Data Interpretation Tables Questions for MAH-CET

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Data Interpretation Tables Questions

Data Interpretation Tables Questions for MAH-CET

Here you can download a free Data Interpretation Tables questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the Data Interpretation Tables of answers for the given questions. These questions will help you to make practice and solve the Data Interpretation Tables questions in the MAH MBA CET exam. Utilize this best PDF practice set which is included answers in detail. Click on the below link to download the Data Interpretation Tables MCQ PDF for MBA-CET 2022 for free.

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Instructions

Read the following Table and answer the given questions

 

 

Question 1: In which of the following subjects was the average percentage of marks obtained by student S the highest?

a) Maths

b) Science

c) History

d) Geography

e) English

1) Answer (C)

Solution:

Total marks obtained by student S in subject :

Maths = 25 + 35 + 40 + 45 + 30 = 175

Science = 31 + 34 + 38 + 27 + 30 = 160

History = 34 + 40 + 36 + 42 + 48 = 200     [MAXIMUM]

Geography = 39 + 37 + 44 + 40 + 30 = 190

English = 31 + 34 + 35 + 45 + 40 = 185

=> Ans – (C)

Question 2: The average percentage of marks obtained by student P in Maths in five periodicals was exactly equal to the average percentage of marks obtained by Student R in the five periodicals in which of the following subjects ?

a) English

b) Geography

c) Science and Geography

d) Maths

e) None of these

2) Answer (B)

Solution:

Total marks obtained by student P in Maths = 40 + 30 + 45 + 20 + 35 = 170

=> Average marks of P in Maths = $\frac{170}{5} = 34$

Marks scored by student R in subject :

Maths = 30 + 36 + 40 + 45 + 40 = 191

Science = 48 + 46 + 31 + 40 + 80 = 245

History = 35 + 45 + 40 + 30 + 35 = 185

Geography = 25 + 35 + 48 + 37 + 25 = 170

=> Average marks of R in Geography = $\frac{170}{5}=34$

=> Ans – (B)

Question 3: what was the total marks of the student’s T in the science in all the periodicals together?

a) 160

b) 180

c) 190

d) 140

e) None of these

3) Answer (C)

Solution:

Total marks of the student’s T in the science in all the periodicals together

= 44 + 36 + 40 + 30 + 40

= 190

=> Ans – (C)

Question 4: What was the average marks of five Subjects of Q in the periodical I ?

a) 32

b) 34

c) 40

d) 30

e) None of these

4) Answer (A)

Solution:

Total marks of five Subjects of Q in the periodical I

= 30 + 25 + 33 + 42 + 30 = 160

=> Required average = $\frac{160}{5}=32$

=> Ans – (A)

Question 5: In which of the following periodicals ,the student P obtained highest percentage of the marks?

a) I

b) II

c) III

d) IV

e) V

5) Answer (E)

Solution:

Total marks obtained by student P :

I = 40 + 30 + 35 + 45 + 24 = 174

II = 30 + 40 + 25 + 47 + 28 = 170

III = 45 + 25 + 15 + 32 + 36 = 153

IV = 20 + 30 + 30 + 39 + 39 = 158

V = 35 + 20 + 40 + 37 + 43 = 175       [MAXIMUM]

=> Ans – (E)

Instructions

Study the following table carefully and answer the given questions

 

Question 6: The total number of candidates qualified from Rural in 1993 and the semi urban in 1990 was exactly equal to the total no of candidates qualified from state -capitals in which of the following years?

a) 1990

b) 1993

c) 1994

d) 1992

e) None of these

6) Answer (E)

Solution:

Number of candidates qualified from Rural in 1993 = 1798

Number of candidates qualified from Semi-Urban in 1990 = 2513

Total = 1798 + 2513 = 4311

Clearly, in none of the year, total number of qualified candidates from state-capitals (Metropolises) is equal to 4311

=> Ans – (E)

Question 7: In 1993 the percentage of candidates qualified to the appeared was 35 from which location?

a) Rural

b) Rural and Metropolises

c) Semi-urban and metropolises

d) Rural and semi urban

e) None of these

7) Answer (A)

Solution:

Percentage of qualified candidates to appeared candidates in 1993 in :

Rural = $\frac{1798}{5032} \times 100 =35.73 \%$

Semi-Urban = $\frac{3528}{9432} \times 100 = 37.40\%$

Metropolises = $\frac{5158}{11247} \times 100 = 45.86\%$

Thus, from rural location the percentage of candidates qualified to the appeared was approximately 35 in 1993

=> Ans – (A)

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Question 8: what was the percentage drop in the Number of semi urban candidates qualified to the appeared from 1991 to 1992?

a) 5

b) 10

c) 15

d) 8

e) 12

8) Answer (A)

Solution:

Percentage of qualified candidates to appeared candidates in Semi-Urban in 1991

= $\frac{2933}{8562} \times 100 = 34.25 \approx 34\%$

Percentage of qualified candidates to appeared candidates in Semi-Urban in 1992

= $\frac{2468}{8139} \times 100 = 30.32 \approx 30\%$

$\therefore$ Percentage drop = 34 – 30 = 4 $\approx$ 5%

=> Ans – (A)

Question 9: In which of the following years was the percentage of candidates qualified to the appeared from the semi-urban area least?

a) 1991

b) 1993

c) 1990

d) 1992

e) None of these

9) Answer (D)

Solution:

Percentage of candidates qualified to the appeared from the semi-urban area in :

(A) : 1991 = $\frac{2933}{8562} \times 100= 34.25\%$

(B) : 1993 = $\frac{3528}{9432} \times 100= 37.40\%$

(C) : 1990 = $\frac{2513}{7894} \times 100= 31.83\%$

(D) : 1992 = $\frac{2468}{8139} \times 100= 30.32\%$     [MINIMUM]

=> Ans – (D)

Question 10: For the candidates from which of the following locations was there continuous increase both in the appeared and the passed?

a) Semi urban

b) State Capitals

c) State capitals and rural

d) Metropolis

e) None of these

10) Answer (E)

Solution:

In Rural location, the number of candidates appeared in 1994 decreased as compared to 1993.

Similarly, in semi-urban and Metropolises location, the number of candidates appeared in 1992 decreased as compared to 1991.

Thus, in none of the locations, there was a continuous increase both in the appeared and the passed.

=> Ans – (E)

Instructions

Study the following Information carefully and answer the questions given below
Number of students appeared ,passed and granted scholarships over the years

 

Question 11: Which of the following seems table most possible reason for substantial increase of number appeared in 1999 ?

a) The increase in the Progressive and in line with the trend from 1994

b) Substantial increase in the Number of scholarships in 1998 as compared to other years

c) Proportion of passed to Appeared increasing progressive?

d) General population increase in the later half of 1999

e) Pass number crossing 200 marks

11) Answer (B)

Solution:

In 1998, the number of scholarships granted was increased as compared to other years and this can be the reason for substantial increase of number appeared in 1999.

=> Ans – (B)

Question 12: In which year is the scholarship Granted to passed Ratio is highest?

a) 1999

b) 1998

c) 1996

d) 1995

e) None of these

12) Answer (B)

Solution:

Ratio of scholarship Granted to passed in :

1994 : $\frac{7}{153}$ = 0.04

1995 : $\frac{7}{151}$ = 0.04

1996 : $\frac{10}{163}$ = 0.06

1997 : $\frac{11}{179}$ = 0.06

1998 : $\frac{15}{182}$ = 0.08   [MAX]

1999 : $\frac{15}{209}$ = 0.07

=> Ans – (B)

Question 13: In which year was the percentage passed to appeared the highest?

a) 1999

b) 1994

c) 1996

d) 1998

e) None of these

13) Answer (C)

Solution:

Ratio of percentage passed to appeared in :

1994 : $\frac{153}{283}$ = 0.54

1995 : $\frac{151}{280}$ = 0.53

1996 : $\frac{163}{291}$ = 0.56    [MAX]

1997 : $\frac{179}{320}$ = 0.55

1998 : $\frac{182}{357}$ = 0.50

1999 : $\frac{209}{422}$ = 0.49

=> Ans – (C)

Question 14: What is the approximate percentage of total passed to total appeared ?

a) 51

b) 55

c) 53

d) 49

e) 57

14) Answer (C)

Solution:

Total appeared students

= 283 + 280 + 291 + 320 + 357 + 422 = 1953

Total passed students

= 153 + 151 + 163 + 179 + 182 + 209 = 1037

=> Required % = $\frac{1037}{1953} \times 100$

= $53.09 \approx 53 \%$

Question 15: What is the approximate Percentage of total Granted Scholarship to the total appeared?

a) 3.5

b) 4.5

c) 4

d) 2.5

e) 2.0

15) Answer (A)

Solution:

Total students who were granted scholarship

= 7 + 7 + 10 + 11 + 15 + 15 = 65

Total appeared students

= 283 + 280 + 291 + 320 + 357 + 422 = 1953

=> Required % = $\frac{65}{1953} \times 100$

= $3.3 \approx 3.5 \%$

Instructions

Read the following table to answer the given questions

Average production of six Machines of the given years in thousands

 

Question 16: For which machine has the less number there been increase in the productin from the previous year?

a) No machine

b) III

c) IV

d) II

e) None of these

16) Answer (A)

Solution:

EndGroup:
StrartGroup:
Instructions: Study the following graph carefully and answer the following questions

Question 17: For which year and the machine has the production highest for the given date?

a) 1999,IV

b) 1998,IV

c) 1998,III

d) 1997,III

e) 1996,IV

17) Answer (B)

Solution:

Production of the machines in thousands :

(A) : 1999,IV = 2050

(B) : 1998,IV = 2070   [MAX]

(C) : 1998,III = 1040

(D) : 1997,III = 1043

(E) : 1996,IV = 1908

=> Ans – (B)

Question 18: Which of the following can be concluded?

a) As the machine becomes older ,the production goes down

b) The production goes down in the initial two or three years then it starts improving

c) All the fluctuations from one year to another year are in the range in the 100

d) Each even numbered machine produces more than odd numbers

e) None of these

18) Answer (E)

Solution:

(A) : Not true for all the machines, eg., Production of machine I increased in 1998 and machine II increased in 1997.

(B) : Not true, as for machine V production increased in the final year.

(C) : Not true, in machine IV, difference in production in 1996 and 1995 is more than 100.

(D) : Not true, as machine II produces less than machine I.

=> Ans – (E)

Question 19: Which machines have shown the least fluctuation in production?

a) I

b) II

c) V

d) VI

e) None of these

19) Answer (B)

Solution:

Fluctuations in production over the years by machine :

I : 6 + 60 + 40 + 60 = 166

II : 2 + 4 + 3 + 0 = 9    [LEAST]

III : 20 + 17 + 3 + 20 = 60

IV : 305 + 195 + 33 + 20 = 553

V : 5 + 10 + 10 + 10 = 35

VI : 140 + 40 + 20 + 20 = 220

=> Ans – (B)

Question 20: How many machines has production lower than 700 for all given years?

a) Nil

b) One

c) Two

d) Three

e) None of these

20) Answer (C)

Solution:

Machine I in the year 1995 and 1996 has production greater than or equal to 700.

Similarly, machine III,IV and VI also has production more than 700 over the years.

Only 2 machines i.e., II and V have production lower than 700 for all the given years.

=> Ans – (C)

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