Question

What number should be subtracted from each of the numbers \[23,30,57\] and \[78\] so that the remainders are in proportion?

Answer 1

Hint: From the given question we have been asked to find a number which when subtracted from the given numbers makes the remainders in proportion. For solving this question first we will assume a number or variable and subtract that variable from the given numbers in the question and then we will use the concept of cross multiplication we will find that number. So, we proceed with our solution as follows.

Complete step by step answer:
Firstly, as mentioned in the above we will assume the number which we have to find as \[x\].
Now we will use the basic mathematical operation which is subtraction and we will subtract this \[x\] from the numbers given the question. So, we get,
\[\Rightarrow 23-x,30-x,57-x,78-x\]
We are given that the remainder numbers are in proportion with each other. So, we can write them as follows.
\[\Rightarrow 23-x:30-x::57-x:78-x\]
Simply,
\[\Rightarrow \dfrac{23-x}{30-x}=\dfrac{57-x}{78-x}\]
From the concept of cross multiplication we will multiply the denominator of one side with the numerator of the other side. So, we get the equation reduced as follows.
\[\Rightarrow \left( 23-x \right)\left( 78-x \right)=\left( 57-x \right)\left( 30-x \right)\]
Now we will simplify this equation using multiplications, addition and subtraction. So, we get the equation simplified as follows.
\[\Rightarrow 1794-23x-78x+{{x}^{2}}=1710-30x-57x+{{x}^{2}}\]
We send all the terms to one side of the equation.
\[\Rightarrow 1794-23x-78x+{{x}^{2}}-1710+30x+57x-{{x}^{2}}=0\]
So, we get,
\[\Rightarrow -14x+84=0\]
\[\Rightarrow 14x=84\]
\[\Rightarrow x=\dfrac{84}{14}\]
\[\Rightarrow x=6\]

Note: Students must be very careful in doing the calculations. Students should know the concept of cross multiplication and also mainly ratio and proportion. We should not do mistakes in cross multiplication like if we write
 As \[\Rightarrow \left( 23-x \right)\left( 30-x \right)=\left( 57-x \right)\left( 78-x \right)\] instead of \[\Rightarrow \left( 23-x \right)\left( 78-x \right)=\left( 57-x \right)\left( 30-x \right)\] then our solution will be wrong.