22 Jun 2017

Practice Aptitude Questions For NICL AO Mains 2017(Quadratic Equation & Data Sufficiency)

Practice Aptitude Questions For NICL AO Mains 2017 (Data Interpretation)
Practice Aptitude Questions For NICL AO Mains 2017 (Quadratic Equation & Data Sufficiency):
Dear Readers, Important Practice Aptitude Questions for NICL AO Mains 2017 Exams was given here with Solutions. Aspirants those who are preparing for Bank exams & all other Competitive examination can use this material.



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Directions (Q. 1-5): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answers.
a) if x > y
b) if x ≥ y
c) if x < y
d) if x ≤ y
e) if x = y or the relationship cannot be established
1.I. (2)^5 + (11)^3/6 = x^3
II. 4y^3 = – (589 ÷ 4) + 5y^3

(a)
(b)
(c)
(d)
(e)

1). I. [(2)^5 + (11)^3]/6 = x^3
(32 + 1331)/6 = x^3
1363/6 = x^3
x^3 = 227.167
x = 6.10
II. 4y^3 = - 589/4 + 5y^3
589/4 = 5y^3 – 4y^3
Y^3 = 147.25
Y = 5.28
x > y.
Answer: A
2.I. 12x^2 + 11x + 12 = 10x^2 + 22x
II. 13y^2 – 18y + 3 = 9y^2 – 10y

(a)
(b)
(c)
(d)
(e)
2).I. 12x^2 + 11x + 12 = 10x^2 + 22x
2x^2 – 11x + 12 = 0
2x^2 – 8x – 3x + 12 = 0
(x – 4) (2x – 3) = 0
x = 4, 3/2
II. 13y^2 – 18y + 3 = 9y^2 – 10y
4y^2 – 8y + 3 = 0
4y^2 – 6y – 2y + 3 = 0
(2y – 3) (2y – 1) = 0
y = 3/2 , 1/2
x ≥ y
Answer: B

3.I. √1225x + √4900 = 0
II. (81)^1/4 y + (343)^1/3 = 0


(a)
(b)
(c)
(d)
(e)
3).35x + 70 = 0
x = -70/35
x = - 2
II. (81)^1/4 y + (343)^1/3 = 0
3y + 7 = 0
3y = – 7
y = – 7/3
y = – 2.33
x > y.
Answer: A

4.I. 18/x^2 + 6/x – 12/x^2 = 8/x^2
II. y^3 + 9.68 + 5.64 = 16.95

(a)
(b)
(c)
(d)
(e)
4).I. 18/x^2 + 6/x + 12/x^2 = 8/x^2
(18 + 6x – 12)/x^2 = 8/x^2
6x + 6 = 8
x = 2/6
x = 0.33
II. y^3 + 9.68 + 5.64 = 16.95
y^3 = 16.95 – 15.32
y^3 = 1.63
y = ∛1.63 = 1.17
x < y
Answer: C
5. I. (x^(7/5) ÷ 9) = 169 ÷ x^(3/5)
II. y^1/4 × y^1/4 × 7 = 273 ÷ y^1/2

(a)
(b)
(c)
(d)
(d)
5).I. x^(7/5) ÷ 9 = 169 ÷ x^(3/5)
x^(7 / 5) ÷ 9 = 169 ÷ x^(3/5)
x^(10/5) = 9 × 169
x^2 = 9 × 169

x = ± (3 × 13)
x = ±39
II. y^1/4 × y^1/4 × 7
= 273/y^1/ 2
y = 273/7
y = 39
x ≤ y
Answer: D

Directions (Q. 6-10): In each of the following questions, a question is followed by information given in three Statements I, II and III. You have to study the question along with the statements and decide the information given in which of the statement(s) is, necessary to answer the question.
6.What is two digit numbers?
I. The difference between the number and the number formed by interchanging the digit is 27.
II. The difference between two digits is 3.
III. The digit at unit's place is less than that at ten’s place by 3.

I and II
I and either II or III
I and III
All of these
None of these
6). Let two digit number be 10x + y.
From I, either x – y = 27/9 = 3
Or, y – x = 27/9 = 3
From II, x – y = 3 or y – x = 3
From III, x – y = 3
Hence, Even by (I) + (II) + (III) we cannot obtain the number.
Answer: E

7.In how many days 10 women can finish the work?
I. 10 men finish the work in 6 days.
II. 10 women and 10 men finish the work in 3 3/7 days.
III. If 10 men work 3 days and after that 10 women are deployed to work for men, the rest work is finished in 4 days.

I and II
Any two of three
I and III
II and III
None of these
7). From I and II, 10 women can finish the work in 1 day
= 7/24 – 1/6 = (7 – 4)/24 = 1/8
10 women can finish the work in 8 days.
From II and III,
Let 10 men can finish the work in x days and 10 women can finish the same work in y days.
Hence,
1/x + 1/y = 7/24 ...(i)
and from III – II,
3/x + 4/y = 1 ...(ii)
from (i) & (ii)
y = 8 days
Again from I and III
3/6 + 4/y = 1
y = 8 days
Answer: B/span>
8. What is the rate of interest Percent per annum?
I. An amount doubles itself in 5 yr on simple interest
II. Difference between the compound interest and the simple interest earned on a certain amount in two years is 400.
III. Simple interest earned per annum is 2000.

Only I
II and III
Any two of three
I or II and III
Only I or both II and III
8).From I, If P = 100
A = 200 and SI = 200 – 100 = 100
Rate = SI × 100/(P × T)
= 100 × 100/(100 × 5)
= 20%
From II and III, Rate
= 400 × 100/2000 × 1
= 20%
Hence, either I alone or II + III will be sufficient.
Answer: E
9.What is the present age of Rakesh?
I. The present age of Rakesh is half of his father's age.
II. After five years the ratio of ages of Rakesh and his father is 6 : 11.
III. Rakesh is younger to his brother by five years.

I and II
I and III
II and III
All of these
Cannot be determined
9).From I, Let present age of Rakesh be x yr and age of his father be 2x yr.
From I and II,
(x + 5)/(2x + 5) = 6/11
12x + 30 = 11x + 55.
x = 25 yr
From I and II, age of Rakesh = 25 yrs.
Hence, only from I & II, age of Rakesh and his father can be obtained.
Answer: A

10.What is the cost of flooring the rectangular hall?
I. Length and the breadth of the hall are in the ratio of 3 : 2
II. Length of the hall is 48 m and cost of flooring is 850 per sq m.
III. Perimeter of the hall is 160 m and cost of flooring is 850 per sq m.

I and II
I and III
Only III
I and either II or III
Any two of the three
10).From I and II.
Length = 3x = 48 m
x = 16
Breadth = 2x = 32 m
Hence, Area of floor = 48 × 32
Cost of flooring = 48 × 32 × 850 = Rs. 1305600
From I and III, 2(l + b) = 160
2(3x + 2x) = 160
10x = 160
x = 16
Length = 3 × 16 = 48 m
Breadth = 2 × 16 = 32 m
Cost of flooring = (48 × 32) × 850 = 1305600
Similarly, from II and III, we can find I = 48 m and b = 32 m and Total cost of flooring = 1305600
Answer: E



        
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