Question

Navdeep is planting lamp posts on divider of the road. The lamp posts are to be erected on a straight line, at a distance interval of 10 meters between consecutive lamp posts. All the lamp posts are kept at origin. So, he errects a lamp post at origin. It is also given that he can carry one lamp post at a time, so, he erects a lamp post and then goes to the origin to fetch another lamp post. In the process of erecting lamp posts, he walks 1.32 kms. Can you tell us how many lamp posts does he erect, if he ends up at the origin, after walking 1.32 kms.
Options

15

14

13

12

Answer 1

Number of Lamp posts = 12   if he walks 1.32 kms

Step-by-step explanation:

Let say there are n lamps

then 1st lamp is to be erected at origin => Walk = 0 m

then for 2nd Lamp erection & return to origin he has to walk  10 * 2 = 20 m

for 3rd Lamp erection & return to origin he has to walk  (10+ 10) * 2 = 40 m

Hence Distances covered are

0  , 20  ,  40 ,...........................................................20(n-1)

a = first term = 0

d = 20

n = number of lamp posts

Total Distance covered = Sum

= (n/2)(0 + 20(n-1))

= 10n(n-1)

walks 1.32 kms = 1.32 * 1000 = 1320 m

=> 10n(n-1) = 1320

=> n(n-1) = 132

=> n² - n - 132 =0

=> n² - 12n + 11n - 132 =0

=> n(n-12) + 11(n-12) = 0

=> (n + 10(n-12) = 0

=> n = 12

Number of Lamp posts = 12

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Answer 2

Answer:

12

Step-by-step explanation:

let the number of lamp posts erected be denoted as $n+1$

when he erects 1 lamp post at the origin, (n) lamp posts remain.

and for each of these, he walks twice the distance for every lamp post until the end.

since the interval distance is 10 m

then

we can form the following relationship

d=(10+10 +20+20 +30+30+ ... +10n+10n)

d=20( 1+2+3+...+n)............................................ (1)

the expression inside the brackets forms an arithmetic progression beginning from 1 to n with increaments of 1.

the sum of an arithmetic progression is given by

$s=\frac{n}{2}(a+l)$

where n is the number of terms ,

          a is the first term

          l is the last term

therefore

$s= \frac{n}{2}(1+n)\\ s=\frac{(n)}{2}*(n+1)$

substituting this in eqn (1)  we get

$d=20(\frac{n(n+1)}{2})\\ 1320=20*\frac{n(n+1)}{2}\\1320=10*n(n+1)\\132=n(n+1)\\n^2+n-132=0\\(n+12)(n-11)=0\\n=11,-12\\n=11$

Therefore the total number of lamp posts erected is 11+1=12