SBI Clerk Quant Previous year 2018 asked Questions With Video Explanations PDF
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SBI Clerk Topic-Wise Important Questions (Download PDF)
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Instructions
What should come in place of the question mark (?) in the following number series?
Question 1: 17, 19, 23, 29, 37, ?
a) 46
b) 49
c) 47
d) 48
e) 45
Question 2: 900, 899, 891, 864, 800, ?
a) 695
b) 685
c) 665
d) 675
e) 655
Question 3: 4, 32, 224, 1344, 6720, ?
a) 26885
b) 26880
c) 26882
d) 26888
e) 26883
Question 4: 56, 54, 58, 50, 66, ?
a) 34
b) 98
c) 38
d) 94
e) 44
Question 5: 655, 637, 622, 610, 601, ?
a) 598
b) 593
c) 595
d) 597
e) 594
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Instructions
approximate value will sonar place of the question mark (1 in ilk. following questions ? (you are not required to calculate the exact value.).
Question 6: 10303.88 $\div$ 55.94 + 62.95 = ?
a) 247
b) 250
c) 260
d) 220
e) None of these
Question 7: $\sqrt{\frac{12321}{36.07}} = ?$
a) 20
b) 18.5
c) 17
d) 17.5
e) None of these
Question 8: 19.03 $\times$ 16.98 $\times$ 13.01 = ?
a) 4000
b) 4100
c) 4200
d) 4250
e) None of these
Question 9: 117% of 459.88 – 162% of 143.02 = ?
a) 290
b) 280
c) 300
d) 306
e) None of these
Question 10: 3/5 $\times$ 4/9 $\times$ 5894.92=?
a) 1527
b) 1572
c) 1752
d) 1725
e) None of these
Instructions
In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer. Give answer :
a:If x < y
b: If x > y
c: If x ≤ y
d: If x ≥ y
e: If relationship between x and y cannot be determined ,
Question 11: I. $15x^{2} + 26x + 8 = 0$
II. $25y^{2} + 15y + 2 = 0$
a) If x < y
b) If x > y
c) If x ≤ y
d) If x ≥ y
e) If relationship between x and y cannot be determined
Question 12: I. $6x^{2} – 19x + 15 = 0$
II. $5y^{2} – 22y + 24 = 0$
a) If x < y
b) If x > y
c) If x ≤ y
d) If x ≥ y
e) If relationship between x and y cannot be determined
Question 13: I. $4x^{2} – 12x + 5 = 0$
II. $4y^{2} – 8y + 3 = 0$
a) If x < y
b) If x > y
c) If x ≤ y
d) If x ≥ y
e) If relationship between x and y cannot be determined
Question 14: I. $10x^{2} + 21x + 8 = 0$
II. $5y^{2} + 19y + 18 = 0$
a) If x < y
b) If x > y
c) If x ≤ y
d) If x ≥ y
e) If relationship between x and y cannot be determined
Question 15: I. $6x^{2} – 5x + 1 = 0$
II. $12y^{2} – 23y + 10 = 0$
a) If x < y
b) If x > y
c) If x ≤ y
d) If x ≥ y
e) If relationship between x and y cannot be determined
Banking Study Material – 15000 Questions
Instructions
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.
a: If quantity I ≥ quantity II
b: If quantity I > quantity II
c: If quantity I < quantity II
d: If quantity I = quantity II or the relationship cannot be established from the information that is given
e: If quantity quantity II
Question 16: Arun and Bhadra are brothers. In how many years from now will Bhadra’s age be 50 years ? •
I. The ratio of the current ages of Arun and Bhadra is 5 : 7 respectively.
II. Bhadra was born 10 years before Arun.
III. 5 years hence, Arun’s age would be three-fourth of Bhadra’s age at that time.
a) Any two of the three
b) Only II and either I or III
c) All I, II and III
d) Only II and III
e) Only I and III
Question 17: A right-angled triangle is inscribed in a given circle. What is the area of the given circle (in cm2) ?
I. The base and height of the triangle (in cm) are both the roots of the equation $x^{2}-23x+120=0$
II. The sum of the base and height of the triangle is 23 cm.
III. The height of the right-angled triangle is greater than the base of the same.
a) III and either omly I or only II
b) ALL I,II and III
c) Only II and III
d) Only I
e) Either I or II
Answers & Solutions:
1) Answer (C)
Let the missing number be x
Here the pattern is
19-17 = 2
23-19 = 4
29-23 = 6
37-29 = 8
As we can see that the difference is geeting increased by 2 every time so ,
x-37 =10,it implies x= 47
2) Answer (D)
Let the missing number be x
900-899 = (1)^2
899-891 = 8 = (2)^3
891-864 = 27 = (3)^3
864-800 = 64 = (4)^3
as we can see the pattern between differences of consecutive numbers is of type (n)^3
so x-800 = (5)^3,this implies that x= 675
3) Answer (B)
Let the missing number be x
32÷4 = 8
224÷32 = 7
1344÷224 = 6
6720÷1344 = 5
As we can see that the division of consecutive numbers is forming a pattern of decreasing numbers by 1, so
x÷6720 = 4, it implies x is 26880
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4) Answer (A)
Let the missing number be x
Here , a pattern is getting formed between consecutive numbers.
54−56 =$(-2)^1$
58−54= $(-2)^2$
50−58= $(-2)^3$
66−50= $(-2)^4$
As we can see that the difference is of the form $(-2)^{n}$
so for missing number x,
x−66= $(-2)^{5}$
This implies that x is 34
5) Answer (C)
Let the missing number be x.
The difference between every two consecutive numbers is forming a pattern.
655-637 = 18
637-622 = 15
622-610 = 12
610-601 = 9
As the difference is getting reduced by 3 every time . So 601-x = 6 which implies that the missing number ,x =595
6) Answer (A)
10303.88$\div$55.94 + 62.95.
~10304$\div$56+63.
=184+63.
=247.
Hence, Option A is correct.
7) Answer (B)
$\sqrt{\frac{12321}{36.07}}$.
~$\sqrt{\frac{12321}{36}}$.
We know that,
$111^{2}=12321, 6^{2}=36$.
~$\sqrt{\frac{12321}{36}}=\frac{111}{6}$.
=$18.5$.
Hence, Option B is correct.
8) Answer (C)
19.03 $\times$ 16.98 $\times$ 13.01.
~19 $\times$ 17 $\times$ 13.
=4199.
~4200.
Hence, Option C is correct.
9) Answer (D)
117% of 459.88 – 162% of 143.02.
~117% of 460-162% of 143.
=$\frac{117\times460}{100}-\frac{162\times143}{100}$.
=$\frac{53820}{100}-\frac{23166}{100}$.
=$538.20-231.66$.
~$306$.
Hence, Option D is correct.
10) Answer (B)
3/5 $\times$ 4/9 $\times$ 5894.92.
~3/5 $\times$ 4/9 $\times$ 5895.
=$3\times4\times131$.
=$1572$.
Hence, Option B is correct.
11) Answer (C)
I. $15x^{2} + 26x + 8 = 0$
=> $15x^2 + 6x + 20x + 8 = 0$
=> $(3x + 4) (5x + 2) = 0$
=> $x = \frac{-4}{3} , \frac{-2}{5}$
II. $25y^{2} + 15y + 2 = 0$
=> $25y^2 + 5y + 10y + 2 = 0$
=> $(5y + 2) (5y + 1) = 0$
=> $y = \frac{-2}{5} , \frac{-1}{5}$
$\therefore x \leq y$
12) Answer (A)
I. $6x^{2} – 19x + 15 = 0$
=> $6x^2 – 9x – 10x + 15 = 0$
=> $(3x – 5) (2x – 3) = 0$
=> $x = \frac{5}{3} , \frac{3}{2}$
II. $5y^{2} – 22y + 24 = 0$
=> $5y^2 – 10y – 12y + 24 = 0$
=> $(5y – 12) (y + 2) = 0$
=> $y = \frac{12}{5} , 2$
$\therefore y > x$
13) Answer (E)
I. $4x^{2} – 12x + 5 = 0$
=> $4x^2 – 2x – 10x + 5 = 0$
=> $(2x – 5) (2x – 1) = 0$
=> $x = \frac{5}{2} , \frac{1}{2}$
II. $4y^{2} – 8y + 3 = 0$
=> $4y^2 – 2y – 6y + 3 = 0$
=> $(2y – 3) (2y – 1) = 0$
=> $y = \frac{3}{2} , \frac{1}{2}$
$\therefore$ No relation can be established.
14) Answer (B)
I. $10x^{2} + 21x + 8 = 0$
=> $10x^2 + 5x + 16x + 8 = 0$
=> $(5x + 8) (2x + 1) = 0$
=> $x = \frac{-8}{5} , \frac{-1}{2}$
II. $5y^{2} + 19y + 18 = 0$
=> $5y^2 + 10y + 9y + 18 = 0$
=> $(5y + 9) (y + 2) = 0$
=> $y = \frac{-9}{5} , -2$
$\therefore x > y$
15) Answer (A)
I. $6x^{2} – 5x + 1 = 0$
=> $6x^2 – 2x – 3x + 1 = 0$
=> $2x (3x – 1) – 1(3x – 1) = 0$
=> $(2x – 1) (3x – 1) = 0$
=> $x = \frac{1}{2} , \frac{1}{3}$
II. $12y^{2} – 23y + 10 = 0$
=> $12y^2 – 8y – 15y + 10 = 0$
=> $4y (3y – 2) – 3 (3y – 2) = 0$
=> $(4y – 3) (3y – 2) = 0$
=> $y = \frac{3}{4} , \frac{2}{3}$
$\therefore y > x$
16) Answer (A)
I & II : Let Arun’s age = $x$ years
=> Bhadra’s age = $x + 10$ years
$\therefore \frac{x}{x + 10} = \frac{5}{7}$
=> $7x = 5x + 50$
=> $7x – 5x = 2x = 50$
=> $x = \frac{50}{2} = 25$
=> Bhadra’s age = $25 + 10 = 35$ years
Thus, I & II are sufficient.
II & III : Let Arun’s age = $x$ years
=> Bhadra’s age = $x + 10$ years
$\therefore \frac{x + 5}{x + 15} = \frac{3}{4}$
=> $4x + 20 = 3x + 45$
=> $4x – 3x = 45 – 20$
=> $x = 25$
=> Bhadra’s age = $25 + 10 = 35$ years
Thus, II & III are sufficient.
I & III : Let Arun’s age = $5x$ years
=> Bhadra’s age = $7x$ years
$\therefore \frac{5x + 5}{7x + 5} = \frac{3}{4}$
=> $20x + 20 = 21x + 15$
=> $21x – 20x = 20 – 15$
=> $x = 5$
=> Bhadra’s age = $7 \times 5 = 35$ years
Thus, I & III are sufficient.
$\therefore$ Any two of the three statements are sufficient.
17) Answer (D)
I : $x^2 – 23x + 120 = 0$
=> $x^2 – 8x – 15x + 120 = 0$
=> $x (x – 8) – 15 (x – 8) = 0$
=> $(x – 8) (x – 15) = 0$
=> $x = 8 , 15$
Thus, base = 8 cm and height = 15 cm (or vice versa)
=> Hypotenuse of right angled triangle = $\sqrt{(8)^2 + (15)^2}$
= $\sqrt{64 + 225} = \sqrt{289} = 17 cm$
Since, triangle is inscribed in circle, => Radius of circle = half of hypotenuse
=> $r = \frac{17}{2} = 8.5$ cm
$\therefore$ Area of circle = $\pi r^2$
= $\frac{22}{7} \times 8.5 \times 8.5 \approx 227 cm^2$
Thus, I alone is sufficient.
Clearly, we cannot find base and height from statements II or III. Thus, they are insufficient.
