As IBPS has released the much-awaited vacancies for the post of Junior Assistants (Clerk) & PO, we have launched 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 IBPS PO/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 PO/Clerk 2019 exam.Q-(1-5) What value will come in place of question mark (?) in the questions given below ?
(1) 188×18×8-8088 = ?
(1) 18984
(2) 19022
(3) 18816
(4) 18626
(5) 18920
(2) 64% of 842 – ?% of 426 =334.4
(1) 34
(2) 38
(3) 42
(4) 46
(5) 48
(3) 9984 ÷
= 384(1) 636
(2) 646
(3) 656
(4) 666
(5) 676
(4) (16)3- (16)2 = ?
(1) 3840
(2) 3620
(3) 3710
(4) 3550
(5) 3320
(5) 26831 + 8348 - 2061 = ? × 58
(1) 569
(2) 573
(3) 575
(4) 571
(5) 582
Q-(6-10) In each of the following questions, two equations (I) and (II) are given. You have to solve them and Give answer-
निम्नलिखित प्रश्नों में दो समीकरण (I) और (II) दिये गये है दोनों समीकरण हल कीजिए और - उत्तर दीजिए
(6) (I) 10x2 -33x+27 = 0
(II) 18y2 - 17y +4 = 0
(1) x > y
(2) x > y
(3) x < y
(4) x < y
(5) x = y or relationship cannot be established
(7) (I) 2x2 + 5x - 12 = 0
(II) 2y2 + 5y - 7 = 0
(1) x > y
(2) x > y
(3) x < y
(4) x < y
(5) x = y or relationship cannot be established
(8) (I) 2x2 + 9x + 10 = 0
(II) 2y2 + 17y + 30 = 0
(1) x > y
(2) x > y
(3) x < y
(4) x < y
(5) x = y or relationship cannot be established
(9) (I) 17x - 57 = 300 -4x
(II) y =
(1) x > y
(2) x > y
(3) x < y
(4) x < y
(5) x = y or relationship cannot be established
(10) (I) 6x2 - 19x + 15 = 0
(II) 4y2 -16y + 15 = 0
(1) x > y
(2) x > y
(3) x < y
(4) x < y
(5) x = y or relationship cannot be established
Answer Key-
- Sol-(1)
2. Sol-(5)
3. Sol-(5)
4. Sol-(1)
? = 4096 – 256 = 3840
5. Sol-(4)
6. Sol-(1)
(I) 10x2-15x-18x+27 = 0
5x(2x - 3) -9(2x-3) = 0
x =
, 
(II) 18y2 - 9y -8y+4 = 0
9y(2y-1) -4(2y-1) = 0
y =
, 
x > y
7. Sol-(5)
(I) 2x2 +8x -3x-12 = 0
2x(x+4) -3(x+4) = 0
x =
,-4(II) 2y2 +7y-2y-7 =0
y(2y+7)-1(2y+7) = 0
y = 1,
So relationship can not be set up.
8. Sol-(2)
(I) 2x2 + 4x +5x+ 10 = 0
2x(x+2) +5 (x+2) = 0
x =
,-2(II) 2y2 + 12y + 5y +30 = 0
2y(y+6) +5(y+6) = 0
y = -6,
x > y9. Sol-(5)
(I) 17x + 4x = 357
21x = 357
x = 17
(II) y = 17
x = y10. Sol-(5)
(I) 6x2 - 10x -9x +15 = 0
2x(3x-5) -3(3x-5) = 0
x =
(II) 4y2 - 10y -6y +15 = 0
2y(2y-5) -3(2y-5) = 0
y =
So, relation can't be established.
