Catherine is now twice as old as Jason but 6 years ago she was 5 times as old as he was. How old is Catherine now?

Question

This is an IQ question.

"Catherine is now twice as old as Jason but 6 years ago she was 5 times as old as he was. How old is Catherine now?"

How to solve such questions? I think their combined age should be given but its not.

Answer 1

It can be helpful to organize the information into a 'now-and-then' chart. Let Jason's age now be $J$.

$$\begin{array} {c|c|c} &\mbox{Jason's age}&\mbox{Catherine's age} \\ \hline \mbox{Now} &J&2J\\ \mbox{6 years ago} &J-6& 2J-6 \end{array}$$

Then $2J-6=5(J-6)$, which you can easily solve.

Answer 2

How to solve such questions?

The approach is to model the given information as some variables and their relationship as equations or inequalities and then try one of the mathematical methods to find a solution, which is then translated back into the terms of the original problem.

Here you have the age of two persons, which one might abstract as variables $C$ and $J$.

Now about the given information:

Catherine is now twice as old as Jason

This can be expressed as $$ C = 2 J \quad (1) $$ if $C$ is Catherine's present age, $J$ is Jason's present age.

$6$ years ago she was $5$ times as old as he was.

Six years ago is $C - 6$ and Jason was $J - 6$ years old. The resulting equation is: $$ C - 6 = 5 \, (J - 6) \quad (2) $$ If you had linear algebra you might recognize this as a inhomogenous system of two linear equations in the unknowns $C$ and $J$. \begin{align} C - 2 J &= 0 \quad (3) \\ C - 5 J &= -24 \quad (4) \end{align} There are systematic ways to solve this.

Solution using algebraic equivalence transformations:

One way is to first solve both variables for one of the variables, this means using equivalence transformation to bring the equations into the form \begin{align} X &= F_1(Y) \\ X &= F_2(Y) \end{align} then equate, $$ F_1(Y) = F_2(Y) $$ and get a value for the other variable (if the system is solvable). $$ X = F_1^{-1}(F_2(Y)) $$ Then that result can be used with one of the original equations where the other variable shows up to solve for that variable too. $$ Y = F_1^{-1}(X) $$ In this case we have $X = C$, $Y = J$ and: \begin{align} C &= 2 J\\ C &= 5 J -24 \end{align} Equating gives: $$ 2J = 5 J - 24 $$ or $$ 24 = 3 J $$ which implies $$ J = 24/3 = 8 $$ Using one of the original equations featuring $C$ we get $$ C = 2 J = 2 \cdot 8 = 16 $$

Graphical Solution:

As these are few variables you can also attempt a graphical solution, meaning: you can try to read the solution (if it exists) from the graph.

(Large Version)

The $x$-coordinate is Catherine's current age $C$, the $y$-coordinate Jason's current age $J$.

The green line $a$ shows all points $(x, y)$ which satisfy equation $(3)$, the blue line $b$ shows all points $(x, y)$ which satisfy equation $(4)$.

If both lines have an intersection, then the points of the intersection set satisfy both equations. They would be solutions. Here the intersection is marked by a big red dot named $P$.

Solution via Methods of Linear Algebra:

More elegant is the formulation using matrices and vectors: $$ A x = b $$ where $$ A = \begin{pmatrix} 1 & -2 \\ 1 & -5 \end{pmatrix} \\ x = \begin{pmatrix} C \\ J \end{pmatrix} \quad b = \begin{pmatrix} 0 \\ -24 \end{pmatrix} $$ For the number of possible solutions of such a system -- no solution, one solution, infinite many solutions -- the determinant of the matrix $A$ is important. Here it is $$ \det A = -3 \ne 0 $$ which means, the system has one solution, the matrix $A$ has an inverse matrix, which can be used to get a solution: $$ x = A^{-1} b $$

You can solve this by hand, using Gaussian elimination, $$ [A \mid b] = \left[ \begin{array}{rr|r} 1 & -2 & 0 \\ 1 & -5 & -24 \end{array} \right] \to \left[ \begin{array}{rr|r} 1 & -2 & 0 \\ 0 & -3 & -24 \end{array} \right] \to \left[ \begin{array}{rr|r} 1 & -2 & 0 \\ 0 & 1 & 8 \end{array} \right] \to \left[ \begin{array}{rr|r} 1 & 0 & 16 \\ 0 & 1 & 8 \end{array} \right] $$ (this calculation is similar to fellow user menag's answer). or using an explicit formula for the inverse of $2 \times 2$ matrices $$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ \det A = a d - b c = 1 \cdot (-5) - (-2) \cdot 1 = -3 \\ A^{-1} = \frac{1}{\det A} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = -\frac{1}{3} \begin{pmatrix} -5 & 2 \\ -1 & 1 \end{pmatrix} $$ which leads to $$ x = A^{-1} b = -\frac{1}{3} \begin{pmatrix} -5 & 2 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 0 \\ -24 \end{pmatrix} = -\frac{1}{3} \begin{pmatrix} -48 \\ -24 \end{pmatrix} = \begin{pmatrix} 16 \\ 8 \end{pmatrix} $$ or some CAS (computer algebra system) like Octave.

Using a Computer Algebra System:

octave> A = [1,1;-2,-5]'
A =

   1  -2
   1  -5

octave> det(A)
ans = -3

octave> inv(A)
ans =

   1.66667  -0.66667
   0.33333  -0.33333

octave> b = [0, -24]'
b =

    0
  -24

octave> x = inv(A)* b
x =

   16
    8

Answer 3

hint: Let $C,J$ be Kathy's and Jason's ages as of now. Thus: $C = 2J , C-6 = 5(J-6)$. Can you finish it off?

Answer 4

First you have to formulate the given information in maths. We write $C$ and $J$ for the ages. Since Catherine is twice as old as Jason we have $$C = 2J.$$ Moreover we know that 6 years ago, Catherine was five times older then Jason. This can be formulated as $$ C - 6 = 5 \cdot (J - 6) = 5J - 30.$$ This equation is equivalent to $$C = 5J - 24.$$ Subtracting the equations yields $$0 = 3J - 24$$ and thus $J = 8$ and $C = 16$.