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InterviewSolution
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. |
If the length of line to be laid each year is in proportion to the estimated cost for material and labour, what fraction of the total length is proposed to be completed by the third year?1). 0.92). 0.73). 0.64). 0.3 |
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Answer» Cost of materials = cost of (cement + steel + BRICKS + other building materials) Cost of materials + Cost of labour for 1st year = 0 + 2.1 = 2.1 lakhs Cost of materials + Cost of labour for 2nd year = 95 + 70 + 15 + 25 + 25 = 230 lakhs Cost of materials + Cost of labour for 3rd year = 80 + 45 + 12 + 18 + 20 = 175 lakhs Cost of materials + Cost of labour for 4th year = 75 + 60 + 16 + 21 + 18 = 190 lakhs Given, length of line to be laid each year is in PROPORTION to the estimated cost for material and labour. ∴ Portion of the total length completed by 3rd year $(= \frac{{2.1 + 230 + 175}}{{2.1 + 230 + 175 + 190}})$ = 0.68 ≈ 0.7 |
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| 2. |
If the cost of materials rises by 5% each year from 1990 onwards, by how much will the estimated cost rise? (approximately)1). Rs. 11.4 lakh2). Rs. 16.4 lakh3). Rs. 21.4 lakh4). Rs. 26.4 lakh |
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Answer» Total cost of materials = cost of (cement + steel + bricks + other BUILDING materials) Estimated cost of materials for 1990 = 80 + 45 + 12 + 18 = 155 lakhs Estimated cost of materials for 1991 = 75 + 60 + 16 + 21 = 172 lakhs Given, cost of materials RISES by 5% each year from 1990 onwards. ∴ Actual cost of materials in 1990 = 155 + 5% of 155 = 162.75 lakhs Actual cost of materials in 1991 = 172 + 5% of 172 = 180.6 lakhs ∴ Increase in estimated cost = (162.75 + 180.6) – (155 + 172) = 16.35 lakhs ≈ 16.4 lakhs |
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| 3. |
1). 20%2). 50%3). 100%4). 200% |
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Answer» <P>Runs scored by P in form of 3’s = 81 – (23 × 1 + 3 × 2 + 4 × 4 + 4 × 6) = 12 Number of 6’s scored by R = 2 ∴ Percentage by which runs scored by P in form of 3’s is greater than number of 6’s scored by R = ((12 – 2)/2) × 100 = 500% |
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| 4. |
By how much is sales of P in 2016 is less than that of S?1). 108002). 110003). 112284). 11616 |
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Answer» <P>Sales of P in 2016 = 12000 × (1 + 10/100) × (1 + 20/100) × (1 + 10/100) = 17424 Sales of S in 2016 = 29040 Sales of P is LESS than S by (29040 – 17424) = 11616 |
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| 5. |
Note : Some values are missing, you must find these values as per given data only.Find the difference between the distribution of bonus in High level management of Account, Administration and security together to the distribution of bonus in Middle level management of Finance, Payroll and security together?1). 1875002). 1365003). 1750304). 12530 |
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Answer» Bonus distribution in HIGH level management of Account, ADMINISTRATION and security together = 2.5 × (8/100) + 6.5 × (15/100) + 4.7 × (15/100) = 1.88 LAKHS Distribution of bonus to Payroll department = 30 – (6.5 + 3.4 + 2.5 + 5.2 + 4.2 + 4.7) = 3.5 lakhs Distribution of bonus in Middle level management of FINANCE, Payroll and security together = 5.2 × (25/100) + 3.5 × (22/100) + 4.7 × (25/100) = 3.245 lakhs Difference = 3.245 – 1.88 = 1.365 lakhs = 136500 |
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| 6. |
1). 20.1%2). 19.62%3). 17.9%4). 18.92% |
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Answer» Total selected SO CANDIDATES for all banks = 300 + 450 + 560 + 360 + 340 + 640 = 2650 The number of selected candidates for Clerk POST by BOB = 520 So, REQUIRED % = 520/2650 × 100 = 19.62% |
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| 7. |
1). 4 ∶ 32). 15 ∶ 113). 18 ∶ 134). 22 ∶ 17 |
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Answer» Total books in SPANISH = 8100 Sum of ratios = 4 + 3 + 2 = 9 No. of medical books in Spanish = (3/9) × 8100 = 2700 No. of other books in Spanish = (2/9) × 8100 = 1800 Total books in French = 33600 Sum of ratios = 7 + 5 + 4 = 16 No. of medical books in French = (5/16) × 33600 = 10500 No. of other books in French = (4/16) × 33600 = 8400 Hence, Total medical books in both languages = 2700 + 10500 = 13200 Total other books in both languages = 1800 + 8400 = 10200 ∴ Required ratio = 13200 ? 10200 = 22 ? 17 |
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| 8. |
What is difference between total number of HR attending seminar on all five days of week and total number of CEO who are attending Seminar on all five days together?1). 22002). 24003). 28804). 2900 |
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Answer» Total no. of HR, ⇒ 2100 + 2000 + 1800 + 1500 + 1200 ⇒ 8600 Let, total no. of employee on Sunday = a According to PROBLEM, ⇒ a × 35/100 = 2100 ⇒ a = 6000 ∴ No. of MD and CEO on Sunday = 6000 - 2100 = 3900 Ratio of MD and CEO on Sunday = 1 ? 2 ∴ No. of CEO on Sunday = 3900 × 2/3 = 2600 Let, total no. of employee on Saturday = b According to problem, $(\Rightarrow b \times 33\frac{1}{3}/100 = 2000)$ ⇒ b × 100/300 = 2000 ⇒ b = 6000 ∴ No. of MD and CEO on Saturday = 6000 - 2000 = 4000 Ratio of MD and CEO on Saturday = 3 ? 7 ∴ No. of CEO on Saturday = 4000 × 7/10 = 2800 Let, total no. of employee on Monday = C According to problem, $(\Rightarrow c \times 37\frac{1}{2}/100 = 1800)$ ⇒ c × 75/200 = 1800 ⇒ c = 4800 ∴ No. of MD and CEO on Monday = 4800 - 1800 = 3000 Ratio of MD and CEO on Monday = 7 ? 8 ∴ No. of CEO on Monday = 3000 × 8/15 = 1600 Let, total no. of employee on Tuesday = d According to problem, ⇒ d × 30/100 = 1500 ⇒ d = 5000 ∴ No. of MD and CEO on Tuesday = 5000 - 1500 = 3500 Ratio of MD and CEO on Tuesday = 3 ? 4 ∴ No. of CEO on Tuesday = 3500 × 4/7 = 2000 Let, total no. of employee on Friday = x According to problem, ⇒ x × 25/100 = 1200 ⇒ x = 4800 ∴ No. of MD and CEO on Friday = 4800 - 1200 = 3600 Ratio of MD and CEO on Friday = 1 ? 1 ∴ No. of CEO on Friday = 3600 × 1/2 = 1800 ∴ Total no. of CEO, ⇒ 2600 + 2800 + 1600 + 2000 + 1800 ⇒ 10800 ∴ Required difference, ⇒ 10800 - 8600 ⇒ 2200 |
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| 9. |
1). Rs. 912, Rs. 15202). Rs. 1216, Rs. 22803). Rs. 1520, Rs. 22804). Rs. 1216, Rs. 3040 |
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Answer» To complete the work X, B’s 1 day work = 1/30 C’s 1 day work = 1/24 D’s 1 day work = 1/12 Now, ratio of their WAGES = ratio of their work efficiencies ⇒ Ratio of their wages = 1/30 ? 1/24 ? 1/12 = 4 ? 5 ? 10 Now, B’s share ? C’s share = 4 ? 5 ⇒ B’s share = (4/5) × 1520 = Rs. 1216 Also, C’s share ? D’s share = 5 ? 10 = 1 ? 2 ⇒ D’s share = 2 × 1520 = Rs. 3040 ∴ B & D will receive Rs. 1216 and Rs. 3040 respectively |
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| 10. |
Note: Some variables are missing, you have to find the missing values on the basis of given information. Calculate simple interest unless specified. All questions are independent from others.If the amount received by Amit is 2 times of the money invested by him then find amount he will receive after 2 years if he invests the same amount of money in compound interest for 2 years compounded half yearly?1). Rs. 94800.552). Rs. 35400.553). Rs. 25850.454). Rs. 15280.55 |
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Answer» Amount invested by Amit i.e. PRINCIPAL = 29000/2 = Rs. 14500 Interest received by Amit = 29000 - 14500 = Rs. 14500 Let the rate of interest for Amit be X ⇒ x = 14500 × 100/(14500 × 5) ⇒ x = 20% ∴ Amount he will receive if compounded half yearly for 2 years = 14500 × (1 + 10/100)4 = 14500 × 11/10 × 11/10 × 11/10 × 11/10 = Rs. 21229.45 |
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| 11. |
Total budget allocated for Healthcare in 2018-19 is 50% more than total budget allocated for Education in same year and sum of both allocated for rural areas in 2018-19 is 7040 cr. Find total budget allocated for Healthcare in 2018-19 is what percent of total budget allocated for all six sectors in 2018-19?1). 23.29%2). 18.73%3). 22.81%4). 20.95% |
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Answer» LET the TOTAL BUDGET allocated for education be x Therefore, total budget allocated for Healthcare = 1.5x (70/100) x + (60/100) × 1.5x = 7040 ⇒ x = 4400 crore Total budget allocated for Healthcare in 2018-19 = 1.5 × 4400 = 6600 crore Total budget allocated for all SECTORS = 31500 crore Percentage = (6600/31500) × 100 = 20.95% |
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| 12. |
What is the respective ratio of selected candidates for the post of PO, Clerk and SO by Syndicate and Dena bank together?1). 5 : 9 : 82). 2 : 5 : 73). 5 : 9 : 74). 15 : 11 :17 |
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Answer» Selected CANDIDATES for the post of PO by Syndicate and Dena bank together = 220 + 280 = 500 Selected candidates for the post of Clerk by Syndicate and Dena bank together = 460 + 440=900 Selected candidates for the post of SO by Syndicate and Dena bank together = 360 + 340 = 700 |
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| 13. |
How many metres of high quality cloth is consumed by A type shirts?1). 8,000 m2). 112,000 m3). 24,000 m4). 30,000 m |
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Answer» % of high QUALITY shirt used in type A = 80% Total shirts MADE of type A = 20000 Each shirt requires 1.5 m length of CLOTH. ∴ Metres of high quality cloth is consumed by A-type shirts = 80% of 20000 × 1.5 ∴ Metres of high quality cloth is consumed by A-type shirts = 24000 m |
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| 14. |
The total of battery production done by HBL during the given period is approximately what percent of that done by Eveready?1). 952). 873). 904). 77 |
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Answer» From the table, The total BATTERY production DONE by HBL in the given periods = 416 lakhs The total battery production done by Eveready in the given periods = 440 lakhs The required percentage = [(416/440) × 100]% = 94.54% ≈ 95% ∴ The total of battery production done by HBL during the given period is APPROXIMATELY 95% of that done by Eveready. |
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| 15. |
Which of the five companies has the highest ratio of battery production done in 2004 to that done in 2002?1). Eveready2). Exide3). Nippo4). Panasonic |
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Answer» From the table,
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| 16. |
1). 382). 503). 404). 43 |
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Answer» Students in standard I are 42, 50, 40, 45, 48 and 52 from all 6 schools. Average no. = TOTAL no. of students/ no. schools Average no. = (42 + 50 + 40 + 45 + 48 + 52)/6 = 277/6 = 46.166 ≈ 46 |
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| 17. |
Initially B works alone for five days then C joins and after three more days A joins and B leaves, in how many was the work completed?1). \(30\frac{{14}}{{39}}\) days2). \(36\frac{{19}}{{39}}\) days3). \(33\frac{{20}}{{39}}\) days4). \(34\frac{{21}}{{39}}\) days |
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Answer» Initially B works alone for FIVE DAYS B’s one day work = $(\frac{1}{{60}})$ ∴ Work done $(= 5 \times \;\frac{1}{{60}} = \frac{1}{{12}})$ ∴ Work left $(= 1 - \frac{1}{{12}} = \frac{{11}}{{12}})$ C joins and B and C work for three days C’s one day work = $(\frac{1}{{80}})$ B and C’s one day work = $(\frac{1}{{60}} + \;\frac{1}{{80}} = \;\frac{7}{{240}})$ ∴ Work done $(= 3 \times \frac{7}{{240}} = \frac{7}{{80}})$ ∴ Work left $(= \frac{{11}}{{12}} - \frac{7}{{80}} = \frac{{199}}{{240}})$ After three more days A joins and B leaves, A and C work together A and C’s one day work = $(\frac{1}{{50}} + \frac{1}{{80}} = \frac{{13}}{{400}})$ Work left = $(\frac{{199}}{{240}})$ ∴ Days TAKEN to complete the remaining work $(= \frac{{199}}{{240}} \times \frac{{400}}{{13}} = \frac{{995}}{{39}})$ ∴ Total number of days $(= 5 + 3 + \;\frac{{995}}{{39}} = 33\frac{{20}}{{39}})$ days |
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| 18. |
Number of students studying in standard IX of state T form what per cent of total number of students from that state? (rounded off to two digits after decimal)1). 16.842). 12.953). 14.724). 15.34 |
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Answer» From the given table, NUMBER of students studying in standard IX of state T = 3.4 LAKHS Total number of students from state T = 3.1 + 3.9 + 4.7 + 4.0 + 3.4 + 4.0 = 23.1 lakhs % of students from state T studying in standard IX out of all the students studying in State T $(= \FRAC{{3.4}}{{23.1}} \times 100\% \;)$ ⇒ % of students from state T studying in standard IX out of all the students studying in State T = 14.7186 % Rounded off to two DIGITS of decimal = 14.72 % |
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| 19. |
The number of green balls is what percentage of total number of balls in basket A and B?1). \(81\frac{{38}}{{127}}\)%2). \(84\frac{{32}}{{127}}\)%3). \(89\frac{{45}}{{127}}\)%4). \(87\frac{{79}}{{127}}\)% |
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Answer» Number of GREEN BALLS = 46 + 12 + 50 + 48 + 58 = 214 Balls in basket A = 28 + 16 + 20 + 46 = 110 Balls in basket B = 52 + 32 + 48 + 12 = 144 Total number of balls in basket A and B = 110 + 144 = 254 Required percentage $(= \frac{{214}}{{254}} \TIMES 100 = 84\frac{{32}}{{127}})$% |
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| 20. |
Consider two situations, in first situation A and E work together for ten days then C joins and after five days E leaves. In second situation B and D work together for six days, then A and E joins and B leaves, after two days A leaves. What is the difference between the total number of days taken in both the situations?1). \(18\frac{7}{{13}}\) days2). \(22\frac{{11}}{{13}}\) days3). \(11\frac{8}{{13}}\) days4). \(15\frac{9}{{13}}\) days |
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Answer» First situation A and E work together for ten days A and E’s one day work $(= \frac{1}{{50}} + \;\frac{1}{{75}} = \frac{1}{{30}})$ Work done $(= 10 \times \frac{1}{{30}} = \frac{1}{3})$ Work left $(= 1 - \frac{1}{3} = \frac{2}{3})$ Then C joins, A, C and E work together for five days A, C and E’s one day work $(= \frac{1}{{50}} + \frac{1}{{80}} + \frac{1}{{75}} = \frac{{11}}{{240}})$ Work done $(= 5 \times \frac{{11}}{{240}} = \frac{{11}}{{48}})$ Work left $(= \frac{2}{3}\;-\;\frac{{11}}{{48}} = \frac{7}{{16}})$ After that A and C work together A and C’s one day work $(= \frac{1}{{50}} + \frac{1}{{80}} = \frac{{13}}{{400}})$ Work left = $(\frac{7}{{16}})$ ∴ Days taken to complete the REMAINING work $(= \frac{7}{{16}} \times \frac{{400}}{{13}} = \frac{{175}}{{13}})$ Total days = $(10 + 5 + \frac{{175}}{{13}} = 28\frac{6}{{13}})$ days Second situation B and D work together for SIX days B and D’s one day work $(= \frac{1}{{60}} + \frac{1}{{150}} = \frac{7}{{300}})$ Work done $(= 6 \times \frac{7}{{300}} = \frac{7}{{50}})$ ∴ Work left $(= 1 - \frac{7}{{50}} = \frac{{43}}{{50}})$ Then A and E joins and B LEAVES, means A, D and E work together A, D and E’s one day work $(= \frac{1}{{50}} + \frac{1}{{150}} + \frac{1}{{75}} = \frac{1}{{25}})$ Work done $(= 2 \times \frac{1}{{25}} = \frac{2}{{25}})$ Work left $(= \frac{{43}}{{50}} - \frac{2}{{25}} = \frac{{39}}{{50}})$ After TWO days A leaves, now D and E work D and E’s one day work $(= \frac{1}{{150}} + \frac{1}{{75}} = \frac{1}{{50}})$ Work left = $(\frac{{39}}{{50}})$ ∴ Days taken to complete the remaining work $(= \frac{{39}}{{50}} \times \frac{{50}}{1} = 39)$ Total days = 6 + 2 + 39 = 47 Difference in days in both situations $(= 47 - \frac{{370}}{{13}} = \frac{{241}}{{13}} = 18\frac{7}{{13}})$ days |
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| 21. |
What is the average of the number of students who passed in Marketing, finance, business, and computer?1). 10002). 3003). 2004). 500 |
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Answer» NUMBER student PASSED in marketing = 500 – 120 = 380 ∴ REQUIRED average now = (380 + 200 + 120 + 100)/4 = 200 |
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| 22. |
What is the growth percentage for Q in year 2016?1). 302). 323). 344). 35 |
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Answer» This gives, 10000 × (1 + 15/100) × (1 + 10/100) × (1 + T/100) = 16951 (1 + T/100) = 1.6951/(1.15 × 1.1) = 1.34 T = 34 So, Q had 34% growth in 2016. |
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| 23. |
Number of boys of standard XI participating in different gamesIf boys of class XI E participating in LUDO together with girls of class XI B and class XI C participating in TT & POLO respectively are selected for a course at the college of sports, what percentage of the students will get this advantage approximately?1). 4.382). 3.513). 10.524). 13.5 |
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Answer» ⇒ Boys of XI E playing ludo = 4 ⇒ GIRLS of XI B playing TT = 25% of 12 = 4 ⇒ Girls of XI C playing POLO = 25% of 8 = 2 ⇒ Number of student selected = 4 + 4 + 2 = 10 ⇒ Number of STUDENTS in the SCHOOL = boys + girls = 228 + 57 = 285 (This is only the total number of student in class XI) ∴ Percentage = 10/285 × 100 = 3.51 |
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| 24. |
If 954 and 2015 schools presented their candidates for 12th and 10th examination in 2016 respectively, on an average, how many students appeared in the respective examination from one school?1). 72, 802). 74, 823). 76, 844). 78, 86 |
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Answer» In 2016, TOTAL no. of students appearing in 12th examination = 37023 + 1811 + 32498 + 737 = 72069 Total no. of schools = 954 Required AVERAGE for 12th examination = 72069/954 = 75.54 ≅ 76 Similarly, Total no. of students appearing in 10th examination = 91172 + 1728 + 74885 + 806 = 168951 Total no. of schools = 2015 ∴ Required average for 10th examination = 168951/2015 = 83.85 ≅ 84 |
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| 25. |
Which day has maximum selling capacity from given options?1). Monday2). Saturday3). Wednesday4). Thursday |
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Answer» Number of PAPERS sold on Monday = 500 + 900 + 520 + 250 + 420 = 2590 Number of papers sold on Wednesday = 250 + 200 + 700 + 100 + 200 = 1450 Number of papers sold on Thursday = 400 + 500 + 830 + 120 + 500 = 2350 Number of papers sold on Friday = 600 + 350 + 400 + 140 + 400 = 1890 Number of papers sold on SATURDAY = 700 + 800 + 300 + 220 + 300 = 2320 From above, Monday has MAXIMUM selling capacity as compare to other options. |
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| 26. |
The registration cost of a mall is what percent of the painting cost if the area of the mall is 48000 sq. ft.1). 75%2). 70%3). 80%4). 65% |
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Answer» Registration cost of MALL = 5% of total LAND VALUE of mall = 1/20 × Area of mall × RATE of 1 sq. ft ⇒ Registration cost of mall = 1/20 × 48000 × 600 = 1440000 = 14.4 lakh Painting Cost = Area of building × Painting Rate = 48000 × 40 = 1920000 = 19.2 lakh ∴ Required % = 14.4/19.2 × 100 = 75% |
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| 27. |
What is the total requirement of cloth?1). 1,50,000 m2). 2,00,000 m3). 2,25,000 m4). 2,50,000 m |
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Answer» From the GIVEN table, Total NUMBER of SHIRTS of type A, B, C, D and E were MADE = 20000 + 30000 + 30000 + 10000 + 10000 ⇒ Total number of shirts of type A, B, C, D and E were made = 100000 Each shirt requires 1.5 m cloth. ∴ $Total cloth requirement = 100000 × 1.5 = 1,50,000 m |
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| 28. |
Which paper is most popular if the difference between the total number of Sakal sold through week and number of Pudhari sold through the week is of 1600?1). Lokmat2). Pudhari3). Loksatta4). Sakal |
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Answer» The paper which is sold out most through the week is most POPULAR. From previous QUESTION, NUMBER of Sakal sold out through week = 3180 Number of Pudhari sold through week = 3180 – 1600 = 1580 Total number of Maharashtra TIMES sold throughout week = 3550 Total number of Loksatta sold throughout week = 2840 Total number of Lokmat sold throughout week = 4150 From above, Lokmat is more popular newspaper amongst all OTHERS. |
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| 29. |
At least 35 students of physics will be awarded on the basis of marks obtained in physics. The minimum qualifying marks in physics eligible for the award would be in the range1). Above 602). 40 to 603). Below 404). Below 50 |
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Answer» From the table, 30 STUDENTS are GETTING MARKS 60 or above and 46 are getting marks 40 or above. ∴ The minimum QUALIFYING mark should LIE in the range 40 to 60. |
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| 30. |
Which bank shows a continuous increase/decrease in loan amount disbursed over 5 years?1). Kotak2). PNB3). SBI4). ICICI |
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Answer» Only Kotak shows a continuous increase in LOAN DISBURSED AMOUNT over 5 years. |
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| 31. |
In the case of which company, the number of battery production in 2003 is the highest percentage of the total battery production by that company during the given period?1). Eveready2). Exide3). Nippo4). Panasonic |
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Answer» From the table,
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| 32. |
1). 1.46%2). 2.31%3). 2.88%4). 3.39% |
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Answer» In 2017, No. of students appeared in 12th = 37872 + 1831 + 33161 + 769 = 73633 No. of students appeared in 10th = 94957 + 1813 + 77765 + 764 = 175299 Total no. of students appeared in 12th and 10th = 73633 + 175299 = 248932 Total no. of students registered in 12th and 10th = 256304 Total no. of students that did not appear = 256304 – 248932 = 7372 ∴ Required PERCENTAGE = (7372/256304) × 100 = 2.88% |
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| 33. |
Total number of MD on Saturday and CEO on Monday together attending seminar is how much percent more or less than total number of MD on Tuesday and CEO on Sunday together attending Seminar? (approximately)1). 31.70% more2). 31.70% less3). 26.11% less4). 26.11% more |
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Answer» Let, total no. of employee on SUNDAY = a According to problem, ⇒ a × 35/100 = 2100 ⇒ a = 6000 ∴ No. of MD and CEO on Sunday = 6000 - 2100 = 3900 Ratio of MD and CEO on Sunday = 1 ? 2 ∴ No. of CEO on Sunday = 3900 × 2/3 = 2600 Let, total no. of employee on Saturday = b According to problem, ⇒ b × 33.33 × 1/100 = 2000 ⇒ b = 6000 ∴ No. of MD and CEO on Saturday = 6000 - 2000 = 4000 Ratio of MD and CEO on Saturday = 3 ? 7 ∴ No. of MD on Saturday = 4000 × 3/10 = 1200 Let, total no. of employee on Monday = C According to problem, ⇒ c × (75/200) = 1800 ⇒ c = 4800 ∴ No. of MD and CEO on Monday = 4800 - 1800 = 3000 Ratio of MD and CEO on Monday = 7 ? 8 ∴ No. of CEO on Monday = 3000 × 8/15 = 1600 Let, total no. of employee on Tuesday = d According to problem, ⇒ d × 30/100 = 1500 ⇒ d = 5000 ∴ No. of MD and CEO on Tuesday = 5000 - 1500 = 3500 Ratio of MD and CEO on Tuesday = 3 ? 4 ∴ No. of MD on Tuesday = 3500 × 3/7 = 1500 ∴ Total number of MD on Saturday and CEO on Monday together, ⇒ 1200 + 1600 ⇒ 2800 ∴ Total number of MD on Tuesday and CEO on Sunday together, ⇒ 1500 + 2600 ⇒ 4100 ∴ Required Percentage, ⇒ (4100 - 2800)/4100 × 100% ⇒ 31.70% less |
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| 34. |
What is the target for CSK?1). 1622). 1633). 1644). 165 |
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Answer» Total runs score by MI = 15 + 0 + 40 + 43 + 22 + 41 + 4 = 165 ∴ Target for CSK = 165 + 1 = 166 |
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| 35. |
Out of total budget allocated for pension in 2017-18 for rural areas, ratio between budget allocated for old person pension to budget allocated for retirement pension is 5 : 12 and budget allocated for retirement pension for rural areas is 3600 cr. Find total budget allocated for pension in 2017-18.1). 6200 crore2). 6600 crore3). 6500 crore4). 6400 crore |
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Answer» Ratio between budget allocated for old PERSON pension to RETIREMENT pension = 5 : 12 Budget allocated for retirement pension for rural AREAS = 3600 crore Budget allocated for old person pension for rural areas = 3600 × 5/12 = 1500 crore Total Budget allocated for pension in rural areas = 3600 + 1500 = 5100 crore Total Budget allocated for pension = (5100 × 100/75) = 6800 crore |
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| 36. |
1). 11%2). 17%3). 21%4). 25% |
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Answer» Total runs scored in the MATCH = 165 Runs scored by Krunal Pandya = 41 ∴ PERCENTAGE of Runs scored by Krunal Pandya = (41/165) × 100 = (4100/165) = 24.84% ≈ 25% |
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| 37. |
How many cities have the average temperature less than 20 °C?1). 02). 13). 24). 3 |
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Answer» Average TEMPERATURE of Faridkot = (22 + 24 + 27 + 29 + 31)/5 = 133/5 = 26.6 °C Average temperature of Nagpur = (21 + 24 + 28 + 35 + 38)/5 = 146/5 = 29.2 °C Average temperature of Mandi = (6 + 8 + 12 + 13 + 16)/5 = 55/5 = 11 °C Average temperature of Delhi = (11 + 15 + 22 + 25 + 27)/5 = 100/5 = 20 °C Average temperature of Jammu = (15 + 16 + 19 + 23 + 24)/5 = 97/5 = 19.4 °C 2 cities have average temperature less than 20 °C |
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| 38. |
Note: Some variables are missing, you have to find the missing values on the basis of given information. Calculate simple interest unless specified. All questions are independent from others.If the interest received by Anita is Rs. 7575 more than interest received by Piku and the rate of interest received by Anita is 8%, then find the interest rate received by Piku every year?1). 8%2). 1.3%3). 2.5%4). 1.5% |
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Answer» Let the rate of INTEREST received by Piku be x% ⇒ 60000 × 8 × 2/100 - x × 3 × 45000/100 = 7575 ⇒ 9600 - 1350x = 7575 ⇒ 9600 - 7575 = 1350x ⇒ x = 2025/1350 = 1.5% |
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| 39. |
If initially C and D work together for eight days and after that A joins them, they all work four days, after which B joins and c leaves, in how many total days was the assignment completed?1). \(25\frac{{10}}{{13}}\) days2). \(29\frac{9}{{13}}\) days3). \(24\frac{{11}}{{13}}\) days4). \(27\frac{{12}}{{13}}\) days |
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Answer» Initially C and D work together for eight DAYS C’s one day work = $(\frac{1}{{80}})$ D’s one day work = $(\frac{1}{{150}})$ C and D’s one day work $(= \frac{1}{{80}} + \frac{1}{{150}} = \frac{{23}}{{1200}})$ ∴ Work done $(= 8 \times \frac{{23}}{{1200}} = \frac{{23}}{{150}})$ ∴ Work left $(= 1 - \frac{{23}}{{150}} = \frac{{127}}{{150}})$ After that A joins them, they all work four days, A, C and D’s one day work $(= \frac{1}{{50}} + \frac{1}{{80}} + \frac{1}{{150}} = \frac{{47}}{{1200}})$ Work done $(= 4 \times \frac{{47}}{{1200}} = \frac{{47}}{{300}})$ Work left $(= \frac{{127}}{{150}} - \frac{{47}}{{300}} = \frac{{69}}{{100}})$ After which B joins and C LEAVES, in how MANY total days was the assignment completed A, B and D’s one day work $(= \frac{1}{{50}} + \frac{1}{{60}} + \frac{1}{{150}} = \frac{{13}}{{300}})$ Work left $(= \frac{{69}}{{100}})$ ∴ Days taken to complete the REMAINING work $(= \frac{{69}}{{100}} \times \frac{{300}}{{13}} = \frac{{207}}{{13}})$ ∴ Total days taken $(= 8 + 4 + \;\frac{{207}}{{13}} = 27\frac{{12}}{{13}})$ days |
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| 40. |
It is found at the end of 1990, that the entire amount estimated for the project has been spent. If for 1991, the actual amount spent was equal to that which was estimated, by what percent (approximately) has the actual expenditure exceeded the estimated expenditure?1). 392). 293). 194). 9 |
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Answer» TOTAL expenditure in the year 1988 = 52.1 lakhs Total expenditure in the year 1989 = 267.5 lakhs Total expenditure in the year 1990 = 196.4 lakhs Total expenditure in the year 1991 = 209.5 lakhs Total estimated expenditure in 4 years = 52.1 + 267.5 + 196.4 + 209.5 = 725.5 lakhs Given, it is FOUND at the end of 1990, that the ENTIRE amount estimated for the PROJECT has been SPENT. ∴ 725.5 lakhs is spent at the end of 1990. Now, for 1991, the actual amount spent was equal to that which was estimated. ∴ % spent over the estimated value $(= \frac{{209.5}}{{725.5}} \times 100\% = 28.8\% \approx 29\%)$ |
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| 41. |
If each person worked alone and did equal amount of work and left, in how many days would the assignment be completed?1). 69.1 days2). 83 days3). 92 days4). 103.75 days |
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Answer» Each person worked alone and did equal amount of work and LEFT that means each person did 20% of work and left A does 20% work in $(= \frac{{20}}{{100}} \times 50 = 10)$ B does 20% work in $(= \frac{{20}}{{100}} \times 60 = 12)$ C does 20% work in $(= \frac{{20}}{{100}} \times 80 = 16)$ D does 20% work in $(= \frac{{20}}{{100}} \times 150 = 30)$ E does 20% work in $(= \frac{{20}}{{100}} \times 75 = 15)$ Total days TAKEN = 10 + 12 + 16 + 30 + 15 = 83 days |
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| 42. |
Number of boys of standard XI participating in different gamesBoys of which of the following classes need to be combined to equal four times the number of girls in class XI B and class XI C ?1). XI D & XI E2). XI A & XI B3). XI A & XI D4). XI B & XI E |
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Answer» Number of girls in XI B + XI C = 24 4 TIMES = 96 ∴ BOYS of XI B and XI E have to be COMBINED |
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| 43. |
1). 10%2). 15%3). 20%4). 25% |
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Answer» No. of BOOKS in German = 22500 No. of books in Japanese = (100 – 82)% of 22500 = 0.18 × 22500 = 4050 No. of journals in Russian = 240 No. of journals in Japanese = (100 + 65)% of 240 = 1.65 × 240 = 396 No. of THESIS in other languages = 10 No. of thesis in Japanese = 40% of 10 = 4 Hence, total collection in Japanese = 4050 + 396 + 4 = 4450 Total collection in other languages = 20320 + 1920 + 10 = 22250 ∴ Required percentage = (4450/22250) × 100 = 20% |
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| 44. |
Which standard has the highest total number of students from all the states together?1). VII2). VIII3). X4). IX |
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Answer» From the given table, Total number of students in lakhs STUDYING in Standard V = 2.5 + 3.2 + 2.3 + 4.8 + 3.1 + 3.3 = 19.2 Standard VI = 2.8 + 2.9 + 3.2 + 2.4 + 3.9 + 4.2 = 19.4 Standard VII = 4.3 + 4.1 + 3.8 + 3.6 + 4.7 + 2.7 = 23.2 Standard VIII = 2.9 + 3.0 + 3.5 + 3.4 + 4.0 + 4.1 = 20.9 Standard IX = 4.1 + 3.6 + 2.9 + 3.4 + 3.4 + 3.6 = 21 Standard X = 4.3 + 2.7 + 3.7 + 4.3 + 4.0 +3.8 = 22.8 Thus standard VII has the HIGHEST total number of students from all the states together. |
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| 45. |
What is the difference between number of students failed in physics and mathematics if pass marks is 40 in each subject.1). 502). 483). 454). 47 |
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Answer» Number of STUDENTS getting 40 or above in MATHEMATICS = 93 ∴ Number of students FAILED in mathematics = 150 - 93 = 57 Number of students getting 40 or above in physics = 46 ∴ Number of students failed in physics = 150 - 46 = 104 ∴ Required difference = 104 - 57 = 47 |
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| 46. |
If 45% in 2017-18 and 35% in 2018-19 of total budget allocated for Railways is allocated for rural areas, then find the difference between total budget allocated for Healthcare & Railways in rural areas in 2017-18 and total budget allocated for Agriculture & Railways in rural areas in 2018-19.1). 2010 crore2). 1990 crore3). 2000 crore4). 2100 crore |
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Answer» Let the TOTAL BUDGET allocated for Railway for RURAL areas in 2017-18 = (45/100) × 9000 = 4050 crore Let the total budget allocated for Railway for rural areas in 2017-18 = (35/100) × 6000 = 2100 crore Budget allocated for Healthcare & Railways in rural in 2017-18 = 4050 + (55/100) × 7200 = 8010 crore Budget allocated for Agriculture & Railways in rural in 2018-19 = 2100 + (75/100) × 5200 = 6000 crore Difference = 8010 – 6000 = 2010 crore |
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| 47. |
Note: Some variables are missing, you have to find the missing values on the basis of given information. Calculate simple interest unless specified. All questions are independent from others.If the interest received by Anita is 20% of the sum invested by her then find how much more money as interest she would have earned if she had invested the money in compound interest?1). Rs. 6002). Rs. 1303). Rs. 2004). Rs. 1500 |
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Answer» Interest received by ANITA = 20 × 60000/100 = Rs. 12000 ⇒ Rate = 12000 × 100/(2 × 60000) = 10% ⇒ Interest for 1st year = 60000 × 10 × 1/100 = Rs. 6000 ⇒ Principal for 2nd year = 60000 + 6000 = Rs. 66000 ⇒ Interest for second year = 66000 × 10 × 1/100 = Rs. 6600 ⇒ TOTAL interest received by Anita = 6600 + 6000 = Rs. 12600 ∴ Anita will receive (12600 - 12000) = Rs. 600 more interest. |
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| 48. |
What is the ratio of the three qualities of dyes in high quality cloth?1). 2 : 3 : 52). 1 : 2 : 53). 7 : 9 : 104). Cannot be determined |
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Answer» From the GIVEN DATA we only know the RELATIONSHIP between the type of SHIRT and CLOTH used and type of shirt and dye used. So we don’t have idea about type of cloth and dye used. |
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| 49. |
If budget allocated for Education in 2017-18 increased by 25% compare to previous year 2016-17 and total Educational budget allocated for rural areas in 2017-18 decreased by 20% compare to previous year 2016-17, then find the ratio between total educational budget allocated for rural areas in 2016-17 to total educational budget allocated in 2016-17.1). 14 : 152). 15 : 163). 17 : 184). 16 : 15 |
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Answer» Budget ALLOCATED for Education in 2017-18 = 6400 crore Total Educational budget allocated for RURAL areas in 2017-18 = (60/100) × 6400 = 3840 crore Budget allocated for Education in 2016-17 = 6400 × (100/125) = 5120 crore Total Educational budget allocated for rural areas in 2016-17 = (100/80) × 3840 = 4800 crore Ratio = 4800/5120 = 15 : 16 |
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| 50. |
1). V2). VI3). I4). IV |
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Answer» The total NUMBERS of students in a standard are:
Hence, the LOWEST no. of students i.e. 277 are in standard I. |
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