A ball dropped from 36 m above the ground rebounds to 1/3rd of the

Question

A ball dropped from 36 m above the ground rebounds to 1/3rd of the height it falls from. If it continues to rebound in this manner, find the total distance the ball can cover?

A. 96
B. 72
C. 54
D. 76
E. 75

GMATinsight · Accepted Answer

Total distance covered by ball  = 36 + 2*(1/3)*36+ 2*(1/3)^2*36+....= 2*36 + 2*(1/3)*36+ 2*(1/3)^2*36+....-36

i.e. Total Distance  = 2* -36

This is the geometric progression with first term, a = 36 and common ratio, r = 1/3

Sum of a Geometric progression with first term a and common ratio r such that  -1≤r≤1 = \frac{a}{(1-r)}

i.e. sum of the series  = \frac{2*36}{(1-1/3)} - 36 = \frac{2*36}{(2/3)} - 36 = 108 - 36 = 72

Answer: Option B

nick1816 · Answer

Distance travelled before 1st rebound = 36

Distance travelled by ball from 1st rebound to 2nd rebound=  2*(\frac{36}{3}) = 24

Distance travelled by ball from 2nd rebound to 3rd rebound = 2*(\frac{12}{3}) = 8

Distance travelled by ball from 3rd rebound to 4th rebound=  2*(\frac{4}{3}) =\frac{ 8}{3}

and so on

Sum of infinite GP = a/1-r. where a is first term and r is common ratio

a=24 and r =1/3

Total distance travelled =  36+    =  36 +    = 72

HarshavardhanR · Answer

1) The ball's first drop is from a height of 36m.
2) The ball now rebounds to 1/3 of the above height (1/3 of 36m = 12m)
3) The ball will again fall from this point (12m)
4) The ball will again rebound to a height of 1/3 x 12m.
5) And so on.

- Distance travelled in 1st fall = 36m
- Distance traveled between 1st fall and 2nd fall (first rebound plus subsequent fall) = (1/3 x 36) + (1/3 x 36)m
- And so on.

So, Overall distance covered = 36 + (2)(36)(1/3) + (2)(36)(1/3^2) + (2)(36)(1/3^3) + .....
=   - 36  (Adding an subtracting 36 to create a proper Geometric Progression) 
=   - 36  (Sum of infinite gp with first term a (2x36) and common ratio r (1/3), when |r| < 1, = a/(1-r)) 
= 108 - 36 = 72

Choice B (72) is the answer.

___
Harsha
Enthu about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com

pratiksha1998 · Answer

Hi,

I wanted to understand what is incorrect in the below approach,

First term of the series is 36, since the ball has a rebound of 1/3rd from the top, from the bottom it will be 2/3, which will be the common ratio.When we apply the formula   \frac{a}{1-r}  , the answer comes out as 108.

Bunuel · Answer

The sequence to sum is:

36 + 24 + 8 + 8/3 + ...

36 is not the first term of the geometric progression. The geometric progression starts from the second term onwards. The ratio of 24 to 36 is 2/3, while the ratio of 8 to 24 and all subsequent consecutive terms is 1/3. So, we have:

36 + (24 + 8 + 8/3 + ...) =

= 36 + 24/(1 - 1/3) =

= 36 + 36 =

= 72.

Answer: B.

luisdicampo · Answer

Deconstructing the Question  
Initial Height ( H ) = 36 m.
Rebound Ratio ( r ) =  1/3 .
Target: Total distance covered by the ball.

Step 1: Understand the Motion  
The ball falls once from 36m.
After that, for every rebound, it travels  up  to the new height and then  down  from that height.
Total Distance = Initial Drop + 2 * (Sum of all rebound heights).

Step 2: Calculate the Geometric Series  
The rebound heights form an infinite Geometric Progression:
1st Rebound:  36 	imes \frac{1}{3} = 12 .
2nd Rebound:  12 	imes \frac{1}{3} = 4 .
Sequence: 12, 4, 4/3, ...

Sum of infinite GP ( S ) where  a=12  and  r=1/3 :
 S = \frac{a}{1 - r} = \frac{12}{1 - 1/3} = \frac{12}{2/3} = 18 .

Step 3: Calculate Total Distance  
Total Distance =  36 + 2(S) 
Total Distance =  36 + 2(18) = 36 + 36 = 72 .

Alternative Method (Short Formula) 
 D = H 	imes \frac{1+r}{1-r} 
 D = 36 	imes \frac{1 + 1/3}{1 - 1/3} = 36 	imes \frac{4/3}{2/3} = 36 	imes 2 = 72 .

Answer:  B

A ball dropped from 36 m above the ground rebounds to 1/3rd of the

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