In a three-digit number, the tens digit is twice the unit’s digit and the hundreds digit is thrice the unit’s digit. If the sum of its all three digits is 12, then find the reversed number.
(a) 642
(b) 462
(c) 246
(d) 416
Answer
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Hint: In this question, we have to first find the number from which we can find the reversed number by writing the unit’s place digit in the hundreds place and hundreds place digit in the unit’s place. As the information of the digits is given in relation to the unit’s digit, we can assume the unit’s digit to be a variable x. Then, we can use the relations between the unit’s digit and the tens and hundreds place digits to find the required number.
Complete step-by-step answer:
It is given that the tens digit is twice the unit’s digit and the hundreds digit is thrice the unit’s digit. Therefore, we if take the unit’s place digit to be x………………………(1.1)
Then,
$\text{Tens digit=2}\times \text{One }\!\!'\!\!\text{ s digit}=2x..........(1.2)$
And
$\text{Hunderd }\!\!'\!\!\text{ s digit=3}\times \text{One }\!\!'\!\!\text{ s digit}=3x..........(1.3)$
It is also given that the sum of all the digits is 12. Therefore, from equations (1.1), (1.2) and (1.3),
$\begin{align}
& \text{One }\!\!'\!\!\text{ s digit+Tens }\!\!'\!\!\text{ s digit}+\text{Hundred }\!\!'\!\!\text{ s digit=12} \\
& \Rightarrow x+2x+3x=12\Rightarrow 6x=12 \\
& \Rightarrow x=\dfrac{12}{6}=2..........(1.4) \\
& \\
\end{align}$
Therefore, from equations (1.1), (1.2), (1.3) and (1.4), we get
\[Ones\text{ }digit=x=2\]
\[Tens\text{ }digit=2x=4\]
\[Hundreds\text{ }digit=3x=6\]
Therefore, the original number should be given by $642$. Therefore, the reversed number which is obtained by writing the digits in reverse order should be equal to $246$ which matches option (c). Hence, option (c) is the correct answer.
Note: We could have directly obtained the reverse number by taking its units place to be 3x, ten’s place to be 2x and hundred’s place to be x by comparing to equations (1.1), (1.2) and (1.3) of the digits of the original number.
Complete step-by-step answer:
It is given that the tens digit is twice the unit’s digit and the hundreds digit is thrice the unit’s digit. Therefore, we if take the unit’s place digit to be x………………………(1.1)
Then,
$\text{Tens digit=2}\times \text{One }\!\!'\!\!\text{ s digit}=2x..........(1.2)$
And
$\text{Hunderd }\!\!'\!\!\text{ s digit=3}\times \text{One }\!\!'\!\!\text{ s digit}=3x..........(1.3)$
It is also given that the sum of all the digits is 12. Therefore, from equations (1.1), (1.2) and (1.3),
$\begin{align}
& \text{One }\!\!'\!\!\text{ s digit+Tens }\!\!'\!\!\text{ s digit}+\text{Hundred }\!\!'\!\!\text{ s digit=12} \\
& \Rightarrow x+2x+3x=12\Rightarrow 6x=12 \\
& \Rightarrow x=\dfrac{12}{6}=2..........(1.4) \\
& \\
\end{align}$
Therefore, from equations (1.1), (1.2), (1.3) and (1.4), we get
\[Ones\text{ }digit=x=2\]
\[Tens\text{ }digit=2x=4\]
\[Hundreds\text{ }digit=3x=6\]
Therefore, the original number should be given by $642$. Therefore, the reversed number which is obtained by writing the digits in reverse order should be equal to $246$ which matches option (c). Hence, option (c) is the correct answer.
Note: We could have directly obtained the reverse number by taking its units place to be 3x, ten’s place to be 2x and hundred’s place to be x by comparing to equations (1.1), (1.2) and (1.3) of the digits of the original number.
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