# 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