 # DI Table Chart Solved examples

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.
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)?
1. 230
2. 23
3. 40
4. 30
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 11-14: 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?
1. B
2. E
3. A
4. D
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.