IBPS PO Exam 2016 – Section wise Full Test-14

IBPS PO Exam 2016 – Section wise Full Test-14:

Dear Readers, IBPS PO 2016 was approaching, for that we have given the Section wise Full Test which consist of all the three sections such as, Aptitude, Reasoning, and English. This Section wise Full Test will be provided on daily basis kindly make use of it.

QUANTITATIVE APTITUDE

Directions (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

d) if x ≤ y

e) if x = y or no relation can be established between x and y.

1). I. x² – 8√3 x + 45 = 0 II. y² – √2 y – 24 = 0

2). I. x – 7√2 x + 24 = 0 II. y – 5√2 y + 12 = 0

3). I. 12 x² – 17 x + 6 = 0 II. 20 y² – 31 y + 12 = 0

4). I. 3 x² – 8 x + 4 = 0 II. 4 y² – 15 y + 9 = 0

5). I. x² – 16 x + 63 = 0 II. y² – 2 y – 35 = 0 

Directions (6 – 10):Study the following information carefully to answer the given questions.

Seven people – P, Q, R, S, T, U and V – are sitting in a straight line with equal distance between each other, but not necessarily in the same order. Some of them are facing north and some are facing south.

Only two people are sitting to the left of V. Only two people sit between V and Q. P sits second to the left of Q. The immediate neighbours of P face opposite directions (ie if one of the neighbours faces south then the other faces north and vice versa). Only one person sits between P and R. U sits third to the left of R. S is not an immediate neighbour of Q. Both the immediate neighbours of R face the same direction (i.e if one neighbour faces south then the other neighbour also faces south and if one neighbour faces north then the other neighbour also faces north). P faces the same direction as R. T faces north. Q sits on the immediate left of T.

6).Who among the following sits exactly between V and the one who is sitting on the immediate left of Q?

c) Other than those given as options

7).Who among the following sits exactly in the middle of the line?

8).Which of the following pairs represents the immediate neighbours of Q?

9).Which of the following pairs represents the people sitting at the two extreme ends of the line?

b) Other than those given as options

10).Who among the following sits second to the right of S?

d) Other than those given as options

Directions (Q. 11-13): Choose the word/group of words which is MOST SIMILAR in meaning to the word/ group of words printed in bold as used in the passage.

11).Inextricably

a)   jointly

b)   partially

c)   completely

d)   inseparably

e)   inadequately

12). Conventional

a)   typical

b)   traditional

c)   ordinary

d)   extraordinary

e)   rare

13). Specious

a)   accurate

b)   truthful

c)   misleading

d)   honest

e)   real

Directions (Q. 14-15): Choose the word/group of words which is MOST OPPOSITE in meaning of the word/group of words printed in bold as used in the passage.

14). Indispensable

a)   additional

b)   vital

c)   crucial

d)   needful

e)   fundamental

15). Rigorous

a)   accurate

b)   proper

c)   rigid

d)   harsh

e)   careless

1)e   2)b   3)d   4)e   5)b   6)a   7)b   8)a   9)e   10)b 11)d   12)b   13)c   14)a   15)e

Solutions:

1). e)I. x² - 8√3x + 45 = 0 => (x + 3√3) (x - 5√3) = 0 => x = 3√3, 5√3

II. y² – √2 y – 24 = 0 => (y + 3√2) (y - 4√2) => y = -3√2, 4√2

Hence the relation cannot be established between x and y

2). b)I. x – 7√2 x + 24 = 0 => (√x - 3√2)( √x - 4√2) = 0;

If √x = 3√2 => x = 18 and if √x = 4√2 => x = 32

II. y – 5√2 y + 12 = 0 => (√y - 2√2)( √y - 3√2) = 0;

If √y = 2√2 => y = 8 and if √y = 3√2 => y = 18

Therefore x ≥ y

3). d) 12x² – 17x + 6 = 0

or 12x² – 9x – 8x + 6 = 0

or 3x(4x – 3) – 2(4x – 3) = 0

or (3x – 2) (4x – 3) = 0

If 3x – 2 = 0 then 3x = 2 => x = 2/3

If 4x – 3 = 0 then x = ¾

II. 20y² – 31y + 12 = 0

or 20y² – 16y – 15y + 12 = 0

or 4y(5y – 4) – 3(5y – 4) = 0

or (4y – 3) (5y – 4) = 0

Therefore y = ¾, 4/5

Hence x ≤ y

4). e) 3x² – 8x + 4 = 0

or 3x² – 6x – 2x + 4 = 0

or (3x – 2) (x – 2) = 0

Therefore x = 2, 2/3

II. 4y² – 15y + 9 = 0

or 4y² – 12y – 3y + 9 = 0

or 4y(y – 3) – 3(y – 3) = 0

or (4y – 3) (y – 3) = 0

Therefore y = ¾, 3

Relation cannot be established between x and y

5). b) x² – 16x + 63 = 0

or x² – 9x – 7x + 63 = 0

or x(x – 9) – 7(x – 9) = 0

or (x – 7) (x – 9) = 0 => x = 7, 9

II. y² – 2y – 35 = 0 => (y + 5)(y – 7) = 0 => y = -5, 7

Hence x ≥ y

(6 – 10):

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