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
​

Answer 1

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.

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

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

$\sf{\implies ad=bc }$

Answer 2

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

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

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

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

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

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

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

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

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