## 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

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

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

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

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

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

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

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

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

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

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