
What must be subtracted from each of the numbers $23,40,57$ and $108$ so that the remainder are in proportions?
Answer
485.1k+ views
Hint: here in this question , the meaning of the remainder is that the numbers we get after subtracting the given number from a particular number. So that the ratio of the first two numbers will be equal to the ratio of the last two numbers. We can easily do it by cross-multiplication method.
Complete step-by-step solution:
Let the number to be subtracted from the given numbers be x.
Now, the remaining numbers are $(23 - x),(40 - x),(57 - x)$ and $(108 - x)$.
It is given that the remaining numbers are in the proportions.
So, $(23 - x):(40 - x)::(57 - x):(108 - x)$
$\dfrac{{(23 - x)}}{{(40 - x)}} = \dfrac{{(57 - x)}}{{(108 - x)}}$
Using the cross-multiplication method,
$(23 - x)(108 - x) = (57 - x)(40 - x)$
Multiply on both sides,
$23*108 - 23x - 108x + {x^2} = 40*57 - 57x - 40x + {x^2}$
$\Rightarrow 2484 - 131x + {x^2} = 2280 - 97x + {x^2}$
Cancel ${x^2}$ from both sides
$2484 - 131x = 2280 - 97x$
Now, transposing $2280$ to the left hand side and $ - 131x$ to the right hand side.
$2484 - 2280 = 131x - 97x$
$\Rightarrow 204 = 34x$
$\Rightarrow \dfrac{{204}}{{34}} = x$
$\Rightarrow x = 6$
Therefore, the required number to be subtracted is from $23,40,57$ and $108$ is $6$.
Note: Do the calculations carefully otherwise the answer would come out to be wrong. You have to subtract the required number from the given number. Don’t do the opposite of it i.e, subtract the given number from the required number.
Complete step-by-step solution:
Let the number to be subtracted from the given numbers be x.
Now, the remaining numbers are $(23 - x),(40 - x),(57 - x)$ and $(108 - x)$.
It is given that the remaining numbers are in the proportions.
So, $(23 - x):(40 - x)::(57 - x):(108 - x)$
$\dfrac{{(23 - x)}}{{(40 - x)}} = \dfrac{{(57 - x)}}{{(108 - x)}}$
Using the cross-multiplication method,
$(23 - x)(108 - x) = (57 - x)(40 - x)$
Multiply on both sides,
$23*108 - 23x - 108x + {x^2} = 40*57 - 57x - 40x + {x^2}$
$\Rightarrow 2484 - 131x + {x^2} = 2280 - 97x + {x^2}$
Cancel ${x^2}$ from both sides
$2484 - 131x = 2280 - 97x$
Now, transposing $2280$ to the left hand side and $ - 131x$ to the right hand side.
$2484 - 2280 = 131x - 97x$
$\Rightarrow 204 = 34x$
$\Rightarrow \dfrac{{204}}{{34}} = x$
$\Rightarrow x = 6$
Therefore, the required number to be subtracted is from $23,40,57$ and $108$ is $6$.
Note: Do the calculations carefully otherwise the answer would come out to be wrong. You have to subtract the required number from the given number. Don’t do the opposite of it i.e, subtract the given number from the required number.
Recently Updated Pages
Why is there a time difference of about 5 hours between class 10 social science CBSE

In cricket, what is a "pink ball" primarily used for?

In cricket, what is the "new ball" phase?

In cricket, what is a "death over"?

What is the "Powerplay" in T20 cricket?

In cricket, what is a "super over"?

Trending doubts
Full Form of IASDMIPSIFSIRSPOLICE class 7 social science CBSE

Convert 200 Million dollars in rupees class 7 maths CBSE

One lakh eight thousand how can we write it in num class 7 maths CBSE

Differentiate between weather and climate How do they class 7 social science CBSE

What are the controls affecting the climate of Ind class 7 social science CBSE

List of coprime numbers from 1 to 100 class 7 maths CBSE


