## Quantitative Aptitude Practice Questions for Bank Exams, RBI Assistants

### Quantitative Aptitude Practice Questions for Bank Exams, RBI Assistants

Directions (1 – 5): What will come in place of “?” in the given series?
1. 1, 2, 33, 290, 2373, ?
a) 22086
b) 27434
c) 28644
d) 29564
e) 30788
2. 3, 4.2, 7.8, 18.6, 51, ?
a) 128.2
b) 132.8
c) 136.4
d) 141.6
e) 148.2
3. 5, 30, 185, 1300, ?, 93650
a) 10400
b) 10405
c) 10505
d) 10550
e) 10600
4. 6, 22, 86, 342, 1366, ?
a) 4832
b) 5264
c) 5462
d) 5064
e) 5678
5. 330, 547, 673, 738, 766, ?
a) 775
b) 780
c) 785
d) 788
e) 792
Directions (6 – 10): In each question, 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 relationship doesn’t exists
6. I.8x2 - 78x + 169 = 0 II.20y2 - 117y + 169 = 0
7. I.(169)1/2x+(289)1/2 = 134 II.(361)1/2y2 – 270 = 1269
8. I.x2-14x +49=0 II.y2-10y-56=0
9. I.8x2+29x-12=0 II.3y2+12y+9=0
10. I.x2-11x+24=0 II.y2-7y+12=0
Solutions:
1. A) The pattern of the series is:
1 × 1 + 13 = 2;
2 × 3 + 33 = 33;
33 × 5 + 53 = 290;
290 × 7 + 73 = 2373
2373 × 9 + 93 = 22086
Hence the required term is 22086
2. E) The pattern of the series is:
3 + (0.4 × 3) = 3 + 1.2 = 4.2
4.2 + (1.2 × 3) = 4.2 + 3.6 = 7.8
7.8 + (3.6 × 3) = 7.8 + 10.8 = 18.6
18.6 + (10.8 × 3) = 18.6 + 32.4 = 51
51 + (32.4 × 3) = 51 + 97.2 = 148.2
Hence the required term is 148.2
3. B) The pattern of the series is
5 × 5 + 5 = 30
30 × 6 + 5 = 185
185 × 7 + 5 = 1300
1300 × 8 + 5 = 10405
10405 × 9 + 5 = 93650
Hence the required term is 10405
4. C) The pattern of the series is
(6 × 4) – 2 = 22
(22 × 4) – 2 = 86
(86 × 4) – 2 = 342
(342 × 4) – 2 = 1366
(1366 × 4) – 2 = 5462
Hence the required term is 5462
5. A) The pattern followed in the series is
330 + 63 + 1 = 547
547 + 53 + 1 = 673
673 + 43 + 1 = 738
738 + 33 + 1 = 766
766 + 23 + 1 = 775
Hence, The required term is 775.
6. A) I.8x2-78x+169=0
Or, 8x2-52x-26x+169 =0
Or, 4x(2x-13)-13(2x-13)=0
Therefore, x=13/2, 13/4
=6.4, 3.25
II.20y2-117y+169 =0
Or, 20y2-65y-52y+169 =0
Or, 5y(4y-13)-13(4y-13)=0
Or, y=13/5,13/4
Therefore, y=2.6, 3.25
Therefore, Hence xy
7. A)  I. (169)1/2x+289=134
Or, 13x+17=134
Or, 13x=117
x=9
II. (361)1/2y2-270=1269
Or, 19y2=1269+270
Or, y2=81
y=±9
Hence, xy
8. E)  I. x2-14x+49=0
Or, x2-7x-7x+49=0
Or,(x-7)(x-7)=0
x=7
II.y2-10y-56=0
Or, y2-14y+4y-56=0
Or,y(y-14)+4(y-14)=0
y=14,-4
Relation can’t be established.
9. E) I. 8x2+29x-12=0
Or, 8x2+32x -3x-12=0
Or,8x(x+4)-3(x+4)=0
x=3/8,-4
II. 3y2+12y+9=0
Or, 3y2+3y+9y+9=0
Or, 3y(y+1)+9(y+1)=0
Or, y=-9/3,-1
= -3,-1
Relationship can’t be established.
10. E)  I. x2-11x+24=0
Or, x2-8x-3x+24=0
Or,x(x-8)-3(x-8)=0
Or,(x-3)(x-8)=0
x=3,8
II. y2-7y+12=0
Or, y2-4y-3y+12=0
Or, y(y-4)-3(y-4)=0
Or, (y-3)(y-4)=0
y=4,3
Hence, relationship can’t be determined.