A two digit number is five times the sum of its digits. If 9

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RRB TC CC Exam (Bangalore) 2007 click here to view all questions

A two digit number is five times the sum of its digits. If 9 is added to the number, the digits interchange. Find the sum of the digits.

A11

B7

C6

D9

quiaz ? Dec 20 '2018 at 14:16

Let a and b the digits we can write the number as "ab" but the number is \(10 \times a+b\)
(for example if a=3 and b=2 the number would be 32 and \(32 =10 \times 3+2\)) 

a two digit number is 5 times the sum of its digit then \(10 \times a+b=5(a+b)\)
when 9 is added to the number the result is the original number with its reversed \(10 \times a+b+9=10 \times b+a \)
then we have the following equations: 
\(10 \times a+b=5(a+b)\)
\(10 \times a+b+9=10 \times b+a \)
then 
\(10a+b=5a+5b\)
\(10a+b+9=10b+a \)
then 
\(5a-4b=0\)

 
\(9a-9b=-9 \)
then 
\(5a-4b=0\)
\(a-b=-1\)
then using 2nd eq a=b-1 
using 1st eq \(5(b-1)-4b=0 \)
then \(5b-5-4b=0\)
then b=5 and a=5-1=4 
the sum of digits is 9.

All DiscussionsClick here to write answer

Let a and b the digits we can write the number as "ab" but the number is \(10 \times a+b\) (for example if a=3 and b=2 the number would be 32 and \(32 =10 \times 3+2\))  a two digit number is 5 times the sum of its digit then \(10 \times a+b=5(a+b)\) when 9 is added to the number the result is the original number with its reversed \(10 \times a+b+9=10 \times b+a \) then we have the following equations:  \(10 \times a+b=5(a+b)\) \(10 \times a+b+9=10 \times b+a \) then  \(10a+b=5a+5b\) \(10a+b+9=10b+a \) then  \(5a-4b=0\)   \(9a-9b=-9 \) then  \(5a-4b=0\) \(a-b=-1\) then using 2nd eq a=b-1  using 1st eq \(5(b-1)-4b=0 \) then \(5b-5-4b=0\) then b=5 and a=5-1=4  the sum of digits is 9. quiaz ? answered Dec 20 '2018 at 14:16 All Comments

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