35. tabulation
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The document provides information about tabulation questions that may appear in competitive exams. It explains that these questions involve analyzing data presented in a table format regarding topics like production over time, imports/exports, employee incomes, etc. The data is arranged systematically in rows and columns with headings. Example questions are provided based on sales data for different battery types sold by a company each year. The questions test the ability to analyze trends, calculate percentages, and compare values across the table. Overall, the document introduces the concept of tabulation questions and provides a sample table and questions to illustrate the type of data analysis skills required to solve them.
![1998
115
85
160
100
145
605
in
eG
K
1. The total sales of all the seven years is the maximum for which battery ?
(a) 4AH
(b) 7AH
(c) 32AH
(d) 35AH
(e) 55AH
2. What is the difference in the number of 35AH batteries sold in 1993 and
1997 ?
(a) 24000
(b) 28000
(c) 35000
(d) 39000
(e) 42000
3. The percentage of 4AH batteries sold to the total number of batteries sold
was maximum in the year:
(a) 1994
. (b) 1995
(c) 1996
(d) 1997
(e) 1998
4. In the case of which battery there was a continuous decrease in sales from
1992 to 1997 ?
(b) 7 AH
(c) 32AH
(d) 35AH
(e) 55AH
(8) 4AH
5. What was the approximate percentage increase in the sales of 55AH
batteries in 1998 compared to that in 1992 ?
(a) 28%
(b) 31%
(c) 33%
(d)34%
(e)37%
eO
nl
Sol. 1. (c) : The total sales (in thousands) of all the seven years for various
batteries are:
For 4AH = 75 + 90 + 96 + 105 + 90 + 105 + 115 = 676
For 7AH = 144 + 126 + 114 + 90 + 75 + 60 + 85 = 694
For 32AH = 114 + 102 + 75 + 150 + 135 + 165 + 160 = 901
For 35 AH= 102 + 84 + 105 + 90 + 75 + 45 + 100 = 601
For 55 AH= 108 + 126 + 135 + 75 + 90 + 120 + 145 = 799.
Clearly, sales are maximum in case of 32AH batteries.
2. (d : Required difference = [(84 - 45) x 1000] = 39000.
3. (d) : The percentages of sales of 4AH batteries to the total sales in different
years are:
For 1992 =(75*100/543)%=13.81%
Th
For 1993=(90*100)/528%=17.05%
For 1994=(96*100/465)%=19.35%
For 1995=(105*100/495)%=20.59%
For 1996=(96*100/465)%=19.35%
For 1997=(105*100/495)%=21.21%
For 1998=(115*100/605)%=19.01%](https://image.slidesharecdn.com/35-131220002035-phpapp02/85/35-tabulation-2-320.jpg)
![eG
K
6. In which subject is the overall percentage the best?
(d) Chemistry (e) Geography
(a) History (b) Maths (c) Physics
Sol. 1.. (e) : Aggregate marks obtained by Sajal
= [(90% of 150) + (60% of 130) + (70% of 120) + (70% of 100) +
(90% of 60) + (70% of 40)] = 135 + 78 + 84 + 70 + 54 + 28 = 449.
.
2. (c) : Aggregate marks obtained by Tarun
= [(65% of 150) + (35% of 130) + (50% of 120) + (77% of 100) + (80% of
60) + (80% of 40)] = 97.5 + 45.5 + 60 + 77 + 48 + 32 = 360.
Total maximum marks (of all the six subjects)
= (150 + 130 + 120 + 100 + 60 + 40) = 600.
Overall percentage of Tarun = 360 x 100 % = 60%.
.
600
3. (b) : Average marks obtained in Physics by all the seven students
in
= 1 [(90% of 120) + (80% of 120) + (70% of 120) + (80% of 120)
7
+ (85% of 120) + (65% of 120) + (50% of 120)]
nl
= 1 [(90 + 80 + 70 +80 + 85 + 65 + 50)% of 120]
7
Th
eO
=1 [520% of 120] = 89.14.
7
4. (b) : From the table it is clear that Sajal and Rohit have 60% or more
marks
in each of the six subjects.
6. (b) : We shall find the overall percentage (for all the seven students) with
respect to each subject.
The overall percentage for any subject is equal to the average of percentages
obtained by all the seven students since the maximum marks for any subject
is the same for all the students.
Therefore, overall percentage for:
(i) Maths = [1(90+100+90+80+80+70+65)]%
7
= [1(575)]% = 82.14%.
7
(ii) Chemistry = [1(50 + 80 + 60 + 65 + 65 + 75 + 35)]%](https://image.slidesharecdn.com/35-131220002035-phpapp02/85/35-tabulation-6-320.jpg)
![= [1 (510)]% = 72.86%.
7
in
eG
K
7
= [1(430)]% = 61.43%.
.
7
(iii) Physics = [1(90 + 80 + 70 + 80 + 85 + 65 + 50)]%
7
=[1 (520)]% = 74.29%.
7
(iv) Geography = [1(60 + 40 + 70 + 80 + 95 + 85 + 77)]%
7
= [1 (507)}_ = 72.43%.
7
(v) History = [1 (70 + 80 + 90+ 60 + 50 + 40 + 80)]%
7
=1 [(470)]% = 67.14%.
7
(vi) Computer Science = [1/7 (80 + 70 + 70 + 60 + 90 + 60 + 80)]%
nl
Clearly; this. percentage is highest for Maths.
ex.4. Study the following table carefully and answer tbe questions given
below:(Bank P.O. 2001)
eO
CLASSIFICATION OF 100 STUDENTS BASED ON THE MARKS
OBTAINED BY THEM IN PHYSICS AND CHEMISTRY IN AN
EXAMINATION
Th
Marks out
40 and
Of 50
above
Subject
physics
9
chemistry 4
(aggregate
Average) 7
30 and
Above
32
20 and
above
80
0 and
above
100
66
10 and
above
92
,.
81,
,21
27
73
87
100
100
1. The number of students scoring less than 40% marks in aggregate is :
(b) 19
(c) 20
(d) 27
(e) 34
(a) 13
2. If at least 60% marks in Physics are required for pursuing higher studies in
Physics,how many students will be eligible to pursue higher studies in](https://image.slidesharecdn.com/35-131220002035-phpapp02/85/35-tabulation-7-320.jpg)
35. tabulation
- 1. Free GK Alerts- JOIN OnlineGK to 9870807070 36. TABULATION in eG K This section comprises of questions in which certain data regarding common disciplines as production over a period of a few years: imports, exports, incomes of employees in a factory, students applying for and qualifying a certain field of study etc. are given in the form of a table. The candidate is required to understand the given information and thereafter answer the given questions on the basis of comparative analysis of the data. Thus, here the data collected by the investigator are arranged in a systematic form in a table called the tabular form. In order to avoid some heads again and again, tables are made consisting of horizontal lines called rows and vertical lines called columns with distinctive heads, known as captions. Units of measurements are given with the captions. SOLVED EXAMPLES eO nl The following table gives the sales of batteries manufactured by a company lit the years. Study the table and answer the questions that follow: (S.B.I.P.O. 1998) NUMBER OF DIFFERENT TYPES OF BATTERIES SOLD BY A COMPANY OVER THE YEARS (NUMBERS _N THOUSANDS) 4AH 1992 75 144 114 102 108 543 1993 90 126 102 84 426 528 Th Year TYPES OF BATTERIES 7AH 32AH 35AH 55AH T0TAL 1994 96 114 75 105 135 525 1995 105 90 150 90 75 510 1996 90 75 135 75 90 465 1997 105 60 165 45 120 495
- 2. 1998 115 85 160 100 145 605 in eG K 1. The total sales of all the seven years is the maximum for which battery ? (a) 4AH (b) 7AH (c) 32AH (d) 35AH (e) 55AH 2. What is the difference in the number of 35AH batteries sold in 1993 and 1997 ? (a) 24000 (b) 28000 (c) 35000 (d) 39000 (e) 42000 3. The percentage of 4AH batteries sold to the total number of batteries sold was maximum in the year: (a) 1994 . (b) 1995 (c) 1996 (d) 1997 (e) 1998 4. In the case of which battery there was a continuous decrease in sales from 1992 to 1997 ? (b) 7 AH (c) 32AH (d) 35AH (e) 55AH (8) 4AH 5. What was the approximate percentage increase in the sales of 55AH batteries in 1998 compared to that in 1992 ? (a) 28% (b) 31% (c) 33% (d)34% (e)37% eO nl Sol. 1. (c) : The total sales (in thousands) of all the seven years for various batteries are: For 4AH = 75 + 90 + 96 + 105 + 90 + 105 + 115 = 676 For 7AH = 144 + 126 + 114 + 90 + 75 + 60 + 85 = 694 For 32AH = 114 + 102 + 75 + 150 + 135 + 165 + 160 = 901 For 35 AH= 102 + 84 + 105 + 90 + 75 + 45 + 100 = 601 For 55 AH= 108 + 126 + 135 + 75 + 90 + 120 + 145 = 799. Clearly, sales are maximum in case of 32AH batteries. 2. (d : Required difference = [(84 - 45) x 1000] = 39000. 3. (d) : The percentages of sales of 4AH batteries to the total sales in different years are: For 1992 =(75*100/543)%=13.81% Th For 1993=(90*100)/528%=17.05% For 1994=(96*100/465)%=19.35% For 1995=(105*100/495)%=20.59% For 1996=(96*100/465)%=19.35% For 1997=(105*100/495)%=21.21% For 1998=(115*100/605)%=19.01%
- 3. Clearly, the percentage is maximum in 1997. 4. (b) : From the table it is clear that the sales of 7AH batteries have been decreasing continuously from 1992 to 1997. eG K 5. (d) : Required Percentage =(145-108)/108)*100 %=34.26%=34%. Ex 2: Study the following table carefully and answer these questions: NUMBER OF CANDIDATES APPEARED AND QUALIFIED IN A COMPETITIVE EXAMINATION FROM DIFFERENT STATES OVER THE YEAR R 1998 App. Qal. 8500 980 9200 1050 8800 1020 9500 1240 1999 App. Qual. 7400 850 8450 920 7800 890 8700 980 2000 App. Qual. 6800 775 9200 980 8750 1010 9700 1200 2001 App. Qual 9500 1125 8800 1020 9750 1250 8950 995 in 1997 App. Qual. M 5200 720 N 7500 840 P 6400 780 Q 8100 950 7800 870 7600 940 9800 1350 7600 945 7990 885 nl 1. Combining the states P and Q, together in 1998, what is the percentage of the candidates qualified to that of the canditates appeared? (c) 12.35% (d) 12.54% (e) 13.50% (8) 10.87% (b) 11.49% eO 2.The percentage of the total number of qualified candidates to the total number appeared candidates among all the five states in 1999 is : (a) 11.49%(b) 11.84% (c)- 12.21% (d) 12.57%(e) 12.7a1 Th 3. What is the percentage of candidates qualified from State N for all the years together, over the candidates appeared from State N during all the years together? (a) 12.36% (b) 12.16% (c) 11.47% (d) 11.15%(e)None of these 4. What is the average of candidates who appeared from State Q during the given yeas? (b) 8760 (c) 8810 (d) 8920 (e) 8990 (8) 8700 5 . In which of the given years the number of candidates appeared from State P has maximum percentage of qualified candidates?
- 4. (8) 1997 (b) 1998 (c) 1999 (d) 2000 (e) 2001 eG K 6. Total number of candidates qualified from all the states together in 1997 is approximately what percentage of the total number of candidates qualified from all the states together in 1998 ? (b) 77% (c) 80% (d) 83% (e) 86% (8) 72% Sol.1.(c)Required Percentage=(1020+1240) *100%=(2260*100)/18300% (8800+9500) =12.35% Required Percentage= (850+920+890+980+1350) *100% (7400+8450+7800+8700+9800) =(4990*100)/42150% =11.84% : Required in (e) Percentage=(84- +1050+920+980+1020)/(7500+9200+8450+9200+8800)*100% =(4810*100)/43150* % nl =11.15% 4. (e) Required average =(8100+9500+8700+9700+8950)/5 =44950/5 eO =8990 5. (e) : The percentages of candidates qualified to candidates appeared from State P during different years are: Th For 1997= 780 * 100% =12.19% 6400 for 1998 = 1020*100 %=11.59% 8800 For 1999 = 890*100 %=11.41%; 7800 For 2000 = 1010* 100 % = 11.54%. 8 750
- 5. For 2001=1250*100 %= 12.82% 9750 :. Maximum percentage is for the year 2001. 6. (c) : Required Percentage =( 720 + 840 + 780 + 950 + 870) . x 100 =80% eG K 980+1050+1020+1240+940 Ex. 3. The following table gives the percentage of marks obtained by seven students in six , different subjects in an examination. Study the table and answer the questions based on it. The numbers in the brackets give the I nl Maths Chemistry Physics GeographyHistory Computer Science (160) (130) (120) (100) (60) (40) 90 50 90 60 70 80 100 80 80 40 80 70 90 60 70 70 90 70 80 65 80 80 60 60 80 65 85 95 50 90 70 75 65 85 40 60 65 35 50 77 80 80 eO (Max. marks) Student Ayush Aman Sajal Rohit Muskan Tanvi Tharun in maximum marks in each subject. (Bank P.O. 2003) Th 1. What was the aggregate of marks obtained by Sajal in all the six subjects? (b) 419 (c) 429 (d) 439 (e) 449 (a) 409 2. What is the overall percentage of Thrun? (c) 60% (d) 63% (e) 64.5% (a) 52.5% (b) 55% 3. What are the average marks obtained by all the seven students in Physics? (rounded off to two digits after decimal) (c) 91.37 (d) 96.11 (e) 103.21 (a) 77.26 (b) 89.14 4. The number of students who obtained 60% and above marks in all the subjects is : (b) 2 (c) 3 (d) None (e) None of these (a) 1
- 6. eG K 6. In which subject is the overall percentage the best? (d) Chemistry (e) Geography (a) History (b) Maths (c) Physics Sol. 1.. (e) : Aggregate marks obtained by Sajal = [(90% of 150) + (60% of 130) + (70% of 120) + (70% of 100) + (90% of 60) + (70% of 40)] = 135 + 78 + 84 + 70 + 54 + 28 = 449. . 2. (c) : Aggregate marks obtained by Tarun = [(65% of 150) + (35% of 130) + (50% of 120) + (77% of 100) + (80% of 60) + (80% of 40)] = 97.5 + 45.5 + 60 + 77 + 48 + 32 = 360. Total maximum marks (of all the six subjects) = (150 + 130 + 120 + 100 + 60 + 40) = 600. Overall percentage of Tarun = 360 x 100 % = 60%. . 600 3. (b) : Average marks obtained in Physics by all the seven students in = 1 [(90% of 120) + (80% of 120) + (70% of 120) + (80% of 120) 7 + (85% of 120) + (65% of 120) + (50% of 120)] nl = 1 [(90 + 80 + 70 +80 + 85 + 65 + 50)% of 120] 7 Th eO =1 [520% of 120] = 89.14. 7 4. (b) : From the table it is clear that Sajal and Rohit have 60% or more marks in each of the six subjects. 6. (b) : We shall find the overall percentage (for all the seven students) with respect to each subject. The overall percentage for any subject is equal to the average of percentages obtained by all the seven students since the maximum marks for any subject is the same for all the students. Therefore, overall percentage for: (i) Maths = [1(90+100+90+80+80+70+65)]% 7 = [1(575)]% = 82.14%. 7 (ii) Chemistry = [1(50 + 80 + 60 + 65 + 65 + 75 + 35)]%
- 7. = [1 (510)]% = 72.86%. 7 in eG K 7 = [1(430)]% = 61.43%. . 7 (iii) Physics = [1(90 + 80 + 70 + 80 + 85 + 65 + 50)]% 7 =[1 (520)]% = 74.29%. 7 (iv) Geography = [1(60 + 40 + 70 + 80 + 95 + 85 + 77)]% 7 = [1 (507)}_ = 72.43%. 7 (v) History = [1 (70 + 80 + 90+ 60 + 50 + 40 + 80)]% 7 =1 [(470)]% = 67.14%. 7 (vi) Computer Science = [1/7 (80 + 70 + 70 + 60 + 90 + 60 + 80)]% nl Clearly; this. percentage is highest for Maths. ex.4. Study the following table carefully and answer tbe questions given below:(Bank P.O. 2001) eO CLASSIFICATION OF 100 STUDENTS BASED ON THE MARKS OBTAINED BY THEM IN PHYSICS AND CHEMISTRY IN AN EXAMINATION Th Marks out 40 and Of 50 above Subject physics 9 chemistry 4 (aggregate Average) 7 30 and Above 32 20 and above 80 0 and above 100 66 10 and above 92 ,. 81, ,21 27 73 87 100 100 1. The number of students scoring less than 40% marks in aggregate is : (b) 19 (c) 20 (d) 27 (e) 34 (a) 13 2. If at least 60% marks in Physics are required for pursuing higher studies in Physics,how many students will be eligible to pursue higher studies in
- 8. eG K Physics? (a) 27 (b) 32 (c) 34 (d)41 (e) 68 3. What is the difference between the number of students passed with 30 as cut-off marks in Chemistry and those passed with :JUas cut-off marks in aggregate? (a) 3 (b) 4 (c) 5 (d) 6 (e) 7 in 4. The percentage of the number of students getting at least 60% marks in Chemistry over those getting at least 40% marks in aggregate, is approximately: (a) 21% (b) 27% (c) 29% (d) 31% (e) 34% 5. If it is known that at least 23 students were eligible for a Symposium on Chemistry the minimum qualifying marks in Chemistry for eligibility to Symposium would lie in the range: (a) 40-50 (b) 30-40 (c) 20-30 (d) Below 20 Th eO nl Sol. 1. (d) : We have 40% of 50 =(40 x 50)= 20. 100 :. Required number = Number of students scoring less than 20 marks in aggregate = 100 - number of students scoring 20 and above marks in aggregate = 100 - 73 = 27. 2. (b) : We have 60% of 50 =(60 x 50) = 30. 100 :. Required number = Number of students scoring 30 and above mark in Physics = 32. . 3. (d) : Required difference = (Number of students scoring 30 and above in mark in Chemistry) (Number of students scoring 30 and . above marks in aggregate) = 27 - 21 = 6. 4. (c) : Number of students getting at least 60% marks in Chemistry = Number of students getting 30 and above marks in Chemistry = 21. Number of students getting at least 40% marks in aggregate = Number of students getting 20 and above marks in aggregate = 73. :. Required Percentage = (21x 100)% = 28.77% ≈29%. 73 6. (c) : Since 66 students get 20 and above marks in Chemistry and out of these 21 students get 30 and above marks, therefore to select top 35 students in Chemistry, the qualifying marks should lie in the range 20-30.