Question

In‌ ‌a‌ ‌3 ‌digit‌ ‌number,‌ ‌units‌ ‌digit,‌ ‌tens‌ ‌digit‌ ‌and‌ ‌hundreds‌ ‌digit‌ ‌are‌ ‌in‌ ‌the‌ ‌ratio‌ ‌\[1:2:3\].‌ ‌If‌ ‌the‌ ‌difference‌ ‌of‌ ‌original‌ ‌number‌ ‌obtained‌ ‌by‌ ‌reversing‌ ‌the‌ ‌digits‌ ‌is‌ ‌\[594\],‌ ‌find‌ ‌the‌ ‌number.‌ ‌

Answer 1

Hint: We have to find the \[3\] digit number which will satisfy the given condition. We solve this question using the concept of formation of numbers. First, we will equate the ratio of the places of the \[3\] digit number to a constant variable and then we could compute the relation for the ratios of the places. Then we form an equation using the considered values and then on simplifying the expressions and putting the values back we will obtain the required \[3\] digit number.

Complete step-by-step solution:
Given:
Let's consider that the place of a \[3\] digit number be \[a\] , \[b\] and \[c\] for the ones place, tens place and the hundreds place respectively.
So, we can write the ratio of the three places as:
\[a:b:c = 1:2:3\]
The expression of the ratio can be also written as:
\[a:b:c = x:2x:3x\]
Where \[x\] is a constant variable.
So, on comparing the two ratios, we can write the value of the variable \[a\] , \[b\] and \[c\] as:
\[a = x\] , \[b = 2x\] and \[c = 3x\]
We also know that a \[3\] digit number can be written as:
\[3 -\text{digit number} = 100 \times a + 10 \times b + 1 \times c\]
Where \[a\] , \[b\] and \[c\] are the digits of the respective place.
So, using the above expression we can write the expression for the \[3\] digit number as:
\[3 - \text{digit number} = 100 \times c + 10 \times b + 1 \times a\]
\[3 -\text{digit number} = 100c + 10b + a\]
Also, the reverse of the number can be written as:
\[\text{Reverse number} = 100a + 10b + c\]
According to the question the difference between the original number and the reverse number is \[594\].
So, we can write the expression as:
\[\left( {100c + 10b + a} \right) - \left( {100a + 10b + c} \right) = 594\]
On further solving, we get the expression as:
\[99c - 99a = 594\]
\[99\left( {c - a} \right) = 594\]
Further, we can write the expression as:
\[c - a = \dfrac{{594}}{{99}}\]
\[c - a = 6\]
Now substituting the values of \[a\] and \[c\] in terms of \[x\], we get the value of the expression as:
\[3x - x = 6\]
\[2x = 6\]
Thus, we get the value of \[x\] as:
\[x = 3\]
Putting the value of \[x\], we get the value of \[a\] , \[b\] and \[c\] as:
As, \[a = x\]
\[a = 3\]
As, \[b = 2x\]
\[b = 2 \times 3\]
\[b = 6\]
As, \[c = 3x\]
\[c = 3 \times 3\]
\[c = 9\]
Thus, we get the required \[3\] digit number as:
\[3-\text{digit number} = 100 \times 9 + 10 \times 6 + 1 \times 3\]
\[3-\text{digit number} = 900 + 60 + 3\]
So, we get the value of \[3\] digit number as:
\[3-\text{digit number} = 963\]
Hence the required \[3\] digit number is \[963\].

Note: The ratio of three numbers can be found out by finding the ratios of two numbers at a time and then the other two and hence we can find the ratio of the three numbers. As in the above question we consider the original ratio to be as:
\[a:b:c = x:2x:3x\]
We can obtain the original ratio as:
\[a:b = x:2x\]
\[\dfrac{a}{b} = \dfrac{x}{{2x}}\]
This can be written as:
\[\dfrac{a}{b} = \dfrac{1}{2}\]
\[a:b = 1:2\]
Similarly, taking the other pair and we will obtain the original ratio.