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)] )+2-3is 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

Answers:
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
Answer: a)

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.
Answer: b)

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
Answer: b)

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
Answer: b)

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

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
Answer: a)

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
Answer: e)

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
Answer: c)

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 =4àhere x
We can have x = 4, y = -5àhere x>y
So we cant determine the relationship
Answer: e)

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

Answer: d)
  

 
 
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