Math, asked by nishaupadhyay259, 5 months ago

What should be subtracted from each of the numbers
58, 76, 73 and 96, so that the remainders are in continued
proportion ?
(a) 9
(b) 7
(c) 4
(d) None of these

Answers

Answered by user0888
37

Answer

 Two ratios are equal then the two are in the continued proportion.

 We subtract x from the four numbers.

\sf{\implies (58-x):(76-x)=(73-x):(96-x)}

 Then the multiplication of the means and extremes are equal.

\sf{\implies (58-x)(96-x)=(76-x)(73-x)}

\sf{\implies x^2-154x+4408=x^2-149x+5548}

\sf{\implies 5x=1140}

\sf{\implies x=228}

 We don't define ratios in negative numbers. This means the solution is rejected. Any number follows the continued proportion.

Learn more

 We don't define ratios if numbers are not positive. Instead, a fraction is used.

 A ratio can be converted into a fraction.

 The means and extremes must equal. \sf{a:b=c:d} can be converted into fractions.

\sf{\implies \dfrac{a}{b} =\dfrac{c}{d} }

\sf{\implies bd\times \dfrac{a}{b} =bd\times \dfrac{c}{d} }

\sf{\implies ad=bc }


user0888: eeeeeeee
Anonymous: fantastic :)
Anonymous: Fabulous
user0888: thanks
IIDarvinceII: Awesome biro ;)
user0888: :D
Answered by Anonymous
7

\: \: \boxed{\boxed{\bf{\mapsto \: \: \: Given \:question}}}

What should be subtracted from each of the numbers

58, 76, 73 and 96, so that the remainders are in continued proportion ?

(a) 9

(b) 7

(c) 4

(d) None of these

_____________________________________________

\: \: \boxed{\boxed{\bf{\mapsto \: \: \: firstly \: let's \: understand \: the \:question}}}

In this question it is asking that what we need to subtract from the number 58 , 76 , 73 and 96 so thatget the remainder as continued proportion

_____________________________________________

\: \: \boxed{\boxed{\bf{\mapsto \: \: \: Solution}}}

Let the unknown value be x

Now according to the question 58 - x , 76 - x , 73 - x and 96 - x

\begin{gathered}\begin{gathered}\\\;\bf{\rightarrow\;\;{ { \red{ 58 - x : 76 - x : :73 - x:  96 - x }} }}\end{gathered}\end{gathered}

Now we know that product of mean = Product of extreme

  • Mean numbers - 76 - x , 73 - x
  • Extreme numbers - 58 - x , 96 - x

\begin{gathered}\begin{gathered}\\\;\bf{\rightarrow\;\;{ { {( 58 - x  )\times (96 - x) =( 76 - x)  \times (73 - x) }} }}\end{gathered}\end{gathered}

so by this we can find the value of x

\begin{gathered}\begin{gathered}\\\;\bf{\rightarrow\;\;{ { {( {x}^{2} - 96 x - 58x + 4408 )= ( {x}^{2}  - 76x - 73x +  5548)  }} }}\end{gathered}\end{gathered}

\begin{gathered}\begin{gathered}\\\;\bf{\rightarrow\;\;{ { {( {x}^{2} - 154x + 4408 )= ( {x}^{2}    - 149x + 5548)  }} }}\end{gathered}\end{gathered}

\begin{gathered}\begin{gathered}\\\;\bf{\rightarrow\;\;{ { {  -149x+ 154x =  5548 - 4408 }} }}\end{gathered}\end{gathered}

 \begin{gathered}\begin{gathered}\\\;\bf{\rightarrow\;\;{ { {  5x =  1140 }} }}\end{gathered}\end{gathered}

 \begin{gathered}\begin{gathered}\\\;\bf{\rightarrow\;\;{ { {  x =   \dfrac{1140}{5}  }} }}\end{gathered}\end{gathered}

 \begin{gathered}\begin{gathered}\\\;\bf{\Longrightarrow\;\; \blue{{ { {  x =   228}}} }}\end{gathered}\end{gathered}

\begin{gathered}\\\;\qquad\qquad\boxed{\boxed{\bf{\purple{Hence,\;\;(d)\; none \; of \; these}}}}\end{gathered}


Anonymous: hope it's help you :)
IIDarvinceII: superb!
Anonymous: thank you dear !
user0888: nice explanation
Anonymous: thank you ! :)
Anonymous: Perfect ❤️
Anonymous: :)
Similar questions