Math, asked by kinjal2003, 9 months ago

pls need the answer urgent​

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Answered by Kidunyoung
0

Answer:

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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