Question

A 10 hectare field is reaped by 2 meb 3 women 4 children together in 10 days if working capablities of amen a women and a child are in the ratio 5:4:2 then 16 hectare field will be reaped by 6 men 4 women and 7 children in

Answer 1

workers)*days/work1= workers2*days/work2

here workers are 2*5+3*4+4*2=30
similarly second workers = 6*5+4*4+7*2=60
and work here is 10 hectare and 16 hectare ,
so now put the values
30*10/10=60*x/16

x=8

Answer 2

$\sf\small\underline\red{Given:-}$

$\tt{\leadsto (2m+3w+4c)=10days}$

$\tt{\leadsto (6m+4w+7c)=?days}$

$\sf{\leadsto Field\:reaped\:_{(2m+3w+4c)}=10hectare}$

$\sf{\leadsto Field\:reaped\:_{(6m+4w+7c)}=16hectare}$

$\sf{\leadsto Ratio\:_{(for\:both)}=5:4:2}$

$\sf\small\underline\blue{To\:Find:-}$

$\tt{\leadsto Work\:done\:_{(6m+4w+7c)}=?days}$

$\sf\small\underline\pink{Solution:-}$

• To solve this question we have to use direct or indirect variations to compare between two conditions. As per the given clue .

$\sf\small\underline\green{Calculation\: begin:-}$

As per the first case - ( i ) :-

$\tt{\leadsto \dfrac{[2m(5) + 3w(4)+4c(2)]×days}{field\:reaped}}$

$\tt{\leadsto \dfrac{(10+12+8)×10}{10}}$

$\tt{\leadsto \dfrac{300}{10}}$

As per the 2nd case - ( ii ) :-

$\tt{\leadsto \dfrac{[6m(5) + 4w(4)+7c(2)]×days}{field\:reaped}}$

$\tt{\leadsto \dfrac{(30+16+14)×D}{16}}$

$\tt{\leadsto \dfrac{60×D}{16}}$

Comparing with case (I) and (ii) :-

$\tt{\leadsto Case-(i)=Case-(ii)}$

$\tt{\leadsto \dfrac{300}{10}=\dfrac{60D}{16}}$

$\tt{\leadsto 60D×10=300×16}$

$\tt{\leadsto 600D=4800}$

$\tt{\leadsto 6D=48}$

$\tt{\leadsto D=8}$

$\sf\large{Hence,}$

$\tt{\leadsto Work\:done\:_{(6m+4w+7c)}=8\:days}$