that :

$\sf \:{ \: x + y = 13}.....eq.(i)$

According to the Question :

$\longrightarrow \sf \: 10y + x = 10x + y + 45$

$\longrightarrow \sf \: 10y + x - 10x - y = 45$

$\longrightarrow \sf \: 9x - 9y = 45$

$\longrightarrow \sf \: 9(x - y)+ 45 = 0$

$\longrightarrow \sf \: x - y = \dfrac{45}{9}$

$\longrightarrow \sf \: x - y = 5......eq.(ii)$

From Equation(i) and Equation(ii) :

$\longrightarrow\sf \: x + y = 13 - x - y = 5$

$\longrightarrow\sf \: x + x - y + y = 13 - 5$

$\longrightarrow \sf \: 2x = 8$

$\longrightarrow \sf \:x = \dfrac{8}{2}$

$\longrightarrow \sf \:x = \large \boxed{ \red{ \sf \: 4}}$

Putting the value of x = 4 in equation(i)

$\longrightarrow \sf \:{ \: x + y = 13}$

$\longrightarrow \sf \:{ \: 4 + y = 13}$

$\longrightarrow \sf y = 13 - 4$

$\longrightarrow \sf \:y = \large \boxed{ \red{ \sf \: 9}}$

Now,

$\star \: \sf Original \: Number =10(4) + 9 = \large\boxed{ \sf \: 49}$

$\star \: \sf Interchanged \: Number =10(9) + 4 = \large \boxed{94}$