8 Oct 2016

Mission IBPS PO 2016 - Practice Quantitative Aptitude Questions with Detailed Solutions

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.


Directions (Q. 1-5):  In the following questions three equations numbered I,II and III are given. Solve the equations and choose the correct option that gives the relation between the variables.
a)    x<y =z
b)    x≤ y<z
c)    x<y>z
d)    x=y>z
e)    None of these

1). I.z2 + z – 56 = 0
     II.7y + 2x = 81
     III.3x + y = 36

2).I.3x + 4y + 5z = 87
    II.3x  + z = 27
    III. x + 3y =29

3). I.x2 + 3x +2 = 0
    II.y2 + 8y + 15 = 0
    III.z2 + 13z + 42 = 0

4).I. x= - √6.25
    II. y 2 + 0.7y – 4.5 = 0
    III.3x + 7z = 7.9

5).I. x + 2y + z =23
    II.3x + 5z = 35
    III.y+2z =15

Directions (Q. 6-10): Find the missing number in the following number series:
6). 410,405,396,378,360,?
a)    351
b)    345
c)    342
d)    350
e)    339

7). 1/9, 4/9, 1 7/9, 7 1/9, 28 4/9, ?
a)    112(1/9)
b)    113(7/9)
c)    115(4/9)
d)    140(5/9)
e)    142(2/9)

8).1685,1712,1837, 2280, 2909, ?
a)    4119
b)    3775
c)    4094
d)    4240
e)    3656

9).3, 0, -4, -17, -74, ?
a)    -309
b)    -302
c)    -363
d)    -290
e)    -377

10). 2025, 2401, 2809, ?, 3721, 4225
a)    3225
b)    3281
c)    3249
d)    3301
e)    3369

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

Solution:
1).From  I,(z-7)(z+8) = 0
=>z = 7, -8
3 × II – 2 × III => 19y =171
=>y = 9
Putting in II, we get x=9
Hence x=y>z
Answer: d)

2).I – II => 4y + 4z = 60
=>y+z = 15(IV)
I – 3III => 5z -5y = 0
=>y=z
Putting in IV, we get y = z = 7.5
Putting in II, x = 6.5
Hence x<y =z
Answer: a)

3).From I, (x+1)(x+2)=0
=>x=-1,-2
From II, (y+3)(y+5)=0
=>Y = -3,-5
From III,(z+6)(z+7)=0
=>z = -6,-7
Hence x>y>z
Answer: e)

4).From I, x = -2.5
Putting in III,-7.5 + 7z = 7.9
=>7z =15.4
=>z=2.2
From II, (y+2.5)(y-1.8)=0
=>y=-2.5,1.8
Hence x≤y<z
Answer: b)

5).II + 2xIII – 1 => 2x + 8z =42
Or x+4z = 21(IV)
3xIV – II =>7z=28
=>z=4
Putting in II,3x + 20 =35
=>x=5
Putting these values in equation I,
We get, 5 + 2y + 4 =23
=>y=7
Hence x<y>z
Answer: c)

6).The next number is obtained by subtracting sum of digits from the number.
Hence the missing number is 360-(3+6+0)=351
Answer: a)

7).The pattern is 1/9,4/9,16/9,64/9,256/9,1024/9.
1024/9 = 113 7/9
Answer: b)

8).1685+33 =1712,1712+53 =1837,1837+73=2280, 2280+93=2909, 2909+113=4240
Answer: d)

9).The pattern is 3×1-3 = 0 ,0×2-4 = -4 ,-4×3-5 = -17,-17×4-6 =-74,-74×5-7=-377
Answer: e)

10).The pattern is 452,492,532,572, 612, 652
Hence missing term =572 = 3249
Answer: c)

  

 


 

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