Quant Test for IBPS PO 2018 Prelim Exam Set – 33

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

  1. I. x^2 – 6x + 135 = 0
    II. y^2 – 30y + 225 = 0

    x=y or relation cannot be established.
    x>=y
    x>y
    y>=x
    y>x
    Option D
    I.X^2 – 6x + 135 = 0
    => x^2 – 15x + 9x + 135 = 0
    => x = +15, –9
    II.Y^2 – 30y + 225 = 0
    =>y^2 – 15y – 15y + 225 = 0
    => y = +15, +15
    x≤y

     

  2. I. 25/√x – 4√x = √x
    II. 2y + (𝑦^2+50)/𝑦 = 5

    x=y or relation cannot be established.
    y>=x
    y>x
    x>y
    x>=y
    Option E
    I.25/ √𝑥 – 4√𝑥 = √𝑥 =>25 – 4x = x
    => 25 = 5x
    =>X = 5
    II.2y+ (y^2+50)/𝑦 = 5y
    =>2y^2 + y2+50 = 5y^2
    => 2y^2 = 50
    =>y = √25
    =>y = ±5
    x≥y

     

  3. I. 8/√𝑥 + 6/√𝑥 = √𝑥
    II. y^3 – (14)^7/2/√𝑦 = 0

    x>y
    y>x
    x>=y
    x=y or relation cannot be established.
    y>=x
    Option D
    I. 8/√𝑥 + 6/√𝑥 = √𝑥
    =>X = 14
    II. Y^3 – (14)^7/2/√𝑦 = 0
    =>𝑦^7/2–(14)^7/2 = 0
    => 𝑦^7/2 = (14)^7/2
    =>y = 14
    x = y

     

  4. I. x^2 – 3481 = 0
    II.y^2 –118y +3481 = 0

    x>=y
    x>y
    y>x
    y>=x
    x=y or relation cannot be established.
    Option D
    I. x^2 – 3481= 0
    =>x^2 = 3481
    =>x = ±59
    II. y^2 – 118y + 3481 = 0
    => y^2 –59y–59y +3481=0
    =>y = +59, +59
    x≤y

     

  5. I. x^3 – 9×2 + 20x = 0
    II. y^3 –14y2 + 48y = 0

    x=y or relation cannot be established.
    y>=x
    x>y
    x>=y
    y>x
    Option A
    I. x^3 – 9×2 + 20x = 0
    =>x(x^2 – 9x + 20) = 0
    =>x^2 – 9x + 20 = 0
    =>x^ 2 – 4x–5x + 20 = 0
    x = 4, 5 and 0
    II. y^3 –14y2 + 48y = 0
    =>y(y^2 –14y + 48) = 0
    =>y^2 –14y + 48 = 0
    =>y^2 –6y –8y+ 48 = 0
    y = 6, 8 and 0
    No relation.

     

  6. 10 years ago the age of Krishna was 𝟏/𝟑 𝐫𝐝 of the age of Banshi. 14 years hence the ratio of ages of Krishna and Banshi will be 5 : 9. Find the ratio of their present ages.
    12 : 29
    13 : 28
    14 : 27
    13 : 29
    11 : 21
    Option D
    Let 10 years ago Krishna’s age = x years.
    Let 10 years ago Banshi’s age = 3x years
    Therefore, [x+10+14]/[3x+10+14] = 5/9 x= 16 years
    So, their present age = 16 + 10 = 26 years & 16 × 3 + 10 = 58 years.
    Ratio = 13 : 29

     

  7. The concentration of glucose in three different mixtures (glucose and alcohol) is 𝟏/𝟐 , 𝟑/𝟓 𝐚𝐧𝐝 𝟒/𝟓 respectively. If 2 litres, 3 litres and 1 litre are taken from these three different vessels and mixed. What is the ratio of glucose and alcohol in the new mixture?
    4:5
    3:2
    7:5
    3:2
    9:7
    Option B
    Concentration of glucose are in ratio
    = 1/2 : 3/5 : 4/5
    Quantity of glucose taken from A = 1 litre out of 2 litre.
    Quantity of glucose taken from B = 3/5 ×3 = 1.8 lt.
    Quantity of glucose taken from C = 0.8 lt.
    So, total glucose taken out from A, B & C, = 3.6 lt.
    So, quantity of alcohol = (2 + 3 + 1) – 3.6 = 2.4 litre.
    Ratio of glucose to alcohol = 3.6/2.4 = 3:2

     

  8. A dealer buys dry fruits at Rs. 100, Rs 80 and Rs. 60 per kilogram. He mixes them in the ratio 4 : 5 : 6 by weight, and sells at a profit of 50%. At what price per kilogram does he sell the dry fruit?
    111
    128
    112
    121
    116
    Option E
    Let he bought 4x kg, 5x kg and 6x kg of dry fruits.
    Total cost price for dealer = (4x×100) + (5x×80) + (6x×60)
    = 400x + 400x + 360x
    = 1160 x
    Total selling price for dealer = 1740x
    Required price per kg = 1740x 15x = 116

     

  9. If Rs.8010 is divided in to three parts such that their amounts after 2 3 and 4 years respectively are equal, the simple interest being at the rate of 2% per annum. Find the difference between the greatest and smallest parts of the sum.
    105
    101
    88
    95
    90
    Option B
    Let the amount divided into three parts x y z
    According to the question, 𝑥 + [𝑥 × 2 × 2]/100 = 𝑦 + [𝑦 × 3 × 2]/100 = 𝑧 + [𝑧 × 4 × 2]/100 𝑥 + 4𝑥/100
    = 𝑦 + 6𝑦/100 = 𝑧 + 8𝑧/100
    = x:y:z = 52:53:54
    x+y+z = 8010
    Required difference = 2 part
    = 8010 × 2/159 = 50.37×2 = 100.74 ==101

     

  10. How many different four letter words can be formed (the words need not be meaningful using the letters of the word “MEDITERRANEAN” such that the first letter is E and the last letter is R?
    45
    60
    48
    59
    50
    Option D
    Case 1: When the two letters are different.
    One has to choose two different letters from the 8 available different choices.
    This can be done in 8 * 7 = 56 ways.
    Case 2: When the two letters are same.
    There are 3 options – the three can be either Ns or Es or As.
    Therefore, 3 ways.
    Total number of possibilities = 56 + 3 = 59

     


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