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Quantitative Aptitude Questions for IBPS/RRB Exam Set- 4

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Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample From all topics ,  which are Important for upcoming IBPS and RRB exams. We have included Some questions that are repeatedly asked in exams !!

Study the following graph carefully and answer the questions given below.
Production of three types of cars by a company over the years(in lakhs).
Graph

  1. What was the percentage drop in the number of C type cars manufactured from 2006 to 2007?
    A. 10
    B. 11
    C. 15
    D. 25
    E. 5
    A.
    Required percentage drop=[(25-22.5)/25]x100=10

  2. What was the difference between the number of B types car manufactured in 2007 and 2008?
    A. 1,00,000
    B. 20,00,000
    C. 10,00,000
    D. 12,50,000
    E. None of these
    E.
    Required difference=(35-27.5)lakh=7,50,000

  3. The total numbers of all the three types cars manufactured was the least in which of the following years?
    A. 2008
    B. 2009
    C. 2006
    D. 2007
    E. 2005
    B.
    Adding all types in all years we can see that 2009 has least numbers of cars manufactured.

  4. In which of the following years was the percentage production of type B to type C cars the maximum?
    A. 2005
    B. 2009
    C. 2008
    D. 2007
    E. 2006
    C.
    Production of B type cars is more than the production of C type cars only in 2006,2007 and 2008.We see the largest difference exists in 2008.

  5. The total production of C type car in 2005 and 2006 together was what percentage of production of B type cars in 2007?
    A. 150
    B. 250
    C. 300
    D. 200
    E. 50
    D.
    Required percentage=[(30+25)/27.5]x100=200

  6. A person cross a stationary bus in 21second.The same bus crosses another person standing over there in 9seconds.What is the ratio of the speed of the bus to the first person ?
    A. 3:4
    B. 3:7
    C. 2:3
    D. 7:3
    E. Cant be determined
    D.
    let the length of the bus be L.
    then speed of the first person
    =L/21units/sec.
    speed of the bus=L/9 units/sec
    Ratio=L/9:L/21
    =7:3

  7. 6 children and 6 women can do an assignment in a duration of 6 days.In how many days 20women can alone complete the given assignment if 9 children alone can complete the assignment in 10 days?
    A. 6.4
    B. 8
    C. Can’t be determined
    D. 7.5
    E. None of these
    E.
    9children alone do in-10days
    6children alone do in- (10×9)/6=15days
    6children in 6days do-6/15=2/5 part
    Rest 3/5 part will be completed by 6women
    in 6days.
    6 women do in 1 day-3/(5×6)=1/10part
    6women alone do in-10days
    =>20women can do in- (6×10)/20=3days

  8. One of the angles of triangle is three-fifth of the sum of the adjacent angles of a parallelogram.Remaining angles of the triangle are in the ratio 5:4.What is the value of the smallest angle of triangle?
    A. 40
    B. 32
    C. 108
    D. 72
    E. None of these
    B.
    One of the angles=(3/5)x180=108(sum of the
    adjacent angles of a parallelogram=180)
    Sum of Remaining angles=180-108=72
    Smallest angle=(4/9)x72=32

  9. Rahul’s salary increases every year by 15% in August.If there is no other increse or reduction in the salary and salary in August 2014 was 30,000 what was his salary(approximately? in August 2011?
    A. 22,685
    B. 19,725
    C. 21,465
    D. 20,895
    E. 23,125
    B.
    Salary in August 2014=30,000
    Salary in August 2011=30,000x
    (1.15×1.15×1.15)=19725.48

  10. 3 peons and 4 sweepers can earn 756 in 7days.11peons and 13 sweepers work for 8 days to earn in 3008. For earning 2480 how much time will be required by 7peons and 9 sweepers?
    A. 8Days
    B. 11Days
    C. 9Days
    D. 7Days
    E. 10Days
    E.
    Let Peons be P and sweepers be S.
    (3P+4S) in one day earn 756/7=108
    (11P+13S) in one day earn 3008/8=376
    By earning 1rs per day we can conclude
    (3P+4S)/108=(11P+13S)/376
    =>P:S=5:3–(i)
    (3P+4S) in 1day earn 108
    =>(3+4x(3/5))P  in 1 day earn 108
    or 27m/5=108
    =>1P in 1 days earns (108×5)/27=20
    thus a S earns 20x(3/5)=12daily
    7P+9S earn (7×20+9×12)=248
    So to earn 2480 to have work 10 days.