Data Interpretation :Pie Chart
Data Interpretation questions for IBPS,SBI,Bank.SSC,UPSC,PSC,RRB, and Insurance exams
Example 1. The pie-chart provided below gives the distribution of land (in a village) under various food crops. Study the pie-chart carefully and answer the questions that follow.
1.Which combination of three crops contribute to 50% of the total area under the food crops ?
(a) Wheat, Barley and Jowan
(c)Rice Wheat and Barley
(b) Rice, Wheat and Jowar
(d) Bajra, Maize and Rice
2. If the total area under jowar was 1.5 million acres, then what was the area (in million acres) under rice ?
(a)6
(b) 7.5
(c) 9
(d) 4.5
3.If the production of wheat is 6 times that of barley, then what is the ratio between the yield per acre of wheat and barley ?
(a) 3:2
(b) 3:1
(c) 12:1
(d) 2:3
4.If the yield per acre of rice was 50% more than that of barley, then the production of barley is what percent of that of rice?
(a) 30%
(b) 33.33%
(c) 35%
(d) 36%
5.If the total area goes up by 5%, and the area under wheat production goes up by 12%, then what will be the angle for wheat in the new pie-chart ?
(a) 62.4°
(b) 76.8°
(c) 80.6°
(d) 84.2°
Sol.1.(c):The total of the central angles corresponding to the three crops which cover 50% of the total area, should be 180°. Now, the total of the central angles for the given combinations are
(i) Wheat, Barley and Jowar = (72° +36° + 18°) = 126°
(ii) Rice, Wheat and Jowar = (72° + 72°+ 18°) = 162°
(iii) Rice, Wheat and Barley. = (72+ 72 + 36°) =180°
(iv) Bajra. Maize and Rice=(18° 45° + 72°)= 135°
Clearly, (iii) is the required combination.
Sol.2(.a): The area under any of the food crops is proportional to the central angle corresponding to that crop.
Let, the area under rice production be x million acres
Then, 18: 72 = 1.5: x => r=[(72×1.5)/18=6
Thus, the area under rice production = 6 million acres.
Sol.3. (b): Let the total production of barley be T tonnes and let Z acres of land be put
under barley production.
Then, the total production of wheat = (6T) tonnes.
Also, area under wheat production = (2Z) acres.
Since, [Area under Wheat production /Area under Barley production]=( 72° / 36°)=2
And therefore ,Area under wheat=2× Area under barley=(2Z) acres.
Now yield per acre for barley =(T/Z) tonnes/acre
There for required ratio =[(3T/Z)/(T/Z)]=3:1
Sol.4.(b): Let Z acres of land be put under barley production
Then, (Area under rice production)/( Area under barley production)= ( 72° / 36°)=2
Therefore, Area under Ric production = 2 x area under barley production = (2Z) acres
Now if p tonnes be the yield per acre of barley then, yield per acre of rice
=(p+50% of p)tonnes=(3p/2 ) tonnes.
Therefore Total production of rice = (yield per acre) x (area under production)
=(3p/2) 2Z = (3pZ)tonnes.
And ,Total production of barley =(pZ) tonnes.
Therefore percentage production of barley to that of rice=[( pZ/3pZ) ×100]%=33.33%
Sol.5.(b) Initially, let t acres be the total area under consideration.
Then, area under wheat production initially was =[(72/360) ×t] acres=(t/5) acres.
Now, if the total area under consideration be increased by 5%, then the new
value of the total area =[(105/100)t] acres.
Also, if the area 'under wheat production be increased by 12%, then the new value of the area under wheat=[(t/5) + (12% of (t/5))]=(112t/500)acres.
Therefore Central angle corresponding to wheat in the new pie-chart
= {[(Area under wheat -new)/(Total area-new)]×360}°
={[(112t/500)/(105t/100) ] ×360 }0=76.8°
Example 2.The Following pie chart show the distribution of students of graduate and post graduate levels in seven different institutes -M, N, P, Q, R, S and T in a town.(Bank Po 2003)
1. How many students of institutes M and S are studying at graduate level ?
(a) 7516
(b) 8463
(C) 9127
(d) 9404
(e)None of the above
2.Number of students studying at Post-graduate level from institutes N and P is:
(a) 5601
(b) 5944
(C) 6669
(d) 7004
(e)None of the above
(d) 2:3
4.If the yield per acre of rice was 50% more than that of barley, then the production of barley is what percent of that of rice?
(a) 30%
(b) 33.33%
(c) 35%
(d) 36%
5.If the total area goes up by 5%, and the area under wheat production goes up by 12%, then what will be the angle for wheat in the new pie-chart ?
(a) 62.4°
(b) 76.8°
(c) 80.6°
(d) 84.2°
Sol.1.(c):The total of the central angles corresponding to the three crops which cover 50% of the total area, should be 180°. Now, the total of the central angles for the given combinations are
(i) Wheat, Barley and Jowar = (72° +36° + 18°) = 126°
(ii) Rice, Wheat and Jowar = (72° + 72°+ 18°) = 162°
(iii) Rice, Wheat and Barley. = (72+ 72 + 36°) =180°
(iv) Bajra. Maize and Rice=(18° 45° + 72°)= 135°
Clearly, (iii) is the required combination.
Sol.2(.a): The area under any of the food crops is proportional to the central angle corresponding to that crop.
Let, the area under rice production be x million acres
Then, 18: 72 = 1.5: x => r=[(72×1.5)/18=6
Thus, the area under rice production = 6 million acres.
Sol.3. (b): Let the total production of barley be T tonnes and let Z acres of land be put
under barley production.
Then, the total production of wheat = (6T) tonnes.
Also, area under wheat production = (2Z) acres.
Since, [Area under Wheat production /Area under Barley production]=( 72° / 36°)=2
And therefore ,Area under wheat=2× Area under barley=(2Z) acres.
Now yield per acre for barley =(T/Z) tonnes/acre
There for required ratio =[(3T/Z)/(T/Z)]=3:1
Sol.4.(b): Let Z acres of land be put under barley production
Then, (Area under rice production)/( Area under barley production)= ( 72° / 36°)=2
Therefore, Area under Ric production = 2 x area under barley production = (2Z) acres
Now if p tonnes be the yield per acre of barley then, yield per acre of rice
=(p+50% of p)tonnes=(3p/2 ) tonnes.
Therefore Total production of rice = (yield per acre) x (area under production)
=(3p/2) 2Z = (3pZ)tonnes.
And ,Total production of barley =(pZ) tonnes.
Therefore percentage production of barley to that of rice=[( pZ/3pZ) ×100]%=33.33%
Sol.5.(b) Initially, let t acres be the total area under consideration.
Then, area under wheat production initially was =[(72/360) ×t] acres=(t/5) acres.
Now, if the total area under consideration be increased by 5%, then the new
value of the total area =[(105/100)t] acres.
Also, if the area 'under wheat production be increased by 12%, then the new value of the area under wheat=[(t/5) + (12% of (t/5))]=(112t/500)acres.
Therefore Central angle corresponding to wheat in the new pie-chart
= {[(Area under wheat -new)/(Total area-new)]×360}°
={[(112t/500)/(105t/100) ] ×360 }0=76.8°
Example 2.The Following pie chart show the distribution of students of graduate and post graduate levels in seven different institutes -M, N, P, Q, R, S and T in a town.(Bank Po 2003)
1. How many students of institutes M and S are studying at graduate level ?
(a) 7516
(b) 8463
(C) 9127
(d) 9404
(e)None of the above
2.Number of students studying at Post-graduate level from institutes N and P is:
(a) 5601
(b) 5944
(C) 6669
(d) 7004
(e)None of the above
3.What is the total number of graduate and post-graduate level students in institute R?
(a) 8320
(b) 7916
(C) 9116
(d) 8099
(e)None of the above
4.What is the ratio between number of students studying at post-graduate and graduate
levels respectively from institute S?
(a) 14:19
(b) 19:21
(C) 17:21
(d) 19:14
(e)None of the above
5.What is the ratio between the number of students studying at post-graduate level from
institute S and the number of students studying at graduate level from institute Q?
(a) 13:19
(b) 21:13
(C) 13:8
(d) 19:13
(e)None of the above
Sol.1.(b) : Students of institute M at graduate level =17% of 27300 = 4641
Students of institute S at graduate level = 14% of 27300 = 3822.
Total number of students at graduate levels in institutes M and S = 4641 + 3822 = 8463
(a) 8320
(b) 7916
(C) 9116
(d) 8099
(e)None of the above
4.What is the ratio between number of students studying at post-graduate and graduate
levels respectively from institute S?
(a) 14:19
(b) 19:21
(C) 17:21
(d) 19:14
(e)None of the above
5.What is the ratio between the number of students studying at post-graduate level from
institute S and the number of students studying at graduate level from institute Q?
(a) 13:19
(b) 21:13
(C) 13:8
(d) 19:13
(e)None of the above
Sol.1.(b) : Students of institute M at graduate level =17% of 27300 = 4641
Students of institute S at graduate level = 14% of 27300 = 3822.
Total number of students at graduate levels in institutes M and S = 4641 + 3822 = 8463
Sol.2. (c) : Required number = (15% of 24700) +1 (12% of 24700)= 3705 + 2964 = 6669
Sol.3. (d) : Required number = (17% of 27300) + (14% of 24700) = 4641 + 3458 =8099
Sol.4. (d) Required ratio = (21% of 24700) / (14% of 27300) =19/14
Sol.5.(d) Required ratio (21% of 24700) / (13% of 27300) =19/13
Example 3. Study the following pi-chart and answer the questions that follows.
1.What is the ratio of car sales together in the months of March and April to that of total car sales in the month of January.
(a) 14:5
(b) 16:5
(c) 7:3
(d) 7:2
(e) None of these
2.What is the percentage change in the car sale in the month of June with respect to previous month.(appxmt value)
(a) 2% increase
(b) 2% decrease
(c) 4% increase
(d) 4% decrease
(e) 6% decrease
3.Number of car sale in April is what percent higher than number of car sales in March
(a) 45%
(b) 38%
(c) 39%
(d) 40%
(e) 32%
4.Total number of sales together in January and February is what percentage of total sales in six months?
( a) 23%
( b) 23.90%
(c) 23.58%
(d) 24.23%
(e) 26.75%
5.Number of sales in July is 25% higher than number of sales in March and number of sales in August is 10% higher than number of sales in July. Then what is the difference between number of sales in June and August?
(a) 34
(b) 24
(c) 33
(d) 23
(e) 13
Sol.1.(b)
Car sales together in the months of March and April=40+96
Car sales in the month of January=30
Required ratio=96:30=16:5
Sol.2.(e)Change in the car sale in the month of June with respect to previous month=34-32=2
Required percentage= (2/34)*100=5.88%(6% appxmt)
Sol.3.(d)Required percentage=[(56-40)/40]*100 %=40%
Sol.4.(c).Required percentage=[(30+20)/212]*100% = 23.58%
Sol.5.(d). Number of sales in July=(125/100) *40 =50
Number of sales in August=(110/100)*50=55
Difference between number of sales in June and August=55-32=23
Sol.3. (d) : Required number = (17% of 27300) + (14% of 24700) = 4641 + 3458 =8099
Sol.4. (d) Required ratio = (21% of 24700) / (14% of 27300) =19/14
Sol.5.(d) Required ratio (21% of 24700) / (13% of 27300) =19/13
Example 3. Study the following pi-chart and answer the questions that follows.
(a) 14:5
(b) 16:5
(c) 7:3
(d) 7:2
(e) None of these
2.What is the percentage change in the car sale in the month of June with respect to previous month.(appxmt value)
(a) 2% increase
(b) 2% decrease
(c) 4% increase
(d) 4% decrease
(e) 6% decrease
3.Number of car sale in April is what percent higher than number of car sales in March
(a) 45%
(b) 38%
(c) 39%
(d) 40%
(e) 32%
4.Total number of sales together in January and February is what percentage of total sales in six months?
( a) 23%
( b) 23.90%
(c) 23.58%
(d) 24.23%
(e) 26.75%
5.Number of sales in July is 25% higher than number of sales in March and number of sales in August is 10% higher than number of sales in July. Then what is the difference between number of sales in June and August?
(a) 34
(b) 24
(c) 33
(d) 23
(e) 13
Sol.1.(b)
Car sales together in the months of March and April=40+96
Car sales in the month of January=30
Required ratio=96:30=16:5
Sol.2.(e)Change in the car sale in the month of June with respect to previous month=34-32=2
Required percentage= (2/34)*100=5.88%(6% appxmt)
Sol.3.(d)Required percentage=[(56-40)/40]*100 %=40%
Sol.4.(c).Required percentage=[(30+20)/212]*100% = 23.58%
Sol.5.(d). Number of sales in July=(125/100) *40 =50
Number of sales in August=(110/100)*50=55
Difference between number of sales in June and August=55-32=23
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