Cable cars at hill stations are one of the major tourist attractions. On a hill station, the length of cable car ride from base point to top most point on the hill is 5000 m. Poles are installed at equal intervals on the way to provide support to the cables on which car moves.
The distance of first pole from base point is 200 m and subsequent poles are installed at equal interval of 150 m. Further, the distance of last pole from the top is 300 m.
Based on above information, answer the following questions using Arithemtic Progression :
(i) Find the distance of 10th pole from the base. [1 Mark]
(ii) Find the distance between 15th pole and 25th pole. [1 Mark]
(iii) (a) Find the time taken by cable car to reach 15th pole from the top if it is moving at the speed of 5 m/sec and coming from top. [2 Marks]
OR
(iii) (b) Find the total number of poles installed along the entire journey. [2 Marks]
Detailed Solution :
Before, I start answering this, I would like to agree that is a little bit challenging case study for students. Most of the students will not be able to imagine the situation. This is why I request you to please look at the image below carefully to understand the situation first.
Click on the image to see Zoomed Version
AP would be
200, 350, 500, 650, . .
Here
a=200
d=a2-a1
d=350-200
d=150
200, 350, 500, 650, . .
Here
a=200
d=a2-a1
d=350-200
d=150
(i) a10=?
an=a+(n-1)d
a10=200+(10-1)150
a10=200+9(150)
a10=200+1350
a10=1550
an=a+(n-1)d
a10=200+(10-1)150
a10=200+9(150)
a10=200+1350
a10=1550
Hence,distance of 10th pole from the base is 1550 m
(ii) a25-a15
= a+24d-(a+14d) [∵an=a+(n-1)d]
=a+24d-a-14d
=0+10d
=10d
=10×150
=1500 m
= a+24d-(a+14d) [∵an=a+(n-1)d]
=a+24d-a-14d
=0+10d
=10d
=10×150
=1500 m
Hence, distance between 15th pole and 25th pole is 1500 m.
(iii) (a)
Distance of 15th Pole from the top
Here the distance of last pole is at a distance of 300 m from the pole and the poles are separated by 150 m distance as usual.
so the distance of 15th pole from top would be
an=a+(n-1)d
a15=300+(15-1)150
a15=300+14(150)
a15=300+2100
a15=2400
Speed of cable car = 5 m/sec
Distance of 15th Pole from the top
Here the distance of last pole is at a distance of 300 m from the pole and the poles are separated by 150 m distance as usual.
so the distance of 15th pole from top would be
an=a+(n-1)d
a15=300+(15-1)150
a15=300+14(150)
a15=300+2100
a15=2400
Speed of cable car = 5 m/sec
Time taken = Distance / speed
= 2400/5
= 480 sec
= 480/60
= 48/6
= 8 minutes
= 2400/5
= 480 sec
= 480/60
= 48/6
= 8 minutes
Hence, time taken by cable car to reach 15th pole from the top is 8 minutes.
OR
(iii) (b)
(iii) (b)
Distance of the last pole from the top
= 5000-300
= 4700
Now AP would be
200, 350, 500, 650, . . 4700
an=4700
a=200
d=150
an=a+(n-1)d
4700=200+(n-1)×150
4700-200=(n-1)×150
4500=(n-1)×150
4500/150=n-1
450/15=n-1
30=n-1
30+1=n
n=31
= 5000-300
= 4700
Now AP would be
200, 350, 500, 650, . . 4700
an=4700
a=200
d=150
an=a+(n-1)d
4700=200+(n-1)×150
4700-200=(n-1)×150
4500=(n-1)×150
4500/150=n-1
450/15=n-1
30=n-1
30+1=n
n=31
Hence, there are 31 poles installed along the entire journey.