Mixed Quantitative Aptitude Questions Set 226

  1. A shopkeeper X sold an article for rs.850 at a profit of 25%. Another shopkeeper Y sold the another article and earned 30% more profit than the profit earned by X. If the selling price of article Y is three-fifth of selling price of article X, then find the profit percent of article Y ?
    64 2/17%
    76 8/17%
    52 3/34%
    52 3/8%
    78%
    Option B
    SP of article X = 850
    CP of article X = 850 * 100/125 = 680
    profit = 170
    profit of article Y = 170*130/100 = 221
    SP of article Y = 850 * 3/5 = 510
    CP = 510 – 221 = 289
    profit percent = 221/289* 100 = 76 8/17%

     

  2. Mixture A contains milk and water in the ratio of 3 : 2 respectively, while mixture B contains water and milk in the ratio of 4 : 3 respectively. If the quantity of water in mixture B is twice the quantity of water in mixture A, then find the quantity of milk in mixture A is what percent quantity of water in mixture B ?
    68%
    60%
    82%
    75%
    56%
    Option D
    Quantity of milk and water in mixture A = 3x and 2x
    according to questions,
    quantity of water in mixture B = 2*2x = 4x
    quantity of milk in mixture B = 4x/4 *3 = 3x
    required percentage = 3x/4x*100 = 75%

     

  3. 340 liters of mixture of milk and water is contained in a container having milk and water mixed in the ratio of 3 : 1 respectively. 76 liters of this mixture is taken out from the container and replaced with (x + 7) liters of milk so that the ratio of water to milk is 6 : 11. What is the value of ‘x’ ?
    42
    48
    38
    35
    52
    Option D
    Quantity of milk in the container = 340 * 3/4 = 255
    quantity of water in the container = 340*1/4 = 85
    76 liters of mixture is taken out and (x + 7) liter of water is added.
    (255 – 57) / 85 – 19 + (x + 7) = 11/6
    198/(73 + x) = 11/6
    11x = 385
    x = 35

     

  4. A can do a piece of work in 30days while B and C together can complete the work in 20 days and A and C together can complete the work in 24 days. If B is worked in two-fifth of his efficiency, then in how many days A and B together complete half of the work ?
    8 days
    6 days
    8 1/3 days
    5 days
    10 days
    Option E
    LCM of 30 , 20 and 24 = 120
    efficiency of A = 120/30 = 4
    efficiency of B and C = 120/20 = 6
    efficiency of A and C together = 120/24 = 5
    efficiency of C = 5- 4 = 1
    efficiency of B 6- 1 = 5
    two-fifth of efficiency B = 5 * 2/5 = 2
    time required to complete half of the work = 60/2 + 4 = 10 days

     

  5. Rohan and Sivan started a business together with initial investments of rs. 1500 and rs.1800 respectively. If after 2.5 years profit share of Rohan and Sivan are rs(x + 80) and (x + 120) respectively, then find the difference between profit of Rohan and Sivan ?
    180
    120
    320
    130
    340
    Option B
    Profit sharing ratio of Rohan and Sivan = (1500 * 2.5) : (1800 * 2.5) = 5 : 6
    x + 80/x + 120 = 5/6
    6x + 480 = 5x + 600
    x = 120

     

  6. Directions : In each of these questions given below two equations. You have to solve both the equations and give answer.
    I. x^2 – 7√3 + 36 = 0
    II. y^2 – 6√2 + 16 = 0
    X > Y
    X < Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation.
    Option E
    I. x^2 – 4√3 – 3√3 + 36 = 0
    x = 4√3 , 3√3
    II. y^2 – 4√2 – 2√2 + 16 = 0
    y = 4√2, 2√2

     

  7. I. x^2 + √2x – 12 = 0
    II. y^2 – 8√3 + 48 = 0
    X > Y
    X < Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation.
    Option B
    I. x^2 + 3√2 – 2√2 – 12 = 0
    x = -3√2, 2√2
    II. y^2 – 4√3 – 4√3 + 48 = 0
    y = 4√3, 4√3

     

  8. I. x^2 – 36 = 0
    II. y^2 – 15y + 36 = 0
    X > Y
    X < Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation.
    Option E
    I. x^2 = 36
    x = 6, -6
    II. y^2 – 12y – 3y + 36 = 0
    y = 12, 3

     

  9. I. y^2 = 144
    II. 2x^2 – 12x + 16 = 0
    X > Y
    X < Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation.
    Option E
    I. y^2 = 144
    y = 12, -12
    II. 2x^2 – 8x – 4x + 16 = 0
    2x = 8, 4
    x = 4, 2

     

  10. I. x^2 – 2x – 195 = 0
    II. y^2 – 24y + 128 = 0
    X > Y
    X < Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation.
    Option E
    I. x^2 – 15x + 13x – 195 = 0
    x = 15, -13
    II. y^2 – 16y – 8y + 128 = 0
    y = 16, 8

     

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