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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
What was the approximate percentage increase in the sales of 55AH batteries in 1998 compared to that in 1992? |
| Answer» Required percentage = (145 - 108) x 100 % 108 = 34.26% 34%. | |
| 2. |
The total sales of all the seven years is the maximum for which battery? |
| Answer» The total sales (in thousands) of all the seven years for various batteries are: For 4AH = 75 + 90 + 96 + 105 + 90 + 105 + 115 = 676 For 7AH = 144 + 126 + 114 + 90 + 75 + 60 + 85 = 694 For 32AH = 114 + 102 + 75 + 150 + 135 + 165 + 160 = 901 For 35AH = 102 + 84 + 105 + 90 + 75 + 45 + 100 = 601 For 55AH = 108 + 126 + 135 + 75 + 90 + 120 + 145 = 799. Clearly, sales are maximum in case of 32AH batteries. | |
| 3. |
What is the difference in the number of 35AH batteries sold in 1993 and 1997? |
| Answer» Required difference = [(84 - 45) x 1000] = 39000. | |
| 4. |
The percentage of 4AH batteries sold to the total number of batteries sold was maximum in the year? |
| Answer» The percentages of sales of 4AH batteries to the total sales in different years are: For 1992 = 75 x 100 % = 13.81%. 543 For 1993 = 90 x 100 % = 17.05%. 528 For 1994 = 96 x 100 % = 18.29%. 525 For 1995 = 105 x 100 % = 20.59%. 510 For 1996 = 96 x 100 % = 19.35%. 465 For 1997 = 105 x 100 % = 21.21%. 495 For 1998 = 115 x 100 % = 19.01%. 605 Clearly, the percentage is maximum in 1997. | |
| 5. |
What is the approximate percentage of total number of candidates selected to the total number of candidates qualified for all five states together during the year 1999? |
| Answer» Required percentage = (82 + 70 + 48 + 65 + 55) % (640 + 560 + 400 + 525 + 350) = 320 x 100 % 2475 = 12.93% 13%. | |
| 6. |
In case of which battery there was a continuous decrease in sales from 1992 to 1997? |
| Answer» From the table it is clear that the sales of 7AH batteries have been decreasing continuously from 1992 to 1997. | |
| 7. |
For which state the average number of candidates selected over the years is the maximum? |
| Answer» The average number of candidates selected over the given period for various states are: For Delhi = 94 + 48 + 82 + 90 + 70 = 384 = 76.8. 5 5 For H.P. = 82 + 65 + 70 + 86 + 75 = 378 = 75.6. 5 5 For U.P. = 78 + 85 + 48 + 70 + 80 = 361 = 72.2. 5 5 For Punjab = 85 + 70 + 65 + 84 + 60 = 364 = 72.8. 5 5 For Haryana = 75 + 75 + 55 + 60 + 75 = 340 = 68. 5 5 Clearly, this average is maximum for Delhi. | |
| 8. |
The percentage of candidates qualified from Punjab over those appeared from Punjab is highestin the year? |
| Answer» The percentages of candidates qualified from Punjab over those appeared from Punjab during different years are: For 1997 = 680 x 100 % = 8.29%. 8200 For 1998 = 600 x 100 % = 8.82%. 6800 For 1999 = 525 x 100 % = 8.08%. 6500 For 2000 = 720 x 100 % = 9.23%. 7800 For 2001 = 485 x 100 % = 8.51%. 5700 Clearly, this percentage is highest for the year 2000. | |
| 9. |
In the year 1997, which state had the lowest percentage of candidates selected over the candidates appeared? |
| Answer» The percentages of candidates selected over the candidates appeared in 1997, for various states are: (i) For Delhi = 94 x 100 % = 1.175%. 8000 (ii) For H.P. = 82 x 100 % = 1.051%. 7800 (iii) For U.P. = 78 x 100 % = 1.040%. 7500 (iv) For Punjab 85 x 100 % = 1.037%. 8200 (v) For Haryana 75 x 100 % = 1.172%. 6400 Clearly, this percentage is lowest for Punjab. | |
| 10. |
The number of candidates selected from Haryana during the period under review is approximately what percent of the number selected from Delhi during this period? |
| Answer» Required percentage = (75 + 75 + 55 + 60 + 75) x 100 % (94 + 48 + 82 + 90 + 70) = 340 x 100 % 384 = 88.54% 88.5% | |
| 11. |
The percentage of candidates selected from U.P over those qualified from U.P is highest in the year? |
| Answer» The percentages of candidates selected from U.P. over those qualified from U.P. during different years are: For 1997 = 78 x 100 % = 10.83%. 720 For 1998 = 85 x 100 % = 13.71%. 620 For 1999 = 48 x 100 % = 12%. 400 For 2000 = 70 x 100 % = 10.77%. 650 For 2001 = 80 x 100 % = 8.42%. 950 Clearly, this percentage is highest for the year 1998. | |
| 12. |
What are the average marks obtained by all the seven students in Physics? (rounded off to two digit after decimal) |
| Answer» Average marks obtained in Physics by all the seven students = 1 x [ (90% of 120) + (80% of 120) + (70% of 120) 7 + (80% of 120) + (85% of 120) + (65% of 120) + (50% of 120) ] = 1 x [ (90 + 80 + 70 + 80 + 85 + 65 + 50)% of 120 ] 7 = 1 x [ 520% of 120 ] 7 = 624 7 = 89.14. | |
| 13. |
The number of students who obtained 60% and above marks in all subjects is? |
| Answer» From the table it is clear that Sajal and Rohit have 60% or more marks in each of the six subjects. | |
| 14. |
What was the aggregate of marks obtained by Sajal in all the six subjects? |
| Answer» Aggregate marks obtained by Sajal = [ (90% of 150) + (60% of 130) + (70% of 120) + (70% of 100) + (90% of 60) + (70% of 40) ] = [ 135 + 78 + 84 + 70 + 54 + 28 ] = 449. | |
| 15. |
In which subject is the overall percentage the best? |
| Answer» We shall find the overall percentage (for all the seven students) with respect to each subject. The overall percentage for any subject is equal to the average of percentages obtained by all the seven students since the maximum marks for any subject is the same for all the students. Therefore, overall percentage for: (i) Maths = 1 x (90 + 100 + 90 + 80 + 80 + 70 + 65) % 7 = 1 x (575) % 7 = 82.14%. (ii) Chemistry = 1 x (50 + 80 + 60 + 65 + 65 + 75 + 35) % 7 = 1 x (430) % 7 = 61.43%. (iii) Physics = 1 x (90 + 80 + 70 + 80 + 85 + 65 + 50) % 7 = 1 x (520) % 7 = 74.29%. (iv) Geography = 1 x (60 + 40 + 70 + 80 + 95 + 85 + 77) % 7 = 1 x (507) % 7 = 72.43%. (v) History = 1 x (70 + 80 + 90 + 60 + 50 + 40 + 80) % 7 = 1 x (470) % 7 = 67.14%. (vi) Comp. Science = 1 x (80 + 70 + 70 + 60 + 90 + 60 + 80) % 7 = 1 x (510) % 7 = 72.86%. Clearly, this percentage is highest for Maths. | |
| 16. |
What is the overall percentage of Tarun? |
| Answer» Aggregate marks obtained by Tarun = [ (65% of 150) + (35% of 130) + (50% of 120) + ((77% of 100) + (80% of 60) + (80% of 40) ] = [ 97.5 + 45.5 + 60 + 77 + 48 + 32 ] = 360. The maximum marks (of all the six subjects) = (150 + 130 + 120 + 100 + 60 + 40) = 600. Overall percentage of Tarun = 360 x 100 % = 60%. 600 | |
| 17. |
Total number of candidates qualified from all the states together in 1997 is approximately what percentage of the total number of candidates qualified from all the states together in 1998? |
| Answer» Required percentage = (720 + 840 + 780 + 950 + 870) x 100 % (980 + 1050 + 1020 + 1240 + 940) = 4160 x 100 % 5230 = 79.54% 80%. | |
| 18. |
What is the average candidates who appeared from State Q during the given years? |
| Answer» Required average = 8100 + 9500 + 8700 + 9700 + 8950 5 = 44950 5 = 8990. | |
| 19. |
In which of the given years the number of candidates appeared from State P has maximum percentage of qualified candidates? |
| Answer» The percentages of candidates qualified to candidates appeared from State P during different years are: For 1997 780 x 100 % = 12.19%. 6400 For 1998 1020 x 100 % = 11.59%. 8800 For 1999 890 x 100 % = 11.41%. 7800 For 2000 1010 x 100 % = 11.54%. 8750 For 2001 1250 x 100 % = 12.82%. 9750 Maximum percentage is for the year 2001. | |
| 20. |
What is the percentage of candidates qualified from State N for all the years together, over the candidates appeared from State N during all the years together? |
| Answer» Required percentage = (840 + 1050 + 920 + 980 + 1020) x 100 % (7500 + 9200 + 8450 + 9200 + 8800) = 4810 x 100 % 43150 = 11.15% | |
| 21. |
Combining the states P and Q together in 1998, what is the percentage of the candidates qualified to that of the candidate appeared? |
| Answer» Required percentage = (1020 + 1240) x 100 % (8800 + 9500) = 2260 x 100 % 18300 = 12.35%. | |
| 22. |
The percentage of total number of qualified candidates to the total number of appeared candidates among all the five states in 1999 is? |
| Answer» Required percentage = (850 + 920 + 890 + 980 + 1350) x 100 % (7400 + 8450 + 7800 + 8700 + 9800) = 4990 x 100 % 42150 = 11.84%. | |
| 23. |
What is the different between the number of students passed with 30 as cut-off marks in Chemistry and those passed with 30 as cut-off marks in aggregate? |
| Answer» Required difference = (No. of students scoring 30 and above marks in Chemistry) - (Number of students scoring 30 and above marks in aggregate) = 27 - 21 = 6. | |
| 24. |
If at least 60% marks in Physics are required for pursuing higher studies in Physics, how many students will be eligible to pursue higher studies in Physics? |
| Answer» We have 60% of 50 = 60 x 50 = 30. 100 Required number = No. of students scoring 30 and above marks in Physics = 32 | |
| 25. |
The percentage of number of students getting at least 60% marks in Chemistry over those getting at least 40% marks in aggregate, is approximately? |
| Answer» Number of students getting at least 60% marks in Chemistry = Number of students getting 30 and above marks in Chemistry = 21. Number of students getting at least 40% marks in aggregate = Number of students getting 20 and above marks in aggregate = 73. Required percentage = 21 x 100 % 73 = 28.77% 29%. | |
| 26. |
The number of students scoring less than 40% marks in aggregate is? |
| Answer» We have 40% of 50 = 40 x 50 = 20. 100 Required number = Number of students scoring less than 20 marks in aggreagate = 100 - Number of students scoring 20 and above marks in aggregate = 100 - 73 = 27. | |
| 27. |
If it is known that at least 23 students were eligible for a Symposium on Chemistry, then the minimum qualifying marks in Chemistry for eligibility to Symposium would lie in the range? |
| Answer» Since 66 students get 20 and above marks in Chemistry and out of these 21 students get 30 and above marks, therefore to select top 35 students in Chemistry, the qualifying marks should lie in the range 20-30. | |
| 28. |
What is the average amount of interest per year which the company had to pay during this period? |
| Answer» Average amount of interest paid by the Company during the given period = Rs. 23.4 + 32.5 + 41.6 + 36.4 + 49.4 lakhs 5 = Rs. 183.3 lakhs 5 = Rs. 36.66 lakhs. | |
| 29. |
The total amount of bonus paid by the company during the given period is approximately what percent of the total amount of salary paid during this period? |
| Answer» Required percentage = (3.00 + 2.52 + 3.84 + 3.68 + 3.96) x 100 % (288 + 342 + 324 + 336 + 420) = 17 x 100 % 1710 1%. | |
| 30. |
Total expenditure on all these items in 1998 was approximately what percent of the total expenditure in 2002? |
| Answer» Required percentage = (288 + 98 + 3.00 + 23.4 + 83) x 100 % (420 + 142 + 3.96 + 49.4 + 98) = 495.4 x 100 % 713.36 69.45%. | |
| 31. |
The total expenditure of the company over these items during the year 2000 is? |
| Answer» Total expenditure of the Company during 2000 = Rs. (324 + 101 + 3.84 + 41.6 + 74) lakhs = Rs. 544.44 lakhs. | |
| 32. |
The ratio between the total expenditure on Taxes for all the years and the total expenditure on Fuel and Transport for all the years respectively is approximately? |
| Answer» Required ratio = (83 + 108 + 74 + 88 + 98) (98 + 112 + 101 + 133 + 142) = 451 586 = 1 1.3 = 10 . 13 | |
| 33. |
If the male population above poverty line for State R is 1.9 million, then the total population of State R is? |
| Answer» Let the total population of State R be x million. Then, population of State R above poverty line = [(100 - 24)% of x] million = 76 x x million 100 And so, male population of State R above poverty line = 2 x 76 x x million 5 100 But, it is given that male population of State R above poverty line = 1.9 million. 2 x 76 x x = 1.9 x = 5 x 100 x 1.9 = 6.25. 5 100 76 x 2 Total population of State R = 6.25 million. | |
| 34. |
What will be the number of females above the poverty line in the State S if it is known that the population of State S is 7 million? |
| Answer» Total population of State S = 7 million. Population above poverty line = [(100 - 19)% of 7] million = (81% of 7) million = 5.67 million. And so, the number of females above poverty line in State S = 3 x 5.67 million 7 = 2.43 million. | |
| 35. |
What will be the male population above poverty line for State P if the female population below poverty line for State P is 2.1 million? |
| Answer» Female population below poverty line for State P = 2.1 million Let the male population below poverty line for State P be x million. Then, 5 : 6 = x : 21 x = 2.1 x 5 = 1.75. 6 Population below poverty line for State P = (2.1 + 1.75) million = 3.85 million. Let the population above poverty line for State P by y million. Since, 35% of the total population of State P is below poverty line, therefore, 65% of the total population of State P is above poverty line i.e., the ratio of population below poverty line to that above poverty line for State P is 35 : 65. 35 : 65 = 3.85 : y y = 65 x 3.85 = 7.15. 35 Population above poverty line for State P = 7.15 million and so, male population above poverty line for State P = 6 x 7.15 million 13 = 3.3 million. | |
| 36. |
If the population of males below poverty line for State Q is 2.4 million and that for State T is 6 million, then the total populations of States Q and T are in the ratio? |
| Answer» For State Q: Male population below poverty line = 2.4 million. Let the female population below poverty line be x million. Then, 3 : 5 = 2.4 : x x = 5 x 2.4 = 4. 3 Total population below poverty line = (2.4 + 4) = 6.4 million. If Nq be the total population of State Q, then, 25% of Nq = 6.4 million Nq = 6.4 x 100 million = 25.6 million. 25 For State T: Male population below poverty line = 6 million. Let the female population below poverty line be y million. Then, 5 : 3 = 6 : y y = 3 x 6 = 3.6. 5 Total population below poverty line = (6 + 3.6) = 9.6 million. If Nt be the total population of State T, then, 15% of Nt = 9.6 million Nt = 9.6 x 100 million = 64 million. 15 Thus, Required ratio = Nq = 25.6 = 0.4 = 2 . Nt 64 5 | |
| 37. |
Which section has the maximum success rate in annual examination? |
| Answer» Total number of students passed in annual exams in a section = [ (No. of students failed in half-yearly but passed in annual exams) + (No. of students passed in both exams) ] in that section Success rate in annual exams in Section A = No. of students of Section A passed in annual exams x 100 % Total number of students in Section A = (14 + 64) x 100 % (28 + 14 + 6 + 64) = 78 x 100 % 112 = 69.64%. Similarly, success rate in annual exams in: Section B (12 + 55) x 100 % = 67 x 100 % = 62.62%. (23 + 12 + 17 + 55) 107 Section C (8 + 46) x 100 % = 54 x 100 % = 67.5%. (17 + 8 + 9 + 46) 80 Section D (13 + 76) x 100 % = 89 x 100 % = 67.94%. (27 + 13 + 15 + 76) 131 Clearly, the success rate in annual examination is maximum for Section A. | |
| 38. |
Which section has the minimum failure rate in half yearly examination? |
| Answer» Total number of failures in half-yearly exams in a section = [ (Number of students failed in both exams) + (Number of students failed in half-yearly but passed in Annual exams) ] in that section Failure rate in half-yearly exams in Section A = Number of students of Section A failed in half-yearly x 100 % Total number of students in Section A = (28 + 14) x 100 % (28 + 14 + 6 + 64) = 42 x 100 % 112 = 37.5%. Similarly, failure rate in half-yearly exams in: Section B (23 + 12) x 100 % = 35 x 100 % = 32.71%. (23 + 12 + 17 + 55) 107 Section C (17 + 8) x 100 % = 25 x 100 % = 31.25%. (17 + 8 + 9 + 46) 80 Section D (27 + 13) x 100 % = 40 x 100 % = 30.53%. (27 + 13 + 15 + 76) 131 Clearly, the failure rate is minimum for Section D. | |
| 39. |
How many students are there in Class IX in the school? |
| Answer» 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. | |
| 40. |
Which section has the maximum pass percentage in at least one of the two examinations? |
| Answer» Pass percentages in at least one of the two examinations for different sections are: For Section A (14 + 6 + 64) x 100 % = 84 x 100 % = 75%. (28 + 14 + 6 + 64) 112 For Section B (12 + 17 + 55) x 100 % = 84 x 100 % = 78.5%. (23 + 12 + 17 + 55) 107 For Section C (8 + 9 + 46) x 100 % = 63 x 100 % = 78.75%. (17 + 8 + 9 + 46) 80 For Section D (13 + 15 + 76) x 100 % = 104 x 100 % = 79.39%. (27 + 13 + 15 + 76) 131 Clearly, the pass percentage is maximum for Section D. | |
| 41. |
During the period between 1995 and 2000, the total number of Operators who left the Company is what percent of total number of Operators who joined the Company? |
| Answer» Total number of Operators who left the Company during 1995 - 2000 = (104 + 120 + 100 + 112 + 144) = 580. Total number of Operators who joined the Company during 1995 - 2000 = (880 + 256 + 240 + 208 + 192 + 248) = 2024. Required Percentage = 580 x 100 % = 28.66% 29%. 2024 | |
| 42. |
What is the difference between the total number of Technicians added to the Company and the total number of Accountants added to the Company during the years 1996 to 2000? |
| Answer» Required difference = (272 + 240 + 236 + 256 + 288) - (200 + 224 + 248 + 272 + 260) = 88. | |
| 43. |
What was the total number of Peons working in the Company in the year 1999? |
| Answer» Total number of Peons working in the Company in 1999 = (820 + 184 + 152 + 196 + 224) - (96 + 88 + 80 + 120) = 1192. | |
| 44. |
What is the pooled average of the total number of employees of all categories in the year 1997? |
| Answer» Total number of employees of various categories working in the Company in 1997 are: Managers = (760 + 280 + 179) - (120 + 92) = 1007. Technicians = (1200 + 272 + 240) - (120 + 128) = 1464. Operators = (880 + 256 + 240) - (104 + 120) = 1152. Accountants = (1160 + 200 + 224) - (100 + 104) = 1380. Peons = (820 + 184 + 152) - (96 + 88) = 972. Pooled average of all the five categories of employees working in the Company in 1997 = 1 x (1007 + 1464 + 1152 + 1380 + 972) 5 = 1 x (5975) 5 = 1195. | |