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Q.

A man takes 20 days to reach the point $B$ under normal circumstances. But, due to the increasingly hostile weather conditions the distance they travel every day reduces by $20%$. In how many days would the man reach the point $B$, taking into consideration weather conditions?

 A.

25

 B.

50

 C.

100

 D.

None of these

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Solution:
Option(D) is correct

Let the total distance to be covered be $d$

On each day under normal weather conditions they travel \(\dfrac{d}{20}\)of the distance.

On $1^{st}$ day they would travel\(=\dfrac{d}{20}\)

On $2^{nd}$ day they would travel \(=0.8\times \dfrac{d}{20}\)

On $3^{rd}$ day they would travel\(=0.8\times \dfrac{0.8\times d}{20}\)

Let he will reach the point $B$ in $n^{th}$ day

\(\Rightarrow \dfrac{d}{20}+0.8\dfrac{d}{20}+0.8^2\dfrac{d}{20}+...0.8^n\dfrac{d}{20}=d\)

\(\Rightarrow \dfrac{d}{20}\times \dfrac{-(0.8)^n+1}{-0.8+1}=d\)

\(\Rightarrow 1-(0.8)n\)

\(=4(0.8)^n\)

\(=-3\)

Since $(0.8)6 n>0$, thus it is never equal to -3

The man will never reach the point B.


(1) Comment(s)


Swathi
 ()

under normal condition assume he walks 100km per day therfore total distance is 100*20days = 2000km

but under bad weather he walks

day1=100km

day2=100-20=80km

day3= 80-20=60km

day4=60-20=40km

day5=40-20=20km

day6=20-20=0km that he doesnt walk atall so he can never reach point b