Question

72 is divided into two parts such twelve times of first when added to the five time of second part makes 472 find each part.​

Answer 1

Answer:

72= 4+(72-4)

=124+5(72-4)=472

=124+360-54=472

74=472-360

74=112

4= 112/7

1st part = 4=16

2nd part = 72-16=56

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Answer 2

Answer:

$\sf{The \ two \ parts \ are \ 16 \ and \ 56 \ respectively.}$

Given:

$\sf{\leadsto{72 \ is \ divided \ into \ two \ parts}}$

$\sf{\leadsto{The \ twelve \ times \ of \ first \ when \ added}}$

$\sf{to \ the \ five \ times \ of \ second \ part}$

$\sf{makes \ 472.}$

To find:

$\sf{Each \ part.}$

Solution:

$\sf{Let \ the \ two \ parts \ be \ x \ and \ y.}$

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

$\sf{\longmapsto{x+y=72...(1)}}$

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

$\sf{\longmapsto{12x+5y=472...(2)}}$

$\sf{Multiply \ equation (1) \ by \ 5, \ we \ get}$

$\sf{\longmapsto{5x+5y=360...(3)}}$

$\sf{Subtract \ equation (3) \ from \ equation (2),}$

$\sf{we \ get}$

$\sf{12x+5y=472}$

$\sf{-}$

$\sf{5x-5x=360}$

__________________

$\sf{7x=112}$

$\sf{\therefore{x=\dfrac{112}{7}}}$

$\boxed{\sf{\therefore{x=16}}}$

$\sf{Substitute \ x=16 \ in \ equation (1), \ we \ get}$

$\sf{16+y=72}$

$\sf{\therefore{y=72-16}}$

$\boxed{\sf{\therefore{y=56}}}$

$\sf\purple{\tt{\therefore{The \ two \ parts \ are \ 16 \ and \ 56 \ respectively.}}}$