Question

Q-14 and Q-16??How To Solve??

Answer 1

sorry I am weak in trigono but can solve 16

Q16
Let days taken by men be 'x'
And let the Days taken by the boys be 'y'

so the equation will be like this
we know that speed=Dist/Time
Here we have to find out speed of each one
and assume distance as no. of men and no. of boys
so the equation would be like this
4/x+6/y=1/5
3/x+4/y=1/7

Let 1/x be a and 1/y be b

so 4a+6b=1/5
3a+4b=1/7

5(4a+6b)=1
20a+30b=1
b=1-20a/30

3a+4(1-20a)/30=1/7
4-100a=(1/7-3a)30
4-100a=30/7-90
4-100a=30-630
-100a=-600-4
a=-604/-100
a=6