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