@
ce
te
xa
m
gr
ou
p
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
1
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
Chapter
Inequality
14
m
gr
ou
p
As we know, if we use the short method, in these types of questions it takes very little time to solve in the examination.
So let us understand the concept.
Quadratic equation: It is a second order polynomial equation with a single variable.
Example: ax 2 + bx + c = 0
There will be two values of x which satisfy the given equation.
Sign Method: Now from the exam's point of view, we can conclude the signs of the roots from the signs of the coefficients.
Case I:
If b = +ve, c = +ve
Example: ax 2 + bx + c = 0, x1 = –ve, x2 = –ve
Case II: If b = –ve, c = –ve,
then one root will be positive(+) (bigger number) and other root will be negative(–) (smaller number)
Example: ax 2 – bx – c = 0, x1 = +ve, x2 = –ve
Case III: If b = +ve, c = –ve
then one root will be –ve (bigger number)
and other root will be +ve (smaller number)
Example: ax 2 + bx – c = 0, x1 = –ve, x2 = +ve
Case IV: If b = –ve, c = +ve
ax 2 – bx + c = 0
x1 = +ve, x2 = +ve
@
ce
te
xa
x’s co-efficient
Constant
x1
x2
(b)
(c)
+
+
–
–
+
–
–
+
–
+
+
+
–
–
+
–
When we see the equation then we can conclude the signs of the roots, so we can find the relation between x and y.
Now we have the signs of the roots of equation and if we remember the table then we can conclude it within 5 seconds.
Now come to the second part:
Suppose we have the equation:
x 2 – 5x + 6 = 0
From the table we can conclude that both the roots of the equation will be +ve.
Now we have to break the constant (6) such that their sum will be 5.
6 = 3 × 2 also, (3 + 2 = 5)
∴ x1 = +3, x2 = +2
Now if there is a value attached to the x's co-efficient, then we have to divide the value to get the roots.
Sol.
2
2x – 11x + 15 = 0
2
∴
2
3×5
6
6 + 5 = 11
5
x1 = + 2 = +3, x2 = + 2 = + 2.5
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
2
2
∴ x1 = –
∴ x1 = –3, x2 = –2
2
y + 3y + 2
@
ce
2 × 1 and, 2 + 1 = 3
2
9
= –7, x2 = – = – 4.5
2
2
ou
II. 4y + 19y + 21 = 0
21
4
2×2
7×3
also, 12 + 7 = 19
∴ y1 = –
12
4
7
= –3, y2 = – = – 1.75
4
∴x<y
6. I. 4x 2 – 29x + 45 = 0 II. 3y 2 – 19y + 28 = 0
2
Sol. (e): I. 4x – 29x + 45 = 0
45
4
2×2
9×5
also, 20 + 9 = 29
xa
te
3 × 2 and, 3 + 2 = 5
14
p
63
2
2×1
9×7
also, 14 + 9 = 23
m
2. I. 5x + 3x – 14 = 0 II. 10y – 3y – 27 = 0
Sol. (e): Here x1 = –ve x2 = +ve
y1 = +ve, y2 = –ve
Now we can't establish the relation between x and
y.
Note: Above two types of questions can be solved
without calculation. So if in the exam we have any
of these cases then we can simply conclude the
relation by the sign method.
3. I. x 2 + 5x+ 6 = 0
II. y 2 + 3y + 2 = 0
2
Sol. (d): x + 5x + 6 = 0
5. I. 2x 2 + 23x + 63 = 0 II. 4y 2 + 19y + 21 = 0
2
Sol. (b):I. 2x + 23x + 63 = 0
gr
Direction (1 – 14): Two equations (I) and (II) are given in
each question. On the basis of these equations you have to
decide the relation between x and y and give 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'.
1. I. x 2 – 3x + 6 = 0
II. 3y 2 + 13y + 12 = 0
Sol. (a): We even don't have to calculate the values.
Here signs of roots of first quadratic equation
(x1 , x2 ) = +ve
and signs of roots of second quadratic equation
(y1 , y2 ) = –ve
∴ x > y (Always)
20
9
∴ x1 = + 4 = +5, x2 = + 4 = +2.25
2
II. 3y – 19y + 28 = 0
28
3
3×1
7×4
also, 12 + 7 = 19
∴ y1 = +
12
3
7
= 4, y2 = + = 2.33
3
∴ No relationship can be estabilished
7. I. 2x 2 – 13x + 21 = 0 II. 5y 2 – 22y + 21 = 0
2
Sol. (c): I. 2x – 13x + 21 = 0
∴ y1 = – 2, y2 = –1
∴ x≤y
2
4. I. 2x + 3x + 1 = 0
Sol. (b): 2x2 + 3x + 1 = 0
2
II. 12y + 7y + 1 = 0
6
2
II. 5y – 22y + 21 = 0
1
∴ x1 = – 2 = –1, x2 = – 2 = – 0.5
2
Now, 12y + 7y + 1 = 0
12 = 4 × 3 and, 4 + 3 = 7
4
1
3
1
∴ y1 = – 12 = – 3 = – 0.33, y2 = – 12 = – 4 = – 0.25
∴y>x
3
7
x1 = + 2 = 3 , x2 = + 2 = 3.5
2 × 1 and, 2 + 1 = 3
2
21
2
2×1
7×3
also, 6 + 7 = 13
21
5
5×1
7×3
also, 15 + 7 = 22
∴ y1 = +
15
5
7
= 3, y2 = + = 1.4
5
x≥y
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
I. 12x 2 + 11x – 56 = 0 II. 4y 2 – 15y + 14 = 0
11. I.
Sol. (d):I. 12x2 + 11x – 56 = 0
1
II. y 3 = 1.63 ⇒ y > 1
21
∴ x1 = – 12 = – 2.67, x2 = + 12 = 1.75
∴y>x
2
4y – 15y + 14 = 0
3
14
7×2
also, 8 + 7 = 15
4
2×2
12. I. x = √2197
II. y 2 = 169
Sol. (c):I. x = 13
II. y 2 = 169 ⇒ y = ± 13
∴x≥y
ou
8
7
4
4
I. 7x – 3y = 13
Sol. (c): I. x = √2304
∴ x = 48 (don't consider – 48 as value of x)
II. 5x + 4y = 40
II. y 2 = 2304 ⇒ y = ± 48
Sol. (b):equation (I) × 4 + equation (II) × 3
∴x≥y
gr
28x − 12y = 52
15x + 12y = 120
14. I.
43x = 172 ⇒ x = 4
1
1
9
√x
6
Sol. (a): I. 35x + 70 = 0 ⇒ x = – 35 = –2
te
@
ce
Direction (1-15): In each of these questions, 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 between x and y cannot be
established.
2
2
II. y + 7y + 12 = 0
2
2.
3.
I. x + 20 = 9x
I. 12x + 3y = 14
II. y 2 + 42 = 13y
II. 4x + 2y = 16
4.
I. x = √625
II. y = √676
5.
2
I. x + 4x + 4 = 0
4
II. y 2 – 8y + 16 = 0
II. y10 – (36)5 = 0
= (36)5 ⇒ y10 = (62 )5 ⇒ y10 = 610
∴y=6
∴x=y
15. I. 7x 2 + 16x – 15 = 0
II. y 2 – 6y – 7 = 0
Sol. (e): I. x1 = –ve x2 = +ve
II. y1 = +ve y2 = –ve
∴ Relation can't be estabilished
II. 3y + 7 = 0 ⇒ y = – 3 = – 2.33
∴ x>y
1
= (x)2
= √x ⇒ x = 6
√x
10
xa
10. I. √1225x + √4900 = 0 II. (81)4 y + (343)3 = 0
m
∴ x<y
7
√x
−
II. y
4 × 7 – 3y = 13 Þ 3y = 15 Þ y = 5
70
15
Sol. (e): I.
∴ put it in equation (I)
I. x + 5x + 6 = 0
II. y 2 = 2304
13. I. x = √2304
∴ y1 = + = 2, y2 = + = 1.75 ; x ≤ y
1.
8
6x = 2 ⇒ x = 3 ⇒ x = 0.33
32
9.
12
+ x − x2 = x2
Sol. (b):I. 18 + 6x – 12 = 8
also, 32 – 21 = 11
II.
6
x2
3
II. y + 9.68 + 5.64 = 16.95
56
8×7
12
4×3
18
p
8.
6.
7.
I. x 2 – 19x + 84 = 0
I. x 3 – 468 = 1729
9
11
I.
9.
I. √784x + 1234 = 1486
√x
= √x
5
8.
√x
+
19
II. y 2 – 25y + 156 = 0
II. y 2 – 1733 + 1564 = 0
II. y –
(2×14) 2
√y
=0
II. √1089y + 2081 = 2345
10. I.
12
√x
–
23
√x
= 5√x
II.
√y
12
–
5√y
12
=
1
√y
11. I. 6x 2 – 49x + 99 = 0 II. 5y 2 + 17y + 14 = 0
12. I.
13. I.
14. I.
15. I.
1
x = (1331)3
2x 2 + 3x + 1 = 0
7x – 3y = 13
2x + 5y = 6
Adda247 Publications
II.
II.
II.
II.
2y 2 – 17y + 36 = 0
12y 2 + 7y + 1 = 0
5x + 4y = 40
5x + 11y = 9
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
Prelims Questions
Directions (1-5):- In each of the following questions, two
equations (I) and (II) are given. Solve the equations and
mark the correct option:
(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.
1. I. x 2 + 5x + 6 = 0
II. y 2 + 9y + 14 = 0
2
2. I. x − 18x + 45 = 0 II. y 2 + 12y − 45 = 0
3. I. 9x 2 + 11x + 2 = 0 II. 8y 2 + 6y + 1 = 0
4. I. 6x² + 5x + 1 = 0
II. 4y 2 – 15y = 4
5. I. x 2 + 3x = 0
II. x 2 + y = 10
15. I. x 2 + 48x + 575 = 0
Directions (6-10):- In each of the following questions, two
equations (I) and (II) are given. Solve the equations and
mark the correct option:
(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.
6. I. x 2 − 25x + 156 = 0
II. y 2 − 29y + 210 = 0
17. I. 3x 2 + 20x + 32 = 0
II. y 2 + 44y + 483=0
two equations (I) and (II) are given. Solve the equations
and mark the correct option:
(c) if x<y
(e) if x = y or no relation can be established between
x and y.
gr
16. I. x 2 + 23x + 132 = 0
II. y 2 + 21y + 110 = 0
II. 5y 2 + 23y + 24 = 0
18. I. x 2 − 29x + 208 = 0
m
te
@
ce
(d) if x ≤y
II. y 2 − 21y + 108 = 0
19. I. x2+30x+224=0
II. y2+35y+306=0
3
20. I. x = √4096
II. y2 =256
Directions (21-25): In each of these questions, two
equations (I) and (II) are given. Solve the equations and
mark the correct option:
Directions (11-15):- In each of the following questions,
two equations (I) and (II) are given. Solve the equations
and mark the correct option:
(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.
11. I. x 2 + 17x + 72 = 0
II. y 2 + 11y + 30 = 0
12. I. 3x 2 − 23x + 40 = 0
II. 5y 2 − 17y + 14 = 0
13. I. x 2 − 26x + 168 = 0
II. y 2 − 29y + 208 = 0
14. I. x3+340=2537
II. y2+23=192
5
(b) if x≥y
ou
(a) if x>y
xa
I. x 2 = 196
II. y = √196
2
I. x + 12x + 35 = 0
II. y 2 + 14y + 48 = 0
9. I. 3x 2 + 23x + 30 = 0
II. y² + 15y + 56 =0
10. I. x 2 + 17x + 72 = 0
II. y 2 + 13y + 42=0
7.
8.
p
Directions (16-20):- In each of the following questions,
21. I. 2x² + 10x + 12 = 0 II. y² + 10x + 25 = 0
(a) x > y
(b) x ≥ y
(c) x < y
(d) x ≤ y
(e) x = y or relation can’t be established.
22. I. x² – 5x + 6 = 0
II. y² + 7y + 6 = 0
(a) x > y
(b) x ≥ y
(c) x < y
(d) x ≤ y
(e) x = y or relation can’t be established.
23. I. x² = 625
II. y = √625
(a) x > y
(b) x ≥ y
(c) x < y
(d) x ≤ y
(e) x = y or relation can’t be established.
24. I. 2x – 3y = 0
II. 4x – 2y = 16
(a) x > y
(b) x ≥ y
(c) x < y
(d) x ≤ y
(e) x = y or relation can’t be established.
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
m
gr
II. y = √289
28. I. x 2 − 25x + 156 = 0
II. y 2 − 32y + 255 = 0
29. I. x2 + 23x + 130 = 0
II. y2 + 30y + 224 = 0
30. I. x 2 − 28x + 195 = 0
II. y 2 − 22y + 117 = 0
Directions (41-45):- Each of the following questions is
provided with 2 statements i.e. Statement I & Statement II.
You have to solve them and mark the correct option.
(a) x>y
(b) x<y
(c) x≤y
(d) x≥y
(e) x=y or no relation can be established.
41. I. x² + 8x + 12 = 0
II. 2y² + 14y + 24 = 0
42. I. x² – x – 30 = 0
II. y² – 15y + 56 = 0
43. I. x² + 31x + 150 = 0
II. y² + 54y + 728 = 0
44. I. x² = 256
ou
Directions (26-30):- In each of the following questions,
two equations (I) and (II) are given. Solve the equations
and mark the correct option:
(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. x 2 + 21x + 108 = 0
II. y 2 + 24y + 143 = 0
27. I. x 2 = 289
II. y= √961
38. I. x 2 + 18x + 77 = 0
II. y 2 + 22y + 117 = 0
39. I. 3 x2+25x+50=0
II. 4y2+23y+33=0
40. I. 2x 2 + 17x + 36 = 0
II. 3y 2 + 20y + 32 = 0
p
36. I. x 2 − 31x + 238 = 0
II. y 2 − 37y + 342 = 0
37. I. x 2 + 215 = 1176
3
25. I. x³ = 1331
II. y = √1331
(a) x > y
(b) x ≥ y
(c) x < y
(d) x ≤ y
(e) x = y or relation can’t be established.
@
ce
te
xa
Directions (31-35):- In each of the following questions,
two equations (I) and (II) are given. Solve the equations
and mark the correct option:
(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.
31. I. 6x 2 + 5x + 1 = 0
II. 2y 2 + 5y + 3 = 0
32. I. x 2 = 4
II. y5 = 32
33. I. x 2 − 11x + 30 = 0
II. y 2 − 15y + 56 = 0
34. I. 3 x2−14x+15=0
II.5y2−14y+8=0
35. I. x 2 + 13x + 42 = 0
II. y 2 + 16y + 63 = 0
Directions (36-40):- In each of the following questions,
two equations (I) and (II) are given. Solve the equations
and mark the correct option:
(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.
6
II. y = √256
45. I. x² – 45x + 506 = 0
II. y² – 9y – 360 = 0
Directions (46-50):- In each of the following questions,
two equations (I) and (II) are given. Solve the equations
and mark the correct option:
(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.
46. I. x 2 − 21x + 110 = 0
II. y 2 − 25y + 156 = 0
47. I. x 2 + 29x + 208 = 0
II. y2 + 35y +306= 0
3
48. I. x = √4096
II. y 2 + 121 = 377
49. I. 3 x2+23x+44=0
II. 4y2+33y+65=0
50. I. x 2 + 41x + 418 = 0
II. y 2 + 47y + 550 = 0
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
I. x 2 + 24x = −119
I. (x + y)2 = 361
I. √x + 4 = √225 − √121
I. 223x + 122y = 791
I. 7x 2 − 44x + 69 = 0
II. 3y 2 + 7 = −10y
II. 92442 = y 2 + 92361
II. y 2 + 329 = 473
II. 122x + 223y = 589
II. 3y 2 − 40y + 133 = 0
ou
p
Directions (16-20): In each of these questions, 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 no relation can be established between
x and y.
16. I. 8x²–10x+3=0
II. 5y²+14y–3=0
17. I. 3x²+13x+12=0
II. y²+9y+20=0
18. I. x²–4x–5=0
II. 7y²–25y–12=0
19. I. x³=216
II. 2y²–25y+78=0
20. I. 5x² + 31x + 48 = 0
II. 3y² + 27y + 42 = 0
Directions (21-25): In each of the following questions two
equations are given. Solve these equations and give
answer:
(a) if x ≥ y, i.e. x is greater than or equal to y
(b) if x > y, i.e. x is greater than y
(c) if x ≤ y, i.e. x is less than or equal to y
(d) if x < y, i.e. x is less than y
(e) x = y or no relation can be established between x and
y
21. I. 6x² + 17x + 5 = 0
II. 2y² + 21y + 49 = 0
22. I. x² - 8x + 15 =0
II. 2y² - 5y - 3 = 0
23. I. 5x² + 11x + 2 = 0
II. 4y² + 13y + 3 = 0
24. I. 4x + 2y = 4
II. 3x + 5y = 3
25. I. 6x² + x- 15 =0
II. 4y² - 24y + 35 = 0
@
ce
te
xa
m
Directions (6-10): In each of these questions, two
equations I. and II. are given. You have to solve both the
equations and answer the following questions.
6. I. x 3 = 2744
II. y 2 = 324
(a) x = y or no relation.
(b) x < y
(c) x ≤ y
(d) x > y
(e) x ≥ y
7. I. (5x − 7)2 = 4 − x(3x − 1)
II. (4y − 3)2 = y(4y − 1) − 1
(a) x = y or no relation.
(b) x < y
(c) x ≤ y
(d) x > y
(e) x ≥ y
8. I. 10x 2 − 29x + 21 = 0
II. 2y 2 − 19y + 45 = 0
(a) x = y or no relation.
(b) x < y
(c) x ≤ y
(d) x > y
(e) x ≥ y
9. I. x 2 + 13x + 42 = 0
II. y 2 + 8y + 12 = 0
(a) x = y or no relation.
(b) x < y
(c) x ≤ y
(d) x > y
(e) x ≥ y
11.
12.
13.
14.
15.
gr
Directions (1-5): In each question two equations
numbered I. and II. are given. You have to solve both the
equations and mark appropriate 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.
1. I. 2x² - 17x + 36 = 0
II. 3y² - 22y + 40 = 0
2. I. x² + 21x + 108 = 0
II. y² + 14y + 48 = 0
3. I. 2x² + 7x – 60 = 0
II. 3y² - 28y + 64 = 0
4. I. x² - 2x – 24 = 0
II. y² + 3y – 40 = 0
5. I. 4x² + 27x + 45 = 0
II. 5y² + 42y + 88 = 0
1
11
x
36x
10. I. 1 = (2 −
)
(a) x = y or no relation.
(b) x < y
(d) x > y
II. (
14y
3
9
+ ) = 13
y
(c) x ≤ y
(e) x ≥ y
Directions (11-15): In each of these questions, two
equations 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 between x and y cannot be
established.
7
Directions (26-30): In each of these questions, two
equations I. and II. are given. You have to solve both the
equations and give answer.
26. I. x 2 − 11x + 30 = 0
II. 56y 2 − 151y + 99 = 0
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or no relation.
27. I. x 2 − 4√3(√3 + 1)x + 48√3 = 0
II. y 2 − 2√5(√5 + 2)y + 40√5 = 0
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or no relation.
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
x
Directions (41-45): Solve the given quadratic equations
and mark the correct option based on your answer—
41. I. x² + 9x = 25x – 63
II. 4y² – 34y + 72 = 0
(a) x = y or no relation can be established between x
and y.
(b) x ≤ y
(c) x < y
(d) x > y
(e) x ≥ y
)=0
7
II. 4 (3y + ) + 37 = 0
y
(a) x > y
(c) x ≥ y
(e) x = y or no relation.
29. I. (x + 15)2 = (y + 19)2
II. x 2 − y 2 = 112
(a) x > y
(c) x ≥ y
(e) x = y or no relation.
1
5
8
y
21
2
2y
=5
(b) x < y
(d) x ≤ y
(b) x < y
(d) x ≤ y
42. I.
II. +
(a) x > y
(c) x ≥ y
(e) x = y or no relation.
x
= –x + 14
II. 30% o f 70y = y² + 90
(a) x ≥ y
(b) x > y
(c) x = y or no relation can be established between x
and y. (d) x ≤ y
(e) x < y
43. I. 6x + 7y = 15
II. 3x + 14y = 19.5
(a) x > y
(b) x = y or no relation can be established between x
and y.
(c) x ≤ y
(d) x < y
(e) x ≥ y
44. I. 7x² + 5x – 18 = 0
II. 3y² + 4y – 20 = 0
(a) x > y
(b) x ≤ y
(c) x = y or no relation can be established between x
and y.
(d) x ≥ y
(e) x < y
45. I. x² + 5x = 5 (2x + 3x)
II. 3y² + 2y = 2(y + 6)
(a) x > y
(b) x ≥ y
(c) x < y
(d) x ≤ y
(e) x = y or no relation can be established between x
and y.
ou
30. I. 3 + x2 = 3x
20% of 225
p
20
x
(b) x < y
(d) x ≤ y
@
ce
te
xa
m
Directions (31-35): In each of these questions, 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 no relation can be established between
x and y.
31. I. 5x 2 − 31x + 30 = 0
II. 8y 2 − 12y + 4 = 0
32. I. 7x 2 − 17x + 6 = 0
II. 5y 2 − 24y + 16 = 0
33. I. 13x 2 + 9x − 4 = 0
II. 2y 2 + y − 3 = 0
34. I. 3x 2 − 60x + 288 = 0
II. 4y 2 − 50y + 156 = 0
35. I. 15x 2 + 10x − 5 = 0
II. 6y 2 + 2y − 4 = 0
gr
1
28. I. 21 + (43 +
Directions (36-40): In each of these questions, 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 or no relation can be established between
x and y.
(d) if y > x
(e) if y ≥ x
36. I. x² – 12x + 32 = 0
II. y² – 20y + 96 = 0
37. I. 2x² – 3x – 20 = 0
II. 2y² + 11y + 15 = 0
38. I. x² – x – 6 = 0
II. y² – 6y + 8 = 0
39. I. x² + 14x - 32 = 0
II. y² – y – 12 = 0
40. I. x² – 9x + 20 = 0
II. 2y² – 12y + 18 = 0
8
Directions (46-50): In the following questions, two
equations I. and II. are given. You have to solve both the
equations and mark the appropriate option.
46. I. 12x 2 − 16x + 5 = 0
II. 30y 2 − 61y + 30 = 0
(a) x < y
(b) x > y
(c) x ≤ y
(d) x ≥ y
(e) x = y or no relation.
47. I. x 2 − 16x + 63 = 0
II. y 2 − 12y + 35 = 0
(a) x < y
(b) x > y
(c) x ≤ y
(d) x ≥ y
(e) x = y or no relation.
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
Directions (51-55): Solve the given quadratic equations
and mark the correct option based on your answer.
(b) x > y
(d) x ≥ y
(a) x > y
(b) x ≥ y
(c) x < y
(d) x ≤ y
(e) x = y or no relation can be established between x and y.
(b) x > y
(d) x ≥ y
51. I. x² + x – 12 = 0
52. I. 6x² – 5x + 1 = 0
II. y² + 2y-15 = 0
II. 3y² + 8y = 3
53. I. 12x² – 7x + 1 = 0
II. 6y² – 5y + 1 = 0
54. I. x² + 7x + 10 = 0
II. 2y² + 5y + 2 = 0
55. I. x² – 2x = 15
II. y² + 5y + 4 = 0
Mains Questions
2. Quantity I : Days after which A and B meet. A and B set
out to meet each other from two places 165 km apart.
A travels 15 km the first day, 14 km second day, 13 km
the third day and so on, B travels 10 km the first, 12 km
the second day, 14 km the third day and so on.
Quantity II: Number of days required to complete the
whole work if A, B and C can complete a piece of work
in 10, 12 and 15 days respectively. A left the work 5
days before the work was completed and B left 2 days
after A had left.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
4. Quantity I: Percentage profit earned by the
shopkeeper if at the time of selling and purchasing he
uses weights 10% less and 20% more per kilogram
respectively and proffesses to all goods at 5% profit.
Quantity II: ‘x’ ; A book was sold for a certain sum and
there was a loss of 20%. Had it been sold for Rs 12
more, there would have been a gain of 30%. ‘x’ would
be value of profit percent if the book were sold for Rs
4.8 more than what it was sold for.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
gr
1. In a two digit number, digit at unit place exceeds, the
digit in its tens place by 2 and the product of the
required number with the sum of its digit is equal to
144.
Quantity I: Value of two digit number
Quantity II: 26
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
3. Quantity I: Present age of Randy, if 10 years are
subtracted from the present age of Randy, then you
would get twelve times of the present age of his
grandson Sandy and Sandy is 19 years younger to
Sundar whose age is 24.
Quantity II: Average age of the remaining persons in
the group if average age of group of 14 persons is 27
years and 9 months. Two persons, each 42 years old,
left the group.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
@
ce
te
xa
m
Directions (1-20): In the given questions, two quantities
are given, one as Quantity I and another as Quantity II. You
have to determine relationship between two quantities
and choose the appropriate option
p
(b) x > y
(d) x ≥ y
ou
48. I. 32x 2 + 44x + 15 = 0
II. 42y 2 + 53y + 15 = 0
(a) x < y
(c) x ≤ y
(e) x = y or no relation.
49. I. (x − 2)2 = x − 2
II. 9y 2 − 36y + 35 = 0
(a) x < y
(c) x ≤ y
(e) x = y or no relation.
50. I. 18x 2 + 39x + 20 = 0
II. 10y 2 + 29y + 21 = 0
(a) x < y
(c) x ≤ y
(e) x = y or no relation.
9
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
p
ou
10. PBA and PDC are two secants. AD is the diameter of the
circle with centre at O. ∠A= 40°, ∠P = 20°
Quantity 1: ∠DBC
Quantity 2: ∠ADB
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
m
6. Quantity I: Overall profit percentage if the cost prices
of two shirts are equal. One shirt is sold for 20% profit
and the other is sold for 10% loss.
Quantity II: Profit % made in selling each meter if the
profit made in selling 20 m of a cloth equals the cost
price of 5 m of that cloth.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
9. P can complete a piece of work in 16 days which Q can
complete in 32 days. P and Q work on alternate days.
Quantity I: Time taken by them to complete the work
if P starts on day 1.
Quantity II: time taken by them to complete the work
if Q starts on day 1.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
gr
5. A group consist of 4 couples in which each of the 4
persons have one wife
Quantity I : Number of ways in which they could be
arranged in a straight line such that the men and
women occupy alternate positions
Quantity II: Eight times the number of ways in which
they be seated around circular table such that men and
women occupy alternate position.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
@
ce
te
xa
7. The largest possible right circular cylinder is cut out
from a wooden cube of edge 7 cm.
Quantity I: volume of the cube left over after cutting
out the cylinder
Quantity II: Surface area of cube remained after
cutting out the cylinder.
Note: compare the magnitudes of both quantities.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
8. Quantity I: Value of y. A vessel contains 2.5 liters of
water and 10 liters of milk. 20% of the contents of the
vessel are removed. To the remaining contents, x liters
of water are added to reverse the ratio of water and
milk. Then y liters of milk are added again to reverse
the ratio of water and milk.
Quantity II: 120 ltr.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
10
11. Quantity I: Height of the tank if the volume of
cylindrical tank is 12320 cubic cm. Its radius and
height are in the ratio of 7 : 10 respectively.
Quantity II: Level kerosene in the jar. A conical vessel
of base radius 2 cm and height 3 cm is filled with
kerosene. This liquid leaks through a hole in the
bottom and collects in a cylindrical jar of radius 2 cm.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
12. I. P2 − 18p + 77 = 0
II. 3q2 − 25q + 28 = 0
Quantity I: Value of P
Quantity II: Value of q
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
17. Quantity I: Time taken by Manoj and Shubham to
complete the work together. When Manoj works alone,
he takes 4.8 hrs more than the time taken by Manoj and
Shubham to complete the work together. When
Shubham works alone, he takes 10.8 hrs more than the
time taken by both of them to complete the work
together.
Quantity II: 7.4 hrs.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
14. If 10 men and 15 women complete a piece of work in 8
days while 12 men and 8 women can complete the
same piece of work in 10 days. If A boy who is 50% less
efficient than the man, can do the same work in 50
days.
Quantity I : Time taken by 2 men, 4 women and 18
boys to complete the work.
Quantity II : Time taken by 9 men, 3 women and 6
boys to complete the same work.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
18. The edge of the cube is 10 cm. Given E is the center of
the semicircle and it is mid-point of the diagonal of the
given cube. (2 marks)
gr
ou
p
13. A man who swim 48m/minute in still water, swims
200m against the current and 200m with the current.
The difference between the time taken by him against
the stream and with the stream is 10 minutes.
Quantity I: speed of current.
Quantity II: Speed of a man who completes 3 rounds of
a circular path of radius 49 m in 14 minutes.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
m
Quantity I: Area of the shaded region
Quantity II: 10 cm2.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
te
xa
15. Babu generally starts from his home at certain time
with a certain speed to pick up his girlfriend from office
at 5 : 00 PM. One day his girlfriend left the office at 3 :
00 PM and starts walking to home with a speed of 40
km/hr and meet Babu in the way who left his home at
his usual time. They reached home 40 min. Earlier than
their usual time.
Quantity I : Speed of boy.
1
Quantity II : 492 2 % of speed of girl.
@
ce
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
2
16. If the cost price of the article is 79 7 % of the mark price
and there is a discount of Rs. 68 on the marked price.
There is a profit of 20% on selling the item.
Quantity I : CP of the article
Quantity II : 1111 Rs.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
11
19. Quantity I: Original duration of flight. In a flight of
3000 km an aircraft was slowed down by bad weather.
Its average speed for the trip was reduced by 100
km/hr. and the time increased by one hour.
Quantity II: Usual time of a man who, when walks at
3
th of his usual pace, reaches his office 20 minutes late.
4
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
20. Wheels of diameters 7 cm and 14 cm start rolling
simultaneously from X and Y, which are 1990.50 cm
apart, towards each other. Both of them make same no.
of revolutions per second. Both of them meet after 10s.
Quantity I: speed of smaller wheel.
Quantity II: 21π cm/s
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I ≤ Quantity II
(e) Quantity I = Quantity II or No relation
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
1
1
23. I. x = √(36)2 × (1296)4
II. x – y + z = 0
II. 2x + 8z = 80
+
x − 11
3x + 7
3x
x+7
+
= 12
x
x − 11
x
= 14 II.
=7
II.
y
−
18y − 5
(18y − 5)
y
y+8
y+8
4y
II.
+
4y − 13
@
ce
29. I.
x+7
x
−
y
=2
= 16
y
4y − 13
te
28. I.
3x
3x + 7
x
+
4y
=9
Direction (1 – 6): In each question two equations
numbered (I) and (II) are given. You should solve both the
equations and mark appropriate answer.
(a) If x=y or no relation can be established
(b) If x>y
(c) If x<y
(d) If x≥y
(e) If x≤y
1. I. x2 = 256
II. y2 – 17y + 16 = 0
2. I. x2 + 20x + 100 =0
II. y2 + 13y + 30 = 0
3. I. 4x² - 8x – 5 = 0
II. 2y² - 11y + 14 = 0
4. I. 6x² + 5x + 1 = 0
II. 20y² + 9y + 1 = 0
12
Direction (36-40): Two equations (I) and (II) are given in
each question. On the basis of these equations you have to
decide the relation between ‘x’ and ‘y’ and give answer
(a) if x > y
(b) if x ≥ y
(c) if x < y
(d) if x ≤ y
(e) if x = y or relation cannot be established.
36. I. 36x 2 + 47 √7x + 105 = 0
xa
Direction (27-29): Two equations (I) and (II) are given in
each question. On the basis of these equations you have to
decide the relation between ‘x’ and ‘y’ and give answer
(a) if x > y
(b) if x ≥ y
(c) if x < y
(d) if x ≤ y
(e) if x = y or Relation cannot be established.
27. I.
p
gr
II. 5x + 6y = 99
m
II. 2y + 3z = 33
III. 6y + 5z = 71
24. I. 8x + 7y = 135
III. 9y + 8z = 121
25. I. (x + y)3 = 1331
III. xy = 28
26. I. x + 3y + 4z = 96
III. 2x + 6y = 120
Direction (30-35): Two equations (I) and (II) are given in
each question. On the basis of these equations you have to
decide the relation between ‘x’ and ‘y’ and give answer
(a) if x > y
(b) if x ≥ y
(c) if x < y
(d) if x ≤ y
(e) if x = y or Relation cannot be established.
30. I. 99x 2 + 149x + 56 = 0
II. 156y 2 + 287y + 132 = 0
31. I. 77x 2 + 58x + 8 = 0
II. 42y 2 + 59y + 20 = 0
32. I. 63x 2 + 172x + 117 = 0
II. 30y 2 + 162y + 216 = 0
33. I. 36x 4 + 369x 2 + 900 = 0
II. 144y 4 + 337y 2 + 144 = 0
34. I. 18x 2 – 13 √7x + 14 = 0
II. 32y 2 – 19√6y + 9 = 0
35. I. x 2 – 82x + 781 = 0
II. y 2 – 5041 = 0
ou
Direction (21-26): Three equations (I), (II) and (III) are
given in each question. On the basis of these equations you
have to decide the relation between ‘x’, ‘y’ and 'z' and give
answer
(a) if x < y = z
(b) if x ≤ y< z
(c) if x < y > z
(d) if x = y > z
(e) if x = y = z or if none of the above relationships is
established.
21. I. 7x + 6y + 4z = 122 II. 4x + 5y + 3z = 88
III. 9x + 2y + z = 78
22. I. 7x + 6y = 110
II. 4x + 3y = 59
III. x + z = 15
II. 35y2 + 20 3y + 63 2y + 36 6 = 0
37. I. 91x 2 + 298x + 187 = 0
II. 247y 2 + 216y – 391 = 0
38. I. 81x 2 – 9x – 2 = 0 II. 56y 2 – 13y – 3 = 0
39. I. 391x 2 +1344x + 1073 = 0
II. 437y 2 +1074y + 589 = 0
40. I. 3216x 2 + 3859x + 481 =0
II. 8132y 2 – 4839y + 978 = 0
5. I. 2x² - 9x + 9 = 0
6. I. 4𝑥 2 − 17𝑥 + 15 = 0
II. 6y² - 17y + 12 = 0
II. 2𝑦 2 − 17𝑦 + 35 = 0
SBI PO Prelims 2020
Direction (7–11): In each question two equations
numbered (I) and (II) are given. You have to solve both the
equations and mark appropriate answer.
(a) If x = y or no relation can be established
(b) If x > y
(c) If x < y
(d) If x ≥ y
(e) If x ≤ y
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
Directions (27-31): In the following questions, calculate
quantity I and quantity II, compare them and answer
according to the following options.
(a) If Quantity I > Quantity II
(b) If Quantity I < Quantity II
(c) If Quantity I ≥ Quantity II
(d) if Quantity I ≤ Quantity II
(e) if Quantity I = Quantity II or no relation can be
established
27. Quantity I. Profit earned on selling an article at Rs.
450 at 20% profit
Quantity II. Cost price of the article which is sold at
Rs.84 on 20% profit
m
Directions (17–21): In the following questions, there are
two equations in x and y. 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 there is no relation between x and y
17. I. 2x2 – 3x + 1 = 0
II. 2y2 – 5y + 3 = 0
18. I. x2 + 21x + 110 = 0
II. y2 + 17y + 72 = 0
2
19. I. x = 4
II. y2 – 6y + 8 = 0
20. I. x² + 9x -22 = 0
II. 2y² - 7y + 6 = 0
21. I. 6x² + 5x + 1 = 0
II. 15y² + 11y + 2 = 0
IBPS Clerk Pre 2020
p
Directions (12-16): In the following two equations
questions numbered (I) and (II) are given. You have to
solve both equations and Give answer
(a) If x > y
(b) If x ≥ y
(c) If y > x
(d) If y ≥ x
(e) If x = y or no relation can be established
12. I. x² - 8x + 15 = 0
II. 2y² - 7y + 5 = 0
13. I. 2x² + x – 28 = 0
II. 2y² - 23y + 56 = 0
14. I. 2x² - 7x – 60 = 0
II. 3y² + 13y + 4 = 0
15. I. x² - 17x – 84 = 0
II. y² + 4y – 117 = 0
2
2
16. I. x = 9
II. (y – 8)2 = 9
IBPS PO Prelims 2020
25. I. x 2 − 20x + 91 = 0
II. y 2 + 16y + 63 = 0
(a) x < y
(b) x > y
(c) x ≤ y
(d) x ≥ y
(e) x = y or no relation.
2
26. I. x − x − 12 = 0
II. y 2 + 5y + 6 = 0
(a) x < y
(b) x > y
(c) x ≤ y
(d) x ≥ y
(e) x = y or no relation.
RRB PO Prelims 2020
ou
II. y3 = 125
II. y2 + 15y + 56 = 0
II. y2 = 64
II. y2 = 2y + 35
II. y2 − 13𝑦 + 40 = 0
SBI Clerk Prelims 2020
gr
7. I. 𝑥 = √25
8. I. x2 + 2x – 35 = 0
9. I. x2 = 81
10. I. 17x2 – 14x – 83 = - 80
11. I. x2 + 4x – 45 = 0
te
xa
28. In a village there are 60% males and rest are females.
30% of total male are illiterate and 25% of total female
are illiterate. Number of illiterate males is 1152.
Quantity I. Literate females in the village.
Quantity II. 1940
@
ce
Directions (22-26): In the given questions, two equations
(I) & (II) are given. You have to solve both the equations
and mark the answer accordingly.
22. I. x 2 + 9x + 20 = 0
II. 8y 2 − 15y + 7 = 0
(a) x < y
(b) x > y
(c) x ≤ y
(d) x ≥ y
(e) x = y or no relation.
23. I. x 2 − 11x + 30 = 0
II. y 2 + 12y + 36 = 0
(a) x < y
(b) x > y
(c) x ≤ y
(d) x ≥ y
(e) x = y or no relation.
24. I. x 2 + 13x + 40 = 0
II. y 2 + 7y + 10 = 0
(a) x < y
(b) x > y
(c) x ≤ y
(d) x ≥ y
(e) x = y or no relation.
13
29. A man invested Rs. P at 12% p.a. on simple interest for
two years.
Quantity I. If at the end of second year he gets
Rs.1200 as interest, then find Rs.P.
Quantity II. Rs.6000
30. Ploughing cost of a rectangular field is Rs.288 at the
rate of Rs.3 per square meter. Length of the field is 4
meters more than the width of field.
Quantity I. Length of rectangular field.
Quantity II. 12 meters.
31. Quantity I. Sum of present ages of Shivam and
Prashant is 32 years and Shivam is 8
years older than Prashant. Find present
age of Prashant.
Quantity II. 15 years.
RRB PO Prelims 2020
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
44. Quantity I:
ou
p
A man invested Rs.5900 for 3 years in a
scheme offering R% p.a. at SI and
received Rs.3186 as interest after 3
years. If the man invested Rs.7900 at
(R+5)% p.a. at SI for 3 years, then find
interest received by man (in Rs.).
Quantity II: A man invested Rs.X at 13% p.a. at CI for
2 years and interest received by him
after 2 years is Rs.2325.96. Find X (in
Rs.).
(a) Quantity I < Quantity II
(b) Quantity I ≤ Quantity II
(c) Quantity I > Quantity II
(d) Quantity I ≥ Quantity II
(e) Quantity I = Quantity II or no relation.
Ratio of CP to MP of an article is 19 : 30.
Shopkeeper allowed 24% discount and
earned 20% profit on selling the article.
If SP of the article is Rs.912, then find
difference between amount of profit
earned and amount of discount allowed
(in Rs.).
Quantity II: Shopkeeper marked an article 70%
above its cost price and he allowed 40%
discount on it. If shopkeeper sold the
article at Rs.183.6, then find sum of
amount of profit earned and amount of
discount allowed(in Rs.).
(a) Quantity I < Quantity II
(b) Quantity I ≤ Quantity II
(c) Quantity I > Quantity II
(d) Quantity I ≥ Quantity II
(e) Quantity I = Quantity II or no relation.
te
xa
m
Directions (37-41): In each of these questions, 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 no relation can be established between x and
y.
37. I. x² - 13x + 40 = 0
II. 2y² - y – 15 = 0
38. I. 5x² + 17x + 6 = 0
II. 2y² + 11y + 12 = 0
39. I. 7x² - 19x + 10 = 0
II. 8y² + 2y – 3 = 0
2
40. I. 𝑥 − 8𝑥 + 15 = 0
II. 𝑦 2 − 3𝑦 + 2 = 0
41. I. 3x² –7x + 4 = 0
II. 2y² – 9y + 10 = 0
SBI Clerk Prelims 2019
43. Quantity I:
gr
Direction (32 – 36): Solve the given quadratic equations
and mark the correct option based on your 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.
32. I. x2 − 14𝑥 + 45 = 0
II. y2 – 18𝑦 + 72 = 0
33. I. x2 + 7𝑥 + 12 = 0
II. y2 +9𝑦 + 20 = 0
34. I. 4x2 – 7𝑥 + 3 = 0
II. 7y2− 17y + 6 = 0
35. I. 2x2 - 19x + 45 =0
II. 2y2 -9y + 4 = 0
36. I. x2 = 144
II. (y +12)2 = 0
SBI PO Prelims 2019
@
ce
Directions (42-46): In the following questions, two
quantities (I) and (II) are given. You have to solve both the
quantities and mark the appropriate answer.
42. Quantity I: In how many ways a committee of 4
members with at least 2 women can be
formed from 8 men and 4 women?
Quantity II: How many 3-digit numbers which are
divisible by 3 can be formed from
0,1,2,3,4,5,6,7,8,9, such that 3-digit
number always ends with an even
number?
(a) Quantity I < Quantity II
(b) Quantity I ≤ Quantity II
(c) Quantity I > Quantity II
(d) Quantity I ≥ Quantity II
(e) Quantity I = Quantity II or no relation.
14
45. Quantity I:
A boat can cover distance of 480 km
each in downstream and in upstream in
total 11 hours. If ratio of speed of boat
in still water to that of stream is 11 : 1,
then find speed of boat in still water (in
km/hr.).
Quantity II: A boat can cover a distance of 350 km
in downstream in 3.5 hours and can
cover a distance of 380 km in upstream
in 5 hours. Find speed of boat in still
water (in km/hr.).
(a) Quantity I < Quantity II
(b) Quantity I ≤ Quantity II
(c) Quantity I > Quantity II
(d) Quantity I ≥ Quantity II
(e) Quantity I = Quantity II or no relation.
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
p
55. I. 9x + 3y = 15
II. 4x + 5y = 14
(a) x = y or no relation can be established
(b) x > y
(c) x ≤ y
(d) x < y
(e) x ≥ y
56. I. 2x 2 − x − 1 = 0
II. 3y 2 − 5y + 2 = 0
(a) x ≤ y
(b) x < y
(c) x = y or no relation can be established
(d) x ≥ y
(e) x > y
IBPS Clerk Prelims 2019
Directions (57-61): Solve the given quadratic equations
and mark the correct option based on your answer—
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or there is no relationship
57. I. x² = 81
II. y² – 18y + 81 = 0
58. I. 4x² - 24x + 32 = 0
II. y² - 8y + 15 = 0
59. I. x² - 21x + 108 = 0
II. y² – 17y + 72 = 0
60. I. x² – 11x + 30 = 0
II. y² - 15y + 56 = 0
61. I. x³ = 512
II. y² = 64
RRB PO Prelims 2019
te
xa
m
Directions (47-51): In the following two equations
questions numbered (I) and (II) are given. You have to
solve both equations and Give answer
(a) If x > y
(b) If x ≥ y
(c) If y > x
(d) If y ≥ x
(e) If x = y or no relation can be established
47. I. x² - 7x + 12 = 0
II. y² - 8y + 12 = 0
48. I. 2x² + x – 28 = 0
II. 2y² - 23y + 56 = 0
49. I. 2x² - 7x – 60 = 0
II. 3y² + 13y + 4 = 0
50. I. x² - 17x – 84 = 0
II. y² + 4y – 117 = 0
2
51. I. x = 81
II. (y-9)2 = 0
IBPS PO Prelims 2019
54. I. 2x 2 + 5x + 3 = 0
II. y 2 + 4y − 12 = 0
(a) x ≤ y
(b) x > y
(c) x = y or no relation can be established
(d) x < y
(e) x ≥ y
ou
B’s present age is 60% more than A’s
present age and ratio of present age of
B to that of C is 5 : 2. D is 8 years
younger than B and D’s present age is
twice of that of C. Find average of
present age of A, B, C & D (in years).
Quantity II: Present age of R is equal to average of
present age of P & Q. 4 years hence, age
of P is twice of age of Q at that time. If R
is 15 years younger than P, then find
age of younger person among P, Q & R.
(a) Quantity I < Quantity II
(b) Quantity I ≤ Quantity II
(c) Quantity I > Quantity II
(d) Quantity I ≥ Quantity II
(e) Quantity I = Quantity II or no relation.
SBI Clerk Mains 2019
gr
46. Quantity I:
@
ce
Directions (52–56): Solve the following quadratic
equation and mark the answer as per instructions.
52. I. x 2 − 2x − 143 = 0
II. y 2 − 169 = 0
(a) x > y
(b) x < y
(c) x ≤ y
(d) x ≥ y
(e) x = y or no relation can be established
53. I. x 2 − 7x − 18 = 0
II. y 2 − 19y + 90 = 0
(a) x ≤ y
(b) x = y or no relation can be established
(c) x > y
(d) x ≥ y
(e) x < y
15
Directions (62-65): In each of the following questions,
two equations (I) and (II) are given. Solve the equations
and mark the correct option:
(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.
62. I. 2x 2 + 11x + 12 = 0
II. 8y2 -22y – 21 =0
63. I. x2 -17x -60=0
II. y2 + 42y +185 =0
2
64. I. x + 41x + 420 =0
II. 6y2 -11y -10 =0
65. I. x2 - 8x - 273 =0
II. y2 +6y -432 =0
RRB PO Mains 2019
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
p
74. Quantity I: Amount obtained after two years on Rs.
2
2450 at 14 % simple interest per annum.
7
Quantity II: Amount obtained after two years on
1
investment of Rs 2450 at 12 2 % 𝑝. 𝑎. compounded
yearly.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I = Quantity II or no relation
(e) Quantity I≤ Quantity II
m
71. Quantity I: A train can cross a pole in 24 sec with a
speed of 75 km/h. Length of train.
Quantity II: A train can cross a pole in 12 sec and a
tunnel in 55.2 sec. If length of tunnel is 1800 m. length
of train.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I = Quantity II or no relation
(e) Quantity I≤ Quantity II
73. Quantity I: value of x
(𝑥 + 3)2 = (𝑥 − 3)2 + 𝑥²
Quantity II: value of y
𝑦 2 − 29𝑦 + 204 = 0
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I = Quantity II or no relation
(e) Quantity I≤ Quantity II
ou
Directions (71–75): In the given questions, two quantities
are given one as ‘Quantity I’ and another as ‘Quantity II’.
You have to determine relationship between two
quantities and choose the appropriate option.
(d) Quantity I = Quantity II or no relation
(e) Quantity I≤ Quantity II
gr
Directions (66-70): Given below are two equations in
each question, which you have to solve and give answer
(a) if𝑥 > 𝑦
(b) if𝑥 ≥ 𝑦
(c) if𝑦 > 𝑥
(d) if𝑦 ≥ 𝑥
(e) if𝑥 = 𝑦 or no relation can be established
66. I. 2𝑥 2 − 5𝑥 + 2 = 0
II. 2𝑦 2 − 9𝑦 + 7 = 0
2
67. I. 3𝑥 + 7𝑥 + 4 = 0
II. 𝑦 2 + 9𝑦 + 20 = 0
68. I. 𝑥 2 − 7𝑥 + 10 = 0
II. 𝑦 2 − 14𝑦 + 45 = 0
69. I. x² – 3x = 4
II. y² + 6y + 8 = 0
70. I. x² – 3x = 10
II. y² + 7y + 10 = 0
RRB Clerk Prelims 2019
xa
75. Quantity I: Actual discount percent. If a shopkeeper
gives 1 article free at every purchase of 4 articles and
also give 20% discount.
Quantity II: Final new profit percent. If selling price
2
36
is increased by 14 7 % then profit percent becomes 19
@
ce
te
72. Quantity I: Marked price of article, if article is marked
at 50% above cost price and on selling the article,
profit earned is 20% and S. P is Rs 1020.
Quantity II: Total cost of fencing a square of side 37.5
meter and cost of wire is Rs 0.17 per centimeter.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
Solutions
1. (c): 𝑥 2 + 5x + 6 = 0
/
6 → 3 × 2 → 3 + 2 = 5 ∴ 𝑥1 = – 3, 𝑥2 = –2
2
𝑦 + 7y + 12 = 0
|
12 → 4 × 3 → 4 + 3 = 7
∴ y = – 4, 𝑦2 = –3; ∴ x ≥ y
16
of initial profit percent.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≥ Quantity II
(d) Quantity I = Quantity II or no relation
(e) Quantity I ≤ Quantity II
RRB Clerk Mains 2019
2. (b): 𝑥 2 – 9x + 20 = 0
|
5 × 4 and, 5 + 4 = 9
∴ 𝑥1 = + 5 𝑥2 = + 4;
𝑦 2 + 42 = 13y
𝑦 2 – 13y + 42 = 0
|
42 → 7 × 6 and, 7 + 6 = 13;
∴ 𝑦1 = +7, 𝑦2 = + 6 ; ∴ y > x
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
3. (b): eqn (1) – (eqn (2) × 3)
10. (a): 5x = –11
12 x + 3y = 14
x=
12 x + 6y = 48
–
34
3
–
= 11.33
Put it in equation 1,
5
5𝑦 2 + 17y + 14 = 0
/
\
5
7×2
and, 10 + 7 = 17
12x + 34 = 14 ⇒ 12x = – 20 ⇒ x = – 3; ∴ x < y
4. (b): x = √625 = 25
−10
5. (b): 𝑥 2 + 4x + 4 = 0
|
2 × 2 and, 2 + 2 = 4
𝑥1 = –2, 𝑥2 = – 2
1
m
13. (b): I. 𝑥1 =
= (2 × 14)
11
2
⇒ (𝑦) = (2 × 14)11/2
∴ y = 28 ∴ x = y
@
ce
9. (a): √784x + 1234 = 1486
28x = 252 ⇒ x = 9
33y = 264 ⇒ y = 8 ∴ x > y
1. (e): I. 𝑥 2 + 5𝑥 + 6 = 0
𝑥 2 + 3𝑥 + 2𝑥 + 6 = 0
(𝑥 + 3)(𝑥 + 2) = 0
𝑥 = −2, −3
17
–1
2
= –0.5
–1
43 x =
172
x = 4 , Put it in equation (i)
3y = 28 – 13 ⇒ y = 5 x < y
15. (b): eqn (1) × 5 – eq (2) × 2
te
8. (e): 9 + 19 = x ⇒ x = 28
= –1, 𝑥2 =
2
–4
14. (b): eq(i) × 4 + eq(ii) × 3.
28x – 12y = 52
15x + 12y = 120
xa
𝑦 2 = 169 ⇒ y = ± 13 ∴ x ≥ y
–2
II. 𝑦1 = 12 = –0.33, 𝑦2 = 𝟒 = –0.25 ∴ y > x
𝑦 2 – 25y + 156 =0
/
13 × 12 and, 13 +12 = 25
∴ 𝑦1 = 13, 𝑦2 = 12 ∴ y ≥ x
7. (c): 𝑥 3 = 1729 + 468
𝑥 3 = 2197 ⇒ x = 13
12. (a): x = (1331)3 = 11
2𝑦 2 – 17y + 36 = 0
/
\
2×1
9×4
and, 9 + 8 = 17
9
8
∴ 𝑦1 = 2 = 4.5, 𝑦2 = 2 = 4 ∴ x > y
gr
6. (d): x2 – 19x + 84 = 0
/
12 × 7 and, 12 + 7 = 19
∴ 𝑥1 = + 12, 𝑥2 = +7
(𝑦)
–7
∴ 𝑦1 = 5 = –2, 𝑦2 = 5 = – 1.4 ∴ x > y
𝑦 2 – 8y + 16 = 0 ⇒ 𝑦1, 𝑦2 = +ve; ∴ y > x
11
2
ou
y = √676 = 26; ∴ x < y
1
5+
2
= – 2.2
y – 5y = 12 ⇒ y = – 3 ∴ x > y
11. (a): 6𝑥 2 –49x + 99 = 0
\
/
2×3
11 × 9
and, 22 + 27 = 49
22 11
27 9
∴ 𝑥1 = 6 = 3 = 3.67, 𝑥2 = 6 = 2 = 4.5
– 3 y = – 34
y=
5
p
–
− 11
10x + 25y = 30
10x + 22y = 18
–
–
–
3 y = 12
y=4
∴ 5x = 9 – 44 , x = –7; x < y
Prelims Solutions
II. 𝑦 2 + 9𝑦 + 14 = 0
𝑦 2 + 7𝑦 + 2𝑦 + 14 = 0
(𝑦 + 2)(𝑦 + 7) = 0
𝑦 = −2, −7
Clearly, no relation can be established
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
2. (b): I. 𝑥 2 − 18𝑥 + 45 = 0
𝑥 2 − 15𝑥 − 3𝑥 + 45 = 0
(𝑥 − 3)(𝑥 − 15) = 0
𝑥 = 3,15
II. 𝑦 2 + 12𝑦 − 45 = 0
𝑦 2 + 15𝑦 − 3𝑦 − 45 = 0
(𝑦 − 3)(𝑦 + 15) = 0
𝑦 = 3, −15
Clearly, 𝑥 ≥ 𝑦
7. (d): I. 𝑥 2 = 196
x =√196
x =±14
II. 𝑦 = √196
y =14
So,x ≤ y
ou
3. (e): I. 9𝑥 2 + 11𝑥 + 2 = 0
9𝑥 2 + 9𝑥 + 2𝑥 + 2 = 0
(9𝑥 + 2)(𝑥 + 1) = 0
9
II. 8𝑦 2 + 6𝑦 + 1 = 0
8𝑦 2 + 4𝑦 + 2𝑦 + 1 = 0
(4𝑦 + 1)(2𝑦 + 1) = 0
1
1
2
4
Clearly, no relation can be established
4. (c): I. 6𝑥² + 5𝑥 + 1 = 0
6𝑥 2 + 3𝑥 + 2𝑥 + 1 = 0
(3𝑥 + 1)(2𝑥 + 1) = 0
3
1
1
𝑦 = −4,4
Clearly, 𝑥 < 𝑦
xa
𝑥 = −3,−2
II. 4𝑦 2 – 15𝑦 = 4
4𝑦 2 − 16𝑦 + 𝑦 − 4 = 0
(4𝑦 + 1)(𝑦 − 4) = 0
@
ce
te
5. (c): I. 𝑥 2 + 3𝑥 = 0
𝑥(𝑥 + 3) = 0
𝑥 = 0, −3
II. 𝑥 2 + 𝑦 = 10
𝑦 = 10 − 𝑥 2
𝑖𝑓 𝑥 = 0, 𝑦 = 10
𝑖𝑓 𝑥 = −3, 𝑦 = 10 − (−3)2 = 1
Clearly, 𝑥 < 𝑦
6. (c): I. 𝑥 2 − 25𝑥 + 156 = 0
x2 -12x -13x +156=0
x(x-12)-13(x-12)=0
(x-12)(x-13)=0
x =12,13
II. 𝑦 2 − 29𝑦 + 210 = 0
y2-14y -15y +210=0
y(y-14) -15(y-14)=0
(y-14)(y-15)=0
y=14,15
So, x<y
18
II. 𝑦² + 15𝑦 + 56 =0
y2 +8y +7y +56 =0
y(y +8)+7(y+8)=0
(y+7)(y+8)=0
y =-7,-8
So, x > y
m
1
9. (a): I. 3𝑥² + 23𝑥 + 30 = 0
3x2 +18x +5x +30 =0
3x(x+6) +5(x+6)=0
(3x+5)(x+6)=0
5
x= -6,−
gr
2
𝑥 = − , −1
𝑦 = − ,−
p
8. (e): I. 𝑥 2 + 12𝑥 + 35 = 0
x2+5x+7x+35=0
x(x+5)+7(x+5)=0
(x+5)(x+7)=0
x= -5,-7
II. 𝑦 2 + 14𝑦 + 48 = 0
y2+ 6y+8y+48 =0
y(y+6)+8(y+6)=0
(y+8)(y+6)=0
y = -8,-6
So, no relation.
10. (c): I. 𝑥 2 + 17𝑥 + 72 = 0
x2+ 8x+9x+72=0
x(x+8)+9(x+8)=0
(x+9)(x+8)=0
x = -8,-9
II. 𝑦 2 + 13𝑦 + 42=0
y2 +6y+7y+42 =0
y(y+6)+7(y+6)=0
(y+6)(y+7)=0
y = -6,-7
So,x<y
11. (c): I. 𝑥 2 + 17𝑥 + 72 = 0
x2 +8x +9x +72=0
x(x+8)+9(x+8)=0
(x+8)(x+9)=0
x = -8,-9
II. 𝑦 2 + 11𝑦 + 30 = 0
y2+5y +6y +30=0
y(y+5) +6(y+5)=0
(y+5)(y+6)=0
y= -5,-6
So,x<y
Adda247 Publications
For More Study Material
Visit: adda247.com
12. (a): I. 3𝑥 2 − 23𝑥 + 40 = 0
3x2 -15x -8x +40=0
3x(x-5)- 8(x-5)=0
(x-5)(3x-8)=0
16. (d): I. 𝑥 2 + 23𝑥 + 132 = 0
x2 +12x +11x +132=0
x(x+12)+11(x+12)=0
(x+11)(x+12)=0
x = -11,-12
II. 𝑦 2 + 21𝑦 + 110 = 0
y2+11y +10y +110=0
y(y+11) +10(y+11)=0
(y+10)(y+11)=0
y= -10,-11
So, x ≤ y
8
x = 5, 3
II. 5𝑦 2 − 17𝑦 + 14 = 0
5y2-10y -7y +14=0
5y(y-2) -7(y-2)=0
(y-2)(5y-7)=0
7
y= 2, 5
ou
So,x>y
17. (e): I. 3𝑥 2 + 20𝑥 + 32 = 0
3x2 +12x +8x +32=0
3x(x+4)+ 8(x+4)=0
(3x+8)(x+4)=0
2
13. (e): I. 𝑥 − 26𝑥 + 168 = 0
x2-12x-14x+168=0
x(x-12)-14(x-12)=0
(x-12)(x-14)=0
x= 12,14
II. 𝑦 2 − 29𝑦 + 208 = 0
y2-13y-16y+208 =0
y(y-13)-16(y-13)=0
(y-13)(y-16)=0
y = 13,16
So, no relation.
3
@
ce
15. (c): I. 𝑥 2 + 48𝑥 + 575 = 0
x2+ 23x+25x+575=0
x(x+23)+25(x+23)=0
(x+23)(x+25)=0
x = -23,-25
II. 𝑦 2 + 44𝑦 + 483=0
y2 +21y+23y+483 =0
y(y+21)+23(y+21)=0
(y+21)(y+23)=0
y = -21,-23
So,x ≤ y
19
gr
8
3
m
II. 5𝑦 2 + 23𝑦 + 24 = 0
5y2+15y +8y +24=0
5y(y+3) +8(y+3)=0
(y+3)(5y+8)=0
te
x =√2197
=13
II. 𝑦² + 23 =192
y2=192-23
= 169
y =±13
So, x ≥ y
x = -4, −
y= -3, −
8
5
So, No relation exists
18. (a): I. 𝑥 2 − 29𝑥 + 208 = 0
x2-13x-16x+208=0
x(x-13)-16(x-13)=0
(x-16)(x-13)=0
x= 16,13
II. 𝑦 2 − 21𝑦 + 108 = 0
y2-9y-12y+108 =0
y(y-9)-12(y-9)=0
(y-12)(y-9)=0
y = 9,12
So, x > y
xa
14. (b): I. x3 +340=2537
x3=2537-340
=2197
p
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
19. (a): I. 𝑥 2 + 30𝑥 + 224 = 0
x2+14x+16x+224=0
x(x+14)+16(x+14)=0
(x+16)(x+14)=0
x= -16,-14
II. 𝑦 2 + 35𝑦 + 306 = 0
y2+18y+17y+306 =0
y(y+18)+17(y+18)=0
(y+18)(y+17)=0
y = -18,-17
So, x > y
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
27. (d): I. x2=289
x = ±17
II. y =√289
y =17
So, x ≤ 𝑦
3
20. (b): I. x=√4096
x =16
II. y2 =256
y = √256
= ±16
So, x ≥ y
ou
21. (a): I. 2x² + 10x + 12 = 0
2x² + 6x + 4x + 12 = 0
(2x + 4)(x + 3)= 0
x= –3, –2
II. y² + 10y + 25 = 0
y² +5y +5y + 25 = 0
(y + 5)(y + 5) = 0
y = -5
∴x>y
22. (a): I. x² – 3x – 2x + 6 =0
(x – 3)(x – 2) = 0
x = +3, + 2
II. y² + 6y + y + 6 = 0
(y + 1)(y + 6) = 0
y = –1, –6
∴x>y
@
ce
25. (e): I. x = +11
y = +11
∴x=y
gr
m
te
24. (a): (I) × 2 – (II)
–6y + 2y = –16
y=4
x=6
x > y.
29. (a): I. 𝑥 2 + 23𝑥 + 130 = 0
x2+13x+10x+130=0
x(x+13)+10(x+13)=0
(x+13)(x+10)=0
x= -13,-10
II. 𝑦 2 + 30𝑦 + 224 = 0
y2+16y+14y+224 =0
y(y+16)+14(y+16)=0
(y+16)(y+14)=0
y = -16,-14
So, x > y
xa
23. (d): I. x = ± 25
II. y = +25
∴x≤y
p
28. (c): I. 𝑥 2 − 25𝑥 + 156 = 0
x2-12x-13x+156=0
x(x-12)-13(x-12)=0
(x-13) (x-12)=0
x= 12, 13
II. 𝑦 2 − 32𝑦 + 255 = 0
y2-15y-17y+255 =0
y(y-15)-17(y-15)=0
(y-15) (y-17)=0
y = 15, 17
So, x < y
26. (e): I. 𝑥 2 + 21𝑥 + 108 = 0
x2 +9x +12x +108=0
x(x+9)+12(x+9)=0
(x+9)(x+12)=0
x = -9,-12
II. 𝑦 2 + 24𝑦 + 143 = 0
y2+11y +13y +143=0
y(y+11) +13(y+11)=0
(y+13)(y+11)=0
y= -13,-11
So, No relation
30. (b): I. 𝑥 2 − 28𝑥 + 195 = 0
x2-13x-15x+195=0
x(x-13)-15(x-13)=0
(x-13)(x-15)=0
x= 13,15
II. 𝑦 2 − 22𝑦 + 117 = 0
y2-13y-9y+117 =0
y(y-13)-9(y-13)=0
(y-13)(y-9)=0
y = 13,9
So, x ≥ y
31. (a): I. 6𝑥 2 + 5𝑥 + 1 = 0
6x2 +3x +2x +1=0
3x(2x+1)+1(2x+1)=0
(3x+1)(2x+1)=0
1 1
x = -3,-2
II. 2𝑦 2 + 5𝑦 + 3 = 0
2y2+2y +3y +3=0
2y(y+1) +3(y+1)=0
(2y+3)(y+1)=0
3
y= - 2,-1
So, x > y
20
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
33. (c): I. 𝑥 2 − 11𝑥 + 30 = 0
x2-5x-6x+30=0
x(x-5)-6(x-5)=0
(x-5) (x-6)=0
x= 5, 6
II. 𝑦 2 − 15𝑦 + 56 = 0
y2-7y-8y+56 =0
y(y-7)-8(y-7)=0
(y-7) (y-8)=0
y = 7, 8
So, x < y
ou
38. (e): I. 𝑥 2 + 18𝑥 + 77 = 0
x2+11x+7x+77=0
x(x+11)+7(x+11)=0
(x+11) (x+7)=0
x= -11, -7
II. 𝑦 2 + 22𝑦 + 117 = 0
y2+9y+13y+117 =0
y(y+9)+13(y+9)=0
(y+9) (y+13)=0
y = -9, -13
So, No relation
gr
34. (e): I. 3𝑥 2 − 14𝑥 + 15 = 0
3x2-9x-5x+15=0
3x(x-3)-5(x-3)=0
(x-3)(3x-5)=0
39. (c): I. 3𝑥 2 + 25𝑥 + 50 = 0
3x2+15x+10x+50=0
3x(x+5)+10(x+5)=0
(3x+10)(x+5)=0
5
@
ce
te
35. (b): I. 𝑥 2 + 13𝑥 + 42 = 0
x2+6x+7x+42=0
x(x+6)+7(x+6)=0
(x+6)(x+7)=0
x= -6, -7
II. 𝑦 2 + 16𝑦 + 63 = 0
y2+9y+7y+63 =0
y(y+9)+7(y+9)=0
(y+9)(y+7)=0
y = -7,-9
So, x ≥ y
xa
5
So, No relation
m
x= 3, 3
II. 5𝑦 2 − 14𝑦 + 8 = 0
5y2-10y-4y+8 =0
5y(y-2)-4(y-2)=0
(y-2)(5y-4)=0
4
y = 2,
36. (c): I. 𝑥 2 − 31𝑥 + 238 = 0
x2 −17x −14x +238=0
x(x-17) −14(x−17)=0
(x-14)(x-17)=0
x = 14,17
II. 𝑦 2 − 37𝑦 + 342 = 0
y2-18y -19y +342=0
y(y-18) -19(y-18)=0
(y-18)(y-19)=0
y= 18,19
So, x < y
21
p
37. (d): I. 𝑥 2 + 215 = 1176
x2 =961
x =± 31
II. y = √961
=31
So, x ≤ 𝑦
32. (d): I. x2=4
x = ±2
II. y5 = 32
y =2
So, x ≤ 𝑦
10
x= -5,- 3
II. 4𝑦 2 + 23𝑦 + 33 = 0
4y2+12y+11y+33 =0
4y(y+3)+11(y+3)=0
(y+3)(4y+11)=0
11
y = -3, - 4
So, x < 𝑦
40. (d): I. 2𝑥 2 + 17𝑥 + 36 = 0
2x2+8x+9x+36=0
2x(x+4)+9(x+4)=0
(x+4)(2x+9)=0
x= -4, -4.5
II. 3𝑦 2 + 20𝑦 + 32 = 0
3y2+12y+8y+32 =0
3y(y+4)+8(y+4)=0
(y+4)(3y+8)=0
y = -4,-2.67
So, x ≤ y
41. (e): I. x² + 6x + 2x + 12 = 0
(x + 6) (x + 2) = 0
x = –2, –6
II. 2y² + 8y + 6y + 24 = 0
(y + 4) (2y + 6) = 0
y = –3, –4
Hence, no relation can be established.
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
47. (a): I. 𝑥 2 + 29𝑥 + 208 = 0
42. (b): I. x² – 6x + 5x – 30 = 0
(x – 6) (x + 5) = 0
x = 6, –5
x2+16x+13x+208=0
II. y² – 8y – 7y + 56 = 0
(x+16) (x+13)=0
x (x+16)+13(x+16)=0
(y – 8) (y – 7) = 0
x= -16, -13
y = 7, 8
II. 𝑦 2 + 35𝑦 + 306 = 0
Hence, y > x
y2+17y+18y+306 =0
(y+18) (y+17)=0
(x + 25) (x + 6) = 0
y = -17, -18
x = –6, –25
(y + 26) (y + 28) = 0
3
48. (b): I. x = √4096
y = –26, –28
x= 16
Hence, x > y
II. y2 + 121 = 377
y2 = 256
gr
44. (c): I. x² = 256
y = ± 16
x = ± √256
So, x ≥ y
x = +16, –16
II. y = √256
49. (e): I. 3𝑥 2 + 23𝑥 + 44 = 0
m
y = 16
Hence y ≥ x
x² – 23x - 22x + 506 = 0
(x – 23) (x - 22) = 0
x = 22, 23
II. y² – 9y – 360 = 0
te
y² – 24y + 15y – 360 = 0
(y – 24) (y + 15) = 0
xa
45. (e): I. x² – 45x + 506 = 0
@
ce
46. (c): I. 𝑥 2 − 21𝑥 + 110 = 0
(3x+11)(x+4)=0
11
x= -4,- 3
4y(y+5)+13(y+5)=0
(y+5)(4y+13)=0
13
y = -5, - 4
So, No relation
50. (b): I. 𝑥 2 + 41𝑥 + 418 = 0
x2+19x+22x+418=0
x(x+19)+22(x+19)=0
x(x-11) −10(x−11)=0
(x+19)(x+22)=0
(x-11)(x-10)=0
II. 𝑦 2 − 25𝑦 + 156 = 0
3x(x+4)+11(x+4)=0
4y2+20y+13y+65 =0
x2 −11x −10x +110=0
x =11,10
3x2+12x+11x+44=0
II. 4𝑦 2 + 33𝑦 + 65 = 0
Hence, no relation.
x= -19, -22
II. 𝑦 2 + 47𝑦 + 550 = 0
y2+22y+25y+550 =0
y2-13y -12y +156=0
y(y+22)+25(y+22)=0
y(y-13) -12(y-13)=0
(y+22)(y+25)=0
(y-13)(y-12)=0
y = -22,-25
y= 13,12
So, x ≥ y
So, x < y
22
ou
So, x > y
II. y² + 26y + 28y + 728 = 0
y = –15, 24
p
y(y+17)+18(y+17)=0
43. (a): I. x² + 25x + 6x + 150 = 0
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
1. (c): I. 2x² - 17x + 36 = 0
2x² - 8x – 9x + 36 = 0
2x(x – 4) – 9 (x – 4) = 0
(2x – 9) (x – 4) = 0
5. (b): I. 4x² + 27x + 45 = 0
4x² + 12x + 15x + 45 = 0
4x(x + 3) + 15 (x + 3) = 0
(4x + 15) (x + 3) = 0
9
x= 2 ,4
𝑥=
II. 3y² - 22y + 40 = 0
3y² - 12y – 10y + 40 = 0
3y(y – 4) – 10 (y – 4) = 0
(y – 4) (3y -10) = 0
4
p
ou
10
−22
𝑦 = −4, 5
𝑥≥𝑦
𝑥>𝑦
2. (a): I. x² + 21x + 108 = 0
x² + 9x + 12x + 108 = 0
x(x + 9) + 12 (x + 9) = 0
(x + 12) (x + 9) = 0
x= -12, -9
II. y² + 14y + 48 = 0
y² + 6y+ 8y+ 48 = 0
y(y + 6) + 8 (y + 6) = 0
(y + 8) (y + 6) = 0
y = -8, -6
y>x
gr
6. (a): I. x³ = 2744
x = 14
II. y² = 324
y = ± 18
So, no relation can be established between x &
y.
xa
m
7. (e): I. (5x – 7)² = 4 – x (3x – 1)
25x² + 49 – 70x = 4 – 3x² + x
⇒ 28x² – 71x + 45 = 0
28x² – 36x – 35x + 45 = 0
4x (7x – 9) –5 (7x – 9) = 0
(7x – 9) (4x – 5) = 0
3. (d): I. 2x² + 7x – 60 = 0
2x² + 15x – 8x – 60 = 0
x(2x + 15) – 4 (2x + 15) = 0
(x – 4) (2x + 15) = 0
te
−15
𝑥 = 4, 2
, −3
II. 5y² + 42y + 88 = 0
5y² + 20y + 22y + 88 = 0
5y (y + 4) + 22 (y + 4) = 0
(5y + 22) (y + 4) = 0
y = 4, 3
@
ce
II. 3y² - 28y + 64 = 0
3y² - 12y- 16y + 64 = 0
3y (y- 4) – 16 (y – 4) = 0
(3y – 16) (y – 4) = 0
4. (e): I. x² - 2x – 24 = 0
x² - 6x + 4x – 24 = 0
x(x – 6) + 4 (x – 6) = 0
(x + 4) (x- 6) = 0
x= 6, -4
II. y² + 3y – 40 = 0
y² + 8y – 5y – 40 = 0
y(y + 8) – 5 (y + 8) = 0
(y – 5) (y + 8) = 0
y = 5, -8
No relation can be established
5
7
4
5
2
II. (4y – 3)² = y (4y – 1) – 1
16y² + 9 – 24y = 4y² – y – 1
⇒ 12y² –23y +10 = 0
12y² – 15y – 8y + 10 = 0
3y (4y – 5) –2 (4y – 5) = 0
(4y – 5) (3y – 2) = 0
⇒ y =4 ,3
𝑦 = 3 ,4
y≥x
9
⇒x= ,
16
23
−15
So, x ≥ y.
8. (b): I. 10x² – 29x + 21 = 0
10x² – 15x – 14x + 21 = 0
5x (2x – 3) –7 (2x – 3) = 0
(2x – 3) (5x – 7) = 0
3
7
⇒ x =2 ,5
II. 2y² – 19y + 45 = 0
2y² – 10y – 9y + 45 = 0
2y (y – 5) –9 (y – 5) = 0
(y – 5) (2y – 9) = 0
9
⇒ y = 5, 2
So, y > x.
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
9. (c): I. x² + 13x + 42 = 0
12. (e):
(𝑥 + 𝑦)2 = 361
⇒ 𝑥 + 𝑦 = ±19 … … (𝑖)
⇒ 92442 = 𝑦 2 + 92361
⇒ 𝑦 2 = 81
⇒ 𝑦 = ±9 … . . (𝑖𝑖)
Using (i) and (ii)
When y=9
⇒ 𝑥 + 𝑦 = ±19
⇒ 𝑥 = 10, −28
When y= - 9
⇒ 𝑥 + 𝑦 = ±19
⇒ 𝑥 = −10, 28
So, no relation can be obtained.
13. (b):
√𝑥 + 4 = √225 − √121
⇒ √𝑥 + 4 = 4
⇒ 𝑥 + 4 = 16
⇒ 𝑥 = 12
And 𝑦 2 = 473 − 329
⇒ 𝑦 = ±12
So, 𝑥 ≥ 𝑦
x² + 7x + 6x + 42 = 0
x (x + 7) +6 (x + 7) = 0
(x + 7) (x + 6) = 0
⇒ x = – 7, –6
II. y² + 8y + 12 = 0
p
y² + 6y + 2y + 12 = 0
y (y + 6) +2 (y + 6) = 0
(y + 6) (y + 2) = 0
1
(2 –
𝑥
2
1= −
1=
11
36𝑥
11
)
𝑥
36𝑥 2
72𝑥 –11
36𝑥²
gr
10. (a): I. 1 =
ou
⇒ y = –6, –2
So, y ≥ x.
36x² = 72x – 11
36x² – 72x + 11 = 0
36x² – 66x – 6x + 11 = 0
6x (6x – 11) –1 (6x – 11) = 0
(6x – 11) (6x – 1) = 0
6
9
,
1
6
II. ( 3 + 𝑦 ) = 13
14𝑦 2 + 27
3𝑦
m
14𝑦
11
= 13
14y² + 27 = 39y
14y² – 39y + 27 = 0
xa
⇒x=
14. (a):
te
14y² – 21y – 18y + 27 = 0
7y (2y – 3) –9 (2y – 3) = 0
7𝑥 2 − 21𝑥 − 23𝑥 + 69 = 0
⇒ (7𝑥 − 23)(𝑥 − 3) = 0
23
⇒ 𝑥 = 3, 7
15. (c):
(2y – 3) (7y – 9) = 0
3
9
@
ce
⇒ y =2 ,7
And 3𝑦 2 − 21𝑦 − 19𝑦 + 133 = 0
⇒ (3𝑦 − 19)(𝑦 − 7) = 0
19
⇒ 𝑦 = 7, 3
So, no relation can be established between x
and y.
11. (c):
𝑥 2 + 24𝑥 + 119 = 0
⇒ x 2 + 7x + 17x + 119 = 0
⇒ (x + 7)(x + 17) = 0
⇒ 𝑥 = −7, −17
And 3𝑦 2 + 10𝑦 + 7 = 0
⇒ 3𝑦 2 + 3𝑦 + 7𝑦 + 7 = 0
⇒ (𝑦 + 1)(3𝑦 + 7) = 0
⇒ 𝑦 = −1, −
7
So, 𝑥 < 𝑦
16. (a): I. 8x²–10x+3=0
8x²–6x–4x+3=0
2x(4x–3)–1(4x–3)=0
(2x–1)(4x–3)=0
1
3
x=2 or 4
II. 5y² + 14y–3=0
5y²+15y–y–3=0
5y(y+3)–1(y+3)=0
(5y–1)(y+3)=0
1
3
y=5 or –3.
So, x<y
24
223𝑥 + 122𝑦 = 791 … … (𝑖)
122𝑥 + 223𝑦 = 589 … … (𝑖𝑖)
Adding equation (i) and (ii)
345(𝑥 + 𝑦) = 1380
𝑥+𝑦 = 4
… … . (𝑖𝑖𝑖)
Subtract equation (ii) from (i)
101(𝑥 − 𝑦) = 202
𝑥 − 𝑦 = 2 … … (𝑖𝑣)
From (iii) and (iv)
𝑥 = 3, 𝑦 = 1
So, 𝑥 > 𝑦
∴
Adda247 Publications
x>y
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
22. (a): I. x² - 8x + 15 = 0
x² - 5x – 3x + 15 = 0
x(x - 5) – 3 (x - 5) = 0
∴ x = 3 or 5
II. 2y² - 5y – 3 = 0
2y² - 6y + y – 3 = 0
2y (y - 3) + 1 (y – 3) = 0
4
x=–3 , –3
II. y²+9y+20=0
y²+5y+4y+20=0
y(y+5) + 4(y+5) =0
y=–5 , –4
∴x>y
𝑦 = 3 𝑜𝑟 −
1
2
∴x≥y
p
17. (a): I. 3x²+13x+12=0
3x²+9x+4x+12=0
3x(x+3) +4(x+3) =0
23. (e): I. 5x² + 11x + 2 = 0
5x² + 10x + x + 2 = 0
5x(x + 2) +1 (x + 2) = 0
19. (d): I. x³=216
x=(216)1/3
x=6
II. 2y²–25y+78=0
2y²–12y–13y+78=0
2y(y–6)–13(y–6) =0
24. (b): I. 4x + 2y = 4
…(i)
II. 3x + 5y = 3
…(ii)
Multiplying (i) by 5 & (ii) by 2 and on solving
x = 1, y = 0
∴x>y
∴ y≥x.
1
x = -2 or − 5
gr
II. 4y² + 13y + 3 = 0
4y² + 12y + y + 3 = 0
4y (y + 3) + 1 (y + 3) = 0
1
∴ y = -3 𝑜𝑟 − 4
m
∴ no relation
xa
13
y= 2 , 6.
ou
18. (e): I. x²–4x–5=0
x²–5x+x–5=0
x(x–5) +1(x–5) =0
x=5 , –1.
II. 7y²–25y–12=0
7y²–28y+3y–12=0
7y(y–4)+3(y–4)=0
(y–4) (7y+3) = 0
y= 4 , –3/7
∴ No relation.
@
ce
te
20. (e): I. 5x² + 31x + 48 = 0
5x² + 15x + 16x + 48 = 0
5x (x + 3) + 16 (x + 3) = 0
x = –3, –16/5
II. 3y² + 27y + 42 = 0
3y² + 21y + 6y + 42 = 0
3y (y + 7) + 6 (y + 7) = 0
y = –7, –2
So, Relation cannot be established
25. (d): I. 6x² + x – 15 = 0
6x² - 9x + 10x – 15=0
3x (2x - 3) +5 (2x – 3)= 0
21. (b): I. 6x² + 17x + 5 = 0
6x² + 2x + 15x + 5 = 0
2x (3x + 1) +5 (3x + 1) = 0
5
1
∴ 𝑥 = − 2 𝑜𝑟 − 3
II. 2y² + 21y + 49 = 0
2y² + 14y + 7y + 49 = 0
2y (y + 7) +7(y + 7) = 0
𝑜𝑟 −
2
5
3
II. 4y² - 24y + 35 = 0
4y² - 14y – 10y + 35 = 0
2y(2y - 7) – 5 (2y - 7) = 0
∴𝑦=
7
2
𝑜𝑟
5
2
∴y>x
26. (a): I. x² – 11x + 30 = 0
x² – 6x – 5x + 30 = 0
x (x – 6) –5 (x – 6)= 0
(x – 6) (x – 5) = 0
x = 6, 5
II. 56y² – 151y + 99 = 0
56y² – 88y – 63y + 99 = 0
8y (7y – 11) –9 (7y – 11) = 0
(7y – 11) (8y – 9) = 0
11 9
7
y = 7 ,8
𝑦 = −7, − 2
So, x > y.
⇒x>y
25
3
∴ 𝑥=
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
1
27. (e): I. x² – 4√3(√3 + 1) 𝑥 + 48√3 = 0
x² – 12x –4√3𝑥 + 48√3 = 0
x² + 15 = 8x
x² – 8x + 15 = 0
x² – 5x – 3x + 15 = 0
x (x – 5) –3 (x – 5) = 0
(𝑥 − 5)(𝑥 – 3) = 0
x = 3, 5
II. y² – 2√5 (√5 + 2) 𝑦 + 40√5 = 0
y² – 10y –4√5𝑦 + 40√5 = 0
y (y – 10) –4√5 (y – 10) = 0
𝑦
II. 2 + 2𝑦 = 5
(y – 10) (y – 4√5) = 0
𝑦 2 + 21
2𝑦
y² + 21 = 10y
y² – 10y + 21 = 0
y² – 7y – 3y + 21 = 0
y (y – 7) –3 (y – 7) = 0
(y – 7) (y – 3) = 0
y = 3, 7
So, no relation between x and y can be
established.
20
28. (c): I. 21 + 𝑥 (43 + 𝑥 ) = 0
20
=0
𝑥²
gr
+
=0
21x² + 43x + 20 = 0
21x² + 28x + 15x + 20 = 0
7x (3x + 4) + 5 (3x + 4) = 0
(3x + 4) (7x + 5) = 0
m
31. (a): I. 5𝑥 2 − 25𝑥 − 6𝑥 + 30 = 0
5𝑥(𝑥 − 5) − 6(𝑥 − 5) = 0
(𝑥 − 5)(5𝑥 − 6) = 0
−4 −5
7
28
12y + 𝑦 + 37 = 0
12𝑦 2 + 28 + 37𝑦
𝑦
=0
xa
x= 3 , 7
II. 4(3𝑦 + 𝑦 ) + 37 = 0
−7 −4
@
ce
y= 4 , 3
te
12y² + 37y + 28 = 0
12y² + 21y + 16y + 28 = 0
3y (4y + 7) +4 (4y + 7) = 0
(4y + 7) (3y + 4) = 0
So, x ≥ y
29. (a): I. (x + 15)² = (y + 19)²
x + 15 = y + 19
x–y=4
… (i)
II. x² – y² = 112
(x + y) (x – y) = 112
…(ii)
From (i) and (ii), we get:
x + y = 28
… (iii)
Now, from (i) and (iii), we get:
x = 16, y = 12
So, x > y
26
=5
ou
y = 10, 4√5
So, no relation can be established between x
and y.
𝑥
𝑥²
21𝑥 2 + 43𝑥 + 20
21
p
x = 12, 4√3
43
8
= 3𝑥
3𝑥²
(x – 12) (x – 4√3) = 0
21 +
8
𝑥 2 + 15
x (x – 12) –4√3 (x – 12) = 0
1
5
30. (e): I. 3 + 𝑥 2 = 3𝑥
6
𝑥 = 5, 5
II. 8𝑦 2 − 8𝑦 − 4𝑦 + 4 = 0
8y(y – 1) –4 (y–1) = 0
1
y = 1, 2
x>y
32. (e): I. 7𝑥 2 − 14𝑥 − 3𝑥 + 6 = 0
7𝑥(𝑥 − 2) − 3(𝑥 − 2) = 0
(7𝑥 − 3)(𝑥 − 2) = 0
𝑥 = 2,
3
7
II. 5𝑦 2 − 20𝑦 − 4𝑦 + 16 = 0
5𝑦(𝑦 − 4) − 4(𝑦 − 4) = 0
4
𝑦 = 4, 5
No relation
33. (e): I. 13𝑥 2 + 13𝑥 − 4𝑥 − 4 = 0
13x (x +1) – 4 (x + 1) = 0
(13𝑥 − 4) (𝑥 + 1) = 0
4
𝑥 = −1, 13
II. 2𝑦 2 + 3𝑦 − 2𝑦 − 3 = 0
𝑦(2𝑦 + 3) − 1(2𝑦 + 3) = 0
(y – 1) (2y + 3) = 0
–3
y = 1, 2
No relation
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
13
39. (c): I. x² + 14x - 32 = 0
x² + 16x -2x - 32 = 0
x (x +16) -2 (x +16) = 0
(x -2) (x + 16) = 0
x = –16, 2
II. y² – y – 12 = 0
y² – 4y + 3y – 12 = 0
y(y – 4) + 3(y – 4) = 0
(y + 3) ( y – 4) = 0
y = – 3, 4
No relation
y=6, 2
x>y
x = –1,
ou
35. (e): I. 15𝑥 2 + 15𝑥– 5𝑥 – 5 = 0
15𝑥(𝑥 + 1) − 5(𝑥 + 1) = 0
(𝑥 + 1)(15𝑥 − 5) = 0
1
3
gr
II. 6𝑦 2 + 6𝑦 − 4𝑦 − 4 = 0
6y (y+1) – 4(y+1) = 0
(y+1) (6y–4)=0
y = -1,
2
3
40. (a): I. x² – 9x + 20 = 0
x² – 5x – 4x + 20 = 0
x (x – 5) –4 (x – 5) = 0
(x – 4) (x – 5) = 0
x = 4, 5
II. 2y² – 12y + 18 = 0
2y² – 6y – 6y + 18 = 0
2y (y – 3) –6 (y – 3) =0
(2y – 6) (y – 3) = 0
y = 3, 3
x>y
te
xa
m
No relation
36. (e): I. x² – 12x + 32 = 0
x² – 8x – 4x + 32 = 0
x (x – 8) – 4 (x – 8) = 0
(x – 8) (x – 4) = 0
x = 8, 4
II. y² – 20y + 96 = 0
y² – 12 y – 8y + 96 = 0
y(y – 12) – 8 (y – 12) = 0
(y – 8) (y – 12) = 0
y = 8, 12
y≥x
@
ce
37. (b): I. 2x² – 3x – 20 = 0
2x² – 8x + 5x – 20 = 0
2x (x – 4) + 5(x – 4) = 0
(x – 4) (2x + 5) = 0
x = 4, –5/2
II. 2y² + 11y + 15 = 0
2y² + 6y + 5y + 15 = 0
2y (y + 3) + 5 (y + 3) =0
(2y + 5) (y + 3) =0
–5
y = 2 , –3
x≥y
38. (c): I. x² – x – 6 = 0
x² – 3x + 2x – 6 = 0
x (x – 3) +2 (x – 3) = 0
(x – 3) (x + 2) = 0
x = 3, –2
27
p
II. y² – 6y + 8 = 0
y² – 2y – 4y + 8 = 0
y (y – 2) – 4 (y – 2) = 0
(y – 2) (y – 4) = 0
y = 2, 4
No relation can be established between x and
y
34. (a): I. 3𝑥 2 − 36𝑥 − 24𝑥 + 288 = 0
3𝑥(𝑥 − 12) − 24(𝑥 − 12) = 0
(𝑥 − 12)(3𝑥 − 24) = 0
𝑥 = 12, 8
II. 4𝑦 2 − 24𝑦 − 26𝑦 + 156 = 0
4𝑦(𝑦 − 6) − 26(𝑦 − 6) = 0
(𝑦 − 6)(4𝑦 − 26) = 0
41. (d): I. x² + 9x = 25x – 63
x² – 16x + 63 = 0
x = 9, 7
II. 4y² – 34y + 72 = 0
4y² – 18y – 16y + 72 = 0
9
𝑦 = 2,4
∴x>y
1
225
42. (c): I. 5 × x = – x + 14
– 45 = x² – 14x
x² – 14x + 45 = 0
x² – 9x – 5x + 45
x = 9, 5
II. 21y = y² + 90
y² – 21y + 90 = 0
y² – 15y – 6y + 90 = 0
y = 15, 6
∴ No relation
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
43. (a): I. 6x + 7y = 15
II. 3x + 14y = 19.5
Solving (i) and (ii)
8
6
7
x>y
x=
44. (c): I. 7x² + 5x – 18 = 0
7x² – 9x + 14x – 18 = 0
x (7x – 9) +2 (7x – 9) = 0
9
II. 3y² + 4y – 20 = 0
3y² + 10y – 6y – 20 = 0
y ( 3y + 10) – 2 (3y + 10) = 0
y=
@
ce
II. 30y² – 61y + 30 = 0
30y² – 36y – 25y + 30 = 0
6y (5y – 6) –5(5y – 6) = 0
(5y – 6) (6y – 5) = 0
6 5
y = 5,6
So, x ≤ y
47. (d): I. x² – 16x + 63 = 0
x² – 9x – 7x + 63 = 0
x (x – 9) –7 (x – 9) = 0
(x – 9) (x – 7) = 0
x = 7, 9
II. y² – 12y + 35 = 0
y² – 7y – 5y + 35 = 0
y (y – 7) – 5 (y – 7) = 0
(y – 7) (y – 5) = 0
y = 5, 7
So, x ≥ y
28
m
xa
2
te
1
−5
6
,
−3
7
gr
45. (e): I. x² + 5x = 25x
x² – 20x = 0
x (x – 20) = 0
x = 0, 20
II. 3y² + 2y = 2y + 12
3y² = 12
y² = 4
y = ±2
∴ No relation.
,
8
49. (e): I. (x – 2)² = x – 2
x² + 4 – 4x = x – 2
x² – 5x + 6 = 0
x² – 3x – 2x + 6 = 0
x (x –3) –2 (x – 3) = 0
(x – 3) (x – 2) = 0
x = 2, 3
II. 9y² - 36y + 35 = 0
9y² – 21y – 15y + 35 = 0
3y (3y – 7) –5 (3y – 7) = 0
(3y – 7) (3y – 5) = 0
∴ No relation
6
−5
ou
10
5
,
So, no relation
y = 2, – 3
x=
4
II. 42y² + 53y + 15 = 0
42y² + 35y + 18y + 15 = 0
7y (6y + 5) +3 (6y + 5) = 0
(6y + 5) (7y + 3) = 0
x = 7, –2
46. (c): I. 12x² – 16x + 5 = 0
12x² – 10x – 6x + 5 = 0
2x (6x – 5) –1 (6x – 5) = 0
(6x – 5) (2x – 1) = 0
−3
p
7
x= , y=
48. (e): I. 32x² + 44x + 15 = 0
32x² + 24x + 20x + 15 = 0
8x (4x + 3) +5 (4x + 3) = 0
(4x + 3) (8x + 5) = 0
7
5
y = 3 ,3
So, no relation
50. (b): I. 18x² + 39x + 20 = 0
18x² + 24x + 15x + 20 = 0
6x (3x + 4) +5 (3x + 4) = 0
(3x + 4) (6x + 5) = 0
−4
−5
x= 3 , 6
II. 10y² + 29y + 21 = 0
10y² + 15y + 14y + 21 = 0
5y (2y + 3) +7 (2y + 3) = 0
(2y + 3) (5y + 7) = 0
y=
−3 −7
2
,
5
So, x > y
51. (e): I. x² + 4x – 3x – 12 = 0
(x + 4) (x – 3) = 0
x = 3, –4
II. y² + 5y -3y -15 = 0
(y + 5) (y -3) = 0
y = –5, 3
⇒ no relation can be established
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
52. (b): I. 6x² – 2x – 3x + 1 = 0
(2x – 1) (3x – 1) = 0
54. (d): I. x² + 2x + 5x + 10 = 0
(x + 2) (x + 5) = 0
x = –2, –5
II. 2y² + 4y + y + 2 = 0
(y + 2) (2y + 1) = 0
1 1
x= ,
2 3
II. 3y² + 9y – y – 3 = 0
3y (y + 3) – 1 (y + 3) = 0
(3y – 1) (y + 3) = 0
y =– 3,
1
y = − 2, – 2
1
⇒x≤y
3
p
⇒x≥y
55. (e): I. x² – 5x + 3x – 15 = 0
(x – 5) (x + 3) = 0
x = 5, –3
II. y² + 4y + y + 4 = 0
(y + 4) (y + 1) = 0
y = –1, –4
⇒ No relation can be established between x &
y.
ou
53. (d): I. 12x² – 3x – 4x + 1 = 0
(3x – 1) (4x – 1) = 0
1 1
x = 3,4
II. 6y² – 2y – 3y + 1 = 0
(2y – 1) (3y – 1) = 0
1 1
gr
𝑦 = 2,3
⇒y≥x
Mains Solutions
m
Days
Quantity II > Quantity I
xa
3. (a): Quantity I:
Let present age of Randy = x
x−10
= 24 − 19
te
1. (b): Quantity I
Let the number be 10x +y
Acc. to question
y=x+2
and
(10x +y) (x+ y) = 144
(10x + x+ 2) (x+ x + 2) =144
(11x+ 2) (x +1) = 72
11x² + 13x + 2= 72
11x² + 13x – 70 = 0
11x² + 35x – 22x – 70 = 0
On solving x = 2
Number is 24
Quantity II > Quantity I
@
ce
2. (b): Quantity I
Let they meet after ‘n’ days
Applying Arithmetic progression
n
n
[2 × 15 + (n − 1)(−1)] + [2 × 10 + (n −
2
2
1)2] = 165
n
[30 − n + 1 + 20 + 2n − 2] = 165
2
n² + 49n – 330 = 0
n = –55, +6
so, they will meet in 6 days
Quantity II
Let required no. of days
12
x – 10 = 5×12
x = 70 years
Quantity II:
Required average
=
111
−2×42
4
14×
609
12
203
= 24 = 8
777
−84
= 2 12
= 25.375 year
Quantity I > Quantity II
4.
(a): Quantity I:
Let C.P of 100 gm = 100 Rs
So, he purchases 120 gm in 100 Rs
105
And sell 90 gm in = 100 × 100 RS
So, % profit
=
=
S.P.−C.P.
C.P.
21 5
−
18 6
5
6
105 100
−
× 100 = 90100120 × 100
× 100 =
36
21−15
18
5
6
120
× 100
= 90 × 100 = 40% profit
Quantity II:
50% → 12 Rs
So, 100→ 24 Rs
So, 80% → 19.2
29
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
(e): Quantity I:
Let first we arrange all 4 men in 4! Ways then we
arrange 4 women in 4P4 ways at 4 places either left
of the man or right of the man.
= 4! × 4P4 + 4! × 4P4 = 2 × 576 = 1152
Quantity II:
Let first we arrange 4 men in 3! Ways, then 4
women can be arranged in 4 places in 4P4 ways
= 3! × 4P4 = 144 = 144 ×8 = 1152
6.
(b):Quantity I: Let C.P. of both shirts be Rs. 100
Total C.P. = Rs. 200
Total S.P. of both shirts = 1.2 × 100 + 0.9 × 100 =
Rs. 210
200
day
1
Total days = 21 2
Quantity II > quantity I
10. (b):
× 100 = 5%
gr
Overall profit =
(210−200)
5
x
4
Profit%
(5−4)
4
× 100 = 25%
Quantity II > Quantity I
7 2
(b):Quantity I: Volume of cube left = 73 − π (2) × 7
= 343 −
22
7
×
49×7
4
te
7.
xa
y
m
Quantity II: Let C.P. of one-metre cloth be Rs. x
And S.P. of one-metre cloth be Rs. y
Then,
20y − 20x = 5x
⇒ 20y = 25x
⇒ =
= 343-269.5 = 73.5 cm3
Quantity II: Surface area of cube left
7 2
7
= 6 × 72 − 2. π (2) + 2π (2) 7
@
ce
= 294 − 77 + 154 = 371 cm2
Quantity II > Quantity I
8.
1
(e): Quantity I: Liters of milk removed =5 ×10=2 ltr.
Remaining milk = 8 ltr
1
Liters of water removed = 5 × 2.5 = 0.5 ltr.
Remaining water = 2 ltr
So milk : water = 4:1
2+x
8
4
=1
⇒ x = 30
32
8+y
1
= 4⇒ y = 128 − 8 = 120 ltr.
Quantity II : 120 ltr
Quantity I = Quantity II.
30
(b):Let, total units of work be 32 units
Then P does 2 units per day.
& Q does 1 unit per day.
Quantity I:
3 units are done in 2 days.
30 units are done in 20 days.
On 21st day P does 2 units and work gets
completed.
Quantity II:
3 units are done in 2 days
30 units are done in 20 days.
On 21st day Q does 1unit work.
1
P completes the remaining one unit in another 2
p
5.
9.
ou
There will be 0% profit if the book were sold for
Rs.4.8 more
Quantity I > Quantity II
∠ABD = 90° (angle in semicircle)
∠DBP = 90°
∠BCD = ∠DAB = 40° (angle subtended by same
arc in same segment)
∠BCD + ∠CPB + ∠CBD + ∠DBP = 180°
⇒ 40° + 20° + ∠𝐷𝐵𝐶 + 90° = 180°
⇒ ∠𝐷𝐵𝐶 = 30°
∠𝐴𝐷𝐵 = 180° − ∠𝐷𝐴𝐵 − ∠𝐴𝐵𝐷 = 180° − 40° −
90° = 50°
Quantity II > Quantity I
11. (a): Quantity I: 12320 = 𝜋(7𝑥)2 × 10𝑥
22
⇒ 12320 = 7 × 49𝑥 3 × 10
⇒ 𝑥3 = 8 ⇒ 𝑥 = 2
Height = 10x = 20 cm.
1
Quantity II: 3 𝜋 × 22 × 3 = 𝜋 × 22 × ℎ
⇒ h = 1 cm.
Level of kerosene in jar = 1 cm.
Quantity I > Quantity 2
12. (c): Quantity I: 𝑝 2 − 18𝑝 + 77 = 0
⇒ 𝑝 2 − 11𝑝 − 7𝑝 + 77 = 0
⇒ (𝑝 − 11)(𝑝 − 7) = 0
⇒ 𝑝 = 11, 7
Quantity II: 3𝑞 2 − 25𝑞 + 28 = 0
⇒ 3𝑞2 − 21𝑞 − 4𝑞 + 28 = 0
⇒ (3𝑞 − 4)(𝑞 − 7) = 0
4
⇒ 𝑞 = 7,
3
Quantity I ≥ Quantity II
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
13. (b):Quantity I: Let, speed of current be x m/minute
200
200
= 48+𝑥 + 10
48−𝑥
⇒ 𝑥 = 32𝑚/𝑚𝑖𝑛.
Quantity II:
22
7
3×2× ×49
14
also,
𝑛(𝑥 + 10.8) = (𝑚 + 𝑛)𝑥
or, 10.8𝑛 = 𝑚𝑥
or, 𝑚𝑥 = 10.8 …………….(ii)
Dividing (i) by (ii),
= 66𝑚/𝑚𝑖𝑛
𝑚
Quantity II > Quantity I.
𝑛
=
3
2
14. (e): Let m → men
w→ 𝑤𝑜𝑚𝑒𝑛
𝑏 →boy
(10𝑚 + 15𝑤) 8 = (12𝑚 + 8𝑤)10
80𝑚 + 120𝑤 = 120𝑚 + 80𝑤
40𝑚 = 40𝑤
And, m = w = 2B
Quantity I
→ 2𝑚 + 4𝑤 + 18𝑏 → 2𝑚 + 4𝑚 + 9𝑚 → 15m
Quantity II
→ 9𝑚 + 3𝑤 + 6𝑏 ⇒ 9𝑚 + 3𝑚 + 3𝑚 → 15𝑚
∴ Quantity I = Quantity II
Putting this in equation (i)
15. (a):
Quantity I < Quantity II
4.8
𝑥
𝑥
= 10.8
𝑥2 =
ou
p
or,
𝑥 2 = 4.8 ×
𝑇ℎ𝑖𝑠 𝑠𝑡𝑒𝑝 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑙𝑦 𝑎𝑝𝑝𝑙𝑖𝑒𝑑
𝑡𝑜 𝑡ℎ𝑒𝑠𝑒 𝑡𝑦𝑝𝑒 𝑜𝑓 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠
10.8 [
]
(𝑆ℎ𝑜𝑟𝑡 − 𝑐𝑢𝑡 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ)
48×108
2
100
12×4×12×9
or, 𝑥 =
10
= 7.2 hrs.
gr
or, 𝑥 =
10×10
12×3×2
m
18. (b):Quantity I:
@
ce
te
xa
Babu returns home 40 minute earlier so he saves
20 min on going and 20 min in returning
So girlfriend travels 100 min before meeting
point
Because babu does not go G to A to A and A to O.
Time
Speed
Babu
20
1
5
×40 200 km/hr
:
:
Girlfriend 100
5
I
×40 40 km/hr
Quantity I → 200 km/h
Quantity II → 197 km/h
∴ Quantity I> Quantity II
16. (b):CP
MP
SP
555x 700x
(700x – 68)
120
∴ 100 × 555𝑥 = 700𝑥 − 68
666𝑥 = 700𝑥 − 68
34x = 68
x=2
∴ Quantity I → CP = 2 × 555 = 1110 Rs.
∴ Quantity I < Quantity II
17. (b):Let, Manoj and Shubham take ‘𝑥’ hrs. to complete
the work working together.
Let, Manoj does ‘𝑚’ units per hour.
And Shubham does ‘𝑛’ units per hour.
ATQ, 𝑚(𝑥 + 4.8) = (𝑚 + 𝑛)𝑥
or, 4.8𝑚 = 𝑛𝑥 …………..(i)
31
Draw EF || BC
Now, E and F are mid-points of AC and AB
respectively.
AF = 5 cm, EF = 5 cm
Radius of semi-circle = 5 cm
1
Area of shaded region = 4 Area of circle – Area of
∆AFE
1
22
1
= 4 × 7 × 5 × 5 − 2 × 5 × 5 ≈ 7.14 cm2
Quantity I < Quantity II
19. (a): Quantity 1:
Let original duration and speed be ‘𝑡’ hr. and ‘𝑠’
km/hr. respectively.
ATQ, 𝑠𝑡 = 3000 ………..(i)
Also, (𝑠 − 1000)(𝑡 + 1) = 3000 ………..(ii)
⇒ 𝑠𝑡 + 𝑠 − 100𝑡 − 100 = 3000
⇒ 3000 + 𝑠 − 100𝑡 − 100 = 3000
⇒ 𝑠 − 100𝑡 = 100
Putting ‘𝑠’ from (i),
3000
𝑡
− 100𝑡 = 100
⇒ 100𝑡 2 + 100𝑡 − 3000 = 0
⇒ 𝑡 2 + 𝑡 − 30 = 0
⇒ (𝑡 + 6)(𝑡 − 5) = 0
⇒ 𝑡 = 5 hr.
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
Quantity 2:
3
New speed = 4th of original
1
By equation (ii) × 3 – equation (iii)
6y + 9z = 99
6y + 5z = 71
−
−
−
4
Hence, new time = th of original
1
3
20
1
23. (e): I. x =√(36)2 × (1296)4 = √6 × 6 = 6
1
i.e., th of original time = =
3
60
3
or, Original time = 1 hr.
Quantity I > Quantity II
4z = 28 ⇒ z = 7
from equation (ii)
2y + 3 × 7 = 33 ⇒ y = 6
Hence, none of the above relationships is
established.
7
20. (e): Perimeter of smaller wheel = 2𝜋. = 7𝜋 cm
ou
... (i)
... (ii)
... (iii)
24. (d):by equation (i) × 5 – (ii) × 8
40x + 35y
= 675
40x + 48y = 792
−
−
−
−13y = −117
⇒ y=9
from equation (i)
8x + 7 × 9 = 135 ⇒ 8x = 135 – 63 =72 ⇒ x = 9
from equation (iii)
9 × 9 + 8z = 121 ⇒ 8z = 121 – 81 = 40 ⇒ z = 5
Clearly x = y > z
m
... (iv)
25. (e): I. (𝑥 + 𝑦)3 = 1331 ⇒ x + y = 11 ⇒ y = 11 – x
put it in equation (iii)
x(11– x) = 28 ⇒ 11x – 𝑥 2 = 28
𝑥 2 – 11x + 28 = 0 ⇒ 𝑥 2 – 7x –4x + 28 = 0
x(x – 7) – 4(x – 7) = 0 ⇒ (x – 7) (x – 4) = 0
So, x = 7, 4
from equation (i)
y = 4, 7
from equation (ii)
7 – 4 + z = 0, z = – 3
or, 4 – 7 + z = 0, z = 3
Hence. no relation can be established.
xa
21. (a): 7x + 6y + 4z = 122
4x + 5y + 3z = 88
9x + 2y + z = 78
By equation (iii) × 3 – equation (ii)
27x + 6y + 3z = 234
4x + 5y + 3z = 88
−
−
−
−
23x + y = 146
By equation (iii) × 4 – equation (i)
36x + 8y + 4z = 312
7x + 6y + 4z = 122
−
−
−
−
29x + 2y = 190
By equation (iv) × 2 – equation (v)
46x + 2y = 292
29x + 2y = 190
−
−
−
17x = 102
x=6
from eqn. (iv)
23 × 6 + y = 146 ⇒ y = 146 – 138 = 8
From equation (iii)
9 × 6 + 2 × 8 + z = 78
⇒ 54 + 16 + z = 78
z = 78 – 70 = 8; ⇒ x = 6, y = 8, z = 8
Hence, x < y = z
p
2
gr
14
Perimeter of larger wheel= 2𝜋. = 14𝜋 𝑐𝑚
2
Let , both take ‘𝑥’ revolutions per second,
Then, (7𝜋 + 14𝜋)10𝑥 = 1990.50 − 10.5
198
198×7
or, 𝑥 = 21𝜋 = 21×22 = 3
speed of smaller wheel = 7𝜋 × 3 = 21𝜋 cm./s.
Quantity I = Quantity II
@
ce
te
... (v)
22. (c): By equation (ii) × 2 – equation (i)
8x + 6y = 118
7x + 6y = 110
−
−
−
x
= 8
from equation (i), 7 × 8 + 6y = 110
⇒ 6y = 110 – 56 = 54 ⇒ y = 9
from equation (iii)
8 + z = 15 ⇒ z = 7; clearly, x < y > z
32
26. (c): By equation (i) × 2 – (ii)
2x + 6y + 8z = 96  2
2x + 8z = 80
−
−
−
6y = 112
112
6
By equation (iii)
2x + 6 ×= 120 ⇒ 2x = 8 ⇒ x = 4
By equation (ii)
8z = 80 – 8 ⇒ 8z = 72 ⇒ z = 9
Hence x < y > z
y=
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
3𝑥
3𝑥 + 7
–
3𝑥 + 7
3𝑥
31. (b):I. 77𝑥 2 + 58x + 8 = 0
\
/
11 × 7
4×2
and, 44 + 14 = 58
44
14
4
2
∴x=– ,– ⇒ x=– ,–
77
77
7
11
II. 42𝑦 2 + 59y + 20 = 0
\
/
7×6
5×4
also, 35 + 24 = 59
35
24
5
4
∴y=– ,– ⇒y=– ,–
42
42
6
7
Hence, x ≥ y
= 14
– 49 – 42x = 14 × 9𝑥 2 + 14 × 21x
– 49 – 42x = 126𝑥 2 + 294x
126𝑥 2 + 336x + 49 = 0
Hence x = –a , –b
both roots of x are – ve
𝑦
18𝑦 − 5
II. 18𝑦 − 5 − 𝑦 = 2
𝑦 2 − 324𝑦 2 − 25 + 180𝑦
18𝑦 2 − 5𝑦
2
=2
28. (e): I.
𝑥 2 +𝑥 2 +49+14𝑥
2
𝑥 2 +7𝑥
p
𝑦 – 324𝑦 – 25 + 180y = 36𝑦 2 – 10y
359𝑦 2 – 190 y + 25 = 0
Hence y = + c, +d
both roots of y are + ve
⇒ y>x
= 12
2𝑥 + 49 + 14x = 12𝑥 2 + 84x
10𝑥 2 + 70x – 49 = 0 ⇒ x = – a, + b
II.
𝑦 2 +𝑦 2 +64+16𝑦
𝑦 2 +8𝑦
2
= 16
𝑥
𝑥 − 11
2
+
𝑥 − 11
𝑥
Þ
𝑥 2 + 𝑥 2 + 121 − 22𝑥
𝑥 2 − 11𝑥
2
m
2𝑦 + 64 + 16y = 16𝑦 2 + 128 y
14𝑦 2 + 112y – 64 = 0 ⇒ y = + c, – d
It is clear that relation cannot be established.
29. (e): I.
32. (a): I. 63𝑥 2 + 172x + 117 = 0
\
/
9×7
13 × 9
also, 91 + 81 = 172
13
9
91
81
x=– ,– ⇒x=– ,–
9
7
63
63
2
II. 30y + 162y + 216 = 0
\
/
6×5
36 × 3 × 2
also, 90 + 72 = 162
90
72
y = – 30, – 30 ⇒ y = – 3, – 2.4
Hence x > y
ou
2
gr
27. (c): I.
=7
xa
2𝑥 + 121 – 22x = 7𝑥 – 77x
5𝑥 2 – 55x – 121 = 0
Hence x = – a, +b
4𝑦
4𝑦 − 13
II. 4𝑦 − 13 + 4𝑦 = 9
33. (e): I. 36x 4 + 369x 2 + 900 = 0
let x 2 = p
36p2 + 369p + 900 = 0
\
/
12 × 3
75 × 12
also, 144 + 225 = 369
–144 –225
p=
,
2
te
16𝑦 2 + 16𝑦 2 + 169 – 104y
= (16𝑦 2 – 52y) × 9
32𝑦 2 + 169 – 104y = 144𝑦 2 – 468y
112𝑦 2 – 364y – 169 = 0
Hence y = + c, – d
Hence Relation cannot be established.
@
ce
30. (a): I. 99𝑥 + 149x + 56= 0
99𝑥 2 + 77x + 72x + 56 = 0
11x (9x + 7) + 8 (9x + 7) = 0
(9x + 7) (11x + 8) = 0
9x + 7 = 0 11x + 8 = 0
−7
−8
x=
x=
9
11
II. 156𝑦 2 + 287y + 132 = 0
156𝑦 2 + 143y + 144y+ 132 = 0
13y (12y + 11) + 12 (12y + 11) = 0
(12y + 11) (13y + 12) = 0
12y + 11 = 0 13y + 12 = 0
−11
−12
y=
y=
12
13
Hence x > y
33
36
36
2 –144 2 –225
so, x = 36 , x = 36
imaginary roots
established.
⇒
relation
cannot
be
34. (e): I. 18x 2 –13 √7x + 14 = 0
\
/
9×2
2×7
9√7 4√7
also, – 9√7 – 4√7 = –13√7 ⇒ x = 18 , 18
II. 32y 2 – 19√6y + 9 = 0
\
/
16 × 2
3×3
also, –16√6 – 3√6 = –19√6 ⇒ y =
Relation cannot be established.
16 √6 3√6
32
32
35. (e): I. x 2 – 82x + 781 = 0
\
/
1
71 × 11
also, –71 – 11 = – 82
x = 71, 11
II. y 2 – 5041 = 0 ⇒ y 2 = 5041 ⇒ y = +71, –71
Hence, relation cannot be established.
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
also, – 221 + 437 = 216
221
437
y = 247, − 247
Hence, relation cannot be established.
x=
–27√7 –20√7
,
36
36
II. 35y 2 + 20√3y + 63√2y + 36√6 = 0
5y (7y + 4√3) + 9√2 (7y + 4√3) = 0
(5y + 9√2) (7y + 4√3) = 0
y=
−9√2
5
,y=
–4 √3
7
Hence, relation cannot be established.
437
gr
40. (c): y > x (according to sign method).
xa
1. (a): I. x = ±16
II. y2 – 16y – y + 16 = 0
y (y – 16) −1(𝑦 − 16) = 0
(y −1) (𝑦 − 16)
y = 1 & 16
So, no relation can be established
𝑥 = −2,2
II. 2y² - 4y – 7y + 14 = 0
2y (y – 2) -7 (y – 2) = 0
(2y – 7) (y – 2) = 0
7
1
1
2
3
𝑥 = − ,−
II. 20y² + 5y + 4y + 1 = 0
(4y + 1) (5y + 1) = 0
1
1
𝑦 = −4,−5
So, y > x
5. (d): I. 2x² - 9x + 9 = 0
2x² - 6x – 3x + 9 = 0
(2x-3) (x-3) = 0
3
𝑥 = 2,3
II. 6y² - 17y + 12 = 0
6y² - 8y – 9y + 12 = 0
(2y-3) (3y-4) = 0
3 4
𝑦 = 2,3
So, x ≥ y
6. (c): I. 4x² – 17x + 15 = 0
4x² – 12x – 5x + 15 = 0
4x (x – 3) –5 (x – 3) = 0
(x – 3) (4x – 5) = 0
𝑦 = 2, 2
34
So, no relation can be established between
x & y.
4. (c): I. 6x² + 3x + 2x + 1 = 0
(3x + 1) (2x + 1) = 0
te
@
ce
3. (a): I. 4x² - 8x – 5 = 0
4x² - 10x + 2x – 5 = 0
2x (2x – 5) + 1 (2x – 5) = 0
(2x + 1) (2x – 5) = 0
1 5
437
Hence relation cannot be established,
/
17 × 23
2. (e): I. x2 + 10x + 10x + 100 =0
x (x + 10) + 10(x +10) = 0
(x + 10) (x + 10) = 0
x = - 10
II. y2 + 10x + 3y + 30 = 0
y (y + 10) + 3(y + 10) =0
(y + 10) (y + 3)
y = −10 − 3
So, x≤y
II. 437y 2 + 1074y + 589 = 0
\
/
19 × 23
19 × 31
also, 361 + 713 = 1074
–361 –713
y=
,
m
\
13 × 19
39. (e): I. 391x 2 + 1344x + 1073 = 0
\
/
17 × 23
29 × 37
also, 493 + 851 = 1344
–493 –851
x = 391 , 391
ou
37. (e): I. 91x 2 + 298x + 187 = 0
\
/
7 × 13
17 × 11
also, 77 + 221 = 298
–77 –221
x= ,
91 91
II. 247y2 + 216y – 391 = 0
38. (e): According to sign method.
Relation cannot be established.
p
36. (e): I. 36x 2 + 47√7x + 105 = 0
\
/
9×4
7×3×5
also, 27√7 + 20√7 = 47√7
x = 3,
Adda247 Publications
5
4
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
II. 2y² - 7y + 5 = 0
2y² - 2y – 5y + 5 = 0
(y - 1) (2y - 5) = 0
5
y = 1, 2
x>y
II. 2y² – 17y + 35 = 0
2y² – 10y – 7y + 35 = 0
2y (y – 5) –7 (y – 5) = 0
(y – 5) (2y – 7) = 0
7
y = 5, 2
13. (d): I. 2x² + x – 28 = 0
2x² + 8x – 7x – 28 = 0
2x (x + 4) – 7 (x + 4) = 0
(2x – 7) (x + 4)= 0
7
x = −4, 2
II. 2y² - 23y + 56 = 0
2y² - 16y – 7y + 56 = 0
2y(y – 8) – 7(y – 8) = 0
(2y – 7) (y – 8) = 0
7
y = 2,8
y≥x
So, y > x
p
7. (a): I. x = 5
II. y = 5
So, x=y
ou
8. (d): I. x2 + 7x – 5x – 35 =0
x (x + 7) – 5 (x + 7) =0
(x + 7) (x – 5) = 0
x = −7, 5
II. y2 + 7y + 8y + 56 = 0
y(y + 7) + 8(y + 7) = 0
(y + 7) (y + 8) = 0
y = − 8, −7
So, x≥y
gr
14. (e): I. 2x² - 7x – 60 = 0
2x² - 15x + 8x – 60 = 0
x (2x – 15 ) + 4 (2x – 15) = 0
(x + 4) (2x − 15) = 0
15
x = −4, 2
II. 3y² + 13y + 4 = 0
3y² + 12y + y + 4 = 0
3y (y + 4) + 1 (y + 4) = 0
(3y + 1) (y + 4) = 0
1
y = − 3 , −4
No relation between x and y
3
x = − 17 , 1
xa
10. (a): I. 17x2 – 14x – 3 = 0
17x2 – 17x + 3x – 3 = 0
17x (x – 1) + 3(x – 1) = 0
(17x + 3) (x – 1) = 0
m
9. (a): I. x = ± 9
II. y = ± 8
So, no relation can be established
te
II. y2 – 2y – 35 = 0
y2 – 7y + 5y – 35 = 0
y(y −7) + 5(𝑦 − 7) = 0
y = 7, - 5
So, no relation can be established
@
ce
11. (e): I. x2 + 9x – 5x – 45 = 0
x(x + 9) – 5(x + 9) = 0
(x – 5) (x + 9) = 0
x = 5, - 9
II. y2 – 5y – 8y + 40 = 0
y(y – 5) – 8(y – 5) = 0
(y – 5) (y – 8) = 0
y = 5, 8
So, x≤y
12. (a): I. x² - 8x + 15 = 0
x² - 3x – 5x + 15 = 0
(x – 3) (x – 5) = 0
x = 3, 5
35
15. (e): I. x² - 17x – 84 = 0
x² +4x – 21x – 84 = 0
(x + 4) (x – 21) = 0
x = -4, 21
II. y² + 4y – 117 = 0
y² - 9y + 13y – 117 = 0
(y – 9) (y + 13) = 0
y = 9, -13
No relation between x and y
16. (c): I. x2 = 92
x=9
II. (y – 8)2 = 32
y = 11
y>x
17. (d): I. 2x2 – 2x - x + 1 = 0
2x(x – 1) – 1(x – 1)
1
x = 2,1
II. 2y2 – 2y – 3y + 3 = 0
2y(y – 1) – 3(y – 1) = 0
y=
3
2
,1
x≤y
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
II. 𝑦 2 + 12𝑦 + 36 = 0
𝑦 2 + 6𝑦 + 6𝑦 + 36 = 0
𝑦(𝑦 + 6) + 6(𝑦 + 6) = 0
(𝑦 + 6)(𝑦 + 6) = 0
𝑦 = −6
So, 𝑥 > 𝑦.
18. (b): I. x2 + 10x + 11x + 110 = 0
x(x + 10) + 11(x + 10) = 0
x = - 10, - 11
II. y2 + 9y + 8y + 72 = 0
y(y + 9) + 8(y + 9)
y = - 9, - 8
x<y
24. (c): I. 𝑥 2 + 13𝑥 + 40 = 0
𝑥 2 + 8𝑥 + 5𝑥 + 40 = 0
𝑥(𝑥 + 8) + 5(𝑥 + 8) = 0
(𝑥 + 8)(𝑥 + 5) = 0
𝑥 = −8, −5
II. 𝑦 2 + 7𝑦 + 10 = 0
𝑦 2 + 5𝑦 + 2𝑦 + 10 = 0
𝑦(𝑦 + 5) + 2(𝑦 + 5) = 0
(𝑦 + 5)(𝑦 + 2) = 0
𝑦 = −2, −5
So, 𝑥 ≤ 𝑦.
ou
p
19. (d): I. x = ±2
II. y2 – 2y – 4y + 8 = 0
y(y – 2) – 4(y – 2) = 0
y = 2, 4
x≤y
20. (e): I. 𝑥 2 + 9𝑥 − 22 = 0
⇒ x² + 11x – 2x – 22 = 0
⇒ (x + 11) (x – 2) = 0
⇒ x = – 11, 2
II. 2y² – 7y + 6 = 0
⇒ 2y² – 4y – 3y + 6 = 0
⇒ 2y(y–2)–3(y–2) =0
⇒ (y–2) (2y–3) = 0
gr
25. (b): I. 𝑥 2 − 20𝑥 + 91 = 0
𝑥 2 − 13𝑥 − 7𝑥 + 91 = 0
𝑥(𝑥 − 13) − 7(𝑥 − 13) = 0
(𝑥 − 13)(𝑥 − 7) = 0
𝑥 = 7, 13
II. 𝑦 2 + 16𝑦 + 63 = 0
𝑦 2 + 9𝑦 + 7𝑦 + 63 = 0
𝑦(𝑦 + 9) + 7(𝑦 + 9) = 0
(𝑦 + 9)(𝑦 + 7) = 0
𝑦 = −7, −9
So, 𝑥 > 𝑦.
3
⇒ y = 2, 2
m
No relation
II. 15y² + 11y + 2 = 0
15y² + 6y + 5y + 2 = 0
3y (5y + 2) + 1 (5y + 2) = 0
2
1
te
∴ y = -5 𝑜𝑟 − 3
∴ no relation can be established
@
ce
22. (a): I. 𝑥 2 + 9𝑥 + 20 = 0
𝑥 2 + 5𝑥 + 4𝑥 + 20 = 0
𝑥(𝑥 + 5) + 4(𝑥 + 5) = 0
(𝑥 + 5)(𝑥 + 4) = 0
𝑥 = −4, −5
II. 8𝑦 2 − 15𝑦 + 7 = 0
8𝑦 2 − 8𝑦 − 7𝑦 + 7 = 0
8𝑦(𝑦 − 1) − 7(𝑦 − 1) = 0
(𝑦 − 1)(8𝑦 − 7) = 0
7
𝑦 = 1, 8
So, 𝑥 < 𝑦.
23. (b): I. 𝑥 2 − 11𝑥 + 30 = 0
𝑥 2 − 6𝑥 − 5𝑥 + 30 = 0
𝑥(𝑥 − 6) − 5(𝑥 − 6) = 0
(𝑥 − 6)(𝑥 − 5) = 0
𝑥 = 5, 6
36
26. (e): I. 𝑥 2 − 𝑥 − 12 = 0
𝑥 2 − 4𝑥 + 3𝑥 − 12 = 0
𝑥(𝑥 − 4) + 3(𝑥 − 4) = 0
(𝑥 − 4)(𝑥 + 3) = 0
𝑥 = 4, −3
II. 𝑦 2 + 5𝑦 + 6 = 0
𝑦 2 + 3𝑦 + 2𝑦 + 6 = 0
𝑦(𝑦 + 3) + 2(𝑦 + 3) = 0
(𝑦 + 3)(𝑦 + 2) = 0
𝑦 = −2, −3
So, 𝑛𝑜 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛.
xa
21. (e): I. 6x² + 5x + 1 = 0
6x² + 3x + 2x + 1 = 0
3x (2x + 1) +1 (2x + 1) = 0
1
1
x = -2 or − 3
27. (a): Quantity I:
20
Required profit = 450 × 120 = Rs.75
Quantity II:
100
Required cost price = 84 × 120 = Rs.70
So, Quantity I > Quantity II.
28. (b): Quantity I:
100
40
Required female = 1152 × 30 × 60 ×
1920
Quantity II:
1940
So, Quantity I < Quantity II.
Adda247 Publications
100−25
100
=
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
29. (b): Quantity I:
ATQ,
𝑃×12×2
100
34. (e): I. 4x2 – 4𝑥 − 3𝑥 + 3 = 0
4x(x −1) − 3 (𝑥 − 1) = 0
x = 1, 3/4
II. 7y2− 17y + 6 = 0
7y2− 14y- 3y + 6 = 0
7y (y – 2) − 3 (𝑦 − 2) = 0
y = 2 , 3/7
So, no relation can be established between x
and y.
35. (a): I. 2x2 - 10x -9x + 45 = 0
2x(x - 5) - 9(x - 5) = 0
x = 5, 9/2
II. 2y2 - 8y - y + 4 = 0
2y(y -4) -1(y -4) = 0
y =4,1/2
So, x >y
= 1200
𝑃 = 5,000 Rs.
Quantity II:
Rs.6,000
So, Quantity I < Quantity II.
Area of a rectangular field =
288
3
p
30. (e): Let breadth of the field be x m.
So, length of the field = (𝑥 + 4) m
= 96 m2
ou
ATQ, 𝑥(𝑥 + 4) = 96
𝑥 2 + 4𝑥 − 96 = 0
𝑥 2 + 12𝑥 − 8𝑥 − 96 = 0
𝑥(𝑥 + 12) − 8(𝑥 + 12) = 0
(𝑥 + 12)(𝑥 − 8) = 0
𝑥 = 8, −12
Quantity I:
Length of rectangular field = 12m
Quantity II: 12 m
So, Quantity I = Quantity II.
m
37. (a): I. x² -13x + 40= 0
x² - 5x – 8x + 40 = 0
x (x -5) – 8 (x - 5) = 0
x = 5, 8
II. 2y² - y – 15 = 0
2y² - 6y + 5y – 15 = 0
2y (y - 3) + 5 (y - 3) = 0
y=3, -5/2
x>y
xa
31. (b): Quantity I:
Let present age of Prashant be x years.
So, present age of Shivam = (𝑥 + 8) years
𝑥 + 8 + 𝑥 = 32
𝑥 = 12 years
Quantity II:
15 years
So, Quantity I < Quantity II.
gr
36. (b): I. x2= 144
x = −12, +12
II. y = −12
So, x≥y
@
ce
te
32. (e): I . x2 − 14𝑥 + 45 = 0
x2 − 9𝑥 − 5𝑥 + 45 = 0
x(x −9) − 5(𝑥 − 9) = 0
x = 9, 5
II. y2 – 12𝑦 − 6𝑦 + 72 = 0
y(y – 12) − 6(𝑦 − 12) = 0
y = 12, 6
no relation can be established between x and
y.
33. (b): I. x2 + 7𝑥 + 12 = 0
x2 +4𝑥 + 3𝑥 + 12 = 0
x(x + 4) + 3(𝑥 + 4) = 0
(x +4) (x +3) = 0
x = −3, −4
II. y2 + 5𝑦 + 4𝑦 + 20 = 0
y(y + 5) + 4(𝑦 + 5) = 0
y = -5, -4
So , x≥y
38. (e): I. 5x² + 17x + 6 = 0
5x² + 15x + 2x + 6 = 0
5x (x + 3) +2(x + 3) = 0
2
x = −3, − 5
II. 2y² + 11y + 12 = 0
2y² + 8y + 3y + 12 = 0
2y (y + 4) + 3 (y + 4) = 0
3
y = −4, − 2
No relation
39. (a): I. 7x² - 19x + 10 = 0
7x² - 14x – 5x + 10 = 0
7x (x - 2) – 5 (x - 2) = 0
5
x = 2, 7
II. 8y² + 2y – 3 = 0
8y² + 6y – 4y – 3 = 0
2y (4y + 3) – 1 (4y + 3) = 0
y=
−3 1
4
,
2
x >y
37
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
40. (a): I. x 2 − 8x + 15 = 0
⇒ x 2 − 5x − 3x + 15 = 0
⇒ x(x − 5) − 3(x − 5) = 0
⇒ (x − 3)(x − 5) = 0
∴ x = 3 or 5
II. y 2 − 3y + 2 = 0
⇒ y 2 − 2y − y + 2 = 0
⇒ y(y − 2) − 1(y − 2) = 0
⇒ (y − 1)(y − 2) = 0
∴ y = 1 or 2
∴x>y
24
20
Required difference = 30𝑥 × 100 − 19𝑥 × 100
= 7.2𝑥 − 3.8𝑥 = Rs.136
Quantity II:
Let cost price of the article be Rs.100x
So, marked price of the article = 100𝑥 ×
170
100
= Rs.170x
60
And, selling price of the article = 170𝑥 × 100
ou
41. (c): I. 3x² –7x + 4 = 0
⇒ 3x²– 4x - 3x +4 = 0
⇒ (3x – 4) (x -1) = 0
4
x = 3 or 1
p
= Rs.102x
ATQ,
102𝑥 = 183.6
⇒ 𝑥 = 1.8
40
Required sum = 170𝑥 × 100 + (102𝑥 − 100𝑥)
= 68𝑥 + 2𝑥 = Rs.126
So, Quantity I > Quantity II.
II. 2y² -9y + 10 = 0
⇒ 2y² - 4y - 5y + 10 = 0
⇒ (2y - 5) (y -2) =0
5
⇒ y = 2 or 2
y>x
480
480
11𝑥−𝑥
+ 11𝑥+𝑥 = 11
⇒𝑥 =8
So, speed of boat in still water = 11x = 88 km/hr.
Quantity II:
Let speed of boat in still water & speed of stream
be ‘a km/hr.’ and ‘b km/hr.’ respectively.
ATQ,
= 150
So, Quantity I > Quantity II.
⇒ 𝑅 = 18%
6
te
43. (a): Quantity I:
ATQ,
5900×𝑅×3
= 3186
100
xa
m
42. (c): Quantity I:
Total number of ways = (8C2 × 4C2) + (8C1 × 4C3) +
(4C4)
= 168 + 32 + 1 = 201
Quantity II:
3-digit numbers which are divisible by 3 and ends
with an even number = (102, 108, 114, -------, 996)
996−102
Required number of 3 – digit numbers=
+1
gr
45. (e): Quantity I:
Let speed of boat in still water & speed of stream
be ‘11x km/hr.’ and ‘x km/hr.’ respectively.
ATQ,
7900×(18+5)×3
@
ce
Required interest =
100
= Rs.5451
Quantity II:
Equivalent rate of interest of 13% p.a. for 2 years
13×13
at CI = 13 + 13 + 100 = 27.69%
ATQ,
𝑋×27.69
= 2325.96
100
⇒ X = Rs.8400
So, Quantity I < Quantity II.
3.5
= (𝑎 + 𝑏)
⇒ (𝑎 + 𝑏) = 100 …(i)
And,
380
5
= (𝑎 − 𝑏)
⇒ (𝑎 − 𝑏) = 76 …(ii)
On solving (i) & (ii), we get:
a = 88 km/hr.
So, Quantity I = Quantity II.
46. (c): Quantity I:
Let A’s present age be 10x years.
So, B’s present age = 10𝑥 ×
160
100
= 16x years
2
And, C’s present age = 16𝑥 × = 6.4x years
5
44. (c): Quantity I:
Let CP & MP of an article be Rs.19x and Rs.30x
respectively.
ATQ,
120
19𝑥 × 100 = 912
⇒ 𝑥 = 40
38
350
And, D’s present age = 2 × 6.4𝑥
= 12.8x years
ATQ,
16𝑥 − 12.8𝑥 = 8
⇒ 𝑥 = 2.5
Hence, required average =
10𝑥+16𝑥+6.4𝑥+12.8𝑥
4
= 11.3x = 28.25 years
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
47. (e): I. x² - 7x + 12 = 0
𝑥 2 − 4𝑥 − 3𝑥 + 12 = 0
(𝑥 − 4)(𝑥 − 3) = 0
𝑥 = 3,4
II. y² - 8y + 12 = 0
𝑦 2 − 6𝑦 − 2𝑦 + 12 = 0
(𝑦 − 6)(𝑦 − 2) = 0
𝑦 = 2,6
No relation can be established
2
y≥x
@
ce
49. (e): I. 2x² - 7x – 60 = 0
2x² - 15x + 8x – 60 = 0
x (2x – 15 ) + 4 (2x – 15) = 0
(x + 4) (2𝑥 − 15) = 0
15
𝑥 = −4, 2
II. 3y² + 13y + 4 = 0
3y² + 12y + y + 4 = 0
3y (y + 4) + 1 (y + 4) = 0
(3y + 1) (y + 4) = 0
1
𝑦 = − , −4
3
No relation between x and y
50. (e): I. x² - 17x – 84 = 0
x² +4x – 21x – 84 = 0
(x + 4) (x – 21) = 0
x = -4, 21
39
p
52. (e): I. x 2 − 13x + 11x − 143 = 0
(x – 13)(x+11) = 0
x = -11, 13
II. y 2 = 169
y = ± 13
clearly, no relation can be established
m
gr
53. (a): I. x 2 − 9x + 2x − 18 = 0
(x – 9) (x + 2) = 0
x = -2, 9
II. y 2 − 10y − 9y + 90 = 0
(y – 10) (y – 9) = 0
y = 9, 10
clearly, x ≤ y
54. (c): I. 2x 2 + 3x + 2x + 3 = 0
(x + 1) (2x + 3) = 0
3
x = -1, -
te
II. 2y² - 23y + 56 = 0
2y² - 16y – 7y + 56 = 0
2y(y – 8) – 7(y – 8) = 0
(2y – 7) (y – 8) = 0
7
𝑦 = 2,8
51. (d): I. 𝑥 2 = 81
𝑥 = ±9
II. (𝑦 − 9)2 = 0
𝑦=9
Clearly, 𝑥 ≤ 𝑦
xa
48. (d): I. 2x² + x – 28 = 0
2x² + 8x – 7x – 28 = 0
2x (x + 4) – 7 (x + 4) = 0
(2x – 7) (x + 4)= 0
7
𝑥 = −4,
II. y² + 4y – 117 = 0
y² - 9y + 13y – 117 = 0
(y – 9) (y + 13) = 0
y = 9, -13
No relation between x and y
ou
Quantity II:
Let present age of P be p years.
So, present age of R = (𝑝 − 15) years
ATQ,
Present age of Q = (2 × (𝑝 − 15)) − 𝑝
= (𝑝 − 30) 𝑦𝑒𝑎𝑟𝑠
Now, (𝑝 + 4) = 2 × (𝑝 − 30 + 4)
⇒ 𝑝 = 56
Hence, present age of R = (𝑝 − 15) years
= 41 years
And, present age of Q = (𝑝 − 30) 𝑦𝑒𝑎𝑟𝑠
= 26 years
So, required age = 26 years
So, Quantity I > Quantity II.
2
II. y 2 + 6y − 2y − 12 = 0
(y – 2) (y + 6) = 0
y = 2, -6
clearly, no relation can be established
55. (d): (II) × 9 – (I) × 4
On solving,
x = 1, y = 2
clearly, x < y
56. (c): I. 2x 2 − 2x + x − 1 = 0
(2x + 1) (x – 1) = 0
1
x=- ,1
2
II. 3y 2 − 3y − 2y + 2 = 0
(3y – 2) (y – 1) = 0
2
y= ,1
3
clearly, no relation
57. (d): x² = 81
x=±9
Y² – 18y + 81 = 0
(y – 9)² = 0
∴ y = 9, 9
∴x≤y
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
II. y2 +42y + 185 =0
y2+37y+5y+185 =0
y(y+37)+5(y+37)=0
(y+5)(y+37)=0
y = -5, - 37
So, x > y
59. (c): x² – 21x + 108 = 0
x² – 9x – 12x + 108 = 0
x(x – 9) – 12 (x – 9) = 0
x = 9, 12
y² – 17y + 72 = 0
∴ y² – 8y – 9y + 72 = 0
y (y – 8) – 9 (y – 8) = 0
∴ y = 8, 9
∴x≥y
2
3
gr
So, x < y
m
65. (e): I. x 2 − 8x − 273 = 0
x2-21x+13x-273=0
x(x-21)+13(x-21)=0
(x+13)(x-21)=0
x= -13, 21
II. y2 +6y – 432 =0
y2+24y-18y-432 =0
y(y+24)-18(y+24)=0
(y-18)(y+24)=0
y = 18, - 24
So, No relation
66. (e):
te
61. (c): x³ = 512
3
x = √512 = 8
y² = 64
y = √64 = ± 8
∴x≥y
2
xa
60. (b): x² – 11x + 30 = 0
x² – 6x – 5x + 30 = 0
∴ x(x – 6) – 5(x – 6) = 0
x = 6, 5
y² – 15y + 56 = 0
y² – 7y – 8y + 56 = 0
y (y – 7) – 8 (y – 7) = 0
∴ y = 7, 8
∴x<y
5
y = ,-
ou
64. (c): I. x2 + 41x + 420 =0
x2+21x+20x+420=0
x(x+21)+20(x+21)=0
(x+20)(x+21)=0
x= -20, - 21
II. 6y2 -11y – 10 =0
6y2-15y+4y-10 =0
3y(2y-5)+2(2y- 5)=0
(2y-5)(3y+2)=0
p
58. (e): 4x² – 24x + 30 = 0
4x² – 16x – 8x + 32 = 0
4x (x – 4) –8 (x–4) = 0
x = 4, 2
y² – 8y + 15 = 0
y² – 5y – 3y + 15 = 0
y(y – 5)–3 (y – 5) = 0
∴ y = 5, 3
∴ No relation exists
@
ce
62. (c): I. 2x 2 + 11x + 12 = 0
2x2 + 8x + 3x + 12 = 0
2x (x + 4) + 3(x + 4) = 0
(x + 4) (2x + 3) = 0
3
x= -4, -
67. (a):
2
II. 8y2 -22y – 21 =0
8y2+6y-28y-21 =0
2y(4y+3)-7(4y+3)=0
(2y-7)(4y+3)=0
7 3
y = ,2
68. (d):
4
So, x < y
63. (a): I. x 2 − 17x − 60 = 0
x2-20x+3x-60=0
x(x-20)+3(x-20)=0
(x+3)(x-20)=0
x= - 3, 20
40
69. (a): I. x² – 3x – 4=0
𝑥 2 − 4𝑥 + 𝑥 − 4 = 0
(x – 4) (x + 1) = 0
x = 4, –1
Adda247 Publications
For More Study Material
Visit: adda247.com
A Complete Guide on Quantitative Aptitude for Banking & Insurance Examinations
II. y² + 6y + 8 = 0
y² + 2y + 4y + 8 = 0
(y + 2) (y + 4) = 0
y = –2, –4
⇒x>y
73. (e): Quantity I:
(x+3)² = (x–3)²+x²
x²+9+6x=x²+9–6x+x²
x²–12x=0
x(x–12)=0
x=0,12
Quantity II:
y²–29y+204=0
y²–12y–17y+204=0
y(y–12)–17(y–12)=0
(y–12)(y–17)=0
y=12, 17
So, Quantity II ≥ Quantity I
ou
p
70. (b): I. x² – 3x = 10
𝑥 2 − 3𝑥 − 10 = 0
x² – 5x + 2x – 10 = 0
(x – 5) (x + 2) = 0
x = –2, 5
II. y² + 7y + 10 = 0
y² + 5y + 2y + 10 = 0
(y + 5) (y + 2) = 0
y = – 2, –5
⇒x≥y
74. (a): Quantity I:
1
Amount = 2450 + 2450 × 7 × 2
= 2450 + 700 = Rs. 3150
Quantity II:
71. (d): Quantity I:
18
1 2
Amount = 2450 (1 + 8)
× 5 = 500 meter.
12
55.2
= 𝑉 …(ii)
On solving (i) & (ii), we get:
𝑥 = 500 𝑚𝑒𝑡𝑒𝑟
So, Quantity II = quantity I
72. (b): Quantity I: Let C. P = Rs. 100 x
150
te
Then M. P= 100 𝑥 ×
120
100
= Rs. 150x
S.P = 100𝑥 × 100 = Rs. 120x
@
ce
ATQ,
120 x = Rs. 1020
x = Rs. 8.5
So, 150x = Rs. 1275
Quantity II:
Perimeter of field = 37.5 ×4 =150 meter
=15000 cm.
Total cost of fencing = 15000×0.17 = Rs. 2550
So, Quantity I < Quantity II.
41
81
64
=Rs 3100.78.
So, Quantity I > Quantity II
75. (d): Quantity I:
Let total M. R. P of 5 article = Rs. 100x
1 article is free, then S. P for 5 articles
= 100x – 20x = Rs. 80x
Again, he gives 20% discount, S.P become of each
80
1
article = 80𝑥 × 100 × 5 = Rs. 12.8𝑥
xa
Now,
𝑥+1800
= 2450 ×
m
Quantity II:
Let speed of train be ‘V km/hr’ and length of train
be ‘x meter’.
ATQ,
𝑥
= 𝑉 …(i)
gr
length of train =
24×75
Actual Discount percentage=
=
7.2
20
20𝑥–12.8𝑥
20𝑥
× 100
× 100 = 36%
Quantity II:
Let C.P = Rs. x
Let Initial S.P= Rs. 7y
Final S.P = Rs. 8y
ATQ
36 7𝑦–𝑥
19
( 𝑥 ) × 100 =
8𝑦−𝑥
𝑥
× 100
252y – 36x = 152y – 19x
𝑥
100
= 17
𝑦
Let C.P= Rs. 100a
Final S.P= Rs. 136a
136𝑎−100𝑎
Final profit % = 100𝑎 × 100 = 36%
So, Quantity I = Quantity II
Adda247 Publications
For More Study Material
Visit: adda247.com