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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeQ (15) Study the given table carefully and answer the following questions
Total Population = Online Surveyors + Offline Surveyors + Who didn’t SurveyWhat is the average of number of Offline surveyors participated from Streets M, O & P.
Correct
Answer :5. 670
Solution:
Village M:
Total population be M = 800+35M/100 + 370; M = 1800; offline surveyors = 630
Total population be O = 680+30O/100 + 545; O = 1750; offline surveyors = 525
Total population be P = 850+45P/100 + 195; P = 1900; offline surveyors = 855
Average = (630+525+855)/3 = 670Incorrect
Answer :5. 670
Solution:
Village M:
Total population be M = 800+35M/100 + 370; M = 1800; offline surveyors = 630
Total population be O = 680+30O/100 + 545; O = 1750; offline surveyors = 525
Total population be P = 850+45P/100 + 195; P = 1900; offline surveyors = 855
Average = (630+525+855)/3 = 670 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeQ (15) Study the given table carefully and answer the following questions
Total Population = Online Surveyors + Offline Surveyors + Who didn’t Survey
Total population in Street N is same as total population in which Street?Correct
Answer :4. Q
Solution:
By observation sum of online surveyors and who didn’t survey in Street N is same as in Street Q and % of offline surveyors is also same. So total population will be same in two streetsIncorrect
Answer :4. Q
Solution:
By observation sum of online surveyors and who didn’t survey in Street N is same as in Street Q and % of offline surveyors is also same. So total population will be same in two streets 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeQ (15) Study the given table carefully and answer the following questions
Total Population = Online Surveyors + Offline Surveyors + Who didn’t SurveyIf all females and 110 males in street Q didn’t participate in the survey, then total how many males have participated in the survey?
Correct
Answer :2. 1590
Solution:
In street Q total population:
x = 790+40x/100+ 410
x =2000 Offline surveyors =800
Total male surveyors = 790+800 =1590Incorrect
Answer :2. 1590
Solution:
In street Q total population:
x = 790+40x/100+ 410
x =2000 Offline surveyors =800
Total male surveyors = 790+800 =1590 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeQ (15) Study the given table carefully and answer the following questions
Total Population = Online Surveyors + Offline Surveyors + Who didn’t SurveyWhat is the average of total population in all five streets together?
Correct
Answer: 3. 1890
Solution:
1800+2000+1750+1900+2000 = 9450/5 = 1890Incorrect
Answer: 3. 1890
Solution:
1800+2000+1750+1900+2000 = 9450/5 = 1890 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeQ (15) Study the given table carefully and answer the following questions
Total Population = Online Surveyors + Offline Surveyors + Who didn’t SurveyApproximately total number of people who didn’t participate in survey in all streets together forms what percent of the total population in all villages together?
Correct
Answer :1. 21%
Solution:
1800+2000+1750+1900+2000 = 9450/5 = 1890Incorrect
Answer :1. 21%
Solution:
1800+2000+1750+1900+2000 = 9450/5 = 1890 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative Aptitude1, 10, 33, ?, 145
Correct
Answer :3. 76
Solution:
1^3 + 1^2 1 = 1
2^3 + 2^2 2 = 10
3^3 + 3^2 3 = 33
4^3 + 4^2 4 = 76
5^3 + 5^2 5 = 145Incorrect
Answer :3. 76
Solution:
1^3 + 1^2 1 = 1
2^3 + 2^2 2 = 10
3^3 + 3^2 3 = 33
4^3 + 4^2 4 = 76
5^3 + 5^2 5 = 145 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative Aptitude1, 6, 21, ?, 105
Correct
Answer :1. 52
Solution:
11^2+1^3 = 1
22^2+2^3 = 6
33^2+3^3 = 21
44^2+4^3 = 52
55^2+5^3 = 105Incorrect
Answer :1. 52
Solution:
11^2+1^3 = 1
22^2+2^3 = 6
33^2+3^3 = 21
44^2+4^3 = 52
55^2+5^3 = 105 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeThe average age of all the 30 students of a class is 26. The minimum age of a student is 20 and the maximum age of another student in the same class is 32 years. When the two students whose average age was 28 years resticated from the class but later on one of the resticated student was readmitted. Now the average age of the class is:
Correct
Answer :5.Can’t be determined
Solution:
Based on the data average cannot be determinedIncorrect
Answer :5.Can’t be determined
Solution:
Based on the data average cannot be determined 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeRs. 2,00,000 was invested by Vikash in fixed deposit @ 10% per annum at CI. However every year he has to pay 10% tax on the compound interest. How much money does Vikash has after 3 years?
Correct
Answer :4. Rs. 259005.8
Solution:
1st year = Interest = 20,000 After tax = 18,000 Amount = 2,18,000
2nd year = Interest = 21800 After Tax = 19620 Amount = 237620
3rd year = Interest = 23762 After tax = 21385.8 Amount = 259005.8Incorrect
Answer :4. Rs. 259005.8
Solution:
1st year = Interest = 20,000 After tax = 18,000 Amount = 2,18,000
2nd year = Interest = 21800 After Tax = 19620 Amount = 237620
3rd year = Interest = 23762 After tax = 21385.8 Amount = 259005.8 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeIf the two digits of the age of Mahesh are reversed then the new age so obtained is the age of his wife. 1/11 of the sum of their ages is equal to the difference between their ages. If Mahesh is elder than his wife then find the difference their ages?
Correct
Answer :3. 9 years
Solution:
Mahesh age be ‘yx’ = 10y+x
Wife’s age = ‘xy’ = 10x+y
1/11 (10y+x) = (10y+x10xy)
5x =4y
Since x and y are digit integers then x =4 and y = 5
Age of Mahesh = 10*5+4 =54
Wife age = 45
Difference = 5445 = 9Incorrect
Answer :3. 9 years
Solution:
Mahesh age be ‘yx’ = 10y+x
Wife’s age = ‘xy’ = 10x+y
1/11 (10y+x) = (10y+x10xy)
5x =4y
Since x and y are digit integers then x =4 and y = 5
Age of Mahesh = 10*5+4 =54
Wife age = 45
Difference = 5445 = 9
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