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Data Interpretation For IBPS PO And RRB PO / CLERK MAIN

 


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Direction (1 - 4)The table given below shows the number of candidates appeared from a state for five different exams over a period of 4 years.



1.If IBPS conducted two exams PO & Clerk in all the years. The ratio of candidates appeared for PO to those appeared for Clerk in the given years is 19: 18, 71: 54, 27: 26, and 67: 60 respectively. What more/less percent candidates appeared for IBPS Clerk in all the years than candidates appeared for NABARD & RBI together in all the years? (Round off to 2 digits after decimal)


A. 

0.45% more


B. 

0.23% less


C. 

1.79% more


D. 

0.79% less


E. 

0.95% more


2.There are only 800 vacancies announced by LIC in all the given years and only 1875 candidates qualified the interview which is last stage of selection. The ratio of non-qualified candidates in given years is 42: 39: 53: 45 respectively. At least 30% candidates who qualified last stage got selected. What can be the maximum possible vacancy percentage filled in 2019?


A. 

52.625%


B. 

53%


C. 

52.25%


D. 

53.5%


E. 

52%


3.If all candidates appeared for NABARD also appeared for SBI, candidates appeared for LIC qualified the last stage of SBI and candidates appeared for RBI got selected through SBI exam. How many times number of candidates who did not get selected from the last stage is the number of candidates who appeared only for SBI in all the given years? (Round off to one digit after decimal)


A. 

0.3


B. 

0.4


C. 

0.5


D. 

0.2


E. 

0.6


4.In which of the two following exams & year, there is a maximum increase in the number of candidates appeared as compared to previous year?


A) SBI in 2017


B) LIC in 2017


C) IBPS in 2019


D) NABARD in 2018


E) RBI in 2018


A. 

A and D


B. 

A and E


C. 

B and C


D. 

B and D


E. 

C and D



5.Read the following passage carefully and answer the given questions. (1.5 marks)


I). In an institute X, there are four courses i.e. Engineering, Science, Arts and Commerce. Number of candidates enrolled in Science course is 1080 while that in Arts course is 880. The number of girls in Engineering course is 50% more than number of boys in Arts.


II). The number of boys in Commerce is 33.33% more than number of girls in Science. The ratio of number of boys to number of Girls in Commerce course is 4: 3.


III). There are (A) girls in Arts course which is half of number of boys in Science course which is half of number of girls in Engineering course. The ratio of number of boys in all courses to that of number of girls is (B): (C). Number of candidates enrolled in Commerce course is 25% of total candidates enrolled in all the courses together.


(NOTE: Total Candidates = Number of boys + Number of girls)



5.Which of the following is true as per the information provided in the passage?


A. 

A = 300; B > C


B. 

A= 240; B < C


C. 

A =300; B < C


D. 

A = 240; B > C


E. 

None of these


6.If there is another course of Management in which number of boys is 50% of number of girls in Science and total number of candidates in it are 25% less than number of total candidates in Arts. Number of girls in Management is D. Then


A. 

A > D


B. 

A < D


C. 

A ≥ D


D. 

A ≤ D


E. 

A = D


7.If in next year, there is an increase of 10% in number of boys in Engineering & Science both while number of girls remain same. What will be the ratio of number of Engineering candidates to number of Science candidate in next year?


A. 

149: 70


B. 

298: 141


C. 

296: 141


D. 

99: 47


E. 

74: 35


8.If there would be no course of Science but total candidates remain same such that all boys and girls from that course are equally divided in all the three remaining courses. What would be the ratio number of boys in Engineering & Arts together to number of girls in Arts & Commerce together?


A. 

57: 53


B. 

None of these


C. 

2: 1


D. 

55: 31


E. 

56: 31


9.What is the percentage of number of boys in Science & Commerce together as compared to number of girls in Engineering & Arts together?


A. 

106.67 %


B. 

105 %


C. 

110 %


D. 

103.25 %


E. 

107.5 %


10.Karan travels from X at a speed of (V + 6) kmph to reach a place Y ‘D’ km far from his starting point to save some journey time. If he travels at ‘V’ kmph, he takes T hour to complete the journey. What is necessary to determine the actual speed of Karan?


Statement I: If he travels one side at his usual speed while returns to his starting position with (V + 6) kmph, time taken is (2T - 1) hours.


Statement II: Ajay who is coming from Y at a speed of 50% more than Karan’s actual speed towards X meets Karan after covering 3D/5 km. (both start at same time)


Statement III: Distance from X to Y is given by (D2 - 100D – 2400 = 0).


A. 

Statements I & II together are required to answer.


B. 

Statements II & III together are required to answer.


C. 

None of the given statements is sufficient to answer, more data is required.


D. 

Statements I & III together are required to answer.


E. 

Any two of the three statements are sufficient to answer.


SOLUTIONS:


1.Answer: D


Candidates appeared for IBPS Clerk in all the years = 1850/37 × 18 + 2000/125 × 54 + 2120/53 × 26 + 2540/127 × 60


= 900 + 864 + 1040 + 1200 = 4004


Candidates appeared for NABARD in all the years = 680 + 755 + 844 + 867


= 3146


Candidates appeared for RBI in all the years = 225 + 241 + 236 + 188 = 890


Candidates appeared for NABARD & RBI together in all the years = 3146 + 890


= 4036


Difference of IBPS Clerk candidates & NABARD and RBI together = 4004 – 4036


= -32


∴more candidates appeared for NABARD & RBI together than IBPS Clerk


Required % = -32/4036 × 100 = 0.79%


2.Answer: A


Total candidates appeared for LIC in all the years = 530 + 612 + 792 + 836


= 2770


Candidates not qualified in all the years = 2770 - 1875 = 895


Candidates not qualified in 2016 = 895/179 × 42 = 210


Candidates not qualified in 2017 = 895/179 × 39 = 195


Candidates not qualified in 2018 = 895/179 × 53 = 265


Candidates not qualified in 2019 = 895/179 × 45 = 225


In order for maximum selection in 2019, there should be minimum selection in 2016, 2017 & 2018 together = 30% of (selection in 2016, 2017 & 2018)


=30/100 × [(530 - 210) + (612 - 195) + (792 - 265)] = 379.2 ≈ 379


Maximum possible selection in 2019 = 800 - 379 = 421


Required % = 421/800 × 100 = 52.625%


3.Answer: B


Candidates who appeared for only SBI = (1720 - 680) + (1840 - 755) + (2115-844) + (2230 - 867) = 4759


Candidates who did not get selected through SBI = (530 - 225) + (612 - 241) + (792 - 236) + (836 - 188) = 1880


Required answer = 1880/4759 = 0.4


4.Answer: C


A) SBI in 2017 = (1840 - 1720)/1720 × 100 = 6.98% ≈ 7%


B) LIC in 2017 = ((612 - 530))/530 × 100 = 15.47% ≈ 15%


C) IBPS in 2019 = (2540 - 2120)/2120 × 100 = 19.81% ≈ 20%


D) NABARD in 2018 = (844 - 755)/755 × 100 = 11.79% ≈ 12%


E) RBI in 2018 = (236 - 241)/241 × 100 = -2.07% ≈ -2% (decrease in number of candidates)


5.From I,

Number of Candidates in Science course = 1080

Number of candidates in Arts course = 880

Let number of boys in Arts course = a

Number of girls in Engineering course = a + 50/100 × a = 1.5a

From II,

Let number of girls in Science course = 3b

Number of boys in Commerce course = 3b + 33.33/100 × 3b = 4b

Number of girls in Commerce course = ¾ × 4b = 3b

From III,

Number of boys in Science course = ½ ×1.5a = 0.75a

Number of girls in Arts course = ½ × 0.75a = 0.375a

Now, 880 = a + 0.375a

a = 640

1080 = 0.75a + 3b

b = 200

Total candidates in Commerce course = 4b + 3b = 7b = 1400

Total candidates enrolled in all the courses together = 1400/25 × 100 = 5600

Total candidates in Engineering course = 5600 - (1080 + 880 + 1400) = 2240

Number of boys in Engineering course = 2240 - 1.5a = 1280

Then,

Engineering:

Total = 2240

Boys = 1280

Girls = 1.5a = 960

Science:

Total = 1080

Boys = 0.75 a = 480

Girls = 3b = 600

Art:

Total = 880

Boys = a = 640

Girls = 0.375 a = 240

Commerce:

Total = 1400

Boys = 4b = 800

Girls = 3b = 600

Answer: D


A = number of girls in Arts = 240


B/C = (total number of boys)/(total number of girls)


= (1280 + 480 + 640 + 800)/(960 + 600 + 240 + 600) = 3200/2400 = 4∶ 3


A = 240; B > C


6.Answer: B


Number of boys in Management course = 50/100 × 600 = 300


Total candidates in Management = 880 - 25/100 × 880 = 660


Number of girls in Management course = D= 660 – 300 = 360


Clearly, A < D


7.Answer: C


Number of candidates in Engineering in next year = 110/100 × 1280 + 960 = 2368


Number of candidates in Science in next year = 110/100 × 480 + 600 = 1128


Required ratio = 2368∶ 1128 = 296∶ 141


8.Answer: E


Number of boys divided = 480/3 = 160


Number of girls divided = 600/3 = 200


Number of boys in Engineering & Arts together = 1280 + 160 + 640 + 160 = 2240


Number of girls in Arts & Commerce together = 240 + 200 + 600 + 200 = 1240


Required ratio = 2240∶ 1240 = 56∶ 31


9.Answer: A


Number of boys in Science & Commerce together = 480 + 800 = 1280


Number of girls in Engineering & Arts together = 960 + 240 = 1200


Required % = 1280/1200 × 100 = 106.67%


10.Answer: D


Distance, D = VT (km)


Let he saves ‘x’ hours by travelling at increased speed


D = (V + 6)(T - x) = VT


D/(V + 6) = T - x; D/V = T


From I,


D/V + D/(V + 6)=2T - 1


T + T - x = 2T - 1


x = 1 hour


This statement alone is not sufficient to answer the question.


From II,


Speed of Ajay = 150/100 × V = 1.5V kmph


Since meeting time is same


(3D⁄5)/1.5V = (D - 3D/5)/V


Nothing can be find from statement II. So,


This statement alone is not sufficient to answer the question.


From III,


Distance, (D2 - 100D – 2400 = 0)


On solving, D = 120 km (another value is -20 which is not possible)


This statement alone is not sufficient to answer the question.


From I & III,


120/(V + 6) = 120/V - 1


120V = 120V + 720 - V2 - 6V


V2 + 6V – 720 = 0


V = 24 kmph (neglecting negative value)


So, Statement I & III together are required to answer the question.


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