Quantitative Aptitude Quiz – March, 2015 – Set 6



Directions (Q. 1-5):
In each of the following number series a wrong number is given. Find out the
wrong number.


1. 161 186 213 242
273 308 341
(a) 186
(b) 213
(c) 242
(d) 273
(e) 308
2. 24 56 110 192 309
464 666
(a) 56
(b) 110
(c) 192
(d) 309
(e) 464
3. 20 34 52 82 124
180
(a) 20
(b) 34
(c) 52
(d) 82
(e) 124
4. 5 14 65 99 197 254
401
(a) 99
(b) 14
(c) 65
(d) 197
(e) 254
5. 28 84 24 51 20 83
(a) 84
(b) 24
(c) 28
(d) 51
(e) 20
Direction (Q. 6-10):
In each of the following questions two equations I and II are given. You have
to solve both the equations and give answer:
(a) x < y
(b) x > y
(c) x ≥ y
(d) x ≤ y
(e) x = y or the
relationship cannot be established.
6.
I. 20×2
– 31x + 12 = 0
II. 20y2 +
y – 12 = 0
7.
I. 2×2
– 27x + 91 = 0
II. 2y2
+ y – 136 = 0
8.
I. 2x -13√x + 21 =
0
II. 2y -15√y
+ 28 = 0
9.
I. x2
= 3136
II. y2
= 1764
10.
I. x2
– 20x + 91 = 0
II. y2
– 6y – 91 = 0

Solutions:

1.
132 – 2 x (1+3) = 161
142 – 2 x (1 + 4) = 186
152 – 2 x (1+5) = 213
2.
33 – 1×3= 24
43 – 2×4 = 56
53 – 3×5 = 110
63 – 4 × 6 = 192
3.
14 + (2×3) = 20
20 + (3 × 4) = 32
32 + (4 × 5) = 52
52 + (5 × 6) = 82
4.
1 × 2 + 3 = 5
4 × 5 – 6 = 14
7 × 8 + 9 = 65
10 × 11 – 12 = 98
13 × 14 + 15 = 197
5.
48 ÷ 2 + 3 = 27
27 × 3 + 3 = 84
84 ÷ 4 + 3 = 24
24 × 2 + 3 = 51
51 ÷ 3 + 3 = 20
20 × 4 + 3 = 83
6.
I. 20 x2 – 31 x + 12 = 0
(4 x – 3) (5x – 4) = 0
x = 3/4 , 4/5
II. 20 y2 + y – 12 = 0
or, (4y – 3) (5y + 4) = 0
y = 3/4 , 4/5
So,  x ≥ y
7.
I. 2×2 – 27x + 91 = 0
or, (x– 7) (2x – 13) = 0
So, x = 7, 13/2
II. 2y2 + y –136 = 0
or, (y –8) (2y + 17)= 0
So, y = 8, –17/2
8.
I. 2x – 13√x + 21 = 0
(√x – 3) ( 2√x –7) = 0
So, x = 9, 49/4
II. 2y – 15√y + 28 = 0
or, (2√y –7) (√y –4) = 0
So, y = 49/4 , 16
Hence, x ≤ y
9.
I. x2 = 3136
So,  x = ±56
II. y2 = 1764
So,  y = ±42
10.
I. x2 – 20 x + 91 = (x –7) (x – 13)= 0 x = 7, 13
II. y2 – 6 y – 91 = (y –13) (y + 7) = 0 y = 13,
–7

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