## 14 Sep 2016

### Mission IBPS PO 2016 - Practice Quantitative Aptitude Questions[Answers Updated]

Mission IBPS PO 2016 - Practice Quantitative Aptitude Questions:
Dear Readers, Important Practice Aptitude Questions for IBPS PO and Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.
1). (17 + 2√7)2 = a + b√7. The pair (a,b) is:
a)   317,68
b)   68,317
c)   327,34
d)   317,34
e)   68,34

2). The value closest to
[(23 – 1)(33 – 1)(43 – 1)…(10003 – 1)] /[ (23 +1)(33 + 1)(43 + 1)…(10003 + 1)]
a)   0.60
b)   0.65
c)   0.70
d)   0.75
e)   0.80

3). Let P = 7/2÷5/2×3/2
Let Q = 7/2÷5/2 of 3/2
Evaluate: P/Q ÷ 5.25
a)   6/5
b)   3/7
c)   8/3
d)   1/8
e)   3/8

4). ([(2n+4 – (2)2n] /[(2)2n+3) ] )+2-3 is equal to
a)   2n-1
b)   1
c)   7/8
d)   1/8
e)   5/8

5). (3 + 4 -6 ÷ 2 + 2) + ((9 ÷ 3 + 6 × 5) ÷ 11)) × ((4 + 5 – 6) + (18 – 3 × 4)) ÷ 9
a)   6
b)   9
c)   12
d)   18
e)   None of this

Directions (6-10): In the following questions two questions numbered I and II are given. You have to solve both the equations and choose the correct option.
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

6). I. 5x2 – 18x + 9 =0
II.           20y2 – 13y + 2=0

7). I. 3/√x + 4/√x = √x
II. y3 – [(7)7/2/√Y] = 0

8). I. 9x – 5.45 = 54.55+4x
II. (√y + 155) – 6 = 7

9). I. x2 + x - 20 =0
II.           y2 – y - 30=0

10). x2 – 365 = 364
II.           y - √324 = √81

1)a   2)b   3)b   4)b   5)b   6)a   7)e   8)c   9)e   10)d

Solutions:
1). (17 + 2√7)2 =  17 × 17 + 2 √7 × 2 √7+2 ×17 × 2 √7
= 289 + 28 + 68√7
= 317 + 68√7
So a = 317 , b = 68

2). a3-1 = (a - 1)(a2+a+1)
a3+1 = (a + 1)(a2-a+1)
also, note, (a2-a+1) = (a+1)2 – (a+1)+1
Numerator = (2 - 1)(22+2+1). (3-1)(32+3+1)……… (1000-1)
(10002+1000+1)
Denominator = (2+1)(22-2+1). (3+1)(32-3+1)……… (1000+1)
(10002-1000+1)
Note that: (22+2+1) = (32-3+1), (32+3+1) = (42-4+1),and so-on
So the required value = 2×(10002+1000+1) / (1000×1001×3) = 0.66
The given value is closest to 0.65.

3). P = (7/2) ÷ (5/2) × (3/2) = (7/2) × (2/5) ×(3/2)= 21/10
Q = (7/2) ÷ (5/2) of  (3/2) = (7/2) ÷ (5/2 × 3/2)= (7/2) × (4/15) = 14/ 15
P/Q = (21/20)/(14/15) = (21 × 15)/(10×14) = 9/4
5.25 = 21/4
Required value = (9/ 4)/ (21/4) = 3/7

4). Consider the first term:
Numerator = 2n × 24 – 2n × 2 = 2n × 14
Denominator = (2)2n+3 = 2n× 16
So, the first term becomes (14 × 2n)/ (16 × 2n) = 14/16 = 7/8
Now, 2-3 = 1/8
So the required value  = 7/8 + 1/8 = 1

5). (3 + 4 – 3 + 2) + ((3+30) ÷ 11)) × (3+6) ÷ 9
= (6 + 3) × 1 = 9

6). Consider equation I:
5x2 – 18 + 9 = 0
5x2 – 15x -3x + 9 = 0
5x(x-3) – 3(x-3) = 0
(x-3) (5x-3) = 0
X = 3 or 3/5
Consider equation II:
20y2 – 13y +2 = 0
20y2 – 8y -5y + 2 = 0
4y(5y-2) – 1(5y-2) = 0
(5y – 2) (4y – 1) = 0
Y = 2/5 or 1/5
Hence, x > y

7). Consider equation I:
7/√x = √x or x = 7
Consider equation II:
Y3 * √y – 77/2 = 0
Y7/2 = 77/2
Y = 7
Hence, x = y

8). Consider equation I:
9x – 5.45 = 54.55 + 4x
5x = 60
X = 12
Consider equation II:
√(y+155) = 6 + 7 = 13
Y + 155 = 169
Y = 14
Hence,  x < y

9). Consider equation I:
x2 + x – 20 =0
x2 + 5x -4x -20 = 0
x(x + 5) – 4(x+5) = 0
(x + 5)(x – 4) = 0
X = -5 or 4
Consider equation II:
Y2 – y – 30 = 0
y2 – 6y + 5y – 30 = 0
y(y - 6) + 5(y – 6) = 0
(y-6)(y+5) = 0
Y = 6 or -5
We can have y = 6 and x =4à here x<y
We can have x = 4, y = -5à here x>y
So we cant determine the relationship

10). Consider equation I:
X2 – 365 = 364
X= 365 + 364 = 729
X = ±27
Consider equation II:
Y = 18 + 9 = 27
So, x ≤ y