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IBPS PO Prelims Quantitative Aptitude (Day-04)

Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for IBPS PO Prelims 2020 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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Missing number series

Direction (1-5): What value should come in the place of (?) in the following number series?

1) 4080, 3360, 2730, 2184, 1716, ?

A) 1320

B) 1850

C) 1560

D) 1210

E) 1080

2) 3, 21, 33, 39, 39, ?

A) 52

B) 48

C) 33

D) 28

E) 22

3) 19, 17, 30, 82, 312, ?

A) 1528

B) 1236

C) 1328

D) 1124

E) 1456

4) 94, 95, 98, 104, 114, ?

A) 110

B) 116

C) 119

D) 124

E) 129

5) 2430, 1620, 1080, 720, ?, 320

A) 640

B) 560

C) 480

D) 680

E) 520

Direction (6-10): Following question contains two equations as I and II. You have to solve both equations and determine the relationship between them and give answer as,

a) If x > y

b) If x ≥ y

c) If x = y or relationship can’t be determined.

d) If x < y

e) If x ≤ y

6)

I) x2+ 2x – 195 = 0

II)y2+ y – 182 = 0

7)

I)x2– 25x + 136 = 0

II)y2– y – 56 = 0

8)

I) 2x2+ 16x + 30 = 0

II)y2+ 6y + 8 = 0

9)

I)x2+ 5x + 6 = 0

II)y2– 2y + 1 = 0

10)

I) x2– 11x + 30 = 0

II)y2– 21y + 90 = 0

Answers :

Directions (1-5) :

1) Answer: A

163 – 16 = 4080

153 – 15 = 3360

143 – 14 = 2730

133 – 13 = 2184

123 – 12 = 1716

113 – 11 = 1320

2) Answer: C

3 + (3 * 6) = 21

21 + (2 * 6) = 33

33 + (1 * 6) = 39

39 + (0 * 6) = 39

39 + (-1 * 6) = 33

3) Answer: A

19 * 1 – 2 = 17

17 * 2 – 4 = 30

30 * 3 – 8 = 82

82 * 4 – 16 = 312

312 * 5 – 32 = 1528

4) Answer: E

94 + 1 = 95

95 + 1 + 2 = 98

98 + 1 + 2 + 3 = 104

104 + 1 + 2 + 3 + 4 = 114

114 + 1 + 2 + 3 + 4 + 5 = 129

5) Answer: C

2430 ÷ 1.5 = 1620

1620 ÷ 1.5 = 1080

1080 ÷ 1.5 = 720

720 ÷ 1.5 = 480

480 ÷ 1.5 = 320

Directions (6-10):

6) Answer: C

x2 + 2x – 195 = 0

x2 + 15x – 13x – 195 = 0

x(x + 15) – 13(x + 15) = 0

(x – 13)(x + 15) = 0

X = 13, -15

y2 + y – 182 = 0

y2 + 14y – 13y – 182 = 0

y(y + 14) – 13(y + 14) = 0

(y – 13)(y + 14) = 0

Y = 13, -14

Relationship between x and y cannot be established.

7) Answer: B

x2 – 25x + 136 = 0

x2 – 17x – 8x + 136 = 0

x(x – 17) – 8(x – 17) = 0

(x – 8)(x – 17) = 0

x = 8, 17

y2 – y – 56 = 0

y2 – 8y + 7y – 56 = 0

y(y – 8) + 7(y – 8) = 0

(y + 7)(y – 8) = 0

y = -7, 8

Hence, x ≥ y

8) Answer: C

2x2 + 16x + 30 = 0

2x2 + 10x + 6x + 30 = 0

2x(x + 5) + 6(x + 5) = 0

(2x + 6)(x + 5) = 0

x = -3, -5

y2 + 6y + 8 = 0

y2 + 4y + 2y + 8 = 0

y(y + 4) + 2(y + 4) = 0

(y + 2)(y + 4) = 0

y = -2, -4

Relationship between x and y cannot be established.

9) Answer: D

x2 + 5x + 6 = 0

x2 + 3x + 2x + 6 = 0

x(x + 3) + 2(x + 3) = 0

(x + 2)(x + 3) = 0

x = -2, -3

y2 – 2y + 1 = 0

y2 – y – y + 1 = 0

y(y – 1) – 1(y – 1) = 0

(y – 1)(y – 1) = 0

y = 1, 1

Hence, x < Y

10) Answer: E

x2 – 11x + 30 = 0

x2 – 6x – 5x + 30 = 0

x(x – 6) – 5(x – 6) = 0

(x – 5)(x – 6) = 0

x = 5, 6

y2 – 21y + 90 = 0

y2 – 15y – 6y + 90 = 0

y(y – 15) – 6(y – 15) = 0

(y – 6)(y – 15) = 0

y = 6, 15

Hence, x ≤ y

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