Question

Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages was 48.

Answer 1

Hint: First of all, let the ages of two friends be A and B respectively. Then according to the given information, form 2 equations to solve for A and B. Check if you can find the values of A and B or not.

Complete step-by-step answer:

Here, we are given that the sum of the ages of two friends is 20 years. Also, four years ago, the product of their ages was 48. We have to find their present ages and check if the above conditions are possible or not.
Let us consider the age of the first friend as A years and let us consider the age of the second friend as B years. We are given that the sum of the ages of two friends is 20 years, so we get,
(Age of First friend) + (Age of Second friend) = 20
By substituting the value of LHS in the above equation, we get,

& A+B=20 \\\ & \Rightarrow B=20-A....\left( i \right) \\\ \end{aligned}$$ Now, four years ago, the age of the first friend was = (A – 4) years. Also, four years ago, the age of the second friend was = (B – 4) years. We are given that four years ago, the product of their ages was 46. So, we get, (Age of First Friend 4 years ago) x (Age of Second Friend 4 years ago) = 48 By substituting the value of LHS in the above equation, we get, (A – 4) (B – 4) = 48 By substituting the value of B = 20 – A from equation (i) to the above equation, we get, = (A – 4) (20 – A – 4) = 48 = (A – 4) (– A + 16) = 48 By simplifying the above equation, we get, $$=\left( -{{A}^{2}}+20A-64 \right)=48$$ $$\Rightarrow {{A}^{2}}-20A+64+48=0$$ $$\Rightarrow {{A}^{2}}-20A+112=0$$ We know that for any quadratic equation, $$a{{x}^{2}}+bc+c$$, we only get real roots if $$D={{b}^{2}}-4ac\ge 0$$. So, now we also find D for the above equation, i.e. $${{A}^{2}}-20A+112=0$$. We get, a = 1, b = - 20, c = 112. So, we can substitute in $$D={{b}^{2}}-4ac$$, and we get, $$={{\left( -20 \right)}^{2}}-4\left( 1 \right)\left( 112 \right)$$ $$=400-448$$ $$=-48$$ Since we get D < 0, therefore real roots of the equation do not exist. Hence, we can say that the condition given in the question is not possible. Note: For these types of questions, students are advised to always take the present age as reference variable and calculate all other ages with respect to the present age. Also, take special care of signs while solving the equation (A – 4)( - A + 16). Also, students must note that for any quadratic equation $$a{{x}^{2}}+bx+c$$, check the value of $$D={{b}^{2}}-4ac$$ first and then only find the roots of the equation.