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प्रश्न
Below is a list of 10 tallest buildings in India.
This list ranks buildings in India that stand at least 150 m (492 ft.) tall, based on standard height measurement. This includes spires and architectural details but does not include antenna marks. Following data is given as per the available information till 2009. Since new buildings are always under construction, go on-line to check new taller buildings.
Use the information given in the table about sky scrapers to answer the following questions:
| Name | City | Height | Floors | Year |
| Planet | Mumbai | 181 m | 51 | 2009 |
| UB Tower | Bengaluru | 184 m | 20 | 2006 |
| Ashok Towers | Mumbai | 193 m | 49 | 2009 |
| The Imperial I | Mumbai | 249 m | 60 | 2009 |
| The Imperial II | Mumbai | 249 m | 60 | 2009 |
| RNA Mirage | Mumbai | 180 m | 40 | 2009 |
| Oberoi Woods Tower I | Mumbai | 170 m | 40 | 2009 |
| Oberoi Woods Tower II | Mumbai | 170 m | 40 | 2009 |
| Oberoi Woods Tower III | Mumbai | 170 m | 40 | 2009 |
| MVRDC | Mumbai | 156 m | 35 | 2002 |
(a) Find the height of each storey of the three tallest buildings and write them in the following table:
| Building | Height | Number of storeys | Height of each storey |
(b) The average height of one storey for the buildings given in (a) is ______.
(c) Which city in this list has the largest percentage of skyscrapers? What is the percentage?
(d) What is the range of data?
(e) Find the median of the data.
(f) Draw a bar graph for given data.
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उत्तर
a. Clearly, Imperial I, Imperial II and Ashok Towers are three tallest building.
| Building | Height | Number of storeys | Height of each storey |
| Imperial I | 249 m | 60 | 294/60 = 4.15 |
| Imperial II | 249 m | 60 | 249/60 = 4.15 |
| Ashok Towers | 193 m | 49 | 193/49 = 3.94 |
b. Average height of each storey of the building given in (a)
= `(["Sum of heights of each storey of three tallest building"])/3`
= `(4.15 + 4.15 + 3.94)/3`
= `12.24/3`
= 4.08
c. We can clearly see from the data, Mumbai has maximum number of skyscrapers from the list given. It has 9 skyscrapers out of the list of 10 buildings given.
∴ Required percentage = `9/10 xx 100 = 90%`
d. Range of data = Maximum height – Minimum height
= 249 – 156
= 93
e. Arranging the data in ascending order, we get 156, 170, 170, 170, 180, 181, 184, 193, 249, 249. Since, there are ten observations, median will be the mean of 5th and 6th observations.
n = 10 ...(Even)
∴ Median = `(n/2 "th observation" + (n/2 + 1) "th observation")/2`
= `((10/2) "th observation" + (10/2 + 1) "th observation")/2`
= `(5 "th observation" + 6"th observation")/2`
= `(180 + 181)/2`
= 180.5
f. A bar graph is as shown below:
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