Tech Mahindra Data Interpretation Quiz 1

After observing the given line graph, we can make the following deductions

Solubility of Potassium Chlorate at 30^{0}C is 0.15 while at 80^{0}C it is 0.4 The increase in % of solubility of Potassium Chlorate= (0.4 – 0.15) 100/0.15 = 166.66%.

The solubility of Potassium Nitrate in one litter at 30^{0}C = 0.38 kg Therefore solubility of Potassium Nitrate in 10 lt. would be = 3.8 kg., Approx = 4 kg.

The solubility of Sodium Nitrate, Potassium Chloride, Sodium Chloride is almost constant as we see the graph between 15^{0}C & 25^{0}C. Also, it can also be seen from graph that solubility of Sodium Chlorate is maximum.

The Solubility of Potassium Chloride at 36^{0}C = 0.4 kg./lt.
Therefore in 100 lt. the amount of Potassium Chloride that can be dissolved = 40 kg.
Number of moles = \frac{40}{0.075}= 533.
Therefore approx. 540 moles can be dissolved in 100 lt. of water at 36^{0}C.

Total sales of branches B1, B3 and B5 for both the years(in thousand numbers)

Required Percentage = [(70 + 80)/(95 + 110) * 100]% = (150/205 * 100)% = 73.17%

Average sales of all the six branches(in thousand numbers) for the year 2006

Required Ratio = (75 + 65)/(85 + 95) = 140/180 = 7/9

Average sales of branches B1, B3 and B6 in 2000 = 1/3 * (80 + 95 + 70) = 245/3