The sum of the digits of the two digit no. Is 5. If the digits are reversed,the no is reduced by 27. Find the no.
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Let digit at unit place be x and at tens be y so no formed is 10y+x . When reversed digit at unit place is y so no formed is 10x +y ATQ 10y+x-10x-y =27 9y-9x=27 so y-x=3 also x+y=5 on adding y=4 so x=1 so no is 41
Step-by-step explanation:
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Let unit digit be x
then the tens digit is 5-x
hence the no. is 10(5-x)+x
on reversing the digits the new number=
10x+(5-x)
A/Q
⇒10(5-x)+x=10x+5-x+27
⇒50-10x+x=10x+5-x+27
⇒50-9x=9x+32
⇒50-32=9x+9x
⇒18=18x
⇒1=x
hence units digit=x=1
and tens digit=5-x=5-1=4
hence the no.=10(5-x)+x=10(5-1)+1=50-10+1=41
then the tens digit is 5-x
hence the no. is 10(5-x)+x
on reversing the digits the new number=
10x+(5-x)
A/Q
⇒10(5-x)+x=10x+5-x+27
⇒50-10x+x=10x+5-x+27
⇒50-9x=9x+32
⇒50-32=9x+9x
⇒18=18x
⇒1=x
hence units digit=x=1
and tens digit=5-x=5-1=4
hence the no.=10(5-x)+x=10(5-1)+1=50-10+1=41
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