# Problem based on ages of two persons.

Bibhu said to Bigyata, "I was twice as old as you were when I was as old as you are.". If the sum of their ages is 35 years, find their present ages.

My Attempt:

Let the present ages of Bibhu and Bigyata be $x$ years and $y$ years respectively. From the second statement of the question: $x+y=35$.

$y=35-x$.

• In your attempt: Case 2? Where's case 1? Oct 13, 2016 at 2:05
• @suomynonA, I have stated that I did not understand the first condition given in the question, which is the case 1.
– pi-π
Oct 13, 2016 at 2:08
• That's just confusing...can you clarify in your question by explaining and removing the case 2 part? Oct 13, 2016 at 2:09
• You mean, $y=35-x$? Where does $50$ comes from? Oct 13, 2016 at 2:10
• @user354073 I made an answer; if you need any more clarification just comment. Oct 13, 2016 at 2:17

Hint:

The equations will be $$x+y=35$$ and $$x-(x-y)=2(y-(x-y))$$ where $x$ is Bibhu's age and $y$ is Bigyata's age.

Here is some explanation for how to get the second equation:

Left hand side of equation: The year when Bibhu was as old as Bigyata would be Bibhu's age minus the difference between the ages.

Right hand side of equation: Since you go back $x-y$ years ago for Bibhu, you also have to do that for Bigyata. You multiply this by 2 because Bibhu says that at that time she was twice as old as Bigyata.

Clearly $x\ge y$.

Second case simplifies to $y=2(y-(x-y))=2(2y-x)=4y-2x$ i.e. $3y=2x$.

Think about it, when Bhibhu was the same age as Bigyata is now i.e. $y$ , (means they are talking about $x-y$ years before), then Bigyata was $y-(x-y)$ years old and twice that is $2(y-(x-y))$.

Therefore, $y=35-\frac{3y}{2}$ or $\frac{5y}{2}=35$.

Thus $y=14$ and $x=21$

$x + y = 35$ and k years before bhibhu was as old as bhigyata is now. $x-k =y , x-k = 2 ( y- k)$ solving these equations will give you $k = 7, x= 21, y=14$

Maybe the problem is easier to comprehend if you consider another variable $$\theta = number\ of\ years\ to\ go\ back\ in\ time\ for\ Bibhu\ to\ be\ twice\ as\ old\ as\ Bigyata$$

We have now 3 equations,

1. "Bibhu ($x$) was twice as old as Bigyata ($y$) $\theta$ years ago": $$x-\theta=2(y-\theta)$$

2. "sum of their age is 35": $$x+y=35$$

3. and we know that $\theta$ years ago Bibhu was Bigyata's age: $$x=y+\theta$$thus $$\theta=x-y$$

Replacing $\theta$ in the equation 1. we get $$x=\frac{3}{2}y$$ replacing $x$ in 2., $y=14$, and $x=21$

An easier way to think about it is if you take 4 variables, $x_1, x_2, y_1, y_2$ with $x, y$ representing Bigyata and Bhibhu respectively and the index $1, 2$ meaning then and now respectively. With that the first statement becomes:

$(I) 2x_1 = y_1 \land y_1 = x_2$

we also have $(II) x_2 + y_2 = 35$

and $(III)x_2 - x_1 = y_2 - y_1$, i.e. the time difference between then and now

It now become easy to solve for any variable, e.g. for $x_1$ we have from $(II): 35=x_2+y_2=^{(I)}y_1+y_2=^{(III)}x_2-x_1+2y_1=^{(I)}5x_1$

Let their present ages be $x$ & $y$. It is given $x+y=35$--(equation 1) and we have to find out $x$ & $y$. Let the years before from the current year be denoted as $'t'$. $x-t=2(y-t)$ (OR) $x-2y= -t$ --> I was twice as old as you were. (eqn 2) $x-t=y$ (OR) $x-y =t$ --> when I was as old as you are. (eqn 3) From eqns 2 & 3 we can calculate the value of $y$ which is $2t$ (eqn 4) Substitute the value of $2t$ for y in eqn $3$ which gives $x-t=2t$ or $x=3t$ (eqn 5). Sutstitute these $x$ & $y$ values in eqn $1$ which gives $5t=35$ or $t=7.$ From this we can find out the values of x&y using eqns 4&5.So the answer is $21$ & $14$.