Answer
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Hint: First assume the three-digit number in the expanded form as $100x + 10y + z$, where $x$ is the value at the once place, $y$ is the value at tens place and $z$ is the value at hundreds place. Then, apply the given conditions to approach the required result.
Complete step-by-step answer:
Given that the number consists of three digits having ones, tens, and hundreds place.
Let the three-digit number is $100x + 10y + z$in the expanded form.
The first condition is given that the right-hand digit is one's place digit being zero, therefore the value of $z$ is zero. That is,
Therefore the original number should be:
$100x + 10y + 0$
$ \Rightarrow 100x + 10y$
If the left hand and middle digit are interchanged, means if tens place and hundreds place is interchanged then the number is diminished by 180.
First, change the tens place and hundreds place then the number should be,
$ \Rightarrow 100y + 10x$
Then the difference between the original number and the changed number should be 180.
$ \Rightarrow \left( {100x + 10y} \right) - \left( {100y + 10x} \right) = 180$
Solve the above equation:
$ \Rightarrow 100x + 10y - 100y - 10x = 180$
$ \Rightarrow 90x - 90y = 180$
$ \Rightarrow 90\left( {x - y} \right) = 180$
$ \Rightarrow x - y = \dfrac{{180}}{{90}}$
$ \Rightarrow x - y = 2$ …(1)
Now, the second condition is given that if the left-hand digit be halved and middle and right-hand digit interchanged then the number is diminished by 454.
When the hundreds place digit be halved then it is given as $\dfrac{{100x}}{2}$ and tens place and ones place digit interchanged then the number should be,
$ \Rightarrow \dfrac{{100x}}{2} + 0 + y$
$ \Rightarrow 50x + y$
As given then the difference between the original number and the changed number should be 454.
$ \Rightarrow 100x + 10y - \left( {50x + y} \right) = 454$
Solve the above equation:
$ \Rightarrow 100x + 10y - 50x - y = 454$
$ \Rightarrow 50x + 9y = 454$ …(2)
Now we have two equations that follow the given conditions.
$x - y = 2$ and
$50x + 9y = 454$
Find the value of \[x\]and$y$ from the above equations:
Multiply (1) with 50 and we get,
$50x - 50y = 100$ …(3)
Now subtract (2) by (3) using the horizontal method of subtraction.
$ \Rightarrow 50x + 9y - \left( {50x - 50y} \right) = 454 - 100$
$ \Rightarrow 50x + 9y - 50x + 50y = 354$
$ \Rightarrow 59y = 354$
$ \Rightarrow y = \dfrac{{354}}{{59}}$
$ \Rightarrow y = 6$
So, the value $y$ is $6$.
Now, substitute 6 as the value of y in the (1).
$x - y = 2$
$x - 6 = 2$
$x = 8$
We have the original number:
\[ \Rightarrow 100x + 10y\]
Now, substitute 8 as the value of$x$ and $6$ as the value of $y$ in the original number:
\[ \Rightarrow 100\left( 8 \right) + 10\left( 6 \right)\]
$ \Rightarrow 800 + 60$
$ \Rightarrow 860$
Therefore the original number is 860.
Note: Any digit can be expressed in the expanded form, therefore the digit $860$ can also be expressed in expanded form as:
$
860 = 800 + 60 + 0 \\
= 8 \times 100 + 6 \times 10 + 0 \\
$
Complete step-by-step answer:
Given that the number consists of three digits having ones, tens, and hundreds place.
Let the three-digit number is $100x + 10y + z$in the expanded form.
The first condition is given that the right-hand digit is one's place digit being zero, therefore the value of $z$ is zero. That is,
Therefore the original number should be:
$100x + 10y + 0$
$ \Rightarrow 100x + 10y$
If the left hand and middle digit are interchanged, means if tens place and hundreds place is interchanged then the number is diminished by 180.
First, change the tens place and hundreds place then the number should be,
$ \Rightarrow 100y + 10x$
Then the difference between the original number and the changed number should be 180.
$ \Rightarrow \left( {100x + 10y} \right) - \left( {100y + 10x} \right) = 180$
Solve the above equation:
$ \Rightarrow 100x + 10y - 100y - 10x = 180$
$ \Rightarrow 90x - 90y = 180$
$ \Rightarrow 90\left( {x - y} \right) = 180$
$ \Rightarrow x - y = \dfrac{{180}}{{90}}$
$ \Rightarrow x - y = 2$ …(1)
Now, the second condition is given that if the left-hand digit be halved and middle and right-hand digit interchanged then the number is diminished by 454.
When the hundreds place digit be halved then it is given as $\dfrac{{100x}}{2}$ and tens place and ones place digit interchanged then the number should be,
$ \Rightarrow \dfrac{{100x}}{2} + 0 + y$
$ \Rightarrow 50x + y$
As given then the difference between the original number and the changed number should be 454.
$ \Rightarrow 100x + 10y - \left( {50x + y} \right) = 454$
Solve the above equation:
$ \Rightarrow 100x + 10y - 50x - y = 454$
$ \Rightarrow 50x + 9y = 454$ …(2)
Now we have two equations that follow the given conditions.
$x - y = 2$ and
$50x + 9y = 454$
Find the value of \[x\]and$y$ from the above equations:
Multiply (1) with 50 and we get,
$50x - 50y = 100$ …(3)
Now subtract (2) by (3) using the horizontal method of subtraction.
$ \Rightarrow 50x + 9y - \left( {50x - 50y} \right) = 454 - 100$
$ \Rightarrow 50x + 9y - 50x + 50y = 354$
$ \Rightarrow 59y = 354$
$ \Rightarrow y = \dfrac{{354}}{{59}}$
$ \Rightarrow y = 6$
So, the value $y$ is $6$.
Now, substitute 6 as the value of y in the (1).
$x - y = 2$
$x - 6 = 2$
$x = 8$
We have the original number:
\[ \Rightarrow 100x + 10y\]
Now, substitute 8 as the value of$x$ and $6$ as the value of $y$ in the original number:
\[ \Rightarrow 100\left( 8 \right) + 10\left( 6 \right)\]
$ \Rightarrow 800 + 60$
$ \Rightarrow 860$
Therefore the original number is 860.
Note: Any digit can be expressed in the expanded form, therefore the digit $860$ can also be expressed in expanded form as:
$
860 = 800 + 60 + 0 \\
= 8 \times 100 + 6 \times 10 + 0 \\
$
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