Mensa Minds Problem

[lantheman]

On an exercise, a soldier marches for four days. On the first dya, he covers 25% of the journey. On the second day, he covers one-third of hte remaining distance, on the third, 25% of the remaining journey and, on the final day he covers half of the remaining distance. he still has 15 miles left to travel. How far has he walked in total?

Here is what I did:
15x2=30
30x4=120
120x3=360
360x4=1440

Total miles traveled = 1440 miles

This seems like too much

Where did I go wrong?

 

[lantheman]

Please help me with this!

Thanks,

 

[lookagain]

On an exercise, a soldier marches for four days. On the first dya, he covers 25% of the journey. On the second day, he covers one-third of hte remaining distance, on the third, 25% of the remaining journey and, on the final day he covers half of the remaining distance. he still has 15 miles left to travel. How far has he walked in total?

Here is what I did:
15x2=30
30x4=120
120x3=360
360x4=1440

Total miles traveled = 1440 miles

This seems like too much

Where did I go wrong?

The fractions below represent \(\displaystyle \ 1 \ minus \ the \ fraction \\)(or decimal) given in the problem.


\(\displaystyle 15 \div \frac{1}{2} = 30\)

\(\displaystyle 30 \div \frac{3}{4} = 40\)

\(\displaystyle 40 \div \frac{2}{3} = 60\)

\(\displaystyle 60 \div \frac{3}{4} = 80\)


Then \(\displaystyle 80\) miles is the length of the entire journey.

\(\displaystyle 80\) miles, less the \(\displaystyle 15\) miles still to walk, leaves

\(\displaystyle 65\) miles that he has walked (marched) in total.

-----------------------------------------------------------------------------------------------

\(\displaystyle * * *\) This is not how I worked this problem. I used algebra instead,
and I understood what was happening in each step.
I don't have a total grasp on this method.

I can post the algebraic solution I used if someone wants me to.

 

[soroban]

Hello, lantheman!

Here's my algebraic approach . . . with lots of baby-talk.

On an exercise, a soldier marches for four days.
On the first day, he covers 25% of the journey.
On the second day, he covers one-third of hte remaining distance.
On the third, 25% of the remaining journey
On the final day, he covers half of the remaining distance.
He still has 15 miles left to travel.
How far has he walked in total?

Let \(\displaystyle x\) = the distance of the entire journey (in miles).

\(\displaystyle \text{Day 1: He covers }\tfrac{1}{4}x\text{ miles.}\)
. . . . . \(\displaystyle \text{He has }\tfrac{3}{4}x\text{ miles remaining.}\)

\(\displaystyle \text{Day 2: He covers }\tfrac{1}{3}\text{ of }\tfrac{3}{4}x \:=\:\tfrac{1}{4}x\text{ miles.}\)
. . . . . \(\displaystyle \text{He has covered a total of: }\,\tfrac{1}{4}x + \tfrac{1}{4}x \:=\:\tfrac{1}{2}x\text{ miles.}\)
. . . . . \(\displaystyle \text{He has }\tfrac{1}{2}x\text{ miles remaining.}\)

\(\displaystyle \text{Day 3: He covers }\tfrac{1}{4}\text{ of }\tfrac{1}{2}x \:=\:\tfrac{1}{8}x\text{ miles.}\)
. . . . . \(\displaystyle \text{He has covered a total of: }\,\tfrac{1}{2}x + \tfrac{1}{8}x \:=\:\tfrac{5}{8}x\text{ miles.}\)
. . . . . \(\displaystyle \text{He has }\tfrac{3}{8}x\text{ miles remaining.}\)

\(\displaystyle \text{Day 4: He covers }\tfrac{1}{2}\text{ of }\tfrac{3}{8}x \:=\:\tfrac{3}{16}x\text{ miles.}\)
. . . . . \(\displaystyle \text{He has covered a total of: }\,\tfrac{5}{8}x + \tfrac{3}{16}x \:=\:\tfrac{13}{16}x\text{ miles.}\)
. . . . . \(\displaystyle \text{He has }\tfrac{3}{16}x\text{ miles remaining.}\)


\(\displaystyle \text{But we are told that, at this point, he has 15 miles remaining.}\)

\(\displaystyle \text{Hence, we have the equation: }\:\tfrac{3}{16}x \:=\:15 \;\;\;\Rightarrow\;\;\;x \:=\:80\)

\(\displaystyle \text{That is, the entire exercise was an 80-mile hike.}\)


\(\displaystyle \text{During the four days, he walked: }\:80-15 \:=\:65\text{ miles.}\)

 

[Denis]

\(\displaystyle \text{Day 1: He covers }\tfrac{1}{4}x\text{ miles.}\)
. . . . . \(\displaystyle \text{He has }\tfrac{3}{4}x\text{ miles remaining.}\)

\(\displaystyle \text{Day 2: He covers }\tfrac{1}{3}\text{ of }\tfrac{3}{4}x \:=\:\tfrac{1}{4}x\text{ miles.}\)
. . . . . \(\displaystyle \text{He has covered a total of: }\,\tfrac{1}{4}x + \tfrac{1}{4}x \:=\:\tfrac{1}{2}x\text{ miles.}\)
. . . . . \(\displaystyle \text{He has }\tfrac{1}{2}x\text{ miles remaining.}\)

\(\displaystyle \text{Day 3: He covers }\tfrac{1}{4}\text{ of }\tfrac{1}{2}x \:=\:\tfrac{1}{8}x\text{ miles.}\)
. . . . . \(\displaystyle \text{He has covered a total of: }\,\tfrac{1}{2}x + \tfrac{1}{8}x \:=\:\tfrac{5}{8}x\text{ miles.}\)
. . . . . \(\displaystyle \text{He has }\tfrac{3}{8}x\text{ miles remaining.}\)

...or you can stop after Day3, since we know 30 miles are left:
3x/8 = 30
x = 80