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Quantitative Aptitude Questions for IBPS PO Mains Exam Set – 36

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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 exams. We have included Some questions that are repeatedly asked in exams.

  1. A’s age is 4 years more than B’s age. A’s age is also 8 years less than C. 7 years hence ratio of B’s age to C’s age will be 7 : 13. Find the ratio of present ages of A, B, and C?
    A. 11 : 19 : 7
    B. 17 : 9 : 11
    C. 11 : 7 : 19
    D. 11 : 12 : 13
    C. 11 : 7 : 19
    Explanation:
    A = B+4
    A = C-8
    B+7 / C+ 7 = 7/13
    Or (A-4)+7 / (A+8)+7 = 7/13
    A+3 / A+15 = 7/13
    A = 11
    So B = A – 4 = 11 – 4 = 7
    C = A+ 8 = 11 + 8 = 19

  2. The sum of ages of A and B is 50. 7 years ago, A’s age was 3 more than twice B’s age then. Find the difference of ages of A and B.
    A. 12
    B. 14
    C. 13
    D. 15
    B. 14
    Explanation:
    A+B = 50
    A – 7 = 2(B-7) + 3
    Solve the two equation to find A and B
    A = 32, B = 18
    Diff = 32 – 18 = 14

  3. A gave 40% of amount to B. From this B gave 20% to C. From remaining B spends 40% on food and 20% of transport. Now he is left with Rs 912. Find the amount received by B.
    A. Rs 2890
    B. Rs 2090
    C. Rs 2270
    D. Rs 2850
    D. Rs 2850
    Explanation:
    Let A has Rs x
    B gets 40/100 * x = 2x/5
    after giving 20% to C, B is left with 80% of 2x/5 = 8x/25
    now he spends 40% and 20%, total 60%, now amount left is 40% of 8x/25 = 16x/125
    so, 16x/125 = 912
    x=7125
    B got 40% of x = 40/100 * 7125 = 2850

  4. 2 men or 3 women or 4 children can do a work in 52 days. In how many days 1 man and 1 woman and 1 child can complete the same work?
    A. 48
    B. 50
    C. 42
    D. 46
    A. 48
    Explanation:
    For this when we are given – a men or b women or c children work in D days. Number of days taken by 1M + 1W + 1C = a*b*c*D / (a*b + b*c + c*a)
    So, days = 2*3*4*52 / (2*3 +3*4 + 4*2) = 48

  5. Four coins are tossed. Find the probability of at most 2 heads.
    A. 5/19
    B. 6/17
    C. 5/16
    D. 11/16
    D. 11/16
    Explanation:
    There are total 24 outcomes
    For atmost 2 heads (11 pairs):
    TTTT, HTTT, THTT, TTHT, TTTH, HHTT, HTHT, HTTH, THHT, THTH, TTHH
    So prob. = 11/16

  6. There are 3 red, 4 white and 6 blue balls. 2 balls are drawn at random. Find the probability that at least 1 is blue.
    A. 19/26
    B. 19/21
    C. 13/24
    D. 14/19
    A. 19/26
    Explanation:
    Total 13 balls, ways of choosing 2 are 13C2 = 13*12/2*1 = 78
    For 1 blue: 6C1 * 7C1
    = 42
    For both blue: 6C2
    = 6*5/2 = 15
    Total prob. = 42+15 / 78 = 57/78 = 19/26

  7. A travels a distance f 50 km in 2 hrs and 30 min. How much faster, in km/hr muct he travel to make such a trip in 5/6 hr less time?
    A. 10
    B. 20
    C. 30
    D. 40
    A. 10
    Explanation:
    Time required = 2hrs 30 min – 50 min = 1 hr 40 min = 1 2/3 hrs
    Required speed = 50 * 3/5 = 30 km/hr
    Original speed = 50 * 2/5 = 20 km/hr
    Diff = 30 – 20 = 10 km/hr

  8. A is travelling on his cycle and has calculated to reach point P at 2 PM if he travels at 10 km/hr. He will reach there at 12 noon if he travels at 15 km/hr. At what speed must he travel to reach P at 1 PM.
    A. 13 km/hr
    B. 12 km/hr
    C. 12.5 km/hr
    D. 14 km/hr
    B. 12 km/hr
    Explanation:
    Let distance = d, time to reach at 1 PM = t
    Then time to reach at 2 PM = t+1 = d/10
    Time to reach at 12 noon = t-1 = d/15
    Solve the 2 equations, d = 60 km, t = 5 hrs
    So speed to reach at 1 PM = 60/5 = 12

  9. The volume of a cone is 1584 cm3 and its height is 42 cm. Find the curved surface area of the cone.
    A. 565.7√2 cm2
    B. 565 cm2
    C. 560.67√2 cm2
    D. 548.34√2 cm2
    A. 565.7√2 cm2
    Explanation:
    Vo. of cylinder is = 1/3 * ᴨr2h = 1584
    h = 42
    Solve and find r, r = 6 cm
    Now length, l = √(h2+r2
    = √ 422+62 = 30√2
    Curved surface area = ᴨrl = 22/7 * 6 * 30√2

  10. A can do a work in 10 days, B in 15 days and C in 12 days. All started the work, but A leaves after 2 days, further 3 days after B left the work. In how much time C completes the remaining work?
    A. 2/5 day
    B. 4/5 day
    C. 1/5 day
    D. 3/5 day
    D. 3/5 day
    Explanation:
    After 2 days, B left after further 3 days, so B worked for 5 days. Let C worked for total x days.
    1/10 * 2 + 1/15 * 5 + 1/12 * x = 1
    Solve, x = 28/5
    This gives that C worked for a total of 28/5 days, in which 5 days he worked before he was left by B. So remaining days to complete remaining work = 28/5 – 5 = 3/5th day