Quantitative Aptitude Questions for SBI PO / NIACL AO / BOB PO 2017

It is time to pace up your preparation with New Pattern Questions of Quantitative Aptitude for SBI PO Prelims and NIACL Assistant Prelims 2017. These Quant questions will also help you in preparing for BOB PO and NICL AO 2017 recruitment examination.

So here we are with the most important questions of Quantitative Aptitude for SBI PO 2017. 

Directions (Q.1-5): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer,

(a) if x<y

(b) if x≤y

(c) if x = y or no relation can be established between x and y

(d) if x>y

(e) if x≥y

 
 
 

Directions (Q.6-10): What should come in place of question-mark (?) in following number series problems?

Q6. 35, 63, 99, 143, ?

(a) 175

(b) 185

(c) 195

(d) 205

(e) None of these

Q7. 26, 105, 400, 1185, ?, 2355

(a) 2360

(b) 2350

(c) 2355

(d) 2340

(e) None of these

Q8. 83, 87,183, 565, ?, 11461

(a) 2270

(b) 2275

(c) 2280

(d) 2290

(e) None of these

Q9. 7, 23, 55, 109, 191, ?

(a) 307

(b) 317

(c) 333

(d) 343

(e) None of these

Q10. 81, 82, 42, 15, ?

(a) 3.75

(b) 4.75

(c) 5.75

(d) 5.25

(e) None of these

Directions (Q.11-15): What approximate value should come in place of the question mark (?) in following questions?

(a) 49

(b) 81

(c) 64

(d) 16

(e) 25

(a) 95

(b) –95

(c) 105

(d) –105

(e) –115

(a) 615 

(b) 645

(c) 675

(d) 715

(e) 725

(a) 62400

(b) 64000

(c) 60400

(d) 64200

(e) 61600

(a) 760

(b) 800

(c) 690

(d) 870

(e) 780

Solutions:

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IBPS SO Exam 2017 - Day 2 (Quadratic Equations)

Dear Aspirants,

We have introduced  "the 12 days study plan to crack IBPS SO Exam 2017". This plan is specifically dedicated to Quantitative Aptitude & English Language and cover all types of questions asked in Bank SO exams. This will help you to prepare in a more efficient and systematic way. So, follow this post to keep updated with the plan.

Directions (1 – 10) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
(a) if x > y
(b) if x ≥ y
(c) if x < y
(d) if x ≤ y
(e) if x = y or no relation can be established between ‘x’ and y.

1. I. 4x + 7y = 42
   II. 3x – 11y = – l

2. I. 9x^2 – 29x + 22 = 0
   II. y^2 – 7y + 12 = 0

3. I. 3x^2 – 4x – 32 = 0
   II. 2y^2 – 17y + 36 = 0

4. I. 3x^2 – 19x – 14 = 0
   II. 2y^2 + 5y + 3 = 0

5. I. x^2 + 14x + 49 = 0
   II.y^2 + 9y = 0

6. I.  30x^2 + 11x + 1 = 0
   II. 42y^2 + 13y + 1 = 0

7. I.   x^2 – x – √2x + √2= 0
   II.  y^2 – 3y + 2 = 0

8. I.   x^2 – 2x – √5x + 2√5 = 0
   II.  y^2 – √3y – √2y + √6 = 0

9. I.   9x^2 + 3x – 2 = 0
   II.  8y^2 + 6y + 1 = 0

10. I.  3x² + 11x + 6 = 0
     II. 3y² + 10y + 8 = 0



ANSWERS :


1. a
x = 7 ; y = 2

2. c
x = 2, 11/9
y = 3, 4

3. d
x = 4, -8/3
y = 4, 9/2

4. a
x = 7, -2/3
y = -3/2, -1

5. e
x = -7
y = 0, -9

6. d
x = -0.16, -0.19
y = -0.14, -0.16

7. e
x = 1, 1.414
y = 1, 2

8. a
x = 2, 2.23
y = 1.414, 1.732

9. e
x = 0.33, -0.66
y = -0.25, -0.5

10. e
x = -0.66, -3

y = -1.33, -2 

Important Note:

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IPPB Scale-1 Officer Prelims 2017 - Day 3 (Quadratic Equations)

Dear Aspirants,

We have introduced  "the 17 days study plan to crack IPPB Scale-1 Prelims 2017". This plan is specifically dedicated to Quantitative Aptitude and cover all types of questions asked in Bank PO exams. This will help you to prepare in a more efficient and systematic way. So, follow this post to keep updated with the plan.

Directions(1-15): In each question two equations are given, find x and y and give the answer:

A. x > y

B. x < y

C. x ≥ y

D. x ≤ y

E. x = y or relation can not be established.

1. a) 3x² - 22x + 7 = 0
    b) y² - 15y + 56 = 0

Sol.
a) 3x² - 21x - x + 7 = 0
    3x(x-7) -1(x-7) = 0
    (3x-1)(x-7) = 0
    x = 7, 1/3.  

b) y² -7y - 8y + 56 = 0
    y(y-7) - 8(y-7) = 0
    (y-8)(y-7) = 0
    y = 8, 7.

Thus, y≥x.

2. a) 2x² - 17x + 36 = 0
    b) 2y² - 19y + 44 = 0

Sol.
a) 2x² - 8x - 9x + 36 = 0
    2x(x-4) - 9(x-4)  = 0
    (2x-9)(x-4) = 0
    x = 9/2, 4.

b) 2y² - 8y - 11y + 44 = 0
    2y(y-4) - 11(y-4) = 0
    (2y-11)(y-4) = 0
    y = 11/2, 4.

Thus, no relation.

3. a) x - √169 = 0
    b) y² - 169 = 0

Sol.
a) x = 13
b) y = ±√169 = ±13.

Thus, x≥y.

4. a) 3x² + 20x + 25 = 0
    b) 3y² + 14y + 8 = 0

Sol.
a) 3x² + 15x + 5x + 25 = 0
    3x(x+5) + 5(x+5) = 0
    (3x+5)(x+5) = 0
    x = -5/3, -5.

b) 3y² + 12y + 2y + 8 = 0
    3y(y+4) + 2(y+4) = 0
    (3y+2)(y+4) = 0
    y = -2/3, -4.

Thus, no relation.

5. a) 3x² + 5x + 2 = 0
    b) 3y² + 18y + 24 = 0

Sol.
a) 3x² + 3x + 2x + 2 = 0
    3x(x+1) + 2(x+1) = 0
    (3x+2)(x+1) = 0
    x = -2/3, -1.
b) 3(y² + 6y + 8) = 0
    y² + 6y + 8 = 0
    y² + 4y + 2y + 8 = 0
    y(y+4) + 2(y+4) = 0
    y = -2, -4.

Thus, x>y.

6. a) 6x² + 31x + 35 = 0
    b) 2y² + 3y + 1 = 0

Sol.
a) 6x² + 21x + 10x + 35 = 0
    3x(2x+7) + 5(2x+7) = 0
    (3x+5)(2x+7) = 0
    x = -5/3, -7/2.

b) 2y² + 2y + y + 1 = 0
    2y(y+1) + 1(y+1) = 0
    (2y+1)(y+1) = 0
    y = -1/2, -1.

Thus, x<y.

7. a) 2x² + 9x + 10 = 0
    b) 4y² + 28y + 45 = 0

Sol.
a) 2x² + 4x + 5x + 10 = 0
    2x(x+2) + 5(x+2) = 0
    (2x+5)(x+2) = 0
    x = -5/2, -2.

b) 4y² +18y + 10y + 45 = 0
    2y(2y+9) + 5(2y+9) = 0
    (2y+5)(2y+9) = 0
    y = -5/2, -9/2.

Thus, x≥y.

8. a) 15x² - 11x - 12 = 0
    b) 20y² - 49y + 30 = 0

Sol.
a) 15x² - 20x + 9x - 12 = 0
    5x(3x-4) +3(3x-4) = 0
    (5x+3)(3x-4) = 0
    x = -3/5, 4/3.

b) 20y² -25y - 24y + 30 = 0
    5y(4y-5) -6(4y-5) = 0
    (5y-6)(4y-5) = 0
    y = 6/5, 5/4.

Thus, no relation.

9. a) 2x² - 15 = 7x
    b) 17y = -7 - 6y²

Sol.
a) 2x² - 10x + 3x - 15 = 0
    2x(x-5) +3(x-5) = 0
    (2x+3)(x-5) = 0
    x = -3/2, 5.

b) 6y² + 3y + 14y + 7 = 0
    3y(2y+1) +7(2y+1) = 0
    (3y+7)(2y+1) = 0
    y = -7/3, -1/2.

Thus, no relation.

10. a) 3x² - 19x - 14 = 0
      b) 2y² +15y + 13 = 0

Sol.
a) 3x² - 21x - 2x - 14 = 0
    3x(x-7) -2(x-7) = 0
    (3x-2)(x-7)
    x = 2/3, 7.

b) 2y² +2y +13y + 13 = 0
    2y(y+1) +13(y+1) = 0
    (2y+13)(y+1) = 0
    y = -13/2, -1.

Thus, x>y.

11. a) (x³ -13x + 12)/(x-1) = 0
      b) (y³ + 5y² - 2y - 24)/(y-2) = 0

Sol.
a) (x³ - x - 12x + 12)/(x-1) = 0
    [x(x²-1) -12(x-1)]/(x-1) = 0
    [(x-1)(x+1)(x-12)]/(x-1) = 0
    x² + x - 12 = 0
Solving, x = 3, -4

b) y³ + 5y² - 2y - 24/(y-2) = 0
    y³ + 7y² - 2y² -14y + 12y - 24/(y-2) = 0
    (y-2)(y² + 7y + 12)/(y-2) = 0
    Solving, y = -4, -3.

Thus, x≥y. 

12. a) y = 2x + 1
      b) 2y = 3x - 1

Sol. Solving the two using substitution method,
x = -3
y = -5

Thus, x>y.

13. a) 9x² - 29x + 22 = 0

      b) y² - 7y + 12 = 0

Sol. 

a) 9x² - 18x - 11x + 22 = 0

    9x(x-2) - 11(x-2) = 0

    (9x-11)(x-2) = 0

    x = 11/9, 2.

b) y² - 4y - 3y + 12 = 0

    y(y-4) - 3(y-4) = 0

    (y-4)(y-3) = 0

    y = 3, 4.

Thus, x<y.

14. a) 3x² - 4x - 32 = 0

      b) 12y² - 109y + 247 = 0

Sol. 

a) 3x² - 12x + 8x - 32 = 0

    3x(x-4) + 8(x-4) = 0

    (3x+8)(x-4) = 0

    x = -8/3, 4.

b) 12y² - 52y - 57y + 247 = 0

    4y(3y - 13) - 19(y - 13) = 0

    (4y-19)(3y-13) = 0

    y = 19/4, 13/3.

Thus, x<y. 

15. a) 4x + 7y = 42

      b) 3x - 11y = -1

Sol.  Solving the two equations using the substitution method,

x = 7

y = 2

Thus, x>y.

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