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

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

& \Rightarrow G.M=H.M+4 \\\ & \Rightarrow H.M=G.M-4 \\\ \end{aligned}$$ We know that the relation between A.M, G.M and H.M as $$\Rightarrow G.{{M}^{2}}=A.M\times H.M$$ By substituting the required values in above equation we get $$\begin{aligned} & \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{aligned}$$ Now, by substituting the value of G.M in A.M we get $$\begin{aligned} & \Rightarrow A.M=G.M+5 \\\ & \Rightarrow A.M=20+5=25 \\\ \end{aligned}$$ Let us assume that the two numbers as $$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{aligned} & \Rightarrow \dfrac{x+y}{2}=25 \\\ & \Rightarrow x+y=50........equation(i) \\\ \end{aligned}$$ We know that the formula of G.M for two numbers $$a,b$$ that is $$G.M=\sqrt{ab}$$ By using the above formula we get $$\begin{aligned} & \Rightarrow 25=\sqrt{xy} \\\ & \Rightarrow xy=400 \\\ \end{aligned}$$ We know that the formula of algebra that is $$\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{aligned} & \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{aligned}$$ Now, by adding equation (i) and equation (ii) we get $$\begin{aligned} & \Rightarrow x+y+x-y=50+30 \\\ & \Rightarrow 2x=80 \\\ & \Rightarrow x=40 \\\ \end{aligned}$$ Now, by substituting $$x=40$$ in equation (i) we get $$\begin{aligned} & \Rightarrow 40+y=50 \\\ & \Rightarrow y=10 \\\ \end{aligned}$$ Therefore we can conclude that the required positive numbers are 10 and 40 **So, the correct answer is “Option A”.** **Note:** Students may make mistakes in solving the problem. That is they may choose the difficult way of solving. We have the mathematical equations of given two conditions as $$\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{aligned} & \Rightarrow \dfrac{a+b}{2}=\sqrt{ab}+5 \\\ & \Rightarrow \sqrt{ab}=\dfrac{2ab}{a+b}+4 \\\ \end{aligned}$$ 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.