# NIACL AO Prelims – Quantitative Aptitude Questions Day- 80

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Dear Readers, Bank Exam Race for the Year 2019 is already started, To enrich your preparation here we have providing new series of Practice Questions on Quantitative Aptitude – Section. Candidates those who are preparing for NIACL AO Prelims 2019 Exams can practice these questions daily and make your preparation effective.

NIACL AO Prelims – Quantitative Aptitude Questions Day- 80

Directions (Q. 1 – 5): What approximate value should come in place of question mark (?) in the following questions?

1) (5/7) of 342 + (2/3) of (7/15) of 1889 =? % of 700

a) 150

b) 90

c) 120

d) 210

e) 250

2) [48.78 ÷ 7 × 5] ÷ 70 = 18.01 ×?

a) 3/47

b) 2/35

c) 4/57

d) 1/36

e) 5/74

3) √? + 27.65 = (375.39 + 72.01) ÷ 49.78 + (18.78 × 2.75)

a) 1225

b) 1444

c) 1681

d) 961

e) 2025

4) 3289 ÷ 14 × 5.01 + 15 % of 500 + 27 % of 299 =?

a) 1330

b) 1860

c) 2150

d) 2740

e) 1570

5) 36 % of 499.8 + 52 % of 301.25 = 4 ×?

a) 75

b) 66

c) 92

d) 127

e) 84

Directions (6-10): In the following questions, two quantities are given as Quantity I and Quantity II. By finding these quantities gives corresponding answer.

6) Quantity I: ‘x’ : 4x2+24x= -27

Quantity II: ‘y’ : 4y2+33y+35=0

a) Quantity I > Quantity II

b) Quantity I < Quantity II

c) Quantity I ≥ Quantity II

d) Quantity I ≤ Quantity II

e) Quantity I = Quantity II or No relation

7) Quantity I: A circle and a rectangle have the same area. Length and breadth of the rectangle is 77cm and 50cm respectively. What is the circumference of circle?

Quantity II: Diameter of a circle is 56 cm and perimeter of a rectangle is same as circumference of the circle. Find out the area of the rectangle, if the length of the rectangle is three times the breadth of the rectangle.

a) Quantity II> Quantity I

b) Quantity II< Quantity I

c) Quantity I ≥ Quantity II

d) Quantity I ≤ Quantity II

e) Quantity I = Quantity II or No relation

8) Quantity I: ‘X’: 5, 9, 36, ?, 404, 525

Quantity II: ‘Y’: 11, 20, ?, 224, 1115, 6684

a) Quantity II > Quantity I

b) Quantity II < Quantity I

c) Quantity I ≥ Quantity II

d) Quantity I ≤ Quantity II

e) Quantity I = Quantity II or No relation

9) Quantity I: ‘X’:  The cost of 5 bags and 12 notes is Rs. 3100 and the cost of 7 bags and 18 notes is Rs. 4400. What is the cost of 3 notes?

Quantity II: ‘Y’:  Y2= 22500

a) Quantity II> Quantity I

b) Quantity II < Quantity I

c) Quantity I ≥ Quantity II

d) Quantity I ≤ Quantity II

e) Quantity I = Quantity II or No relation

10) Quantity I: “A”: X, Y, and Z can finish a work in 5, 6, and 10 Days respectively. They start working together. If Y stops after 2 days, how long would it take X and Z to finish the remaining work.

Quantity II: “B”: B3= 27/1331

a) Quantity II> Quantity I

b) Quantity II< Quantity I

c) Quantity I ≥ Quantity II

d) Quantity I ≤ Quantity II

e) Quantity I = Quantity II or No relation

Directions (1-5):

(5/7) of 343 + (2/3) of (7/15) of 1890 = x % of 700

(5/7) * 343 + (2/3) * (7/15) * 1890 = (x/100)*700

245 + 588 = 7x

833 = 7x

X = 833/7

X = 119 = 120 (Approximately)

[49 ÷ 7 × 5] ÷ 70 = 18 ×?

(7*5)/70 = 18x

35/70 = 18x

X = (1/2)*(1/18) = 1/36

√x + 28 = (375 + 72) ÷ 50 + (19 × 3)

√x + 28 = 447/50 + 57

√x = 9 + 57 – 28

√x = 38

X = 382 = 1444

3290 ÷ 14 × 5 + 15 % of 500 + 27 % of 300 = x

(3290/14)*5 + (15/100)*500 + (27/100)*300 = x

X = 1175 + 75 + 81

X = 1331= 1330 (Approximately)

36 % of 500 + 52 % of 300 = 4 ×?

(36/100)*500 + (52/100)*300 = 4x

4x = 180 + 156

4x = 336

X = (336/4) = 84

X = 84

Directions (6-10):

Quantity I:

4x2+24x= -27

4x2+24x+27=0

4x2+18x+6x+27=0

2x(2x+9)+3(2x+9)=0

X= – 3/2 , – 9/2 = -1.5, -4.5

Quantity II:

4y2+33y+35=0

4y2+28y+5y+35=0

4y(y+7)+5(y+7)=0

y= – 5/4 , – 7 = -1.2, -7

No Relation

Quantity I:

Area of circle= Area of Rectangle

πr2= (l × b)

22/7× r2= 77×50

r2= 25×49

r=35 cm

Circumference of circle= 2πr

= 2 × 22/7 × 35

= 220 cm

Quantity II:

Perimeter of rectangle= Circumference of circle

2(l + b) = 2πr

2(l + b) = 2 × 22/7 × 56/2

l + b= 88cm

According to the question,

3b+b= 88

b=22

Area of rectangle= l × b = 22×66= 1452cm2

Quantity II> Quantity I

Quantity I:

5, 9, 36, ?, 404, 525

5 + 22 = 9

9 + 33 = 36

36 + 52 = 61

61 + 73 = 404

404 + 112 = 525

Quantity II:

11, 20, ?, 224, 1115, 6684

11 * 2 – 2 = 20

20 * 3 – 3 = 57

57 * 4 – 4 = 224

224 * 5 – 5 = 1115

1115 * 6 – 6 = 6684

Quantity II < Quantity I

Quantity I:

According to the question:

5b + 12n = 3100—– (i)

7b + 18n = 4400 —– (ii)

After solving equation (i) and (ii),

We get, b= 500 and n= 50

Cost of 3 notes = 50×3= 150

Quantity II:

Y2= 22500

Y= ±150

Quantity I ≥ Quantity II

Quantity I:

According to the question,

X + Y + Z= 1/5 + 1/6 + 1/10

= (12+10+6)/60 = 28/60 = 7/15

Their work for 2 Days= 14/15

Remaining work = 1 – 14/15 = 1/15

Required Time to finish the work by X and Z= (1/15)/(1/5+ 1/10) = 1/15 × 10/3 = 2/9 Days

Quantity II:

B3 = 27/1331 => B =3/11

Quantity II> Quantity I