Chapter-38 (LINE-GRAPHS)



                                             38. LINE-GRAPHS

  This section comprises of question in which the data collected in a particular discipline
are represented by specific points together by straight lines. The points are plotted on a two-dimensional plane taking one parameter on the horizontal axis and the other on the vertical axis. The candidate is required to analyse the given information and thereafter answer the given  questions on the basis of the analysis of data.

                                                SOLVED EXAMPLES

Ex. 1. In a school the periodical examination are held every second month. In a session during Apr. 2001 – Mar. 2002, a student of Class IX appeared for each of the periodical exams. The aggregate marks obtained by him in each periodical exam are represented in the line-graph given below. Study the graph and answer the questions based on it.                                                                        (S.B.I.P.O 2003)        

       MARKS OBTAINED BY A STUDENT IN SIX PERIODICAL EXAMS HELD IN                  

          EVERY TWO MONTHS DURING THE YEAR IN THE SESSION 2001-02 


               Maximum Total Marks In each Periodical Exam = 500


       
      
  1. The total number of marks obtained in Feb. 02 is what percent of the total marks obtained in Apr. 01?
(a) 110%      (b) 112.5%    (c) 115%    (d) 116.5%    (e) 117.5%
  1. What are the average marks obtained by the student in all the periodical exams of during the session.
 (a) 373        (b) 379           (c) 381       (d) 385           (e) 389
  1. what is the percentage of marks obtained by the student in the periodical exams of Aug. 01 and Oct. 01 taken together?
             (a) 73.25% (b) 75.5%       (c) 77%     (d) 78.75%     (e) 79.5%
  1. In which periodical exams there is a fall in percentage of marks as compared to the previous periodical exams?
             (a) None      (b) Jun. 01      (c) Oct. 01  (d) Feb. 01   (e) None of these
  1. In which periodical exams did the student obtain the highest percentage increase in marks over the previous periodical exams?
 (a) Jun. 01    (b) Aug. 01   (c) Oct. 01 (d) Dec. 01   (e) Feb. 02

Sol. Here it is clear from the graph that the student obtained 360, 365, 370, 385, 400 and 405 marks in periodical exams held in Apr. 01, Jun. 01, Aug. 01, Oct. 01, Dec. 01 and  Feb. 02 respectively.

  1. (b) : Required percentage = [(405/360)*100] % = 112.5 %
  2. (c) : Average marks obtained in all the periodical exams.
                = (1/6)*[360+370+385+400+404] = 380.83 » 381.
  1. (d) : Required percentage = [(370+385)/(500+500) * 100] % = [(755/1000)*100]% =75.5%
  2. (a) : As is clear from graph, the total marks obtained in periodical exams, go on increasing. Since, the maximum marks for all the periodical exams are same , it implies that the percentage of marks also goes on increasing. Thus, in none of the periodical exams, there is a fall  in percentage of marks compared to the previous exam.
  3. (c) : Percentage increases in marks in various periodical exams compared to the previous exams are:
For Jun. 01 = [(365-360)/360 * 100 ] % = 1.39 %
For Aug. 01 = [(370-365)/365 * 100 ] % = 1.37 %
For Oct. 01 = [(385-370)/370 * 100 ] % = 4.05%
For Dec. 01 = [(400-385)/385 * 100 ] % = 3.90 %
For Feb. 02 = [(405-400)/400 * 100 ] % = 1.25 %

Ex. 2. The following line- graph the ratio of the amounts of imports by a Company to the amount of exports from that Company over the period from 1995 to 2001. The questions given below are based on this graph.                                                                                                                               (S.B.I.P.O 2001)

               

                            Ratio of value of Import to Export by a Company over the Years



  1. In how many of the given years were the exports more than the imports?
a. 1                  b. 2                   c. 3                 d. 4

2.The imports were minimum proportionate to the exports of the Company in the year:
  a..1995    b.1996             c.1997             d.2001

3.If the imports of the Company in 1996 was Rs.272 crores, the exports from the Company in 1996 was:
a. Rs.370 crores           b.Rs.320 crores           c.Rs.280 crores
d.Rs.275 crores           e.Rs.264 crores

4.What was the percentage increase in imports from 1997 to 1998?
a. 72    b.56     c.28     d.None of these          e.Data inadequate

5.If the imports in 1998 was Rs.250 crores and the total exports in the years 1998 and 1999 together was Rs.500 crores, then the imports in 1999 was:
a.Rs.250 crores            b.Rs.300 crores           c.Rs 357 crores
d.Rs 420 crores           e.None of these

Sol: 1.  d : The exports are more than the imports implies that the ratio of value of imports to exports is less than 1.
Now, this ratio is less than 1 in the years 1995,1996,1997 and 2000.
Thus, there are four such years.

2.   c:  The imports are minimum proportionate to the exports implies that the ratio of the    value of imports to exports has the minimum value.
Now, this ratio has a minimum value of 0.35 in 1997, i.e., the imports are minimum proportionate to the exports in 1997.

3.   b:  Ratio of imports to exports in the years 1996=0.85.
Let the exports in 1996=Rs.320 crores.
Then,272/x =0.85  implies x=272/.85 = 320.

Exports in 1996 = Rs.320 crores.
4.        e: The graph gives  only the ratio of imports to exports for different years. To find the percentage increase in imports from 1997 to 1998, we require more details such as the value of imports or exports during these years. Hence, the data is inadequate to answer this question.

5.    d: The ratio of imports to exports for the years 1998 and 1999 are 1.25 and 1.40 respectively.
Let the exports in the year 1998 = Rs. x crores
Then, the exports in the year 1999=Rs(500-x) crores.
1.25=250/x implies x=250/1.25=200
Thus the exports in the year 1999=Rs. (500-200)crores=Rs.300 crores
Let the imports in the year 1999=Rs. y crores
Then, 1.4=y/300 implies y=(300*1.4)=420.
Imports in  the year 1999=Rs.420 crores.

Ex.3.Study the following line-graph and answer the question based on it.

 

Number of vehicle Manufactured by Two Companies over the Years

                        (Numbers in thousands)

         
1.What is the difference between the total productions of the two Companies in the given years?
a. 19000          b. 22000          c.26000           d.28000           e.29000

2.What is the difference between the numbers of vehicles manufactured by Company Y in 2000 and 2001?
a.50000           b.42000           c.33000           d.21000           e.13000

3.What is the average number of vehicles manufactured by Company X over the given period? (rounded off to the nearest integer)
a.119333         b.113666         c.112778         d.111223         e. None of these

4.In which of the following years, the difference between the productions of Companies X and Y was the maximum among the given years?
a.1997             b.1998             c.1999                         d.2000             e.2001

5.The production of Company Y in 2000 was approximately what percent of the production of Company X in the same year?
a.173.              b.164               c.132               d.97                 e.61

Sol:  From the line-graph it is clear that the productions of Company X in the years 1997,1998,1999,2000,2001 and 2002 are 119000,99000,141000,78000,120000and 159000 respectively and those of Company Y are 139000,120000,100000,128000,107000 and 148000 respectively.

     1. (c) :   Total production of Company X from 1997 to 2002                            
                  = 119000+99000+141000+78000+120000+159000 = 716000
              and total production of Company Y from 1997 to 2002
                   =139000+120000+100000+128000+107000+148000=742000                                            
               Difference=742000-716000=26000.

  1. (d) : Require difference  = 128000-107000 = 21000.
  2. (a) : Average number of vehicles manufactured by Company X
               = (91/6)* (119000 + 99000 + 141000 + 78000 + 120000 + 159000) = 119333.

  1. (d) : The difference between the production of Companies X and Y in various years are.
              For 1997 = (139000 – 119000) = 20000;
              For 1998 = (120000 – 99000) = 21000;
              For 1999 = (141000 – 100000) = 41000;
              For 2000 = (128000 – 78000) = 50000;
              For 2001 = (120000 – 107000) = 13000;
              For 2003 = (159000 – 148000) = 11000;
              Clearly, maximum difference was in 2000.


  1. (b) : Required   percentage = [( 128000/78000)* 100] % = 164 %.

Ex. 4. The following line-graph gives the percent profit earned by two Companies X and Y  during the period 1996 – 2001. Study the line – graph and answer the questions that are based on on it.

      Percentage Profit Earned by Two Companies X and Y over the Given years
                        % profit/ loss = [(Income – Expenditure) / Expenditure] * 100




1.      If the expenditure of Company Y in 1997 was Rs. 220 crores, what was its income in 1997?
(a). Rs. 312 crores   (b). Rs. 297 crores  (c) Rs. 283 crores (d) Rs. 275 crores (e)Rs.261 crores

2.If the incomes of the two companies were equal in 1999,then what was the ratio of expenditure of Company X to that of company Y in 1999 ?
(a) 6:5  (b) 5:6  (c) 11:6  (d) 16:15  (e) 15:16                                                 

3.The incomes of the companies X and Y in 2000 were in the ratio of 3:4 respectively.What was the respective ratio of their expenditures in 2000?
(a) 7:22  (b) 14:19  (c) 15:22  (d)27:35   (e) 33:40

4.If the expenditure ofcompanies X and Y in 1996 were equal and the total income of the two companies in 1996 was Rs.342 crores, what was the total profit of the twocompanies together in 1996 ? (Profit = Income – Expenditure)
(a)    Rs.240crores  (b) Rs.171crores  (c) Rs.120crores  (d) Rs.102crores  (e)None of these.

5.The expenditure ofcompany X in the year 1998 was Rs.200 crores and the income Company X in 1998 was the same as its expenditure in 2001 was:
(a)    Rs.465crores  (b)Rs.385crores   (c)Rs.295crores    (d)Rs.255crores

Sol.1.(b) : Profit percent ofcompany Y in 1997=35.
Let the income of company Y in 1997 be Rs.x crores
Then,35 = x-220  X  100  Þ x =297
                   220
\Income of company Yin 1997 = Rs.297crores

     2.(d): Let the incomes of the twocompanies X and Yin 1999 be Rs.x and let the Expenditures of companies X and Y in 1999 be E1 and E2 respectively
            Then, for Company X we have:

50=  x-E1   x  100      Þ    50    = x     -1       Þ  x = 150  E1
          E1                            100     E1                          100

Also, for the Company Y we have:

60 =  x-E2   *100 =>60  = x  -1 =>x =160 E2
           E2                                                                                                      100     E2              100        
From (i) and (ii),we get
150 E1 =160 E2  =>E1=160 =16(Required ratio)
100         100           E2  150    15

3.(c):Let the incomes in 2000 of companies X and Y be 3x and 4x respectively.And let the expenditure in 2000 of  companies X and Y be E1 and E2 respectively.
Then, for company X we have:
65=3x-E1 *100  =>65 = 3x   -1 =>E1=3x  *(100)
        E1                  100   E1                            165
For company Y we have:
50 =4x-E2  *100 => 50 = 4x  -1 =>E2 = 4x* (100)
         E2                   100   E2                            150
From (i)and(ii) we get:
E1 = 3x*(100/165) =3* 150 =15(Required ratio)
E2    4x*(100/150)    4*165    22


4.(d):Let the expenditures of each of the Companies X and Y in 1996 be Rs.xcrores.And let the income of Company X in 1996 be Rs.zcrores so that the income of Company Y in 1996 =Rs.(342-z)crores.
Then,for company X we have:
40= z-x *100 => 40 = z   -1 => x  = 100z
         x                100   x                    140
Also for company Y we have:
45= (342-z)-x   *100 => 45 =  (342-z)  -1  =>x = (342 –z)* 100
            x                         100        x                              145
From(i)and (ii) we get:
100z =  (342-z) *100 =>z = 168
140                          145

Substituting z=168 in (i),we get: x=120
\ Total expenditure of companies X and Y in 1996=2x=Rs.240crores.
      Total income of companies X and Y in 1996=Rs.342 crores.
\Total profit =Rs.(342-240)crores =Rs.102 crores


5.(a): Let the income of company X in 1998 be Rs.x crores.
                                                                                    Then,55= x-200  *100 => x = 310.
                                      200
\Expenditure of Company X in 2001= Income of company X in 1998 = Rs.310crores
Let  the income of company X in 2001 be Rs.z crores
Then,50 = z-310  *100 =>z = 465.
                   310
\Income of company X in 2001 = Rs.465 crores.

Chapter-37 (PIE-CHARTS)



37. PIE-CHARTS

IMPORTANT FACTS AND FORMULAE

The pie-chart or a pie-graph is a method of representing a given numerical data in the form of sectors of a circle.
The sectors of the circle are constructed in such a way that the area of each sector is proportional to the corresponding value of the component of the data.
From geometry, we know that the area of a circle is proportional to the central angle.
So, the central angle of each sector must be proportional to the corresponding value of the component.
Since the sum of all the central angle is 360°, we have          
                                                         Value of the component                      °  
Central angle of the component =        ------------------------ * 360                                             
                                                                     Total value                                            

SOLVED EXAMPLES

The procedure of solving problems based on pie-charts will be clear from the following solved examples.

Example 1. The following pie-chart shows the sources of funds to be collected by the National Highways Authority of India (NHAI) for its Phase II projects. Study the pie-chart and answer the questions that follow.
                  
                     SOURCES OF FUNDS TO BE ARRANGED BY  NHAI
                        FOR PHASE II PROJECTS (IN CRORES RS.)







Total funds to be arranged for Projects (Phase II) =Rs.57,600 crores.
1. Near about 20% of the funds are to be arranged through:
(a) SPVS                     (b) External Assistance
(c) Annuity                  (d) Market Borrowing

2. The central angle corresponding to Market Borrowing is:
(a) 52°                                     (b) 137.8°
(c) 187.2°                                (d) 192.4°

3. The approximate ratio of the funds to be arranged through Toll and that through Market Borrowing is:
(a) 2:9                                      (b) 1:6
(c) 3:11                                    (d) 2:5

4. If  NHAI could receive a total of Rs. 9695 crores as External Assistance, by what percent (approximately) should it increase the Market Borrowings to arrange for the shortage of funds ?
(a) 4.5%                                   (b) 7.5%
(c) 6%                                      (d) 8%

5.If the toll is to be collected through an outsourced agency by allowing a maximum 10% commission, how much amount should be permitted to be collected by the outsourced agency, so that the project is supported with Rs. 4910 crores ?
(a) Rs.6213 crores                   (b) Rs. 5827 crores
(c) Rs. 5401 crores                  (d) Rs. 5216 crores
                                     

SOLUTION

1. (b): 20% of the total funds to be arranged = Rs.(20% of 57600) crores
                                                                       = Rs.11520 crores Rs.11486 crores.

2. (c):Central angle corresponding to Market Borrowing =   29952
                                                                                                --------  * 360°
                                                                                                57600
                                                                                          = 187.2°
3. (b):                             4910      1        1
         Required ratio =  ------- = ---- = ----
                                     29952    6.1      6  

4. (c):Shortage of funds arranged through External Assistance
                              =Rs.(11486-9695) crores =Rs. 1791 crores.
        therefore, Increase required in Market Borrowings =Rs. 1791 crores.
       
                                                             1791    
     Percentage increase required =     -------- * 100   % =  5.98 % = 6%
                                                            29952
5. (c):Amount permitted  = (Funds required from Toll for projects of Phase II ) +
                                                                                    (10 % of these funds)            
                                                =Rs. 4910 crores + Rs. (10% of 4910) crores            
                                                =Rs. (4910 + 491) crores = Rs. 5401 crores.

Example 2. 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.

    DISTRIBUTION OF AREAS (IN ACRES) UNDER VARIOUS FOOD CROPS


1.Which combination of three crops contribute to 50% of the total area under the food crops ?
(a) Wheat, Barley and Jowar                                           (b)Rice, Wheat and Jowar
(c) Rice, Wheat and Barley                                             (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 %         (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°

                                                                      SOLUTIONS
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.

2.(a):   The area under any of the food crops is proportional to the angle corresponding to that crop.
Let the area under the rice production be x million acres.
Then, 18:72 = 1.5:x Þ x=(72*15/18)=6
Thus, the area under rice production be = 6 million acres.

3.(b): Let the total production of barley be T tones and let Z acres of land be put under barley    production.
Then, the total production of wheat =(6T) tones.
Also,area under wheat production = (2Z) acres.

\Area Under Wheat Production     72°    
   ------------------------------------- = ----   =2 
            Area Under Barley Production    36°    
            And therefore,Area under wheat = 2*Area under Barley = (2Z)acres

Now, yield per acre for wheat = (6T/2Z) tones/acre = (3T/Z) tones/acre
And yield per acre for barley = (T/Z) tones/acre.
                                              3T/Z
             \Required ratio  = -------- = 3:1.  
                                               T/Z
        4. (b):Let Z acres of land be put under barley production.
                                               Area Under Rice Production         72 °
                              Then,       -----------------------------------   =   ----  =  2.
                                               Area Under Barley Production       36°
\Area under rice production = 2 * area under barley production = (2Z) acres.
Now,if p tones be the yield per acre of barley then ,yield per acre of rice
                        =(p+50% of p) tones =(3/2 p) tones.
\Total production of rice = (yield per acre) * (area under production)
                                          =  (3/2 p)*2Z=(3pZ) tones.
And,Total production of barley = (pz) tones.
\Percentage production of barley to that rice = (pZ/3pZ *100)%= 33 1/3% .

5.(b): Initially,let  t  be the total area under considerations.
The 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%,
                                                                       t                      t
then the new value of area under wheat =     --  +(12% of   --)   acres = (112t/500)acres.
                                                                       5                     5
 \Central angle corresponding to wheat in the pie-chart
                                 Area  Under Wheat (new)           °        (112t/500)         °
                         =      ------------------------------- *360    =     ------------*360      =76.8°
                                   Total area (new)                                 (105t/100)
                                                                        
Example 3.The following pie-charts show the distribution of students of graduate and post graduate levels in seven different institute-M,N,P,Q,R,S and T in a town.

DISTRIBUTION OF STUDENTS AT GRADUATE AND POST-GRADUATE     LEVELS  IN SEVEN INSTITUTES-M,N,P,Q,R,S AND T.

             Total     Number of students of                               Total     Number of students of  
                               graduate level                                                  post graduate level







1.How many students of institutes M and S are studying at graduate level?
    (a) 7516          (b) 8463         (c) 9127      (d) 9404

2.Total number of students studying at post -graduate level from institutes N and P is:
     (a) 5601          (b) 5944          (c) 6669       (d) 7004

3.What is the total number of graduate and post-graduate level students in institute R?
      (a) 8320          (b) 7916        (c) 9116         (d) 8372
      
      4 .What is the ratio between the 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

5.What is the ratio between the number of students studying 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

                                                            SOLUTION
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 students at graduate level in institutes M and S = 4641+3822=8463                                                                               

2.(c):Required number = (15% of 24700) + (12% of 24700) = 3705 + 2964 = 6669.

3.(d):Required number = (18% of 27300) + (14% of 24700) = 4914 + 3458 = 8372.

4.(d):Required ratio =  (21% of 24700) = 21 * 24700 = 19
                                     (14% of 27300)    14 * 27300     14

5.(d):Required ratio = (21% of 24700) = 21 * 24700 = 19
                                    ( 13% of 27300)    13 * 27300    13

Example 4.Study the following pie-chart and the table and the answer the questions based on them.
                  
PROPORTION OF POPULATION OF SEVEN VILLAGES IN 1997

                                                                                                              
                                 




Oval:        R 16%     Z 11%

S 11%                     Y15 %

                             X 16%     
                                
T 21%         V 10%     
                                                                                        
village
% population below   poverty
X
38
Y
52
Z
42
R
51
S
49
T
46
V
58



1.Find the population of villages S if the population of village X below poverty line in 1997 is 12160.
    (a). 18500          (b) 20500          (c) 22000          (d) 26000

2.The ratio of the population of the village T below poverty line to that of village Z below poverty line in 1997 is:
     (a) 11:23           (b) 13:11          (c) 23:11            (d) 11:13

3.If the population of village R in 1997 is 32000,then what will be the population of village Y below poverty line in that year?
      (a) 14100          (b)15600          (c) 16500          (d) 17000

4.If in 1998, the population of villages Y and V increases by 10% each and the percentage of population below poverty line remains unchanged for all the villages, then find the population of village V below poverty line in 1998,given that the population of village Y in 1997 was 30000.
       (a) 11250          (b) 12760           (c) 13140           (d) 13780

5.If in 1998,the population of village R increases by 10% while that of village Z reduces by 5% compared that in 1997 and the percentage of population below poverty line remains unchanged for all the village,then find the approximate ratio of population of village R below poverty line for the year 1999.
        (a) 2:1                (b)  3:2              (c) 4:3              (d) 5:4

                                                                  SOLUTION
1.(c):Let the population of village X be x
         Then,38% of x=12160  Þ  x = 12160 * 100      =3200
                                                                 38
         Now ,if s be the population village S,then
                                                              11 * 32000
         16:11 = 32000 : s           Þ  s=                                 = 22000.
                                                                     16
2.(c): Let N be the total population of all the seven villages.
Then ,population of village T below poverty line = 46% of (21% of N) and population of   village Z below poverty line = 42% of (11% of N)


       \Required ratio   =   46% of (21% of N)   =      46 * 21   =    23
                                         42% of (11% of N)           42 * 11           11     

3.(b): Population of village R = 32000(given)
          Let the population of village Y be y.
          Then, 16:15  =  32000 : y     Þ    y  = 15 * 32000   =30000
                                                                             16
4.(b): Population of village Y in 1997  =  30000(given) .
          Let the population village V in 1997 be v.
             Then, 15:10 = 30000:v       Þ  v  = 30000 * 10       = 20000.
                                                                              15                          
           Now population of village V in 1998 = 20000 + (10% 0f 20000) = 20000.
           \Population of village V below poverty line in 1998  =  58% of 22000 = 12760.

5.(a) : Let the total population of all the seven villages in 1997 be N.
           Then,population of village R in 1997 = 16% of N =  16/ 100 N
            And population of village Z in 1997 = 11% of N =11/100 N

            \Population of village R in 1999 = {16/100N+(10% of 16/100 N)}=1760/10000 N
           and population of village Z in 1999 = {11/100 N-(5% of 11/100 N)} = 1045/10000 N.
           Now,population of village R below poverty line for 1999 = 51% of(1760/10000 N)
           And population of village Z below poverty line 1999 = 42% of (1045/10000 n)
                                                    51% of (1760/10000 N)      51 * 1760        2
Required ratio =           -------------------------- =  -------------   =  ------    
                                                     42% of (1045/10000 N)    42 * 1045