IBPS Clerk Pre 10.11.2018




Directions (1-5): What should come in place of question mark (?) in the following questions? 

Q1.
50
60
55
75
80
Solution:


Q2.
137.5
125.5
117.5
112.5
107.5
Solution:


Q3. 22.5 × 12 + (11)² – √(?) = (19)²
800
750
825
900
950
Solution:
270 + 121 – √(?) = 361 √(?)=30 ? = 900

Q4. 624/?+3.5×6+27=(6)²+440/2
5
3
4
8
7
Solution:


Q5.
44
46
56
54
64
Solution:


Directions (6-10): What will come in the place of the question mark (?) in the following number series? 

Q6. 125, 189, 314, 530, 873, ?
1385
1265
1525
1375
1785
Solution:


Q7. 150, 300, 60, 120, 24, 48, ?
16.4
7.6
9.6
12
30
Solution:
Pattern is
150 × 2 = 300
300 ÷ 5 = 60
60 × 2 = 120
120 ÷ 5 = 24
24 × 2 = 48
48 ÷ 5 = 9.6

Q8. 106, 184, 267, 357, 456, ?
486
566
626
766
546
Solution:


Q9. 142, 119, 100, 83, 70, ?
69
49
42
59
53
Solution:
Pattern is
142 – 23 = 119
119 – 19 = 100
100 – 17 = 83
83 – 13 = 70
70 – 11 = 59

Q10. 120, 60, 90, 225, ?
782.5
787.5
752.5
875
855.5
Solution:
Pattern is
 ×0.5, ×1.5, ×2.5, ×3.5, ×4.5 ….
∴ 225 × 3.5 = 787.5

Directions (11-15): Solve the given quadratic equations and mark the correct option based on your answer— 

Q11. (i) x² – x – 132 = 0 
         (ii) y² – 22y + 112 = 0
x > y
x ≥ y
x < y
x ≤ y
x = y or no relation can be established between x and y.
Solution:
X² – 12x + 11x – 132 = 0
x(x – 12) + 11 (x – 12) = 0
(x + 11) (x – 12) = 0
x = –11, 12
y² – 14y – 8y + 112 = 0
y(y – 14) – 8(y – 14) = 0
(y – 8) (y – 14) = 0
y = 8, 14
No relation can be established between x & y.

Q12. (i) x² – 8x + 15 = 0 
         (ii) y² – 2y – 3 = 0
x > y
x ≥ y
x < y
x ≤ y
x = y or no relation can be established between x and y.
Solution:
x² – 3x – 5x + 15 = 0
(x – 3) (x – 5) = 0
x = 3, 5 y² – 3y + y – 3 = 0
y(y – 3) + 1 (y – 3) = 0
(y + 1) (y – 3) =0
y = –1, 3
x ≥ y

Q13. (i) x² + 5√3 x+18=0 
         (ii) y²-√3 y-18=0
x > y
x ≥ y
x < y
x ≤ y
x = y or no relation can be established between x and y.
Solution:
x²+2√3 x+3√3 x+18=0
x(x+2√3)+3√3 (x+2√3)=0
(x+3√3)(x+2√3)=0
x=-3√3,-2√3
y²+2√(3) y-3√(3 ) y-18=0
(y+2√3)(y-3√3)=0
y= -2√(3,) 3√3
y ≥ x

Q14. (i) 2x² – x – 3 = 0 
         (ii) 4y² – 24y + 35 = 0
x > y
x ≥ y
x < y
x ≤ y
x = y or no relation can be established between x and y.
Solution:
2x² – x – 3 = 0
(x + 1) (2x – 3) = 0
x= -1, 3/2 4y² – 24y + 35 = 0
4y² – 14y – 10y + 35 = 0
y=5/2, 7/2
y > x

Q15. (i) 2x² – 7x + 6 = 0 
         (ii) 2y² – y– 1= 0
x > y
x ≥ y
x < y
x ≤ y
x = y or no relation can be established between x and y.
Solution:
2x²– 3x – 4x + 6 = 0
x(2x – 3) – 2 (2x – 3) = 0
(x – 2) (2x – 3) = 0
x=2, 3/2
2y² – 2y + y – 1 = 0
2y(y – 1) + 1 (y – 1) = 0
(2y + 1) (y – 1) = 0
y = -1/2 , 1
x > y

               

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