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IBPS Clerk Main: Quants Day 14

By
AC Team 1
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Hello Aspirants,

Welcome to Online Quant Section in AffairsCloud.com. We are starting IBPS Clerk course 2015 and we are creating sample questions in Quantitative Aptitude section, type of which will be asked in IBPS Clerk Main Exam.

Stratus – IBPS Clerk Course 2015 

Stratus - IBPS Clerk - Daily Test - Quants
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  1. There are 3 red balls, 4 blue balls, 5 white balls. Three balls are drawn at random. What is the probability that there are at least 2 red balls?
    A) 21/58
    B) 7/55
    C) 9/52
    D) 12/55
    E) None of these
    B. 7/55
    Explanation:
    Case 1 : 2 red balls:
    Prob = 3C2 × 9C1 / 12C3
    Case 2 : 3 red balls:
    Prob = 3C3 / 12C3
    Add the two cases.

  2. A man starts from point P at 6:00 AM and reaches point Q at 9:00 AM. Another man starts from point Q at 7:00 AM and reaches point P at 11:00 AM. At what time both men will meet?
    A) 8:00 AM
    B) 8:08 AM
    C) 8:34 AM
    D) 9:18 AM
    E) 7:42 AM
    B) 8:08 AM
    Explanation:
    Let the distance from A to B is ‘d’ km.
    First train reaches destination in 3 hrs, so speed of first train = d/3 km/hr.
    Speed of second train = d/4 km/hr
    Speed of first train = d/3 km/hr, so after 1 hr, i.e. at 7AM, the first train has covered d/3 km.
    The distance left = (d – d/3) km = 2d/3 km. relative speed = d/3 + d/4.
    So time they will meet at = 7:00 AM + (2d/3) / (d/3 + d/4)
    = 7:00 AM + 1 (1/7)*60 = 8:08 AM

  3. 80 litres of a mixture contains 20% water and the rest milk. What amount of water must be added so that the resulting mixture contains only 50% milk?
    A) 40 litres
    B) 50 litres
    C) 52 litres
    D) 48 litres
    E) None of these
    D) 48 litres
    Explanation:
    Mixture contains 20% water.
    The water which will be added will be 100% water
    The new mixture contains 50% milk, this gives 50% water
    So by allegation rule
    Mixture                     Water
    20                              100
    .                 50
    50                               30
    This gives 50 : 30 = 5 : 3
    Let x litres of water to be added to 80 litres mixture.
    So 80/x = 5/3
    x = 48

  4. 3 liquids A, B, and C are present in the ratio 3 : 4 : 5. After 12 litres of the mixture taken out and 8 litres of A is poured in, the ratio of A to C becomes 7 : 5. Find the original quantity of the mixture?
    A) 48 litres
    B) 36 litres
    C) 39 litres
    D) 60 litres
    E) 32 litres
    B) 36 litres
    Explanation:
    Let A = 3x, B = 4x, C = 5x
    When 12 litres taken out, left out quantities are
    A = 3x – 3/(3+4+5) * 12 = 3x – 3
    B = 4x – 4/(3+4+5) * 12 = 4x – 4
    C = 5x – 5/(3+4+5) * 12 = 5x – 5
    Now after 8 litres of A poured in, A becomes (3x – 3) + 8 = 3x + 5
    Now given, A : C = 7 : 5. So
    3x+5 / 5x-5 = 7/5
    Solve, x = 3
    Total initial quantity = 3x + 4x + 5x = 12x = 12*3 = 36

  5. 2/5th of work is completed in 10 days by 20 men working 6 hours each day. In how many days the remaining work will be completed by 10 women working 8 hours each day such that 4 women do as much work as is done by 2 men?
    A) 22
    B) 42
    C) 38
    D) 45
    E) 50
    D) 45
    Explanation:
    4 w = 2m, so 1w = 1/2 m
    And then 10 w = 5 m, so we have to find the number of days for 5 men to complete remaining work (1 – 2/5) = 3/5th.
    M1*D1*H1*W2 = M2*D2*H2*W1
    20*10*6*(3/5) = 5*D2*8*(2/5)
    Solve, D2 = 45 days

  6. 2x2 + 13x + 15 = 0, 2y2 – 5y – 12 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    D) x ≤ y
    Explanation:
    2x2 + 13x + 15 = 0
    2x2 + 10x + 3x + 15 = 0
    Gives x = -5, -3/2
    2y2 – 5y – 12 = 0
    2y2 – 8y + 3y – 12 = 0
    Gives y = -3/2, 4
    Put on number line
    -5               -3/2                       4

  7. 2x2 + 5x + 2 = 0, 3y2 + 5y – 2 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    E) x = y or relationship cannot be determined
    Explanation:
    2x2 + 5x + 2 = 0
    2x2 + 4x + x + 2 = 0
    Gives x = -2, -1/2
    3y2 + 5y – 2 = 0
    3y2 + 6y – y – 2 = 0
    Gives y = -2, 1/3
    Put on number line
    -2            -1/2                   1/3

  8. 5x2 – 7x – 6 = 0, 5y2 + 23y + 12 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    C) x ≥ y
    Explanation:
    5x2 – 7x – 6 = 0
    5x2 – 10x + 3x – 6 = 0
    Gives, x = -3/5, 2
    5y2 + 23y + 12 = 0
    5y2 + 20y + 3y + 12 = 0
    Gives y = -4, -3/5
    Put on number line
    -4           -3/5               2

  9. 3x2 – 4x – 4 = 0, 2y2 – 11y + 15 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    B) x < y
    Explanation:
    3x2 – 4x – 4 = 0
    3x2 – 6x + 2x – 4 = 0
    Gives, x = 2, -2/3
    2y2 – 11y + 15 = 0
    2y2 – 6y – 5y + 15 = 0
    Gives, y = 5/2, 3
    Put on number line
    -2/3              2                  5/2                3

  10. x2 + x – 12 = 0, 2y2 – 9y + 10 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    E) x = y or relationship cannot be determined
    Explanation:
    x2 + x – 12 = 0
    Gives, x = -4, 3
    2y2 – 9y + 10 = 0
    2y2 – 9y – 5y + 10 = 0
    Gives, y = 2, 5/2
    Put on number line
    -4            2               5/2                  3





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