DIRECTIONS for questions 1 to 5: Read the table given below and answer the questions accordingly.
Shares traded in Chandigarh, Punjab, Haryana (In Rupees)

Chandigarh 
Punjab 
Haryana 
Total 
Name of the company 
High 
Low 
High 
Low 
High 
Low 
High 
Low 
Pepsodent 
438 
396 
420 
395 
452.5 
413.75 
1310.5 
1204.75 
ZARA 
148 
140 
189.5 
180 
190 
180 
527.5 
500 
Bagh Bakri Chai 
28 
22.75 
28.25 
23 
29 
24.5 
85.25 
70.25 
RCOM 
190.5 
177 
189 
177.5 
191.25 
177.5 
570.75 
532 
Birla 
160 
143 
158 
140.5 
161.25 
102.5 
479.25 
386 
Example 1. The average for the high rates of Chandigarh for all the companies was
1. Rs.192.9
2. Rs.202.9
3. Rs.169.9
4. Rs.178
Sol: Option (1)
The average for all the high involved with the Chandigarh region was
(438+148+28+190.5+160)/5 which gives 192.9.
Example 2. Total of low value of the Punjab region exceeds the total of the low value of the region of Haryana by how much?
1. 17.75
2. 15.67
3. 15.57
4. None of these
Sol: Option(1)
Total value for the low of the Punjab region for the given companies is 395+180+23+177.5+140.5=916 whereas the total for the low of the Haryana region is 413.75+180+24.5+177.5+102.5=898.25. So it exceeds by 916 – 898.25=17.75.
Example 3. Low for RCOM for all the given states/UT form what percent of the low for ZARA for the same?
1.98.4%
2.101%
3.106.4%
4. None of these
Sol: Option(3)
The total of lows for the RCOM for all the given states/ UT are given to be 532 and the ZARA has the total for their lows to be 500.Thus ATQ (532/500)*100 = 106.4%
Example 4. High for Chandigarh for BIRLA form what percent of the High for BIRLA for Punjab and Haryana combined?
1. 50%
2. 44%
3. 40%
4. 55%
Sol: Option(1)
High for the company "BIRLA" in Chandigarh was 160 and Total of high in Punjab and Haryana happen to be 319.25.
So, ATQ the percentage required will be (160/319.25)*100=50% approx.
Example 5. Total of high for all the companies in all the States/UT are more than the lows for all the states/UT by how much?
1. 260
2. 170
3. 270
4. 280.25
Sol: Option(4)
The total of high for all the given places is 2973.25 whereas the low totals to 2693. So the given data differ by 280.25.
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DIRECTIONS for questions 6  10: Study the following table carefully and answer the questions given below.
Number of candidates appeared (App.) and percentage of candidates qualified (Qual.) under different disciplines over the years.
Year 
German 
French 
Chinese 
Latin 
Greek 

App. 
Qual% 
App. 
Qual% 
App. 
Qual% 
App. 
Qual% 
App. 
Qual% 
2011 
842 
29 
908 
21 
1928 
40 
579 
45 
843 
42 
2012 
1019 
27 
878 
28 
2028 
38 
608 
38 
719 
36 
2013 
985 
31 
1156 
31 
2536 
42 
492 
42 
645 
41 
2014 
1215 
28 
1290 
32 
2113 
45 
714 
55 
720 
39 
2015 
1429 
34 
1075 
24 
1725 
36 
801 
48 
586 
48 
2016 
1128 
24 
1416 
35 
1820 
39 
726 
51 
620 
35 
Example 6. Approximately, what was the percentage increase in the number of candidates who appeared under the German stream from 2013 to 2014(approx)?
Sol: Option(2)
Number of students in the year 2013 were 985 and they increased to 1215, so there was a net increase of 230.Talking of percentage there was an increase of 230 in the initial value 985, so the increase will be (230/985)*100 = 23.35
Example 7. How many students were qualified for the Chinese stream in the year 2014?
1. 450
2. 950
3. 650
4. 900
Sol: Option(2)
The actual no. of candidates that were qualified under Chinese stream = 45% of 2113 = 950.
Example 8. According to the given data which were the 2 years under which the number of students who qualified in the Greek stream is the same(approx.)?
1. 2013, 2014
2. 2014, 2015
3. 2012, 2013
4. 2011 & 2016
Sol: Option(2)
In the year 2014 i.e. (39% * 720) = 280(approx). For 2015 i.e. (48% * 586) = 281(approx). They are approximately equal.
Example 9. The number of students who qualified in the year 2011 under the streams Greek and French combined was
1. 505
2. 545
3. 530
4. 560
Sol: Option(2)
Students who qualified under the stream Greek in 2011 (42% of 843) = 354(approx)
Students who qualified under the stream French in 2011 (21% of 908) = 191(approx)
Total for the years 354+191=545
Example 10. Which will be the pair of years in German whose total equals to the number of students who appeared in the Chinese discipline in 2014?
1. 2016, 2015
2. 2013, 2016
3. 2016, 2014
4. 2016, 2012
Sol: Option(2)
In the year 2013 the number of candidates who were in German = 985. For the year 2016, the total no of candidates in German happens to be = 1128
So Total = 985 + 1128 = 2113. Students who appeared under the Chinese stream in 2014 = 2113
DIRECTIONS for questions 1114: Study the following graph carefully and answer the questions given below:
Production (in Lakhs) of Widgets & Percentage defect over the years in five factories A, B, C, D, E.
Year 
A 
B 
C 
D 
E 

Widgets 
% Defect 
Widgets 
% Defect 
Widgets 
% Defect 
Widgets . 
% Defect 
Widgets 
% Defect 
2001 
76 
5 
58 
11 
39 
5 
59 
9 
28 
8 
2002 
82 
6 
46 
9 
37 
9 
62 
8 
36 
4 
2003 
65 
8 
49 
8 
45 
6 
47 
12 
42 
15 
2004 
70 
12 
52 
12 
42 
13 
54 
4 
31 
9 
2005 
85 
9 
64 
14 
38 
11 
57 
7 
49 
11 
2006 
80 
11 
54 
10 
40 
8 
68 
5 
38 
7 
Example 11. The average production for the given data will be maximum for which factory?
Sol: Option(3)
If we look closely then we will find that the production of factory A happens to be maximum and thus its average will also be highest. We can also verify this by calculating the number of widgets individually and dividing it by the number of given years i.e. 6.
Example 12. What was the difference in defected widgets of factory C for the year 2006 and defected widgets of factory E for the year 2003?
1. 2800
2. 4000
3. 3500
4. 3100
Sol: Option(4)
Reqd. No. = (15% of 42000)  (8% of 40000) = 6300  3200 = 3100
Example 13. The average number of widgets that were produced in the year 2002 formed what percent of the widgets that were produced in the year 2006?
1. 94%
2. 90%
3. 85%
4. 99%
Sol: Option(1)
The total of the widgets that were produced in the year 2002 was 263(Lakhs) and the ones produced in the year 2006 were 280(Lakhs). The percentage value is (263/280) x 100 = 94 % (approx).
Note: We have taken the total in place of average as the result in both the cases will be same. The reason is in case of average, we have to divide both numerator and denominator by 6 which will not make any difference in final result.
Example 14. What was the difference in the number of widgets of factory C which were free from any defect in the years 2004 and 2006?
1. 300
2. 150
3. 260
4. 400
Sol: Option(3)
Required difference = (87% of 42000) – (92% of 40000) = 36540 – 36800 = 260
So the difference is 260.