
What number must be added to each term of the ratio \[9:16\] to make the ratio \[2:3\]?
Answer
552.9k+ views
Hint:To find the term needed to add to \[9:16\] to make \[2:3\], we assume that the terms given are \[x\] in both the denominator and the numerator with the ratio \[9:16\] added to which is equal to \[2:3\] . After placing the values in the ratio we find the value of \[x\] by cross multiplying the ratios.
Complete step by step solution:
Let us assume that the number needed to be added is taken as \[x\].
Now to find the term\ratio needed i.e. \[2:3\] to add the unknown term of \[x\] to the numerator and denominator of the previous ratio of \[9:16\].
The value of the ratio on the LHS of the equation down below is equal to the ratio on the RHS when \[x\] is added in both the numerator and denominator of the LHS given.
\[\Rightarrow \dfrac{9+x}{16+x}=\dfrac{2}{3}\]
Cross multiplying the value of the ratio of the LHS and RHS, we get the value of the unknown variable of \[x\] as:
\[\Rightarrow 3\left( 9+x \right)=2\left( 16+x \right)\]
\[\Rightarrow 27+3x=32+2x\]
\[\Rightarrow x=5\]
Therefore, the value needed to be added with the ratio of \[9:16\] to get \[2:3\] is \[5\].
Note: Student may go wrong if they try to add a single number instead of adding two number in both the denominator and numerator as given below:
\[\Rightarrow \dfrac{9+x}{16+x}=\dfrac{2}{3}\] correct form
\[\Rightarrow \dfrac{9}{16}+x=\dfrac{2}{3}\] Incorrect form
Complete step by step solution:
Let us assume that the number needed to be added is taken as \[x\].
Now to find the term\ratio needed i.e. \[2:3\] to add the unknown term of \[x\] to the numerator and denominator of the previous ratio of \[9:16\].
The value of the ratio on the LHS of the equation down below is equal to the ratio on the RHS when \[x\] is added in both the numerator and denominator of the LHS given.
\[\Rightarrow \dfrac{9+x}{16+x}=\dfrac{2}{3}\]
Cross multiplying the value of the ratio of the LHS and RHS, we get the value of the unknown variable of \[x\] as:
\[\Rightarrow 3\left( 9+x \right)=2\left( 16+x \right)\]
\[\Rightarrow 27+3x=32+2x\]
\[\Rightarrow x=5\]
Therefore, the value needed to be added with the ratio of \[9:16\] to get \[2:3\] is \[5\].
Note: Student may go wrong if they try to add a single number instead of adding two number in both the denominator and numerator as given below:
\[\Rightarrow \dfrac{9+x}{16+x}=\dfrac{2}{3}\] correct form
\[\Rightarrow \dfrac{9}{16}+x=\dfrac{2}{3}\] Incorrect form
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