
A number of two digits has 3 for its unit’s digit, and the sum of digits is \[\dfrac{1}{7}\] of the number itself. The number is
A. 43
B. 53
C. 63
D. 73
Answer
580.2k+ views
Hint: As the units digit is given as 3, so the number can be represented as \[{\rm{1}}0{\rm{x}} + {\rm{3}}\], where, x is ten's digit and its digits are x and 3. Hence, according to the question, we can say that, \[x + 3 = \dfrac{1}{7}\left( {10x + 3} \right)\] and from this solve for x and thus, find the two digit number.
Complete step-by-step answer:
In the question, we are given certain features about a two-digit number which is that '3' must be in the unit’s digit and the sum of the digits is \[\dfrac{1}{7}\] of the number itself.
We have to find a two digit number, so we can suppose it to be \[10x + y\] where x represents tens place and y represents the unit place.
Here, it is said that the number 3 is fixed for the units place, so, the number can be represented as \[10x + 3\] .
Now, we have to find the value of x such that the sum of the digits \[ \Rightarrow \left( {x + 3} \right){\rm{ is }}\dfrac{1}{7}\] of the number which is \[10x + 3\] which can be represented in form of equation as,
\[x + 3 = \dfrac{1}{7}\left( {10x + 3} \right)\]
Now, on cross multiplication we get,
\[7\left( {x + 3} \right) = 10x + 3\]
Now, further on simplifying we get,
\[7x + 21 = 10x + 3\]
On subtracting 7x from both the sides, we get,
\[\begin{array}{l}7x + 21 - 7x = 10x + 3 - 7x\\ \Rightarrow 21 = 3x + 3\end{array}\]
Now, subtracting 3 from both the sides, we get,
\[\begin{array}{l}21 - 3 = 3x + 3 - 3\\ \Rightarrow 18 = 3x\end{array}\]
So, the value of x is \[\dfrac{{18}}{3} \Rightarrow 6\]
The number represented as \[10x + 3.\] We found the value of x as 6, thus, the number after putting x as 6, we get, \[10 \times 6 + 3 \Rightarrow 63\]
Thus, the two digit number is 63.
Hence, the correct option is C.
Note: In the question, it is given that, if the number is divided by 7 its quotient will be the sum of the digits, so from this we can conclude that the number is divisible by 7. We can check which one of the options is divided by 7 and tick the correct option.
Complete step-by-step answer:
In the question, we are given certain features about a two-digit number which is that '3' must be in the unit’s digit and the sum of the digits is \[\dfrac{1}{7}\] of the number itself.
We have to find a two digit number, so we can suppose it to be \[10x + y\] where x represents tens place and y represents the unit place.
Here, it is said that the number 3 is fixed for the units place, so, the number can be represented as \[10x + 3\] .
Now, we have to find the value of x such that the sum of the digits \[ \Rightarrow \left( {x + 3} \right){\rm{ is }}\dfrac{1}{7}\] of the number which is \[10x + 3\] which can be represented in form of equation as,
\[x + 3 = \dfrac{1}{7}\left( {10x + 3} \right)\]
Now, on cross multiplication we get,
\[7\left( {x + 3} \right) = 10x + 3\]
Now, further on simplifying we get,
\[7x + 21 = 10x + 3\]
On subtracting 7x from both the sides, we get,
\[\begin{array}{l}7x + 21 - 7x = 10x + 3 - 7x\\ \Rightarrow 21 = 3x + 3\end{array}\]
Now, subtracting 3 from both the sides, we get,
\[\begin{array}{l}21 - 3 = 3x + 3 - 3\\ \Rightarrow 18 = 3x\end{array}\]
So, the value of x is \[\dfrac{{18}}{3} \Rightarrow 6\]
The number represented as \[10x + 3.\] We found the value of x as 6, thus, the number after putting x as 6, we get, \[10 \times 6 + 3 \Rightarrow 63\]
Thus, the two digit number is 63.
Hence, the correct option is C.
Note: In the question, it is given that, if the number is divided by 7 its quotient will be the sum of the digits, so from this we can conclude that the number is divisible by 7. We can check which one of the options is divided by 7 and tick the correct option.
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