Question

Rahim sets out to cross a forest. On the first day, he completes 1/10th of the journey. On the second day, he covers 2/3rd of the distance travelled the first day. He continues in this manner, alternating the days in which he travels 1/10th of the distance still to be covered, with days on which he travels 2/3 of the total distance already covered. At the end of seventh day, he finds that 22 ½ km more will see the end of his journey. How wide is the forest?

Answer 1

looks hard >.<

Answer 2

1/10->first day
2/3*1/10->second day
so1/10+1/15=1/6

Answer 3

1/6-1=5/6 remaining

Answer 4

5/6*1/10=1/12,1/12-5/6=3/4-->3rd day

Answer 5

3/4*2/3=1/2,1/2-3/4=1/4 -->4th day

Answer 6

1/4*1/10=1/40,1/4-1/40=9/40-->5th day

Answer 7

9/40*2/3=3/20,3/20-9/40=3/40-->6th day

Answer 8

3/40*1/10=3/400,3/400-3/40=27/400-->7th day

Answer 9

something went wrong?

Answer 10

remaining should be calculated carefully.always make mistake

Answer 11

@ganeshie8 u know very well to calculate remaining

Answer 12

if remaining is correct then we get the answer i think

Answer 13

i think 4th day is 2/3*1/12=1/18,1/18-3/4=2/3

Answer 14

im getting 120km

Answer 15

total width of forest = 120km

Answer 16

yes u are right!:)

Answer 17

huh....

Answer 18

u take a long time to reply

Answer 19

say the forest is \(\large x\) km :

\[
\begin{array}{c|c|c}
\text{day}&\text{distance covered}& \text{total distance covered} & \text{distance remaining} \\
\hline \\
1&\color{red}{\dfrac{1}{10}(x)}&\dfrac{1}{10}(x) &\dfrac{9}{10}(x) \\
\hline \\
2&\color{red}{\dfrac{2}{3} . \dfrac{1}{10}(x)}& \dfrac{1}{6}(x) &\dfrac{5}{6}(x) \\
\hline \\


\end{array}
\]

Answer 20

lets fill that table ^^

Answer 21

o-o its not loaded fully

Answer 22

Oh must be a poblem with latex, il take a screenshot and attach

Answer 23

http://prntscr.com/4dcv5k

Answer 24

yes i am done that upto 2nd day earlier

Answer 25

3rd day balance=59/60 ahhh

Answer 26

13/15 ahh

Answer 27

say the forest is \(\large x\) km :

\[
\begin{array}{c}
\text{day}&\text{distance covered}& \text{total distance covered} & \text{distance remaining} \\
\hline \\
1&\color{red}{\dfrac{1}{10}(x)}&\dfrac{1}{10}(x) &\dfrac{9}{10}(x) \\
\hline \\
2&\color{red}{\dfrac{2}{3} . \dfrac{1}{10}(x)}& \dfrac{1}{6}(x) &\dfrac{5}{6}(x) \\
\hline \\
3&\color{red}{\dfrac{1}{10} . \dfrac{5}{6}(x)}& \dfrac{3}{12}(x) &\dfrac{9}{12}(x) \\
\hline \\
4&\color{red}{\dfrac{2}{3} . \dfrac{3}{12}(x)}& \dfrac{5}{12}(x) &\dfrac{7}{12}(x) \\
\hline \\

\end{array}
\]

Answer 28

first four days ^^

Answer 29

here is the screenshot
http://prntscr.com/4dcy9j

Answer 30

same i have done ya instead of 9/12 i made a mistake of 3/4 .then it will be correct .see at the top

Answer 31

5/6*1/10=1/12,1/12-5/6=3/4-->3rd day

Answer 32

9/12 is same as 3/4 :)

Answer 33

how u got 5/12 in total distance column

Answer 34

3/4*2/3=1/2,1/2-3/4=1/4 -->4th day

Answer 35

oh i went wrong

Answer 36

oh i multiplied the balance into 2/3 instead 3/12 of 3rd day

Answer 37

@ganeshie8 can you please help me on my questions i tagged you

Answer 38

5 1/10*7/12 63/120 57/120

Answer 39

is it right? for 5th day

Answer 40

fast @ganeshie8 .always i make mistake in calculation........

Answer 41

15/80x=45/2

Answer 42

\[
\begin{array}{c}
\text{day}&\text{distance covered}& \text{total distance covered} & \text{distance remaining} \\
\hline \\
1&\color{red}{\dfrac{1}{10}(x)}&\dfrac{1}{10}(x) &\dfrac{9}{10}(x) \\
\hline \\
2&\color{red}{\dfrac{2}{3} . \dfrac{1}{10}(x)}& \dfrac{1}{6}(x) &\dfrac{5}{6}(x) \\
\hline \\
3&\color{red}{\dfrac{1}{10} . \dfrac{5}{6}(x)}& \boxed{\dfrac{3}{12}(x)} &\dfrac{9}{12}(x) \\
\hline \\
4& \boxed{\color{red}{\dfrac{2}{3} . \dfrac{3}{12}(x)}} & \dfrac{5}{12}(x) &\dfrac{7}{12}(x) \\
\hline \\
5& \boxed{\color{red}{\dfrac{1}{10} . \dfrac{7}{12}(x)}} & \dfrac{57}{120}(x) &\dfrac{63}{120}(x) \\
\hline \\

6& \boxed{\color{red}{\dfrac{2}{3} . \dfrac{57}{120}(x)}} & \dfrac{95}{120}(x) &\dfrac{25}{120}(x) \\
\hline \\
7& \boxed{\color{red}{\dfrac{1}{10} . \dfrac{25}{120}(x)}} & \dfrac{975}{1200}(x) &\dfrac{225}{1200}(x) \\
\end{array}

\]

Answer 43

you're correct :)
http://prntscr.com/4dd48s

Answer 44

thank you :)