Question

sum of the digits of a two digit. no. is 7 .when we interchange the digits it is found that a resulting new no. is greater than the original no .by 45 .find the new no.????​

Answer 1

Step-by-step explanation:

y+x

10x+y

10y+x

x+y=7.....(given).....eq(1)

10y+x=10x+y+45.......eq(2)

7y+7x=45

10y+x=10x+y+45

9y-9x=45

y-x=45

solving eq1 and eq2

x=1,y=6

orignal no =16

Number =61

Hope it's help you

Answer 2

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

Sum of the digits of a two digit number is 7.When we interchange the digits it is found that a resulting new number is greater than original number by 45.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The new number.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Let the tens place digit be r

Let the ones place digit be m

$\boxed{\bf{The\:original\:number=10r+m}}}}$

$\boxed{\bf{The\:reversed\:number=10m+r}}}}$

A/q

$\longrightarrow\rm{r+m=7}\\\\\longrightarrow\rm{m=7-r....................(1)}$

&

$\longrightarrow\rm{10m+r=10r+m+45}\\\\\longrightarrow\rm{10m-m+r-10r=45}\\\\\longrightarrow\rm{9m-9r=45}\\\\\longrightarrow\rm{9(m-r)=45}\\\\\longrightarrow\rm{m-r=\cancel{\dfrac{45}{9} }}\\\\\longrightarrow\rm{m-r=5}\\\\\longrightarrow\rm{7-r-r=5\:\:\:[from(1)]}\\\\\longrightarrow\rm{7-2r=5}\\\\\longrightarrow\rm{-2r=5-7}\\\\\longrightarrow\rm{-2r=-2}\\\\\longrightarrow\rm{r=\cancel{-2/-2}}\\\\\longrightarrow\rm{\orange{r=1}}$

Putting the value of r in equation (1),we get;

$\longrightarrow\rm{m=7-1}\\\\\longrightarrow\rm{\orange{m=6}}$

Thus;

$\underbrace{\sf{The\:original\:number\:is\:10r+m=[10(1)+6]=10+6=\boxed{\bf{16}}}}}$