write a two digit number,the sum of two digits of which is 14 and if 29 is substracted from the number,the two digits will be equal.Let us form the simultaneous equation by solving them and let us see what will be the 2 digit number.
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Step-by-step explanation:
Given:–
In a two digit number,the sum of two digits of which is 14 and if 29 is substracted from the number,the two digits will be equal.
To find:–
What is the two digit number?
Solution:–
Let the digit at 10s place in the two digits number = X
The place value of X = 10× X = 10X
Let the digit at ones place in the two digits number = Y
The place value of Y = Y×1 = Y
Then , The two digits number = 10X+Y
Given that .
Condition –1:
The sum of two digits = 14
X+Y = 14 ————(1)
Condition –2:–
If 29 is subtracted from the number then the two digits will be equal
=> (10X+Y)-29 =10Z+Z (Z is some digit)
=>10X+Y-29 = 11Z
=>10X +Y = 11Z+29——–(2)
On Subtracting (1) from (2) then
10X +Y = 11Z +29
X+Y = 14
(-)
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9X +0 = 11Z +29-14
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=>9X = 11Z+29-14
=>9X = 11Z +15
=>X = (11Z+15)/9
If Z = 1 then X = (11+15)/9 = 26/9
The digit can not be a fraction
If Z = 2 then X = (22+15)/9 = 37/9
If we continue like this ..
If Z = 6 then X = (66+15)/9 = 81/9 = 9
So we get a natural number
Therefore, X = 9
On Substituting the value of X in (1) then
=>9+Y = 14
=>Y = 14-9
=>Y = 5
X = 9 and Y = 5
Answer:–
The required two digit number is 95
Check:–
Sum of the digits = 9+5 = 14
If 29 subtracted from it = 95-29 = 66
The both digits are equal.
Verified the given relations
Answer:
[tex]\huge\ \sf{\red {[[«\: คꈤ \mathfrak Sฬєя \: » ]]}}[/tex]
The required two digit number is 95
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[tex]\sf \purple{Given:}[/tex]
[tex]\sf In \: a \: two \:digit \:number, \:the \: sum \: of \: two \: digits [/tex] [tex]\sf of \:which \:is \:14 \:and \: if \:29 \:is \:substracted[/tex] [tex]\sf \:from \: the \:number, \:the \: two \:digits \: will \: be \: equal.[/tex]
[tex]\sf\orange{To \:find:}[/tex]
[tex]\sf the \: two \:digit \: number[/tex]
[tex]\huge{\underline{\mathtt{\red{S}\pink{O}\green{L}\blue{U}\purple{T}\orange{I}\red{O}{N}}}}[/tex]
[tex]\sf Let \:the \:digit \: at \: 10s \: place \: in \:the \: two \: digits \: number = X[/tex]
[tex]\sf The \: place \:value \:of \:X = 10 \:× \:X = 10X[/tex]
[tex]\sf Let \:the \:digit \:at \:ones \: place \: in \: the \: two \: digits \: number = Y[/tex]
[tex]\sf The \:place \: value \:of Y = Y×1 = Y[/tex]
[tex]\sf Then , \: The \: two \:digits \: number = 10X+Y[/tex]
[tex]\sf\blue{Given \:that:}[/tex]
[tex]\sf Case -1:[/tex]
[tex]\sf The \:sum \:of \: two \: digits = 14[/tex]
[tex]\sf X+Y = 14 ————(1)[/tex]
[tex]\sf Case -2:-[/tex]
[tex]\sf If \:29 \:is \: subtracted \:from \:the \:number \: then \: the \: two \:digits \:will \:be \: equal[/tex]
[tex]\sf ⟹ (10X+Y)-29 =10D+D (D \:is \: some \: digit)[/tex]
[tex]\sf ⟹ 10X+Y-29 = 11D[/tex]
[tex]\sf ⟹ 10X +Y = 11D+29——–(2) [/tex]
[tex]\sf By \:Subtracting \: (1) \: from \:(2) \:we \: get [/tex]
[tex]\sf 10X +Y = 11D +29[/tex]
[tex]\sf X+Y = 14[/tex]
(-)
_______________
[tex]\sf 9X +0 = 11D +29-14[/tex]
________________
[tex]\sf ⟹ 9X = 11D+29-14[/tex]
[tex]\sf ⟹ 9X = 11D +15[/tex]
⟹ [tex]\ X = \dfrac{11D+15}{9}[/tex]
If D = 1 then [tex]\ X= {\dfrac{11+15}{9} = \dfrac{26}{9}}[/tex]
[tex]\sf The \:digit \: can \:not \:be \: a \: fraction [/tex]
If D = 2 then [tex]\ X= {\dfrac{22+15}{9} = \dfrac{37}{9}}[/tex]
[tex]\sf If \: we \: continue \: like \: this …[/tex]
If D = 6 then [tex]\ X= {\dfrac{66+15}{9} = \dfrac{81}{9}}= 9[/tex]
[tex]\sf So \: we \:get \: a \: natural \: number [/tex]
[tex]\sf Therefore, \: X = 9[/tex]
[tex]\sf putting \: the \:value \: of \: X \: in \: (1) ,we \: get[/tex]
[tex]\sf ⟹ 9+Y = 14[/tex]
[tex]\sf ⟹ Y = 14-9[/tex]
[tex]\sf ⟹Y = 5[/tex]
[tex]\sf X = 9 \:and Y = 5[/tex]
[tex]\sf The \: required \:two \:digit \:number \:is \:95[/tex]