the sum of the digits of two digit number is 9 if the digits are reversed the new number demonized by 9 equals 4 times of original number find the original number.
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the sum of the digits of two digit number is 9 if the digits are reversed the new number demonized by 9 equals 4 times of original number find the original number.
please give answer first
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LetTen's place digit = x
units place digit = (9-x)
the original number = 10x + (9 - x )
=10x + 9 - x
=9x + 9 ------(1)
If digits are reversed the
new number = 10(9 - x ) + x
= 90 - 10x + x
= 90 - 9x ------(2)
relation given between (1) and (2) is
(2) - 9 = 4 × (1)
90 - 9x - 9 = 4 ( 9x + 9 )
81 - 9x = 36x + 36
divide each term with 9 we get
9 - x = 4x + 4
9 - 4 = 4x + x
5 = 5x
therefore
x = 1
put x=1 in (1)
Required number = 9x + 9
= 9 × 1 + 9
= 9 + 9
= 18
Hope this will helps you.
Answer:
let the numerator be x
so, the denomination is x+2
fraction formed = x/x+2
according to the question:
=>x/x+2+3=1/2
=> x/5 = 1/2
transposing 5 to rhs
=> x=5/2
hence, the rational no. is 5/2