Question

the sum of two digit no is 15. if the no formedby reserving the digit is less than the orignal no by 27 then the orignal no is​

Answer 1

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Let the number =10x+y

Reversed number=10y+x

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A.T.Q

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x+y=15 ..........i)

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10y+x=10x+y-27

10y-y+x-10x=-27

9y-9x=-27

y-x=-3..............ii)

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From equation i)&ii),we get

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x+y=15

-x+y=-3

2y=12

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y=6

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putting the value of 'y' in equation i),we get

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x=9

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ORIGINAL NUMBER=10x+y

=96

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$\bf\bold\orange{MARK\:AS \:Brainliest}$

Answer 2

Answer:

$\sf{The \ original \ number \ is \ 96.}$

Given:

$\sf{\leadsto{The \ sum \ of \ two \ digits \ number \ is}}$

$\sf{15.}$

$\sf{\leadsto{The \ number \ formed \ by \ reversing \ the}}$

$\sf{digits \ is \ less \ than \ the \ original \ number \ by}$

$\sf{27.}$

To find:

$\sf{The \ original \ number.}$

Solution:

$\sf{Let \ the \ unit's \ place \ of \ the \ two \ digit \ number}$

$\sf{be \ x \ and \ the \ ten's \ place \ be \ y.}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{y+x=15}$

$\sf{\therefore{x+y=15...(1)}}$

$\sf{\longmapsto{Original \ number=10y+x}}$

$\sf{\longmapsto{Number \ with \ reversed \ digits=10x+y}}$

$\sf{According \ to \ the \ second \ condition. }$

$\sf{10y+x=10x+y+27}$

$\sf{\therefore{10y-y+x-10y=27}}$

$\sf{\therefore{9y-9x=27}}$

$\sf{\therefore{9(y-x)=27}}$

$\sf{\therefore{y-x=\dfrac{27}{9}}}$

$\sf{\therefore{y-x=3}}$

$\sf{\therefore{-x+y=3...(2)}}$

$\sf{Add \ equations \ (1) \ and \ (2), \ we \ get}$

$\sf{x+y=15}$

$\sf{+}$

$\sf{-x+y=3}$

_______________

$\sf{2y=18}$

$\sf{\therefore{y=\dfrac{18}{2}}}$

$\boxed{\sf{\therefore{y=9}}}$

$\sf{Substitute \ y=9 \ in \ equation \ (1), \ we \ get}$

$\sf{x+9=15}$

$\sf{\therefore{x=15-9}}$

$\boxed{\sf{\therefore{x=6}}}$

$\sf{\longmapsto{Original \ number=10y+x}}$

$\sf{\longmapsto{Original \ number=10(9)+6}}$

$\sf{\longmapsto{Original \ number=90+6}}$

$\sf{\longmapsto{Original \ number=96}}$

$\sf\purple{\tt{\therefore{The \ original \ number \ is \ 96.}}}$