Data Interpretation Questions Solution Bank PO Clerk Set 19

Hello Friends, We are sharing some important Data Interpretation Questions Solutions Bank PO Clerk IBPS exams 2017 2018. These Data Interpretation Questions Solutions are very important for IBPS Clerk, SBI PO exam and IBPS PO & other upcoming exams like for SSC, CAT, XAT, NMAT, MAT & other exams 2017.

Data Interpretation Questions Solutions Bank PO Clerk provided by Bank4Study.com will help you a lot in your goal to crack the test. You will be taught to solve these questions better by solving many Data Interpretation problems.

You need to practice as many DI Questions as possible for Competitive Exams to know data interpretation techniques. Data Interpretation Questions Solutions Bank PO Clerk will help you master the concepts of data interpretation better. Once you are sure that you have mastered the concepts involved in data interpretation, then you can easily crack the Quantitative Section of Exams.

Data Interpretation Questions Solution Bank PO Clerk

Directions (Q. 1–5) : Study the following graph to answer the given questions—

The graph depicts the impact on production when three groups of employees were given three different types of incentive schemes during the period February to June.

1. The approximate impact (i.e., increase in production) of ‘No Supervision’ from Jan. to February was approximately what per cent to that of ‘Efficiency Increment’ on production ?

(A) 110

(B) 260

(D) 95

(E) None of these

1. (E) Increase from Jan, to Feb. in ’No Supervision’. = 30 – 15 = 15 thousand

Increase from Jan, to Feb. in ‘Efficiency Increment’ = 25 – 15 = 10 thousand

Reqd. percentage = 15/10 × 100% = 150% Ans.

2. For which of the following pairs of months the total production of all the three groups together is exactly the same ?

(A) January & June

(B) February & April

(D) February & March

(E) None of these

2. (B) Total production of all the three groups in Feb. = 15 + 25 + 30 = 70 thousand

total production of all the three groups in April = 15 + 20 + 35 = 70 thousand Ans.

3. Which of the following is the increase in average production after ‘No Supervision’ scheme was introduced from February ?

(A) 16,000

(B) 10,000

(D) 31,000

(E) None of these

3. (E) Production in Feb. in ‘No supervision’ = 30 thousand

average production in ‘No supervision’ from Feb. to June = (30 + 35 + 35 + 30 + 25)/5 = 31 thousand

Reqd. increase = 31 – 30 = 1 thousand Ans.

4. Which of the incentives showed the most dramatic impact in June ?

(A) No supervision

(B) All the three

(D) Lunch Allowance

(E) Efficiency Increment & No supervision

4. (B) In the month of June the decrease of each is 5000

5. How many items were produced after the introduction of ‘Efficiency Increment’ by that group till June ?

(A) 1,20,000

(B) 1,00,000

(D) 1,50,000

(E) None of these

5. (C) Total production from Feb. to June in ‘Efficiency Increment’ = (25 + 25 + 20 + 20 + 15) thousand = 1,05,000

Directions (Q. 6–10) : Study the following table carefully and answer the questions based on it.

6. What was the total number of defective B type toys in 1995 and defective D types toys in 1993 ?

(A) 13290

(B) 14600

(D) 14260

(E) None of these

6. (B) No. of defective B type of toys in 1995 + No. of defective D type toys in 1993. = (64000 × 14)/100 × (47000 × 12)/100 = 8960 + 5640 = 14600 Ans.

7. The average production in the given years of which of the following types of the toys was highest ?

(A) B

(B) E

(D) C

(E) None of these

7. (C) Average production of ‘A’ type of toys = (76 + 82 + 65 + 70 + 85 + 80)/6 × 1000 = (458 × 1000)/6

Average production of ‘B’ type of toys = (58 + 46 + 49 + 52 + 64 + 54)/6 × 1000 = (323 × 1000)/6

Average production of ‘C’ type of toys = (39 + 37 + 45 + 42 + 38 + 40)/6 × 1000 = (241 × 1000)/6

Average production of ‘D’ type of toys = (59 + 62 + 47 + 54 + 57 + 68)/6 × 1000 = (347 × 1000)/6

Average production of ‘E’ type of toys = (28 + 36 + 42 + 31 + 49 + 38)/6 × 1000 = (224 × 1000)/6

Hence it is clear from above that the average production of ‘A’ type of toys was maximum.

8. What was the average number of defect free toys of all type in 1994 ?

(A) 45680

(B) 42790

(D) 44790

(E) None of these

8. (D) Total number of defect free toys in the year 1994.

= [(70 × (100 – 12))/100 + (52 × (100 – 12 ))/100 + (42(100 – 13))/100 + (54(100 – 4))/100 + (31(100 – 9))/100] × 1000

= [(70 × 88) + (52 × 88) + (42 × 87) + (54 × 96) + (31 × 91)] × 1000/100

= [6160 + 4576 + 3654 + 5184 + 2821] × 10

= 22395 × 10 = 223950 Average = 223950/5 = 44790

9. Among the given years in which years, the average percentage defect of all the five types of toys was lowest ?

(A) 1992

(B) 1995

(D) 1996

(E) None of these

9. (A) Average percentage of all five types of defective toys in 1991 = (5 + 11 + 5 + 9 + 8) × 1/5 = 38 × 1/5

Average percentage of all five types of defective toys in 1992 = (6 + 9 + 9 + 8 + 4) × 1/5 = 36 × 1/5

Average percentage of all five types of defective toys in 1993 = (8 + 8 + 6 + 12 + 15) × 1/5 = 49 × 1/5

Average percentage of all five types of defective toys in 1994 = (12 + 12 + 13 + 4 + 9 ) × 1/5 = 50 × 1/5

Average percentage of all five types of defective toys in 1995 = (9 + 14 + 11 + 7 + 11) × 1/5 = 52 × 1/5

Average percentage of all five types of defective toys in 1996 = (11 + 10 + 8 + 5 + 7) × 1/5 = 41 × 1/5

Hence, average percentage of all five types of defective toys in 1992 was minimum.

10. What was the difference in the numebr of defect-free B type toys between 1992 and 1993 ?

(A) 3220

(B) 7700

(D) 3860

(E) None of these

10. (A) Reqd. difference = [46(100 – 9)/100 – 49(100 – 8)/100] × 1000 = [46 × 91 – 49 × 92] × 1000/100 = (4186 – 4508) × 10 = + 3220