Each stroke of a hammer inserts 30% of the nail, out of the wood, into

Question

Each stroke of a hammer inserts 30% of the nail, out of the wood, into the wood. How many strokes are needed to push 80% of a nail into the wood?

A. 3
B. 4
C. 5
D. 6
E. 7

gmatophobia · Answer

Let the length of the nail = 1 unit

With each strike, 30% of the length gets into the wood. We're asked to find the number of strikes required to push 80% of the nail.

If 80% of the nail is pushed, 20% of the nail will remain.

0.2 = 1  ^n

0.2 =  ^n

n = 2 ; value = 0.49
n = 3 ; value = 0.343
n = 4 ; value = 0.2401
n = 5 ; value = 0.178

Option C

rickyric395 · Answer

According to me answer should be 4.
Let the length be L.
1st try , inside wood length = 0.3L , outside wood length = 0.7L.
2nd try, inside wood length = 0.3(0.7L)   , outside wood length = 0.7*0.7L.
3rd try, inside wood length = 0.3*0.7*(0.7L)   , outside wood length = 0.7*0.7*0.7L.

We can see inside wood length can be written as  0.3*(0.7)^{n-1}  , n is number of attempt.
We want inside length to be 0.8L.
Equating  0.3*(0.7)^{n-1}L =0.8L
We get  n\approx{3.9225}  i.e 4

Bunuel  can you please take a look at it.

Regor60 · Answer

After the 2nd hit, the inside length is not .3*.7. That is the contribution of the 2nd hit to the existing inside length of .3. This mistake is carried through in your example.

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rickyric395 · Answer

Hey  Regor60  , thanks for pointing out, with inside wood length phrase I used, I meant on nth try the subsequent portion will be shoved into the wood. So in totality the portion inside wood would be  0.3L+0.3*0.7L+0.3*(0.7)^{2}L....... = 0.8L

CatNinja7 · Answer

You've made a mistake in the red highlighted portion.

You should treat your series like a GP (Geometric Progression). Each consecutive stroke ADDS to the length inside, whereas you have equated just the final term in the series to 0.8L.

My modified approach:
a = 0.3 (first term)
r = 0.7 (the multiplier)
Sum of a GP = a(1-r^n)/(1-r) (n = number of terms = number of strokes)
=> 0.3*(1- (0.7)^n)/(1-0.7) = 1- (0.7)^n

This is GREATER THAN OR EQUAL TO 0.8, as more than 80% should be inside
=>1- (0.7)^n >= 0.8
=> (0.7)^n <= 0.2
This is true for n=5

Hence  C

RenB · Answer

Hi  gmatophobia 
'If 80% of the nail is pushed, 20% of the nail will remain.'
Here, how do we realise that we have to consider the nail that remains and not the nail that is pushed?
I tried solving considering the nail that is pushed and got stuck

gmatophobia · Answer

Hey RenB We can do the other way around as well. However the math gets a bit tedious. Assume that the length of the nail is 100 cm. We need at least 80 cm of the nail into the wood. Here is the working of the same. https://gmatclub.com/forum/download/file.php?id=82974 Hope this helps. Happy to answer any questions, if not. GMAT-Club-Forum-oyc5jwih.png

Bunuel · Answer

After solving above problem, check similar questions to practice: https://gmatclub.com/forum/if-1-2-of-th ... 66101.html (Official Guide) https://gmatclub.com/forum/if-2-3-of-th ... 28432.html https://gmatclub.com/forum/if-3-4-of-th ... 97300.html https://gmatclub.com/forum/on-january-1 ... 82667.html https://gmatclub.com/forum/after-instal ... 43260.html https://gmatclub.com/forum/if-1-2-of-th ... 65131.html https://gmatclub.com/forum/each-stroke- ... 06881.html Hope it helps.

m7runner · Answer

Probably the most crisp way to get this done - thanks  gmatophobia ! However, if your brain isn't braining on d-day, probably another way to get this done (quick and dirty though, so you'll have to keep your rough work on the scratch pad/whiteboard pretty clear).

→ Assume the length of the nail is 100 units.
→ 1st time: When you hammer it in, 30 units of the nail is in the wood; 70 units is outside the wood.
→ 2nd time: When you hammer again, (30/100)*70 = 21 units of the wood is inside OUT OF 70 units. That means, 70-21 = 49 units is outside the wood.
→ 3rd time: When you hammer again, (30/100)*49 = 14.7 units of the wood is inside OUT of 49 units. Therefore, 49-14.7 = 34.3 units is outside the wood.
→ 4th time: When you hammer again, (30/100)*34.3 = 10.29 units of the wood is inside OUT OF 34.3 units. Therefore, 34.3-10.29 = 24.01 units is outside the wood.

Let's take a deep breath and do a recap: Out of 100 units, 24.01 units is outside the wood on the 4th hammer ( TIME!  - sorry, Lewis Hamilton reference). So when the hammer hit-eth the 5th time, surely 80 units of the nail is inside the wood. Hence, option C is the correct answer choice.

Each stroke of a hammer inserts 30% of the nail, out of the wood, into

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