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Quantitative Aptitude Questions For IBPS Clerk Mains | 12- 01 - 19

Mahendra Guru
Quantitative Aptitude Questions For IBPS Clerk Mains | 12- 01 - 19
Dear Aspirants,

As IBPS has released the official notification of the Common Recruitment Process for selection of personnel for Clerical cadre Posts and the exam is tentatively scheduled to be held December 2018 & January 2019. Looking at the notification, we have now started subject-wise quizzes for the exam. It will include quizzes of all the subjects- Quantitative Aptitude, English, Reasoning and Computer. All these quizzes will be strictly based on the latest pattern of the IBPS Clerk exam and will be beneficial for your preparations. So, keep following the quizzes which will provide you a set of 10 questions daily.

Here, we are providing you important questions of Quantitative Aptitude for IBPS Clerk 2018 exam.

Q1-5 In each of the following questions, two equations are given. You have to solve them and state the correct relationship-

निम्नलिखित प्रश्नों में दो समीकरण (I) और (II) दिये गये है दोनों समीकरण हल कीजिए और सही सम्बन्ध स्थापित कीजिये - 

Q1. (I) 3x2 - 4 x -32 = 0          (II) 3y2 - 19 y - 14 = 0
(1) If x > y 
(2) If x > y 
(3) If x < y 
(4) If x < y 
(5) If x = y or relationship cannot be established 

Q2. (I) 2x2 - 17 x + 36 = 0           (II) 9y2 - 29 y + 22 = 0
(1) If x > y 
(2) If x > y 
(3) If x < y 
(4) If x < y 
(5) If x = y or relationship cannot be established 

Q3. (I) 3x + 5y = 34.5           (II) 4x - 9 y = -1
(1) If x > y 
(2) If x > y 
(3) If x < y 
(4) If x < y 
(5) If x = y or relationship cannot be established 

Q4. (I) (II)
(1) If x > y 
(2) If x > y 
(3) If x < y 
(4) If x < y 
(5) If x = y or relationship cannot be established  

Q4. (I) x2 + 16 x + 64 = 0         (II) y2 +121 y = 0
(1) If x > y 
(2) If x > y 
(3) If x < y 
(4) If x < y 
(5) If x = y or relationship cannot be established 

Q6-10. Read the following pie chart and table carefully and answer the questions asked.
Number of employees in different departments of Compaq and Sony company.
Ratio of males and females in different departments of both companies.
निम्नलिखित पाई-चार्ट और सारणी का ध्यानपूर्वक अध्ययनकर नीचे पूछे गये प्रश्नों के उत्तर दीजिए। 
कॉम्पैक और सोनी के विभिन्न विभागों में पुरुष और महिला का अनुपात 

Q6. What is the average number of males in department D5 and D6 in Compaq?

कॉम्पैक में विभाग D5 और D6 में पुरूषों की औसत संख्या क्या है?
(1) 1602 
(2) 1596 
(3) 1612 
(4) 1628 
(5) 1702 

Q7. The number of females in department D1 in Sony is equal to number of males in which department of Compaq?

सोनी में विभाग D1 में महिलाओं की संख्या कॉम्पैक कम्पनी के किस विभाग में पुरूषों की संख्या के बराबर है?
(1) D2 
(2) D4 
(3) D3 
(4) D6 
(5) None of these 

Q8. What is the respective ratio between the number of females in department D1 in Compaq and that of females in department D4 in Sony?

कॉम्पैक में विभाग D1 में महिलाओं की संख्या और सोनी में विभाग D4 में महिलाओं की संख्या के बीच क्रमशः अनुपात क्या है?
(1) 1 : 2 
(2) 3 : 2 
(3) 3 : 1 
(4) 2 : 3 
(5) None of these 

Q9. What is the difference between the number of females in department D5 of both companies?

दोनों कम्पनियों के विभाग D5 में महिलाओं की संख्या के बीच अन्तर क्या है?
(1) 170 
(2) 160 
(3) 168 
(4) 162 
(5) 165 

Q10. What is the total number of females in department D3 of both companies?

दोनों कम्पनियों के विभाग D3 में महिलाओं की कुल संख्या क्या है?
(1) 2880 
(2) 2940 
(3) 2950 
(4) 2910 
(5) 2980 

Answer Key-
Q1. (5) 
(I) 3x2 - 12x + 8x - 32 = 0
3x (x- 4) +8 (x-4) = 0
x = 4,-

(II) 3y2 - 21y + 2y - 14 = 0
3y (y - 7) +2 (y - 7) = 0
Relationship Cannot Be Established.

Q2. (1) 
(I) 2x2 - 8x - 9x + 36 = 0
2x (x- 4) -9 (x - 4) = 0
(II) 9y2 - 18y - 11y +22 = 0
9y (y - 2) -11 (y - 2) = 0
: x y

Q3. (1)
From Eq. (I) and (II)
y = 3
x = 6.5 
.'. x y

Q4. (3)
(I)3x + 7 = 0
x =

(II)6y3 = 972
y3 = 162
< y

Q5. (5)
(I) x2 + 8x +8x + 64 = 0
x (x + 8) +8 (x + 8) = 0
x = - 8, -8

(II) y2 + 121 y = 0
y (y + 121) = 0
y = 0 , -121 
Relationship can not be established.

Q6. (1)
Average number =
= 1602

Q7. (3)
D3

Q8. (2)
Ratio = 1500 : 1000 = 3 : 2

Q9. (5)
Difference = 525 -360 = 165

Q10. (4)
Total number = 1440+1470
= 2910

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