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# A bicycle rider coasts down the hill, travelling 4ft in the first seco

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Math Expert
Joined: 02 Sep 2009
Posts: 64314
A bicycle rider coasts down the hill, travelling 4ft in the first seco  [#permalink]

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30 Jul 2019, 22:16
00:00

Difficulty:

55% (hard)

Question Stats:

60% (01:51) correct 40% (01:51) wrong based on 121 sessions

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A bicycle rider coasts down the hill, travelling 4ft in the first second. In each succeeding second, he travels 5ft farther than in the preceeding second. If the rider reaches the bottom of the hill in 11 seconds, find the total distance travelled.

A. 54
B. 304
C. 309
D. 314
E. 319

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Director
Joined: 16 Jan 2019
Posts: 614
Location: India
Concentration: General Management
WE: Sales (Other)
Re: A bicycle rider coasts down the hill, travelling 4ft in the first seco  [#permalink]

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30 Jul 2019, 22:31
1
1
We have an arithmetic sequence with first term $$a_1=4$$ and common difference $$d=5$$

The 11th term $$a_{11}=a_1+(11-1)*d = 4+50=54$$

Sum of terms $$= \frac{n}{2}(a_1+a_{11}) = \frac{11}{2}(4+54) = 319$$

Intern
Joined: 24 Jul 2017
Posts: 2
GMAT 1: 460 Q43 V12
GMAT 2: 700 Q50 V34
A bicycle rider coasts down the hill, travelling 4ft in the first seco  [#permalink]

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30 Jul 2019, 22:39
1
Distance traveled at 11th sec = Distance traveled at 1st sec + (11-1) * 5 = 54
Sum = 11/2 * (1st + 11th) = 11/2 * (4+54) = 319
CEO
Joined: 03 Jun 2019
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GMAT 1: 690 Q50 V34
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A bicycle rider coasts down the hill, travelling 4ft in the first seco  [#permalink]

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30 Jul 2019, 23:43
Bunuel wrote:
A bicycle rider coasts down the hill, travelling 4ft in the first second. In each succeeding second, he travels 5ft farther than in the preceeding second. If the rider reaches the bottom of the hill in 11 seconds, find the total distance travelled.

A. 54
B. 304
C. 309
D. 314
E. 319

Given: A bicycle rider coasts down the hill, travelling 4ft in the first second. In each succeeding second, he travels 5ft farther than in the preceeding second. The rider reaches the bottom of the hill in 11 seconds.

Asked: Find the total distance travelled.

The distance travelled in first second = 4 m = 5-1 ft
The distance travelled in second second = 4 + 5 ft = 5*2 -1 ft = 9ft
In general
The distance travelled in nth second = 5n -1 ft
Total distance travelle in n seconds = 5n(n+1)/2 - n ft
Sum of 11 terms of the sequence = 5*11*12/2 - 11 = 30*11 -11 = 29*11 = 319 ft
Total distance travelled.= 319 ft

IMO E
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Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com
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Joined: 13 Mar 2019
Posts: 27
Re: A bicycle rider coasts down the hill, travelling 4ft in the first seco  [#permalink]

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08 Aug 2019, 19:44
1
Check the pattern -
!st sec - 4ft.
2nd - 4+5
3rd - 4+2*5
....
11th - 4+10.5

So Total = 44+(1+2+3+4.....10)*5 = 319.
Manager
Joined: 09 Nov 2015
Posts: 193
Re: A bicycle rider coasts down the hill, travelling 4ft in the first seco  [#permalink]

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09 Aug 2019, 22:10
1
I guess we could shave off a few seconds by eliminating the step of calculating the distance traveled in the 11th second and directly applying the formula for 'Sum of an Arithmetic Series':
Sn=(n/2){2a+(n-1)d} where 'n' is the number of terms, 'a' is the 1st term and 'd' is the Common Difference (difference between one term and the next).
Total distance traveled = (11/2){2*4+(11-1)*5} = 319 ANS: E

Any time saving, however small, helps!
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 10669
Location: United States (CA)
Re: A bicycle rider coasts down the hill, travelling 4ft in the first seco  [#permalink]

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12 Aug 2019, 18:30
Bunuel wrote:
A bicycle rider coasts down the hill, travelling 4ft in the first second. In each succeeding second, he travels 5ft farther than in the preceeding second. If the rider reaches the bottom of the hill in 11 seconds, find the total distance travelled.

A. 54
B. 304
C. 309
D. 314
E. 319

We can use the formula sum = avg x quantity

avg = (4 + 4 + 5 x 10)/2 = 58/2 = 29

Thus, the distance traveled was 29 x 11 = 319 ft.

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Re: A bicycle rider coasts down the hill, travelling 4ft in the first seco   [#permalink] 12 Aug 2019, 18:30