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IBPS Clerk Main: Quants Day 6

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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. 2x2 + 3x – 35 = 0, 2y2 – 3y – 9 = 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 + 3x – 35 = 0
    2x2 + 10x – 7x – 35 = 0
    Gives x = -5, 7/2
    2y2 – 3y – 9 = 0
    2y2 – 6y + 3y – 9 = 0
    Gives y = -3/2 3
    Put on number line
    -5     -3/2     3      7/2

  2. 3x2 + 10x + 7 = 0, 2y2 – 5y = 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 + 10x + 7 = 0
    3x2 + 3x + 7x + 7 = 0
    Gives x = -7/3, -1
    2y2 – 5y = 0
    y(2y-5) = 0
    gives y = 0, 5/2
    put on number line
    -7/3      -1       0        5/2

  3. 3x2 – 7x – 20 = 0, 6y2 + 31y + 35 = 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:

    3x2 –7x – 20 = 0
    3x2 – 12x + 5x – 20 = 0
    Gives, x = -5/3, 4
    6y2 + 31y + 35 = 0
    6y2 + 10y + 21y + 35 = 0
    Gives y = -7/2, -5/3
    Put on number line
    -7/2        -5/3         4

  4. 3x2 + 4x – 32 = 0, y2 + y – 12 = 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:

    3x2 + 4x – 32 = 0
    3x2 + 12x – 8x – 32 = 0
    Gives, x = -4, 8/3
    y2 + y – 12 = 0
    gives, y = -4, 3
    put on number line
    -4        8/3       3

  5. 3x2 + 7x + 4 = 0, 3y2 + y – 2 = 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:

    3x2 + 7x + 4 = 0
    3x2 + 3x + 4x + 4 = 0
    Gives, x = -4/3, -1
    3y2 + y – 2 = 0
    3y2 + 3y – 2y – 2 = 0
    Gives, y = -1, 2/3
    Put on number line
    -4/3       -1       2/3

  6. A, B, and C can complete a work in 20/11 hours. Time taken by C to complete the same work is 9/2 times the time taken by A and B together to complete that work. Also time taken by A to complete the work is 5/6 times the time taken by B and C together to complete that work. What is the time in which B can alone complete the work?
    A) 3 1/3 hours
    B) 6 hours
    C) 6 2/3 hours
    D) 5 hours
    E) None of these
    C) 6 2/3 hours
    Explanation:

    (1/A) + (1/B) + (1/C) = 11/201/C = (2/9) * [(1/A) + (1/B)]

    1/A = (6/5) * [(1/B) + (1/C)]
    From 2nd and 3rd equation put values of 1/A and 1/C in equation 1 to find value of B


  7. The average of money with A, B, C, and D is Rs 200. If the money with A increases by 100% and that with B increases by 200%, then the average money with four increases by Rs 62.50. If the percentage increases of the amounts with A and B gets swapped, average money with A and B will be Rs 275. What is the amount of money that A had originally?
    A) Rs 200
    B) Rs 175
    C) Rs 150
    D) Rs 180
    E) None of these
    C) Rs 150
    Explanation:

    (A+B+C+D)/4 = 200
    So A+B+C+D = 200*4
    Money with A increases by 100%, so with A money becomes (200/100)*A = 2A
    With B money becomes 3B
    (2A+3B+C+D)/4 = 200+62.50
    Or A+2B+(A+B+C+D) = 262.5*4
    So A+2B = 262.5*4 – 200*4
    Or A+2B = 62.5*4
    Also given that if increase in money gets swapped
    (3A+2B)/2 = 275
    Now solve both equations in A and B to find value of A

  8. Some toffees were bought at 10 for Rs 1.50 and an equal number at 20 paise each. On selling all the toffees at 20 for Rs 4, a profit of Rs 10 was made. How many toffees were purchased?
    A) 200
    B) 300
    C) 400
    D) 450
    E) 250
    C) 400
    Explanation:

    Let x toffees purchased. If 10 toffees are for Rs 1.50 then x toffees for Rs 15x/100
    If each toffee is for 20 paise or RS 20/100, then x toffees for Rs 20x/100
    Total CP of 2x toffees = (15x/100) + (20x/100) = 35x/100
    Now 20 toffees sold for Rs 4, so SP of 2x toffees = 4x/10
    Now given that (35x/100) + 10 = 4x/10
    Solve, x = 200
    Total toffees bought = 2x

  9. In a business started by A and B, a profit of Rs 50,000 was made after a year. B being a working partner received 20% of the annual profit as his salary. Had the entire profits divided in the ratio of their investments A would have received Rs 8000 more as his profit share than what he actually got. What is the actual profit share of A?
    A) Rs 35,000
    B) Rs 30,000
    C) Rs 42,000
    D) Rs 32,000
    E) Rs 40,000
    D) Rs 32,000
    Explanation:

    As salary, B got = 20/100 * 50,000 = Rs 10,000
    So remaining profit = 50,000 – 10,000 = 40,000
    Let the investment ratio of A and B is x : y
    Then A will get = [x/(x+y)] * 40,000
    But if entire profits were divided between them
    Then A would have got [x/(x+y)] * 50,000
    Now given that [x/(x+y)] * 50,000 = [[x/(x+y)] * 40,000] + 8000
    Solve, x = 4y
    So the ratio of investments become 4 : 1
    So A will get [4/(4+1)] * 40,000

  10. A bag contains 5 black and 7 green balls. Two balls are drawn one after the other with 1st ball being replaced. What is the probability that both the balls are green?
    A) 36/225
    B) 48/121
    C) 49/144
    D) 40/221
    E) None of these
    C) 49/144
    Explanation:

    7/12 * 7/12