Hello and welcome to ExamPundit. Here is a set Data Interpretation Problems.
Study the following graph carefully and answer the questions
given below.
Percentage growth in population of six states A, B, C, D, E
and F from 2012 to 2013 and 2013 to 2014.
1. Population of
state ‘F’ in 2013 was approximately what percent of its population in 2014 ?
(a) 60
(b) 67
(c) 75
(d) 55
(e) None of these
2. If the population
of state ‘B’ in the year 2012 was 5 lakh, what was approximately its population
in the year 2014?
(a) 9.50 lakh
(b) 8 lakh
(c) 10.50 lakh
(d) 14.50 lakh
(e) None of these
3. If the population
of states C and D in 2013 are in the ratio of 2 : 3 respectively and the population
of state ‘C’ in 2012 was 2.5 lakh,
what was the population of state ‘D’ in 2013 ?
(a) 5.25 lakh
(b) 4.75 lakh
(c) 3.50 lakh
(d) 6 lakh
(e) None of these
4. In 2012 the
population of states B and D are equal and the population of state B in 2014 is
4 lakh, what approx was the population
of state ‘D’ in 2014?
(a) 3 lakh
(b) 3.50 lakh
(c) 6 lakh
(d) 5 lakh
(e) None of these
5. Population of
state ‘E’ in 2012 was what fraction of its population in 2014?
(a) 2/3
(b) 4/5
(c) 5/4
(d) 3/2
(e) None of these
Solutions:
1. Let the population of state ‘F’ in 2013 = 100
Then population of state ‘F’ in 2014 = 150 (Since growth is
50%)
So, Required % = 100/150 × 100 » 67%
2. Population of state ‘B’ in the year 2014 =
5 × 155/100 135/100
»
10.50 lakh
3. C12/D12 = 2/3
and C12 = 2.50 Lakhs
So, C12 = 2.5 x 140/100 =
3.5 Lakhs
So we get, the population of D in 2013 = 3 x 3.50/2
= 5.25 Lakhs
4. Population of state ‘B’ in 2012
= 4 × 100 × 100/155 × 135= Population
of D in 2012
Population of state ‘D’ in 2014 = 1×100×100×160×170/100×100×155×135
= 5 lakhs
5. Suppose population of state E in 2012 = 100
Then population of state E in 2013 = 125
*since 25% growth
And population of state E in 2014 = 125 (120/100)
= 150
*since 20% growth
So, required ratio = 100/150 = 2/3
Regards
Team ExamPundit
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