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nick1816
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A number of saplings are lying at a place by the side of a straight ro Updated on: 01 Jun 2019, 20:22
Originally posted by nick1816 on 01 Jun 2019, 18:03.
Last edited by nick1816 on 01 Jun 2019, 20:22, edited 1 time in total.
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95% (hard)

Question Stats:

46% (03:03) correct 54% (02:43) wrong based on 169 sessions

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A number of saplings are lying at a place by the side of a straight road. These are to be planted in a straight line at a distance interval of 10 meters between two consecutive saplings. John, the forester, can carry only one sapling at a time and has to move back to the original point to get the next sapling. In this manner he covers a total distance of 1320 meters. How many saplings does he plant in the process if he ends at the starting point?

(Note- He plants first sapling at the starting point.)

A. 10
B. 11
C. 12
D. 13
E. 14
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C
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A number of saplings are lying at a place by the side of a straight ro 01 Jun 2019, 19:18
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nick1816
A number of saplings are lying at a place by the side of a straight road. These are to be planted in a straight line at a distance interval of 10 meters between two consecutive saplings. John, the forester, can carry only one sapling at a time and has to move back to the original point to get the next sapling. In this manner he covers a total distance of 1320 meters. How many saplings does he plant in the process if he ends at the starting point?

A. 10
B. 11
C. 12
D. 13
E. 14
Show more

Firstly it is slightly unclear whether he plants a sapling at the start point or starts at 10m from the start point.

Travel time for first sapling = 2*10
Travel time for second sapling = 2*20
Travel time for third sapling = 2*30
.....
Travel time for nth sapling = 2*10n

So total distance = 2*10+2*20+2*30+.....2*10n = 1320
\(2(10+20+30+...10n)=1320........20(1+2+3+...n)=1320........20*\frac{n(n+1)}{2}=1320...........n(n+1)=132=12*11\)
Thus, n = 11.

If he starts planting from 10 m from start point, the answer is 11, but if he plants the first sapling at the start point, then answer is 12.

I would assume that the first sapling is at the start point, so answer is 12.

C
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A number of saplings are lying at a place by the side of a straight ro 02 Jun 2019, 00:00
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nick1816
A number of saplings are lying at a place by the side of a straight road. These are to be planted in a straight line at a distance interval of 10 meters between two consecutive saplings. John, the forester, can carry only one sapling at a time and has to move back to the original point to get the next sapling. In this manner he covers a total distance of 1320 meters. How many saplings does he plant in the process if he ends at the starting point?

(Note- He plants first sapling at the starting point.)

A. 10
B. 11
C. 12
D. 13
E. 14
Show more

solving such word based problems under 2 mins is a task , does not seems to be gmat type question

anyways given that each sapling is planted at distance of 10 mtrs and first is at starting point then for each successive sapling total distance travelled will be in succession of 10 mts
for eg ; 2nd sapling = 10 mtrs
3rd sapling =30 mtrs
4th sapling= 50 mtrs and so on
use formula
sn=n/2 * ( 2*a+(n-1)*d)
here n we can substitute answer options say n=12
and a=1 d=20
sn=12/2*(2*1+220)
we get sn = 1322 which is close to 1320 IMO C ..
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A number of saplings are lying at a place by the side of a straight ro 02 Jun 2019, 16:18
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IMO this question is a GMAT level question and should take less than 2 minutes.(As the distances are making an Arithmetic series)
Anyways, i think you understood the question wrongly, that's why it took you longer than 2 minutes

1. He plants first sapling at the starting point, distance covered by him=0
2. He plants second sapling at a distance of 10 m, distance covered by him to plant 2nd sapling and return back to starting point= 10+10=20
3. He plants third sapling at a distance of 20 m, distance covered by him to plant 3rd sapling and return back to starting point= 20+20=40
and so on

Total distance= 0+20+40 ...... 20(n-1)=1320
20(n-1)(n)/2=1320
(n-1)(n)=132=11*12
n=12

Should take max 1 and half minute!!!!




Archit3110
nick1816
A number of saplings are lying at a place by the side of a straight road. These are to be planted in a straight line at a distance interval of 10 meters between two consecutive saplings. John, the forester, can carry only one sapling at a time and has to move back to the original point to get the next sapling. In this manner he covers a total distance of 1320 meters. How many saplings does he plant in the process if he ends at the starting point?

(Note- He plants first sapling at the starting point.)

A. 10
B. 11
C. 12
D. 13
E. 14

solving such word based problems under 2 mins is a task , does not seems to be gmat type question

anyways given that each sapling is planted at distance of 10 mtrs and first is at starting point then for each successive sapling total distance travelled will be in succession of 10 mts
for eg ; 2nd sapling = 10 mtrs
3rd sapling =30 mtrs
4th sapling= 50 mtrs and so on
use formula
sn=n/2 * ( 2*a+(n-1)*d)
here n we can substitute answer options say n=12
and a=1 d=20
sn=12/2*(2*1+220)
we get sn = 1322 which is close to 1320 IMO C ..
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A number of saplings are lying at a place by the side of a straight ro 27 Jan 2020, 09:08
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nick1816
A number of saplings are lying at a place by the side of a straight road. These are to be planted in a straight line at a distance interval of 10 meters between two consecutive saplings. John, the forester, can carry only one sapling at a time and has to move back to the original point to get the next sapling. In this manner he covers a total distance of 1320 meters. How many saplings does he plant in the process if he ends at the starting point?

(Note- He plants first sapling at the starting point.)

A. 10
B. 11
C. 12
D. 13
E. 14
Show more

First sapling: d=0
F --10m--> Second: d=10
F <--10m-- Second: d=10
F ------20m------> Third: d=20
F <------20m------ Third: d=20
F ---------30m---------> Fourth: d=30
F <---------30m--------- Fourth: d=30

2(10)+2(20)+2(30)+…=1320
2(10+20+30…)=1320
2(Sum of multiples of 10)=1320
2(n/2(2a+m(n-1))=1320
n(2a+m(n-1)=1320
a=0 and m=10
n(0+10n-10)=1320
10n^2-10n-1320=0
n^2-n-132=0
(n-12)(n+11)
n>0=12

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A number of saplings are lying at a place by the side of a straight ro 14 Mar 2026, 00:03
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