Question

A number lies between $300$ and $400$. If the number is added to the number formed by reversing its digits, the sum is $888$ and if the unit's digit and the ten's digit change places, the new number exceeds the original number by $9$. Then the number is:

• A. $354$
• B. $341$
• C. $348$
• D. $345$

Answer 1

The correct option is D $345$

The number lies between 300 and 400.

Thus, we know its a three digit number so let the number be $abc$.

Now, the value of number will be $100a+10b+c$.
On reversing, the number formed will be $100c+10b+a$
Sum $=100a+10b+c+100c+10b+a=888$
$=101a+20b+101c=888$
Now, the other condition is when units and tens places are interchanged.
So, the number formed will be $100a+10c+b$
This number exceeds $100a+10b+c$ by 9
So $100a+10b+c+9=100a+10c+b$
$9c-9b=9$
$c-b=1$
$c=b+1$
Now, as the number lies between 300 and 400, the value of $a$ has to be $3$.
So putting value in first equation we get :
$101a+20b+101c=888$
$101\ast 3+20b+101(b+1)=888$
$303+20b+101b+101=888$
$404+121b=888$
$121b=484$
$b=4$
$c=4+1=5$
Number will be $345$.