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

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

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New post 30 Jul 2019, 22:16
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E

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

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New post 30 Jul 2019, 22:31
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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 \(= \frac{n}{2}(a_1+a_{11}) = \frac{11}{2}(4+54) = 319\)

Answer is (E)
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A bicycle rider coasts down the hill, travelling 4ft in the first seco  [#permalink]

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New post 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
Answer: E
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A bicycle rider coasts down the hill, travelling 4ft in the first seco  [#permalink]

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New post 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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Re: A bicycle rider coasts down the hill, travelling 4ft in the first seco  [#permalink]

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New post 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.
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Re: A bicycle rider coasts down the hill, travelling 4ft in the first seco  [#permalink]

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New post 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!
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Re: A bicycle rider coasts down the hill, travelling 4ft in the first seco  [#permalink]

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New post 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.

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

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

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