One-quarter of the sum of the two numbers is 14 and a half their difference is 5. The numbers are
(a) 23, 30
(b) 33, 43
(c) 23, 33
(d) 25, 35
Answer
600k+ views
Hint: For solving this problem we need to consider the numbers as some variables like \['x’, ‘y'\] and write the equations of the conditions given. They gave two conditions, so we will get two equations and we have two unknowns. So, we can solve for variables \['x’, ‘y'\] to get the two numbers.
Complete step-by-step solution
Let us assume that \['x’, ‘y'\] are the two numbers.
We are given that the quarter of the sum of two numbers is 14.
By applying this condition we can write
\[\begin{align}
& \Rightarrow \dfrac{1}{4}\left( x+y \right)=14 \\
& \Rightarrow x+y=56...equation(i) \\
\end{align}\]
Now the second condition is that the half the difference of numbers is 5.
For applying the difference let us assume that \[x >y\]. By applying the second condition we get
\[\begin{align}
& \Rightarrow \dfrac{1}{2}\left( x-y \right)=5 \\
& \Rightarrow x-y=10.....equation(ii) \\
\end{align}\]
Here, we have two equations with two variables.
So, let us solve these equations to get the numbers. By adding equation (i) and equation (ii) we get
\[\begin{align}
& \Rightarrow \left( x+y \right)+\left( x-y \right)=56+10 \\
& \Rightarrow 2x=66 \\
& \Rightarrow x=33 \\
\end{align}\]
Now, for finding the value of \['y'\] let us take one equation from equation (i) or equation (ii) and substitute value of \['x'\].
By substituting value of \['x'\] in equation (i) we will get
\[\begin{align}
& \Rightarrow 33+y=56 \\
& \Rightarrow y=23 \\
\end{align}\]
So, we got both the values of assumed variables \['x’, ‘y'\] which represent the numbers.
So the numbers are \[33,23\]. So, option (c) is the correct answer.
Note: Students may make mistakes in applying the conditions to mathematical equations. Due to over reading instead of taking \[\dfrac{1}{4}\left( x+y \right)=14\] for first condition they may take \[\left( x+y \right)=\dfrac{1}{4}\times 14\] which results in wrong answer. Reading the conditions and changing them to mathematical equations is important. Also, we need to take care of the calculations part.
Complete step-by-step solution
Let us assume that \['x’, ‘y'\] are the two numbers.
We are given that the quarter of the sum of two numbers is 14.
By applying this condition we can write
\[\begin{align}
& \Rightarrow \dfrac{1}{4}\left( x+y \right)=14 \\
& \Rightarrow x+y=56...equation(i) \\
\end{align}\]
Now the second condition is that the half the difference of numbers is 5.
For applying the difference let us assume that \[x >y\]. By applying the second condition we get
\[\begin{align}
& \Rightarrow \dfrac{1}{2}\left( x-y \right)=5 \\
& \Rightarrow x-y=10.....equation(ii) \\
\end{align}\]
Here, we have two equations with two variables.
So, let us solve these equations to get the numbers. By adding equation (i) and equation (ii) we get
\[\begin{align}
& \Rightarrow \left( x+y \right)+\left( x-y \right)=56+10 \\
& \Rightarrow 2x=66 \\
& \Rightarrow x=33 \\
\end{align}\]
Now, for finding the value of \['y'\] let us take one equation from equation (i) or equation (ii) and substitute value of \['x'\].
By substituting value of \['x'\] in equation (i) we will get
\[\begin{align}
& \Rightarrow 33+y=56 \\
& \Rightarrow y=23 \\
\end{align}\]
So, we got both the values of assumed variables \['x’, ‘y'\] which represent the numbers.
So the numbers are \[33,23\]. So, option (c) is the correct answer.
Note: Students may make mistakes in applying the conditions to mathematical equations. Due to over reading instead of taking \[\dfrac{1}{4}\left( x+y \right)=14\] for first condition they may take \[\left( x+y \right)=\dfrac{1}{4}\times 14\] which results in wrong answer. Reading the conditions and changing them to mathematical equations is important. Also, we need to take care of the calculations part.
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