Quant Test for NIACL 2018 Prelim Exam Set – 12

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

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

    x>=y
    x=y or relation cannot be established.
    x>y
    y>=x
    y>x
    Option A
    6y^2 + 1/2 = 7/2 y
    => 12y^2 – 7 y + 1 = 0
    =>12y62 – 44y – 3y + 1 = 0
    => 4y (3y – 1) – 1 (3y –1) = 0
    => (3y – 1) (4y – 1) = 0
    y = 1/3,1/4
    12x^2 – 10x + 2 = 0
    => 6x^2 – 5x + 1 = 0
    =>3x (2x –1) – 1 (2x –1)
    x = 1/3,1/2
    x>=y

     

  2. I. √x + 6 = √121 – √36
    II. y^2 + 112 = 473

    x=y or relation cannot be established.
    y>=x
    x>=y
    y>x
    x>y
    Option E
    x = 19
    y= + 19, -19
    x > y

     

  3. I. 17 x = (13)^2 + √196 + (5)^2 + 4 x
    II. 9 y – 345 = 4y – 260

    x=y or relation cannot be established.
    x>y
    y>x
    y>=x
    x>=y
    Option C
    17x = 169 + 14 + 25 + 4x
    =>17x = 218 + 4x
    =>13x = 218
    x = 16
    9y – 345 = 4y – 260
    =>5y = 85
    y = 17
    y>x

     

  4. I. (441)^1/2 x^2 – 111 = (15)^2
    II. √121 y^2 + (6)^3 = 260

    x>y
    y>x
    x>=y
    x=y or relation cannot be established.
    y>=x
    Option D
    441 x^ 2 – 111 = (15)^2
    => 21x^2 – 111 = 225
    =>21x^2 = 336
    =>X^2 = 16
    x = + 4, -4
    121 y^2 + (6)^3 = 260
    =>11y^2 + 216 = 260
    =>11y^2 = 44
    => Y^2 = 4
    y = + 2, -2
    Cannot established.

     

  5. I. 17 x + (13)^2 – 114 = (15)^2
    II. √121 y^ 2 + (6)^3 = 260

    y>=x
    x=y or relation cannot be established.
    x>y
    x>=y
    y>x
    Option C
    17x + (13)^2 – 114 = (15)^2
    => 17x + 169 – 114 = 225
    => 17x = 170
    x = 10
    121 y ^2 + (6)^3 = 260
    =>121 y^2 = 260 – 216
    =>11y^2 = 44
    y = + 2, -2
    x>y

     

  6. A train travelling at a speed of 75 meter per hour enters a tunnel 3(1/2) miles long. The train is 1/4 miles 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 min.
    8 min.
    6 min.
    4 min.
    3 min.
    Option E
    Total distance covered = 7/2+1/4 miles
    =154 miles
    Time taken = (15/4*75) hrs. =120 hrs
    =(120*60)min. = 3 min.

     

  7. The simple interest on a sum of money will be Rs.900 after 8 years. If the principal is tripled for the last 4 years, what will be the total interest at the end of the 8th year?
    1210
    1350
    1800
    2000
    1500
    Option C
    Let sum be x
    n=8 yrs
    SI=900
    R=(100×𝑆𝐼)/( 𝑃×𝑛)
    => R=(100×900)/( 𝑥×8) = 11250/𝑥
    For 1st 4 yrs SI= (𝑥 × 11250/𝑥 × 4 × 1/100)
    = 450
    For last 4 yrs.
    SI= (3𝑥 × 11250/𝑥 × 4 × 1/100) = 1350
    Total interest = 1350 + 450
    =1800

     

  8. There are three different mixtures (glucose and alcohol) in which the concentration of glucose 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. Find the ratio of glucose and alcohol in the new mixture?
    3:2
    9:8
    7:5
    2:1
    5:4
    Option A
    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

     

  9. Mr.Pankaj sells kaju katri at Rs. 15 per kg. A kaju katri is madeup of flour and sugar in the ratio of 5 : 3. The ratio of price of sugar and flour is 7 : 3 (per kg). Thus he earns 66(𝟐/𝟑) % profit. What is the cost price of sugar?
    15
    10
    20
    14
    18
    Option D
    C.P. = (15×100×3)/500 = Rs. 9
    Now, flour : sugar = 5 : 3
    In 1 kg = flour = 625 gm.
    Sugar = 375 gm.
    Let price of sugar = 7x,
    Flour = 3x
    According to question, 625/1000 ×3x + 375/1000 ×7x = 9
    x = 2
    So, C.P. of sugar = 7 × 2
    = 14 Rs./kg.

     

  10. A box contains 3 ballons of 1 shape, 4 ballons of 1 shape and 5 ballons of 1 shape. Three ballons of them are drawn at random, what is the probability that all the three are of different shape?
    3/11
    7/9
    5/12
    4/13
    3/11
    Option A
    Total = 3+4+5 = 12
    N(S) = 12c3 = 220
    N(E) = 3c1 * 4c1 * 5c1 = 60
    Required Probability = 60/220 = 3/11

     


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