Question

In a family of husband, wife and a daughter, the sum of the husband’s age, twice the wife’s age and thrice the daughter’s age is 85; while the sum of twice the husband’s age, 4 times the wife’s age and 6 times the daughter’s age is 170. It is also given that the sum of 5 times the husband’s age, ten times the wife's age and 15 times the daughter’s age equals 450. The number of possible solution, in terms of the age of husband, wife and daughter, to this problem is
A. 0
B. 1
C. 2
D. Infinitely many

Answer 1

Hint: In this question we have given we have in a family of husband, wife and a daughter, the sum of the husband’s age, twice the wife’s age and thrice the daughter’s age is 85; while the sum of twice the husband’s age, 4 times the wife’s age and 6 times the daughter’s age is 170. It is also given that the sum of 5 times the husband’s age, ten times the wife's age and 15 times the daughter’s age equals 450. We need to find the number of possible solutions. So for the solution we first need to consider the age of husband, wife and daughter and after that by the given data we can form linear equations which will help us to find the age.
So for the linear equations we need to know one criteria,
so for this let, $a_{1}x+b_{1}y+c_{1}z=d_{1}$ and $a_{2}x+b_{2}y+c_{2}z=d_{2}$ be two linear equations,
Then,
(i) the equations have no solution, if $$\dfrac{a_{1}}{a_{2}} =\dfrac{b_{1}}{b_{2}} =\dfrac{c_{1}}{c_{2}} \neq \dfrac{d_{1}}{d_{2}}$$
(ii) the equations have infinitely many solution, if $$\dfrac{a_{1}}{a_{2}} =\dfrac{b_{1}}{b_{2}} =\dfrac{c_{1}}{c_{2}} =\dfrac{d_{1}}{d_{2}}$$
(iii) For the other conditions the solution will be unique.

Complete step-by-step answer:
Let us consider the age of husband, wife and daughter is h, w, and g respectively.
First statement,
The sum of the husband’s age, twice the wife’s age and thrice the daughter’s age is 85,
$\therefore h+2w+3g=85$.............(1)
Second statement,
The sum of twice the husband’s age, 4 times the wife’s age and 6 times the daughter’s age is 170,
$\therefore 2h+4w+6g=170$...........(2)
And the third statement,
The sum of 5 times the husband’s age, ten times the wife's age and 15 times the daughter’s age equals 450,
$\therefore 5h+10w+15g=450$..........(3)
Now if we compare the equation (1), (2) and (3) with $a_{1}h+b_{1}w+c_{1}g=d_{1}$, $a_{2}h+b_{2}w+c_{2}g=d_{2}$ and $a_{3}h+b_{3}w+c_{3}g=d_{3}$ respectively, then we can write,
$a_{1}=1,\ b_{1}=2,\ c_{1}=3,\ d_{1}=85$
$a_{2}=2,\ b_{2}=4,\ c_{2}=6,\ d_{2}=170$
$a_{3}=5,\ b_{3}=10,\ c_{3}=15,\ d_{3}=450$
Now,
$$\dfrac{a_{1}}{a_{2}} =\dfrac{1}{2} ,\ \dfrac{b_{1}}{b_{2}} =\dfrac{2}{4} =\dfrac{1}{2} ,\ \dfrac{c_{1}}{c_{2}} =\dfrac{3}{6} =\dfrac{1}{2} ,\ \dfrac{d_{1}}{d_{2}} =\dfrac{85}{170} =\dfrac{1}{2}$$
Therefore, we can write,
$$\dfrac{a_{1}}{a_{2}} =\dfrac{b_{1}}{b_{2}} =\dfrac{c_{1}}{c_{2}} =\dfrac{d_{1}}{d_{2}}$$
Which implies, equation (1) and (2) coincide with each other, i.e, equation has infinitely many solution

Similarly for equation (2) and (3),
$$\dfrac{a_{2}}{a_{3}} =\dfrac{2}{5} ,\ \dfrac{b_{2}}{b_{3}} =\dfrac{4}{10} =\dfrac{2}{5} ,\ \dfrac{c_{2}}{c_{3}} =\dfrac{6}{15} =\dfrac{2}{5} ,\ \dfrac{d_{2}}{d_{3}} =\dfrac{170}{450} =\dfrac{2}{5}$$
Therefore,
$$\dfrac{a_{2}}{a_{3}} =\dfrac{b_{2}}{b_{3}} =\dfrac{c_{2}}{c_{3}} =\dfrac{d_{2}}{d_{3}}$$
Which implies, equation (2) and (3) coincide with each other, i.e, equation has infinitely many solution

So if (1) and (2) coincide and (2) and (3) coincide then it implies they all coincide, that means the equations have infinitely many solutions.
Hence the correct option is option D.


Note: To solve this type of equation we have to know that if two linear equations have infinitely many solutions then two equations must coincide(i.e, lies on one another) with each other, and if two linear equations coincide then the ratio of the coefficient of x, y and z is equals to the ratio of constant part.