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Data Interpretation MCQ Questions and Answers - Table Chart

Learn Table Chart Data Interpretation mcq question and answers with easy and logical explanations Page-22 section-1

Directions (1 - 5): Study the following table which shows the amount of money invested (Rupees in crore) in the core infrastructure areas of two districts. A and B of a State, and answer the below five questions.
  District A District B
Core Area 1995 1996 1995 1996
Electricity
Chemical
Thermal
Solar
Nuclear		 815.2
389.5
632.4
468.1
617.9		 1054.2
476.7
565.9
589.6
803.1		 2065.8
745.3
1232.7
1363.5
1674.3		 2365.1
986.4
1026.3
1792.1
2182.1
Total 2923.1 3489.5 7081.6 8352.0
1
If the total investment in district B shows the same rate of increase in 1997, as it had shown from 1995 to 1996, what approximately would be the total investment in B in 1997?
Discuss
Answer & Solution
Difference of 1995 & 1996
= 8352 - 7081.6
= 1270.4
Increase % = $$\frac{1270.4}{7081.6}$$  × 100
                  = 17.9% (Approx)
B investment in 1997
= $$\frac{117.9}{100}$$ × 8352
= Rs. 9850 crore (Approx)
2
The total investment in electricity and thermal energy in 1995, in these two districts A and B formed approximately what percent of the total investment made in that year?
Discuss
Answer & Solution
Investment of A & B in electricity & thermal energy in 1995
= 815.2 + 2065.8 + 632.4 + 1232.7
= 4746.1
Total investment of A & B in 1995
= 2923.1 + 7081.6
= 10004.7
∴ Required percentage
= $$\frac{4746.1}{10004.7}$$  × 100
= 47% (Approx)
3
Approximately how many times was the total investment in 1995 and 1996 in district B was that of total investment of district A in the same years?
Discuss
Answer & Solution
Total investment of A
= 2923.1 + 3489.5
= 6412.6
Total investment of B
= 7081.6 + 8352
= 15433.6
∴ Times = $$\frac{15433.6}{6412.6}$$   = 2.4 (Approx)
4
By approximately what percent was the total investment in the two districts A and B More in 1996 as compared to 1995?
Discuss
Answer & Solution
Total investment of A & B in 1995
= 2923.1 + 7081.6
= Rs. 10004.7
Total investment of A & B in 1996
= 3489.5 + 8352
= Rs. 11841.5
∴ Increase %
$$\eqalign{ & = \frac{11841.5 - 10004.7}{10004.7} \times 100 \cr & = \frac{1836.8}{10004.7} \times 100 \cr & = 18\% \left(\text{Approx} \right) \cr} $$
5
In district B, the investment in which area in 1996 did show the highest percentage increase over the investment in that area in 1995?
Discuss
Answer & Solution
$$\eqalign{ & \text{Chemical %} \cr & = \frac{986.4 - 745.3}{745.3} \times 100 \cr & = \frac{241.1 \times 100}{745.3} \cr & = 32.34\% \cr & \cr & \text{Solar %} \cr & = \frac{1792.1 - 1363.5}{1363.5} \times 100 \cr & = \frac{428.6}{1363.5} \times 100 \cr & = 31.4\% \cr & \cr & \text{Electricity %} \cr & = \frac{2365.1 - 2065.8}{2065.8} \times 100 \cr & = \frac{299.3}{2064.8} \times 100 \cr & = 14.4\% \cr & \cr & \text{Nuclear %} \cr & = \frac{2182.1 - 1674.3}{1674.3} \times 100 \cr & = \frac{507.8}{1674.3} \times 100 \cr & = 30.3\% \cr} $$
So the highest percentage is in Chemical area