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Ragav Ramakrishnan

· started a discussion

· 1 Months ago

?? (Question: 46123.0 )

sum of the digits be..6+9=15??

Question:
Let d be a two digit number. If half of d exceeds one third of d by the sum of the digits of d, then which of the following can be the sum of the digits of d?
Options:
A) 6
B) 8
C) 9
D) 15
E) None of These.
Solution:
Units place be x and tens place be y

\(\therefore \) Number = \(10y+x\)

\(\cfrac{10y+x}{2}-\cfrac{10y+x}{3}=x+y \)

\(\Rightarrow\cfrac{30y+3x-20y-2x}{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,=x+y\)

\(\Rightarrow 10y+x=6(x+y)\)

If \(x+y=9\)

Number= \(6\times9=54\)

Therefore the sum of digits = 9

kanhaiya jha

· commented

· 1 Months ago

Required Number= 54

So, sum of their digits = 5+4 = 9
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