# Data Interpretation - Table Charts - Discussion

### Discussion :: Table Charts - Table Chart 7 (Q.No.2)

A school has four sections A, B, C, D of Class IX students.

The results of half yearly and annual examinations are shown in the table given below.

 Result No. of Students Section A Section B Section C Section D Students failed in both Exams 28 23 17 27 Students failed in half-yearlybut passed in Annual Exams 14 12 8 13 Students passed in half-yearlybut failed in Annual Exams 6 17 9 15 Students passed in both Exams 64 55 46 76

2.

How many students are there in Class IX in the school?

 [A]. 336 [B]. 189 [C]. 335 [D]. 430

Answer: Option D

Explanation:

Since the classification of the students on the basis of their results and sections form independent groups, so the total number of students in the class:

= (28 + 23 + 17 + 27 + 14 + 12 + 8 + 13 + 6 + 17 + 9 + 15 + 64 + 55 + 46 + 76)

= 430.

 Shashwat Shekhar said: (Aug 28, 2012) How do they form independent groups?

 Sai Madhav.K said: (Aug 30, 2012) Students failed in both the exams can be inclusive of students failed in Half yearly but passed in Annual and students failed in Annual but passed in Half yearly right? Its not mentioned in the question that they perform independent operation, then how can we take that criteria into consideration? According to the question the answer is 336. Kindly correct me if I am wrong.

 Ashish Jain said: (Feb 21, 2013) 430 is right one because there may be 4 persons e.g A, B, C and D whose result may vary. A passed in both exams. B failed in both exams. C passed in half yearly and failed in annual. This is diff from B. D failed in half yearly and passed in annual. This is diff from A. Please correct me if I'm wrong.

 Shweta Shinde said: (Jul 26, 2013) Consider 2 venn diagrams for fail and pass students. In fail we have failed in both, failed in annual but not half, Failed in half but not annual, so total failed students are addition of all (28+23+17+27 + 14+12+... +16+17+... ) = x. Now in pass venn diagram we have passed in both, And passed in 1 and failed in other is already considered so add 64+55+... To x n we get 430.

 Abc said: (Dec 1, 2013) According to me the answer would be 335. How do they form independent groups?

 Manav said: (Feb 10, 2014) Students failed in both the exams can't pass any or both of the exam. Student failed in half yearly only can't failed in both exam also pass both the exam also pass annual. So all the 4 groups are independent of each other. NOW short cut to solve this question. Just add all the tens position number only. Adding row wise. 20+20+10+20 (first row over) +10+10+0+10 (second row over) +0+10+0+10 (third row over) +60+50+40+70 = 340. Now see the options only one option is bigger than 340 tick it and enjoy the life.

 Seetha said: (May 10, 2014) I think the correct answer is 336. Total students in the class = (no of students passed in both exams + no of students failed in both exams).

 Darshana said: (Jun 24, 2014) The four group are independent so the answer is sum of all the numbers.

 Prasanna said: (Dec 16, 2016) Why 430? it should be 336 (sum of students passed in both in exams + sum of students failed in both the exam).

 Kusum said: (Feb 4, 2017) I think the answer should be 336.

 Kushi said: (Feb 4, 2017) The answer should be 336. The sum of failed student and the passed one in both exams.

 Suresh said: (Feb 4, 2017) Yes, you are correct @Prasanna.

 Rahul Raghuwanshi said: (Feb 12, 2017) According to me, the answer is 336.

 Gauri Pawar said: (Oct 17, 2017) Answer is 336. Total students = Passed in both exam + failed in both exam = 241 +95 =336.

 Solomon said: (Mar 16, 2018) The answer is 335.

 Akash Singh said: (Feb 16, 2019) HF + AF = 28+23+7+27. HF + AP = 14+12+8+13. HP + AF = 6+17+9+15. HP + AP = 64+55+46+76. Adding all equations:- 2(HP+HF+AP+AF) = 430. HP+ HF = no. student = NS. AP +AF = no. student = NS. 2(NS +NS) = 430. 4*NS = 430.

 Arpith said: (May 31, 2019) To all those who consider 366 is the answer. Please note this. Let's keep it simple.. let's assume the number of students as 4. One student has passed in HY failed in AE. One student has failed in HY and passed in AE. One student has passed in both. One student has failed in both. Now, draw a table with this. If you have to get the class strength you need to add them 'ALL'. Which is (1+1+1+1) = 4. NOT, (students who passed in both) + students who (failed in both), that'll give you (1+1)= 2. Which is wrong.

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