Question

The A.M of two positive numbers exceeds the G.M by 5 and G.M exceeds the H.M by 4. Then the numbers are:
(a) 10, 40
(b) 10, 20
(c) 20, 40
(d) 10, 50

Answer 1

Hint: We solve this problem by using the A.M, G.M and H.M of two numbers. If there are two numbers \[a,b\] then
\[A.M=\dfrac{a+b}{2}\]
\[G.M=\sqrt{ab}\]
\[H.M=\dfrac{2ab}{a+b}\]
We also have the relation between A.M, G.M and H.M as
\[\Rightarrow G.{{M}^{2}}=A.M\times H.M\]
By using these relations and given conditions we find the required numbers. While applying the given condition to mathematical equations we need to keep in mind that
\[A.M\ge G.M\ge H.M\]

Complete step-by-step answer:
We are given that the A.M of two positive numbers exceeds G.M by 5
By converting the above statement into mathematical equation we get
\[\Rightarrow A.M=G.M+5\]
We are also given that the G.M exceeds H.M by 4

By converting the above statement into mathematical equation we get
\[\begin{align}
  & \Rightarrow G.M=H.M+4 \\
 & \Rightarrow H.M=G.M-4 \\
\end{align}\]
We know that the relation between A.M, G.M and H.M as
\[\Rightarrow G.{{M}^{2}}=A.M\times H.M\]
By substituting the required values in above equation we get
\[\begin{align}
  & \Rightarrow G.{{M}^{2}}=\left( G.M+5 \right)\left( G.M-4 \right) \\
 & \Rightarrow G.{{M}^{2}}=G.{{M}^{2}}+G.M-20 \\
 & \Rightarrow G.M=20 \\
\end{align}\]
Now, by substituting the value of G.M in A.M we get
\[\begin{align}
  & \Rightarrow A.M=G.M+5 \\
 & \Rightarrow A.M=20+5=25 \\
\end{align}\]
Let us assume that the two numbers as \[x,y\] such that \[x>y\]
We know that the formula of A.M of two numbers \[a,b\] that is
\[A.M=\dfrac{a+b}{2}\]
By using the above formula we get
\[\begin{align}
  & \Rightarrow \dfrac{x+y}{2}=25 \\
 & \Rightarrow x+y=50........equation(i) \\
\end{align}\]
We know that the formula of G.M for two numbers \[a,b\] that is
\[G.M=\sqrt{ab}\]
By using the above formula we get
\[\begin{align}
  & \Rightarrow 25=\sqrt{xy} \\
 & \Rightarrow xy=400 \\
\end{align}\]
We know that the formula of algebra that is
\[\Rightarrow {{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab\]

By using the above formula to \[x,y\] such that \[x>y\]we get
\[\Rightarrow {{\left( x-y \right)}^{2}}={{\left( x+y \right)}^{2}}-4xy\]
By substituting the required values in above equation we get
\[\begin{align}
  & \Rightarrow {{\left( x-y \right)}^{2}}={{\left( 50 \right)}^{2}}-4\left( 400 \right) \\
 & \Rightarrow {{\left( x-y \right)}^{2}}=2500-1600 \\
 & \Rightarrow x-y=\sqrt{900}=30.....equation(ii) \\
\end{align}\]
Now, by adding equation (i) and equation (ii) we get
\[\begin{align}
  & \Rightarrow x+y+x-y=50+30 \\
 & \Rightarrow 2x=80 \\
 & \Rightarrow x=40 \\
\end{align}\]
Now, by substituting \[x=40\] in equation (i) we get
\[\begin{align}
  & \Rightarrow 40+y=50 \\
 & \Rightarrow y=10 \\
\end{align}\]
Therefore we can conclude that the required positive numbers are 10 and 40

So, the correct answer is “Option A”.

Note: Students may make mistakes in solving the problem. That is they may choose the difficult way of solving.
We have the mathematical equations of given two conditions as
\[\Rightarrow A.M=G.M+5\]
\[\Rightarrow G.M=H.M+4\]
Now, by assuming the two positive numbers as \[x,y\] such that \[x>y\] and using the formulas of A.M, G.M and H.M we get
\[\begin{align}
  & \Rightarrow \dfrac{a+b}{2}=\sqrt{ab}+5 \\
 & \Rightarrow \sqrt{ab}=\dfrac{2ab}{a+b}+4 \\
\end{align}\]
Trying to solve these two equations is very difficult because we need to square the two functions twice which leads to equation of power 4 that cannot be solved easily.
So we need to try to solve for any one of A.M, G.M and H.M to make our solution easy.