39 CHAP

# 39 CHAP - 39.LINE-GRAPHS This section comprises of question...

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39.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%

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2. 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 3. 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% 4. 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 5. 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. 3. (d) : Required percentage = [(370+385)/(500+500) * 100] % = [(755/1000)*100]% =75.5% 4. (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. 5.
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## This note was uploaded on 08/13/2011 for the course FSD 011 taught by Professor Vinh during the Spring '11 term at Beacon FL.

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39 CHAP - 39.LINE-GRAPHS This section comprises of question...

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