Question

The age of the father of two children is twice that of the elder one added to four times that of the younger one. If the geometric mean of the ages of the children is $4\sqrt{3}$ and their harmonic mean is 6, then father’s age is
(a) 36
(b) 40
(c) 50
(d) 56

Answer 1

Hint: Let the ages of the father, elder child and younger child be x,y and z years, respectively. Now use the conditions given in the question along with the formula $\text{Geometric mean}=\sqrt{yz}$ and $\text{Harmonic mean}=\dfrac{2yz}{y+z}$ to form 3 equations. Now solve the equations to get the value of x.

Complete step-by-step answer:
To start with the question, we let the ages of the father, elder child and younger child be x,y and z years, respectively.
Now it is given in the question that the age of the father of two children is twice that of the elder one added to four times that of the younger one. So, if we represent this in form of equation, we get
$x=2y+4z......(i)$
Also, it is given that the geometric mean of the ages of the two children is $4\sqrt{3}$ . We know that $\text{Geometric mean}=\sqrt{yz}$ . Therefore, we can say that
$4\sqrt{3}=\sqrt{yz}$
Squaring both side of the equation will give:
$yz=48.........(ii)$
$\Rightarrow y=\dfrac{48}{z}.........(iii)$
Now as it is given that the harmonic mean of the ages of the two children is 6 . We know that $\text{Harmonic mean}=\dfrac{2yz}{y+z}$ . Therefore, we can say that
$\dfrac{2yz}{y+z}=6$
If we substitute the value of yz from equation (ii), we get
$\dfrac{2\times 48}{y+z}=6$
$\Rightarrow y+z=16$
Now we will use equation (iii) to substitute y. On doing so, we get
$\dfrac{48}{z}+z=16$
$\Rightarrow \dfrac{48+{{z}^{2}}}{z}=16$
$\Rightarrow 48+{{z}^{2}}=16z$
$\Rightarrow {{z}^{2}}-16z+48=0$
$\Rightarrow {{z}^{2}}-12z-4z+48=0$
$\Rightarrow \left( z-4 \right)\left( z-12 \right)=0$
Therefore, the possible values of z are 4 and 12. But if we put the value of z to be 12 in equation (iii), we get y=4 which is not possible as y is the age of the elder child, so z=4 and corresponding to this y=12.
Now we will put y=12 and z=4 in equation (i). On doing so, we get
$x=2\times 12+4\times 4$
$x=40\text{ years}\text{.}$
Therefore, the age of the father is 40 years. Hence, the answer is option (b).

Note: Remember that the formulas of geometric mean and harmonic mean are only valid for positive numbers and in the above question y and z were the age of children, so they were positive for sure. Also, be careful about the substitutions and calculation part, as students generally make an error in the calculation part. If students do not realise that the value of z cannot be 4, they will take z=4 and get y=12. Now, using these values, the age of father would come up to be 56, which is option (d). So, this must be noted.