pls need the answer urgent
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Step-by-step explanation:
CASE I
He walks 4km/h, then he is 20m late.
Let us assume the time as 't¹'
S=D/t¹
4=D/t¹
D=4t¹
t¹=D/4---------equation 1
CASE II
s=5km/h
Time=t²
Distance is same
S=d/t²
t²=d/5------------equation 2
Let the perfect time or required time be T
t¹=T+20( twenty minutes late)
T=t¹-20----eq3
t²=T-10(ten minutes early)
T=t²+10-----eq4
Now from eq 3 and 4
t²+10=t¹-20
t²=t¹-30
Now by substituting values from equation 1 and 4 we have=
D/4-30=D/5
D/4-D/5=30
5D-4D/20=30
D=600
Edit- i have made some minor mistakes by not converting hour into minute but the answer process will be tge same
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