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Question Number 35088 by Raj Singh last updated on 15/May/18

\boldsymbol a \boldsymbol s o n \boldsymbol a n d \boldsymbol f a t h e r \boldsymbol d o \boldsymbol a \boldsymbol w o r k \boldsymbol i n

24 \boldsymbol d a y . \boldsymbol i f \boldsymbol b o t h \boldsymbol w o r k \boldsymbol t o g e t h e r

\boldsymbol a n d \boldsymbol f a t h e r \boldsymbol w o r k \boldsymbol l a s t 6 \boldsymbol d a y \boldsymbol o n l y

\boldsymbol t h e n \boldsymbol t h e y \boldsymbol h o w \boldsymbol m u c h \boldsymbol d o

Commented by Rasheed.Sindhi last updated on 15/May/18

Data seems incomplete .

Commented by tanmay.chaudhury50@gmail.com last updated on 15/May/18

l e t f a t h e r d o \frac{1}{f} p o r t i o n i n a d a y

s o n d o \frac{1}{s} p o r t i o n i n a d a y

s o 24 (\frac{1}{f} + \frac{1}{s}) = 1

\frac{1}{f} + \frac{1}{s} = \frac{1}{24}

i n l a s t 6 d a y f a t h e r d o \frac{6}{f} p o r t i o n

r e m a i n i n g w o r k (1 - \frac{6}{f})

\frac{6}{f} + \frac{6}{s} = \frac{6}{24}

\frac{6}{f} = \frac{1}{4} - \frac{6}{s}

r e m a i n i n g w o r k = (1 - \frac{6}{f})

= (1 - \frac{1}{4} + \frac{6}{s})

= \frac{3}{4} + \frac{6}{s}

s o r e m a i n i n g w o r k > \frac{3}{4}