Quant Test for IBPS PO 2018 Prelim Exam Set – 3

Directions(1-5): Find the relation between x and y and choose a correct option.

  1. I. 3√x– 18/ √x = √x
    II. 2√y + (y−36)/ √y = –√y

    x>=y
    x=y or relation cannot be established.
    y>=x
    x>y
    y>x
    Option B
    3√x– 18/ √x = √x
    =>3x–18 = x
    =>2x = 18
    =>x = 9
    2√y +( y−36)/ √y = –√y
    =>2y+y–36 = –y
    =>4y = 36
    =>y = 9
    x=y

     

  2. I. 6 y^2 + 1/2 = 7/2 y
    II. 12 x^2 + 2 = 10 x

    y>=x
    x>y
    x>=y
    y>x
    x=y or relation cannot be established.
    Option C
    6y^2 + 1/2 = 7/2 y
    12y^2 – 7 y + 1 = 0
    12y^2 – 44y – 3y + 1 = 0
    4y (3y – 1) – 1 (3y –1) = 0
    (3y – 1) (4y – 1) = 0
    y = 1/3, ¼
    12x^2 – 10x + 2 = 0
    6x^2 – 5x + 1 = 0
    3x (2x –1) – 1 (2x –1)= 0
    x = 1/3, ½
    x>=y

     

  3. I. x ^4 – 1000 = 296
    II. y^2 + 330 = 346

    x=y or relation cannot be established.
    x>=y
    x>y
    y>x
    y>=x
    Option A
    x = + 6,-6
    y = + 6, -6
    No relation

     

  4. I. 21x^2 – 26x + 5 = 0
    II. y = √0.1024

    y>x
    x>y
    y>=x
    x=y or relation cannot be established.
    x>=y
    Option D
    x = 0.23, 1
    y = + 0.32
    No relation

     

  5. I. (6^ 3+9^2)/ 11 = x^3
    II. 15y^3= (36×18) + 12y^3

    x>=y
    y>x
    x=y or relation cannot be established.
    y>=x
    x>y
    Option B
    (6^3+9^2)/ 11 = x^3
    =>216+81 11 = x ^3
    =>297 11 = x ^3
    =>x ^3 = 27
    x = +3
    15y^3= (36×18) + 12y^3
    =>15y^3 –12y^3 = (36×18)
    =>3y^3 = (36×18)
    =>3y^3 = 648
    =>y ^3 = 216
    y = +6
    y>x

     

  6. How many 3-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?
    700
    660
    720
    656
    685
    Option C
    The word ‘LOGARITHMS’ has 10 different letters.
    Required answer = 10P3
    =10×9×8=720

     

  7. Compound Interest for two years at 20% per annum would fetch a sum of money Rs. 482, if the interest was payable half- yearly then it was payable annually. What is the sum?
    35000
    22800
    33000
    20000
    25200
    Option D
    Let sum=Rs.x
    C.I. when compounded half yearly = [x(1+10/100)^4−x] = 4641/10000
    C.I. when compounded annually = [x(20/100)^2−x] = 11/25
    4641/10000 x−11/25x = 482
    => x=20000

     

  8. A and B can do a piece of work in 45 and 40 days respectively.They began work together but A leaves after X days and B finished the rest of the work in X+14 days. After how many days did A leave?
    11
    7
    9
    10
    5
    Option C
    Total work in one day = (1/45+1/40) = 17/360
    In X days, 17X/360
    Remaining work = (1-17x/360)
    Now, B can do 1 pt. work in (x+14)/(1-17x/360) days.
    (x+14)/(1-17x/360) = 40
    X = 9 days.

     

  9. A train travelling at a speed of 75 mph enters a tunnel 3(1/2) miles long. The train is ¼ mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?
    5
    3
    7
    4
    2
    Option B
    Total distance covered = (7/2+1/4) miles
    = 154 miles
    Time taken = (154*75) hrs = 120 hrs.
    = (1/20*60) min.= 3 min

     

  10. In a city of an adult population, 45% of men and 25% of women are married. If no man marries more than one woman, and vice versa, What is the percentage of total population of adults who are married ?
    20.25%.
    32.14%.
    28.15%.
    25.23%.
    30.12%.
    Option B
    Let no. of men = x & total population be 100.
    Total women = (100-x)
    Married man = 45x /100 —————(1)
    & married women = 25/100 (100-x) ——— (2)
    From (1) & (2) x = 500/14
    So, Married man = 225 /14
    Married woman = 225/14
    Total married population = men + women = 450/14
    Required % = (450/14)/100 *100
    = 450/14% = 32.14%.

     


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