Crack IBPS Exam 2017 - Quantitative Aptitude Scoring Part (Day-21)

Crack IBPS Exam 2017 - Quantitative Aptitude Scoring Part (Day-1)

Crack IBPS Exam 2017 - Quantitative Aptitude Scoring Part (Day-21):

Dear Readers, Nowadays most of the aspirants are facing huge trouble to increase the overall marks. To score high you need to practice more and more standard questions daily. “Practice does not make perfect, Only Perfect Practice makes perfect”.

Direction (1-10): What value should come in place of question mark (?) in the following questions?
1). (17)^8.8 × (289)^(–1.4) ÷ 1/(17)^-1 = 17 × (17)^?

1). Answer: c)
17 × 17^? = (17)^8.8 × (17^2)^(–1.4) ÷ (17)^1
17 × 17^? = (17)^8.8 × (17)^(–2.8) ÷ 17 = (17)^(8.8 – 2.8 – 1) = (17)^5
17^? = 17^(5 – 1) = 17^4
? = 4

2. 2.4% of 775 + 8.4% of 525 = 30% of?

2). Answer: e)
30 × ? / 100 = 2.4 × 775 / 100 + 8.4 × 525 / 100
30 × ? / 100 = 18.6 + 44.1 = 62.7
? = 62.7 × 100 / 30 = 627/3 = 209

3. {0.00102 ÷ 0.000017} × 17.75 = ?

3). Answer: c)
? = 0.00102 / 0.000017 × 17.75
? = 1020 / 17 × 17.75
? = 60 × 17.75 = 1065

4. (1728)^(2/3) ÷ 1/(5832)^(2/3) = ? × 9

4). Answer: a)
9 × ? = (12^3)^(2/3) ÷ 1/(18^3)^(2/3) =(12)^2 ÷ (18)^(-2)
9 × ? = (12)^2 × (18)^2
? = 12 × 12 × 18 × 18 / 9
? = 5184

5).(1260 ÷ 28) × 6.4 = 45% of?

5). Answer: b)
45 × ? / 100 = 1260 / 28 × 6.4 = 45 × 6.4
? = 6.4 × 100 = 640

6. 85% of 4/7 of 6755 = ? + 1687

6). Answer: c)
? = 85 × 4 × 6755 / (100 × 7) – 1687 = 17 × 193 - 1687
? = 3281 - 1687 = 1594

7. (5568 ÷ 87)^(1/3) + (72 × 2)^(1/2) = (?)^(1/2)

7). Answer: a)
(?)^(1/2) = (5568 / 87)^(1/3) + (144)^(1/2)
(?)^(1/2) = (64)^(1/3) + (12^2)^(1/2) = (4^3)^(1/3) + (122)^(1/2)
(?)^1/2 = 4 + 12 = 16
? = 256

8. √(13^2 + 28 ÷ 4 - (3)^3 +107) = (?)^2

8). Answer: d)
(?)^2 = √(13^2 + 28 ÷ 4 - (3)^3 +107)
= √(169 + 7 - 27 +107)
= √(283 – 27) = √256 = 16
? = √16 = 4
9). Answer: b)
(0.7)^(?+3) = (0.7)^(2 × 4) × (0.7)^(3 × 4) ÷ (0.7)^(4 × 4)
(0.7)^(?+3) = (0.7)^(8 + 12 – 16) = (0.7)^4
? = 4 – 3 = 1

9. (0.49)^4 × (0.343)^4 ÷ (0.2401)^4 = (70 ÷ 100)^(? + 3)

9). Answer: b)
(?)^2 = 1224 × 306 = (18 × 17 × 4) × (18 × 17)
(?)^2 = (18 × 17 × 2)^2
? = 18 × 17 × 2 = 612

10. 45% of √2025 ÷ 0.01 = (?)^2 ÷ 25

10). Answer: e)
√2025 = 45
Now, 45 × 45 / (0.01 × 100) = (?)^2 ÷ 25
2025 = (?)^2 / 25
or. (?)^2 = 50625
? = √50625 = 225

Direction (11-20): What approximate value should come in place of question mark (?) in the following questions?
11. 78.99% of 9875.99 - 38.09% of 6785.05 = 2479.05 + ? of 4895.99

11). Answer: a)
? × 4896 / 100 ≈ 79 × 9876 /100 - 38 × 6785 / 100 - 2479
= 0.79 × 9876 - 0.38 × 6785 - 2479
≈ 7802 - 2578 - 2479
= 7802 - 5057 = 2745
? = 2745 / 4896 × 100 = 56.06 ≈ 56

12. √((4095.99)^(1/3)) × √65535.89 = (?)^2

12). Answer: e)
(?)^2 ≈ √(4096)^(1/3) × √65536
= √16^(3 ×(1/3)) × √16^4
= 4 × 16 × 16 = 256 × 4 = 1024
? = √1024 = 32

13.5030.05 ÷ 42.93 + 24.49% of 5049.93 ÷ 100 = ?

13). Answer: c)
5030.05 ÷ 42.93 + 24.49% of 5049.93 ÷ 100 = ?
? ≈ 5030 ÷ 43 + 24.5% of 5050 ÷ 100
? = 116.9764 + 1237.25 ÷ 100
? ≈ 117 + 13
? = 130

14. 52920 ÷ 3214.05 × 514.13 + 5231.92 = ?

14). Answer: b)
? ≈ 52920 ÷ 3214 × 514 + 5232
? = 16.46 × 514 + 5232
? = 8460.44 + 5232 = 13692.44 ≈ 13695

15. ³√ 6850 × √12540 = ? × 52

15). Answer: a)
³√6850 × √12540 = ? × 52
or, ? ≈ 19 × 112 / 52
? = 40.92 ≈ 41

16.√6398.99 ÷ ³√4099.99 × 24.89 = (?)^³

16). Answer: b)
(?)^3 = √6398.99 ÷ ³√4099.99 × 24.89
or, (?)^3 ≈ 80 ÷ √(16 ×16 × 16) × 25
or, (?)^3 = 80 ÷ 16 × 25
or, (?)^3 = 125 = 53
? = 5

17. (87.65% of 7159.89 - 68.99% of 8939.89) × 6.06 = (?)^2

17). Answer: d)
(?)^2 ≈ (88 × 7160 / 100 - 69 × 8940 /100) × 6
= (88 × 71.60 - 69 × 89.40) × 6
≈ (6300 - 6168) × 6
= 132 × 6 = 792
? = √792 = 28.14 ≈ 28
18). Answer: b)
40 × ? / 100 ≈ 449 × 346 / 64
or,? = 449 × 346 × 100 / (64 × 40)
? = 6068.5 ≈ 6065

18.449.03 × 345.88 ÷ 64 = 40.02% of ?

18). Answer: b)
(50243408)^(1/3) ≈ (50243409)^(1/3) = 369
and, (48627124)^(1/3) ≈ (48627125)^(1/3) = 365
Again, (7529535)^(1/3) ≈ (7529536)^(1/3) =196
So, ? = 369 - 365 + 196
? = 200

19. 37.9% of 638.05 + 25.25% of 4401.9 = ?

19). Answer: d)
37.9% of 638.05 + 25.25% of 4401.9 = ?
or, ? ≈ 38% of 638 + 25% of 4402
= 242.44 + 1100.5 = 1342.94
? ≈ 1345

20. 833.956 - 543.005 - 108.98 = 19.8% of ?

20). Answer: d)
? × 20 / 100 ≈ 834 - 543 - 109 = 182
? = 182 × 5 = 910

Direction (21 – 25): In the following number series only one number is wrong. Find out the wrong number.21. 2, 10, 30, 68, ?

21). Answer: b)
The series is 1^3 + 1, 2^3 + 2, 3^3 + 3, 4^3 + 4, 5^3 + 5, …

22. 33, 321, 465, 537, 573, ?

22). Answer: b)
The difference between the numbers is + 288, + 144, + 72, + 36, +18, ..

23. 2, 9, ?, 105, 436, 2195

23). Answer: c)
The sequence of the series is ×1 + (1×7), ×2 + (2×6), ×3 + (3×5), ×4 +(4×4),…

24. 4, 6, 14, 44, ?, 892

24). Answer: c)
The series is ×1 + 2, ×2 +2, ×3 + 2, ×4 + 2, ×5 + 2,

25.126, 64, 34, 20, 14, ?

25). Answer: a)
The sequence of the series is ÷2 + 1, ÷2 + 2, ÷2 +3, ÷2 +4, ÷2+5,..

Direction (26-30): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and given answer:
a) If x > y
b) If x ≥ y
c) If x < y
d) If x ≤ y
e) If x = y or no relation can be established between x and y.

26. I. x2 – 16 = 0
II. y^4 – {[(8 x 2)^(9/4)] / √y } = 0

26). Answer: d)
I. x^2 – 16 = 0
X^2 = 16
x = ±4
II. y^4 – {[(8 x 2)^(9/4)] / √y } = 0
Y^4 √y – [(4^2)^(9/4)] = 0
y ^(4+(1/2)) – (4)9/2 = 0
y^(9/2) – 4^(9/2) = 0
y = 4
Hence x ≤ y

27.I. x^2 - 2√3x – 72 = 0
II. y^2 + 9√3x + 60 = 0

27). Answer: b)
I. x^2 - 2√3x – 72 = 0
X^2 - 6√3x + 4√3x – 72 = 0
x (x - 6√3) + 4√3 (x - 6√3) = 0
x = -4√3, 6√3
II. y^2 + 9√3y + 60 = 0
Y^2 + 4√3y + 5√3y + 60 = 0
y(y + 4√3) + 5√3 (y + 4√3) = 0
y = -4√3, -5√3
Hence x ≥ y

28.I. x^6 – [(√(54 × 6))^6.5 / √x] = 0
II. y – (36 × 81)^(1/2) = 0

28). Answer: c)
I. x^6 – [(√(54×6))^6.5 / √x] = 0
X^(6+(1/2)) – (18)^(6.5) = 0
X^(6.5) = 18^(6.5)
x = 18
II. y – (36 × 81)^(1/2) = 0
y – (6 × 9) = 0
y – 54 = 0
y = 54
Hence x < y

29. I. x^2 – 17x – 234 = 0
II. 4y^2 + 32y – 192 = 0

29). Answer: e)
I. x^2 – 17x – 234 = 0
X^2 – 26x + 9x – 234 = 0
x(x - 26) + 9 (x – 26) = 0
(x + 9) (x - 26) = 0
x = -9, 26
II. 4y^2 + 32y – 192 = 0
Y^2 + 8y – 48 = 0
y^2 + 12y – 4y – 48 = 0
y (y + 12) – 4 (y + 12)
y = - 12, 4
Hence relation cannot be determined

30. I. x^2 = 784
II. y^2 – 11y – 102 = 0

30). Answer: e)
I. x^2 = 784
x = ± 28
II. y^2 – 11y – 102 = 0
y^2 – 17y + 6y – 102 = 0
y (y - 17) +6 (y - 17) = 0
y = 17, -6
Hence Relation cannot be determined