The total length of a runway is 1000 m. The elevation at an interval of 100 m is as follows :

Distance(m) 100 200 300 400 500 600 700 800 900 1000

Elevation(m) 100 99.2 98.6 99.2 101.4 97.6 100 101.6 102.4 101

The effective gradient of the runway (in %, round off to two decimals) will be____
0.48

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Solution

The correct option is A 0.48
Effective gradient =

$[\frac{Maximun,difference,in,elevation}{Total,runway,length} \times 100]$

$= [\frac{(102.4 - 97.6)}{1000} \times 100]$

= 0.48 %

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The median of the following data is 525. Find the values of x and y, if the total frequency is 100.

$\begin{matrix} C l a s s I n t e r v a l & F r e q u e n c y 0 - 100 & 2 100 - 200 & 5 200 - 300 & x 300 - 400 & 12 400 - 500 & 17 500 - 600 & 20 600 - 700 & y 700 - 800 & 9 800 - 900 & 7 900 - 1000 & 4 \end{matrix}$ [4 MARKS]

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Distance(m)	100	200	300	400	500	600	700	800	900	1000
Elevation(m)	100	99.2	98.6	99.2	101.4	97.6	100	101.6	102.4	101

The total length of a runway is 1000 m. The elevation at an interval of 100 m is as follows : Distance(m) 100 200 300 400 500 600 700 800 900 1000 Elevation(m) 100 99.2 98.6 99.2 101.4 97.6 100 101.6 102.4 101 The effective gradient of the runway (in %, round off to two decimals) will be____0.48

The correct option is A 0.48Effective gradient = [Maximun,difference,in,elevationTotal,runway,length×100] =[(102.4−97.6)1000×100] = 0.48 %

The correct option is A 0.48
Effective gradient =

$[\frac{Maximun,difference,in,elevation}{Total,runway,length} \times 100]$

$= [\frac{(102.4 - 97.6)}{1000} \times 100]$

= 0.48 %