When Ahmad was same age as Bamon, Cell was $6$. When Bamon was same age as Cell, Ahmet was $26$. What is the Cell's current age?

Question

When Ahmad was same age as Bamon, Cell was $6$. When Bamon was same age as Cell, Ahmet was $26$. What is the Cell's current age?

My attempt:

$$A = \text{Ahmad}, B = \text{Bamon}, C = \text{Cell}$$

$$t= \text{passed time}$$

$$C = 6 -t, A-t = B$$ and

$$A = 20-t, B-t = C$$

Sorry If I'm wrong.

Answer 1

The tag is misleading. It is not a word-problem but linear algebra.

First, let us understand each sentence:

When Ahmad was the same age as Bamon, Cell was 6.

So, $t_1$ years bevor, Cell was $6$ and Ahmad was same age as Bamon. So, you get $$ C-t_1=6\text{ and }A-t_1=B. $$ You can combine these equations. Because $C=6+t_1$ and $t_1=A-B$, you get $C=6+A-B$.

Now,

When Bamon was the same age as Cell, Ahmet was 26.

Here, you can't say, that the passed time is the same as in the first sentence. So, you have to use a time $t_2$ and you can use the same arguments as above to get $$ B-t_2=C\text{ and }A-t_2=26. $$ Again, combine these equations to eliminate $t_2$. Because $C=B-t_2$ and $t_2=A-26$, you get $C=B-(A-26)=B-A+26$.

Together, you have the two equations $$ C=6+A-B\\ C=B-A+26 $$ Next, you can add the two equations and you get: $$ 2C=32 $$ Finally, you can conclude $C=16$.

Answer 2

I will generalize it to help you entirely on these age questions. Let $a,b,c$ be the ages of Ahmad, Bamon, Cell respectively in the present.

When $A$ was the same age as $B$, $C$ was $x$ years old. When $B$ was same age as $C$, $A$ was $y$ years old. What is $C's$ current age?

"When $A$ was the same age as $B$, $C$ was $x$ years old."

$y$ years after this statement (not after the present), $A$ would be $b+y$ years old, $B$'s age would be unknown, $C$'s age would be $x+y$ years old.

"When $B$ was same age as $C$, $A$ was $y$ years old $(y>x)$."

$b$ years after this statement (not after the present), $A$ would be $b+y$ years old, $B$ would be $c+b$ years old, $C$'s age would be unknown.

We have two points in time that is the same (because $A$'s age is the same), so we merge them to conclude that if $A$ is $b+y$ years old, then $B$ is $c+b$ years old and $C$ is $x+y$ years old.

$c$ years before that point (not before the present), $A$ was $b+y-c$ years old, $B$ was $b$ years old and $C$ was $x+y-c$ years old, which happens to be the present because "$B$ was $b$ years old" is true in the present (I will ignore the use of "was" here), this means that:

• In the present, $A$ is $b+y-c$ years old, $B$ is $b$ years old, $C$ is $x+y-c$ years old.

• In the present, $A$ is $a$ years old, $B$ is $b$ years old, $C$ is $c$ years old.

$\Rightarrow x+y-c=c \Rightarrow c=\frac{x+y}{2}$, which $x$ and $y$ are already given.

Hope this is useful, as this is the third time I answered your problems. This answer is pretty much proof-based.