QUANTITATIVE APTITUDE - FOR UPCOMING EXAMINATION

Direction (Q. 1-5): What value should come in the place of question mark (?) in the following number series?

1). 28, 42, 84, 210, 630, ?

a)    2675

b)    2515

c)    2445

d)    2375

e)    2205

2). 13, 15, 22, 50, 113, ?

a)    239

b)    272

c)    287

d)    296

e)    306

3). 8, 17, 38, 85, 186, ?

a)    318

b)    354

c)    397

d)    408

e)    427

4). 5042, 5187, 5334, 5483, 5634, ?

a)    5716

b)    5748

c)    5787

d)    5793

e)    None of these

5). 3, 9, 17, 27, 39, 53, ?

a)    64

b)    69

c)    74

d)    78

e)    81

Direction (Q. 6-10) : In the following questions two equations numbered (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 the relationship can't be established.

6). I. 14x² + 17x - 6 = 0

II. 6y² - 13y + 5 = 0

7). I. x = √7

II.6y² - 7y - 20 = 0

8). I. 3x² + 8x - 35 = 0

II. y² - 2y - 48 = 0

9). I. x² - 23x + 132 = 0

II. y = ₃√1331

10). I. 7x - 5y = 64

II. 4x + 3y = 19

SOLUTION

1).E) Series is × 1.5, ×2, ×2.5, ×3, ×3.5

2). A) Series is + 1³ +1, +2³ -1, +3³ +1, 4³-1

3). C) Series is ×2+1² , ×2+2² , ×2+3² , ×2+4²

4).C) Series is (71)² +1, (72)² +3, (73)² +5

5).B) Series is 1× 2 + 1, 2 × 3 + 3, 3 × 4 + 5, 4 × 5 + 7, 5 × 6 + 9

6). C)  14 x² + 17 x – 6 = 0

14 x2 + 21 x – 4 x – 6 = 0

7 x (2 x + 3) – 2 (2 x + 3) = 0

(2 x + 3) (7 x – 2) = 0

x = -(3/2), 2/7

6 y² – 3y – 10 y + 5 = 0

3 y (2 y –1) – 5 (2 y –1) = 0

(3 y – 5) (2 y – 1) = 0

y = 5/3, 1/2

Hence, x < y

7). A)  x = √7 = 2.645

6 y²  –15 y + 8 y –20 = 0

3y (2y – 5) + 4 (2 y – 5) = 0

(3 y + 4) (2 y – 5) = 0

y = - (4/3), 5/2

 x > y

8).E)  3x² + 15 - 7x - 35 = 0

3x (x + 5) - 7 (x + 5) = 0

(3x - 7) ( x + 5) = 0

x = –5, 7/3

y² – 8 y + 6 y – 48 = 0

y (y –8) + 6 (y –8) = 0

(y + 6) (y –8) = 0

y = – 6 , 8

No relation between x & y

9). B)  x2 - 23x + 132 = 0    y = ₃√1331

 x² -12x - 11x + 132 = 0      y = 11

 x (x - 12) – 11 (x - 12) = 0

(x –11) (x – 12) = 0

 x = 11, 12   x ≥ y

10).A)  equn. (I) × 3 + equn. (II) ×5

(21x – 15y = 192) + (20x +15y=95)

41x = 287

x =7 and y = -3

x > y