Mixed Quantitative Aptitude Questions Set 219

  1. A man deposited a total amount of rs. 36000 partially in scheme A and scheme B. Scheme A offers simple interest at the rate of 14% p.a while scheme B offers compound interest at the rate of 20% p.a compounded annually. If the interest earned from scheme A after 2 years is rs.2160 less than the interest earned from scheme B after 2 years, then what is the amount deposited in scheme B ?
    15000
    17000
    23000
    14000
    17500
    Option B
    Let amount deposited in scheme B = rs. x
    amount deposited in scheme A = rs.(36000 – x)
    Rate of compound interest after 2 years = 20 + 20 + 20 * 20/100 = 44%
    x * 44/100 – (36000 – x) * 14/100 * 2 = 2160
    (44x – 108000 + 28x)/100 = 2160
    72x = 1224000
    x = 17000
    amount deposited in scheme B is rs.17000

     

  2. The income of A and B are in the ratio 4 : 5 respectively. If the expenditure of A and B are rs.14000 and rs.18000 and ratio of saving of A and B is 6 : 7, then what is the saving of B ?
    8000
    9200
    7000
    6800
    8200
    Option C
    Let income of A and B = 4x and 5x respectively
    4x – 14000/5x – 18000 = 6/7
    28x – 98000 = 30x – 108000
    x = 5000
    income of B = 5 * 5000 = 25000
    saving of B = 25000 – 18000 = 7000

     

  3. A takes 5 days less than B to complete a work alone. A and B together can complete the work in 13 7/11 days. Find the time taken by B to complete the whole work ?
    25 days
    30 days
    45 days
    43 days
    40days
    Option B
    Let B alone can complete the whole work = x days
    A alone can complete the whole work = (x – 5) days
    1/x – 5 + 1/x = 11/150
    2x – 5/x^2 – 5x = 11/150
    11x^2 – 55x = 300x – 750
    11x^2 – 355x + 750 = 0
    11x^2 – 330x – 25x + 750 = 0
    x = 30
    B alone can complete the whole work in 30 days

     

  4. Three containers having capacity in the ratio of 4 : 5 : 3 contain mixture of milk and water in the ratio of 2 : 3, 4 : 1 and 3 : 1 respectively. If all three containers poured in a big container, then find the ratio of water to milk in final mixture ?
    28 : 137
    85 : 152
    83 : 157
    134 : 49
    5 : 2
    Option C
    Let total mixture in three container are 400, 500 and 300 respectively
    total quantity of milk in big container = 400 * 2/5 + 500 * 4/5 + 300 * 3/4 = 785
    total quantity of water in big container = 400 * 3/5 + 500 * 1/5 + 300 * 1/4 = 415
    required ratio = 415 : 785 = 83 : 157

     

  5. B and C started a partnership business with the capital of rs.(x + 4000) and rs.(x + 2000) respectively. After one year A joined the business with capital of rs.(x + 12000) while B left the business. If at the end of three years the profit sharing ratio of A , B and C is 20 : 6 : 15, then find the difference between amount invested by A and B in the business ?
    15000
    13000
    14000
    8000
    12000
    Option D
    A : B : C = (x + 12000) * 2 : (x + 4000) * 1 : (x + 2000) * 3
    = 2x + 24000 : x + 4000 : 3x + 6000
    2x + 24000/x + 4000 = 20/6
    20x + 80000 = 12x + 144000
    8x = 64000
    x = 8000
    difference = (8000 + 12000) – (8000 + 4000) = 8000

     

  6. Directions : In these questions two equations is given, solve the equations and answer the questions given below.
    I. x^2 – 42x + 392 = 0
    II. y^2 – 27y + 180 = 0
    X < Y
    X > Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation
    Option E
    I. x^2 – 28x – 14x + 392 = 0
    x = 28 , 14
    II. y^2 – 15y – 12y + 180 = 0
    y = 15, 12

     

  7. I. x^2 – 8x – 65 = 0
    II. 2y^2 – 27y + 91 = 0
    X < Y
    X > Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation
    Option E
    I. x^2 – 13x + 5x – 65 = 0
    x = 13, -5
    II. 2y^2 – 14y – 13y + 91 = 0
    2y = 14, 13
    y = 7, 6.5

     

  8. I. x^3 = 216
    II. 5y^2 + 2y – 185 = 0
    X < Y
    X > Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation
    Option B
    I. x^3 = 216
    x = 6, 6
    II. 5y^2 + 15y – 13y – 37 = 0
    5y = – 15, 13
    y = -3 , 2.6

     

  9. I. x^2 + 17x + 60 = 0
    II. y^2 – 2y – 15 = 0
    X < Y
    X > Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation
    Option A
    I. x^2 + 12x + 5x + 60 = 0
    x = -12, -5
    II. y^2 – 5y + 3y – 15 = 0
    y = 5, -3

     

  10. I. x^2 – 15x + 36 = 0
    II. y^2 – 19y + 60 = 0
    X < Y
    X > Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation
    Option E
    I. x^2 – 12x – 3x + 36 = 0
    x = 12 , 3
    II. y^2 – 15y – 4y + 60 = 0
    y = 15, 4

     

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