Current Affairs PDF Sales

Quantitative Aptitude Questions for IBPS PO Mains Exam Set – 40

AffairsCloud YouTube Channel - Click Here

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. At a certain rate per annum, the simple interest on a sum of money for 1 year is Rs 260 and the compound interest on the same sum for 2 yrs is Rs 540.80. The rate of interest per annum is
    A. 10%
    B. 4%
    C. 8%
    D. 6%
    C. 8%
    Explanation:
    SI for 2 yrs = 2 × 260 = 520
    Difference in CI and SI for 2 yrs = 540.80 – 520 = 20.80
    For 2 yrs, Difference = r×SI / 2×100
    So, 20.80 = r×520/2× 100
    Solve, r = 8%

  2. In how many ways the letters of word COMPLETE be arranged such that all vowels are always together?
    A. 2290
    B. 1290
    C. 2150
    D. 2160
    D. 2160
    Explanation:
    CMPLT OEE
    We have to take OEE together, so in all we have 6 letters {C, M, P, L, T, OEE}
    Ways for these 6 is 6!
    Now for OEE, ways are 3!, and since it contains 2 E’s, so ways are 3!/2!
    Total ways = 6! × 3!/2! = 2160

  3. There are 5 males and 4 females. A committee is to be formed comprising 4 people. How many ways are there to do so such that a particular male is always to be included?
    A. 52
    B. 56
    C. 48
    D. 30
    B. 56
    Explanation:
    Since a particular male is to included always, so there remains 4 males for selection and 4 females, and also since a total of 4 is to selected, and 1 male is already included, so 3 are to be selected now.
    So ways are 8C3 = 56

  4. x2 + 28x + 195 = 0; y2 + 28y – 60 = 0
    A. x ≤ y
    B. x ≥ y
    C. x > y
    D. No relationship
    C. 4
    Explanation:
    x2 + 28x + 195 = 0
    (x+15)(x+13) = 0
    x = -15, -13
    y2 + 28y – 60 = 0
    (y+30)(y-2) = 0
    y = -30, 2
    put on number line
    -30        -15         -13               2
    We can see that no relationship can be established.

  5. A and B together can complete the work in 4 days. A alone starts working and leaves it after working for 3 days, completing only half of the work. In how many days can it be completed if the remaining job is undertaken by B?
    A. 5
    B. 8
    C. 6
    D. 4
    C. 6
    Explanation:
    1/A + 1/B = 1/4
    A works for 3 days and completes half of the work. This gives
    1/A * 3 = 1/2
    A = 6
    So 1/B = 1/4 – 1/6 = 1/12
    So B can complete a job in 12 days, so half of the job will be complete in 6 days.

  6. A man, a woman or a boy can do a piece of work in 3, 4 and 12 days respectively. How many boys must assist one man and one woman to do the work in one day?
    A. 6
    B. 5
    C. 4
    D. 10
    B. 5
    Explanation:
    Let ‘x’ boys. So,
    1/3 + 1/4 + 1/12 * x = 1
    Solve, x = 5

  7. While selling a cooler a shopkeeper gives a discounts of 10% on the marked price. If he gives a discount of 12% he earns Rs 35 as profit. The marked price of cooler is
    A. Rs 2250
    B. Rs 1750
    C. Rs 1625
    D. Rs 1725
    B. Rs 1750
    Explanation:
    35*100 / (-10 – (-12)) = 1750

  8. A man buys a chair and a table for Rs 6000. He sells the chair at a loss of 10% and table at a gain of 10%. He still gains Rs 100 on the whole. The cost price of table is
    A. Rs 3050
    B. Rs 3500
    C. Rs 2850
    D. Rs 4000
    B. Rs 3500

    Explanation:
    Gain% on the whole = 100/6000 * 100 = 5/3%
    By the method of allegation:
    -10                                      10
    .                              5/3
    10 – 5/3 = 25/3                    5/3 – (-10) = 35/3
    Ratio = 25/3 : 35/3 = 5 : 7
    So CP of table = 7/(5+7) * 6000 = 3500

  9. If the compound interest for two successive years is Rs 1100 and Rs 1210 respectively, then what is the rate percent?
    A. 11%
    B. 10%
    C. 12.1%
    D. None of these
    B. 10%
    Explanation:
    R% of 1100 = 1210 – 1100
    R/100 * 1100 = 110
    R = 10%

  10. By selling 8 articles, a tradesman makes a loss equal to the cost price of 4 articles. Find loss%.
    A. 25%
    B. 100/3%
    C. 50%
    D. None of these
    C. 50%
    Explanation:
    C.P of 8 articles – S.P of 8 articles = C.P of 4 articles
    C.P of 4 articles = S.P of 8 articles
    Loss% = (8 – 4)/8 * 100 = 50%