Question

13 In a two digit number the sum of the digits is 7. If the number with the order of the
digits reversed is 28 greater than twice the unit's digit of the original number, find the
number.​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\sf{The \ number \ is \ 43.}$

Given:

$\sf{\mapsto{In \ a \ two \ digit \ number \ the \ sum \ of \ the}}$

$\sf{digits \ is \ 7.}$

$\sf{\mapsto{If \ the \ number \ with \ the \ order \ of \ the}}$

$\sf{digits \ reversed \ is \ 28 \ greater \ than \ twice}$

$\sf{the \ units \ digit \ of \ the \ original \ number.}$

To find:

$\sf{The \ number.}$

Solution:

$\sf{Let \ the \ unit's \ place \ of \ the \ digit \ be \ x}$

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

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

$\sf{\longmapsto{x+y=7...(1)}}$

$\sf{Here,}$

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

$\sf{\leadsto{Number \ with \ reversed \ digit=10x+y}}$

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

$\sf{\longmapsto{10x+y=2x+28}}$

$\sf{\longmapsto{therefore{8x+y=28...(2)}}}$

$\sf{Subtract \ equation (1) \ from \ equation (2)}$

$\sf{we \ get,}$

$\sf{8x+y=28}$

$\sf{-}$

$\sf{x+y=7}$

_________________

$\sf{7x=21}$

$\sf{\therefore{x=\dfrac{21}{7}}}$

$\boxed{\sf{\therefore{x=3}}}$

$\sf{Substitute \ x=3 \ in \ equation \ (1) \ we \ get,}$

$\sf{3+y=7}$

$\sf{\therefore{y=7-3}}$

$\boxed{\sf{\therefore{y=4}}}$

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

$\sf{=10(4)+3}$

$\sf{=40+3}$

$\sf{\leadsto{\therefore{Original \ number=43}}}$

$\sf\purple{\tt{\therefore{The \ number \ is \ 43.}}}$