Quantitative Aptitude Quiz For SBI / RBI Main Exam | 15-05-2020

Dear Readers,

Mahendras has started special quizzes for SBI / RBI Main Exam so that you can practice more and more to crack the examination. This SBI / RBI Main Exam quiz series will mold your preparations in the right direction and the regular practice of these quizzes will be really very helpful in scoring good marks in the Examination. Here we are providing you important question of Quantitative aptitude for SBI / RBI Main Exam.

Q-(1-5) What will come in place of question mark (?) in the given number series?

दी गयी संख्या श्रृंखला में प्रश्नवाचक चिन्ह (?) के स्थान पर क्या आएगा?

(1) 24 25 ? 41 – 8 73

(1) 13

(2) 16

(3) 25

(4) 43

(5) 50

(2) 16 ? 32 16 4 0.5
(1) 32

(2) 20

(3) 24

(4) 28

(5) 18

(3) 9 11 16 26 ? 69

(1) 31

(2) 38

(3) 46

(4) 45

(5) 43

(4) 8 20 35 54 78 ?
(1) 114

(2) 169

(3) 106

(4) 108

(5) 136

(5) 16.8 16.85 16.75 16.9 16.7 ?
(1) 17.5

(2) 17.25

(3) 16.95

(4) 16.25

(5) 18.85

Q-(6-10) In the fol­lowing questions two equations numbered I and II are given. You have to solve both the equations and Give answer:

निम्नलिखित प्रश्नों में दो समीकरण I और II दिए गए हैं। आपको दोनों समीकरणों को हल करना होगा और उत्तर देना होगा:

(6) I. x2 – 1 = 0
II. y2 + 4y + 3 = 0

(1) If x > y

(2) If x < y

(3) If x > y

(4) If x < y

(5) If x = y or relation cannot be established.


(7) I. x2 – 0.10x + 0.24 = 0
II. y2 – 0.14y + 0.48 = 0
(1) If x > y

(2) If x < y

(3) If x > y

(4) If x < y

(5) If x = y or relation cannot be established.


(8) I. (x – 18)2 = 0
II. y2 = 324

(1) If x > y

(2) If x < y

(3) If x > y

(4) If x < y

(5) If x = y or relation cannot be established.


(9) I. 3x2 + 7x = 6
II. 6 (2y2 + 1) = 17y
(1) If x > y

(2) If x < y

(3) If x > y

(4) If x < y

(5) If x = y or relation cannot be established.


(10) A, B and C can independently finish a piece of work in 18 days, ‘x’ days and 27 days
respectively. A and C started working together and after 6 days B replaced both of them. If B could finish the remaining work in 16 days, what is the value of ‘x’?

A, B और C स्वतंत्र रूप से एक काम को क्रमशः18 दिनों, ‘x’ दिनों और 27 दिनों में पूरा कर सकते हैं । A और C ने एक साथ काम करना शुरू किया और 6 दिन बाद B ने दोनों का स्थान ले लिया । यदि B, 16 दिनों में शेष कार्य पूरा कर सका, तो ‘x’ का मान क्या है?

(1) 32

(2) 36

(3) 34

(4) 40

(5) None of these

Answer Key-

Q-(1) Sol-(2)
+ 12, – 32, + 52, – 72, + 92

Q-(2) Sol-(1)
× 2, × 1, ×1/2 , ×1/4, ×1/8

Q-(3) Sol-(5)
9 11 16 26 ? 69
+ 2 + 5 + 10 + 17 + 26
+ 3 + 5 + 7 + 9

Q-(4) Sol-(4)
8 20 35 54 78 ?
+ 12 + 15 + 19 + 24 + 30
+ 3 + 4 + 5 + 6

Q-(5) Sol-(3)
+ 0.05, – 0.1, + 0.15, – 0.2, + 0.25

Q-(6) Sol-(3)
I. x2 – 1 = 0
x = + 1, – 1
II. y2 + 4y + 3 = 0
y2 + 3y + y + 3 = 0
y(y + 3) + 1(y + 3) = 0
(y + 1) (y + 3) = 0
y = – 1, – 3
Hence, x > y

Q-(7) Sol-(4)
I. x2 – 0.10x + 0.24 = 0
x2 – 0.6x – 0.4x + 0.24 = 0
x (x – 0.6) + 0.4 (x – 0.6) = 0
(x + 0.4) (x – 0.6) = 0
x = – 0.4, 0.6
II. y2 – 0.14y + 0.48 = 0
y2 – 0.6y – 0.8y + 0.48 = 0
y (y – 0.6) – 0.8 (y – 0.6) = 0
(y – 0.6) (y – 0.8) = 0
y = 0.6, 0.8
Hence, x < y

Q-(8) Sol-(3)
I. (x – 18)2 = 0
x = 18, 18
II. y2 = 324
y = 18, – 18
Hence, x > y

Q-(9) Sol-(4)
I. 3x2 + 7x = 6
3x2 + 7x – 6 = 0
3x2 + 9x – 2x – 6 = 0
3x (x + 3) – 2(x + 3) = 0
(3x – 2) (x + 3) = 0
x = 2/3, – 3
II. 6 (2y2 + 1) = 17y
12y2 – 17y + 6 = 0
12y2 – 9y – 8y + 6 = 0
3y (4y – 3) – 2(4y – 3) = 0
(3y – 2) (4y – 3) = 0
y = 2/3,3/4
Hence, x < y

Q-(10) Sol-(2)
(A + C)’s 1 day work = 5/54
(A + C)’s 6 day work = 5/9
Remaining work = 1 – 5/9=4/9
According to the question,
16/x=4/9
x = 36 days