The two digit number \[{\text{x}}\] is the digit at unit place and \[{\text{y}}\] is the digit at unit place, then the number is :
\[
{\text{A}}.10y + x \\
{\text{B}}.10x + y \\
{\text{C}}.10y - x \\
{\text{D}}.10 - y \
\]
Answer
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Hint – Use the concept of ones and tens and make an equation, take an example of any two digit number.
It is a two digit number \[{\text{x}}\] is the digit at unit place and \[{\text{y}}\] is the digit at tens place .
Let us take an example of a two digit number \[51\] so here \[\,1\] is at the unit place and \[{\text{5}}\] is digit at the tense place . So here,
$x = 1\,\,\& \,\,y = 5$ ……(i)
We can write \[51\] as
\[51 = 50 + 1 = 10(5) + 1\]
We can write the above equation as \[10y + x\] ……(From (i))
Hence the correct option is A.
Note – In these types of questions we must have to know that, just left of decimal point is one place and left of one's place is a tens place. We should know that when we write the numbers of all these places in addition/subtraction then these numbers get multiplied by the number of their respective names, then aon adding/subtracting we will get the number .
It is a two digit number \[{\text{x}}\] is the digit at unit place and \[{\text{y}}\] is the digit at tens place .
Let us take an example of a two digit number \[51\] so here \[\,1\] is at the unit place and \[{\text{5}}\] is digit at the tense place . So here,
$x = 1\,\,\& \,\,y = 5$ ……(i)
We can write \[51\] as
\[51 = 50 + 1 = 10(5) + 1\]
We can write the above equation as \[10y + x\] ……(From (i))
Hence the correct option is A.
Note – In these types of questions we must have to know that, just left of decimal point is one place and left of one's place is a tens place. We should know that when we write the numbers of all these places in addition/subtraction then these numbers get multiplied by the number of their respective names, then aon adding/subtracting we will get the number .
Last updated date: 22nd Sep 2023
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