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# Quantitative Aptitude Quiz For SBI Clerk | 27-06-19

Mahendra Guru

As SBI has released the much-awaited vacancies for the post of Junior Assistants (Clerk), 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 SBI 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 SBI Clerk 2019 exam.

Q. (1 – 3). What approximate value should come in place of the question mark (?) in the following questions?

निम्नलिखित प्रश्नों में प्रश्नवाचक चिन्ह (?) के स्थान पर लगभग मान क्या आयेगा?

Q.1.

(1) 18
(2) 13
(3) 11
(4) 16
(5) 20

Q.2. 19.068 × 18.345 × ? = 11538.9337

(1) 30
(2) 40
(3) 45
(4) 50
(5) 35

Q.3.

(1) 50
(2) 55
(3) 45
(4) 60
(5) 35

Q. (4 – 5). In each question two equation I & II are given, on the basis of these you have to find out relation between p and q.

निम्नलिखित प्रश्नों में दो समीकरण I और II दिए गये हैं आपको दोनो समीकरण हल करना है और उत्तर दीजिए-

Q.4. (I) p2 + 7p + 12 = 0
(II) q2 – 2q – 8 = 0

(1) p = q or relation cannot be established/या सम्बन्ध स्थापित नहीं किया जा सकता
(2) p > q
(3) q > p
(4) p ≥ q
(5) q ≥ p

Q.5. (I) p2 – 3p + 2 = 0
(II) q2 – 7q = – 12

(1) p = q or relation cannot be established/या सम्बन्ध स्थापित नहीं किया जा सकता
(2) p > q
(3) q > p
(4) p ≥ q
(5) q ≥ p

Q. (6 – 10). In each of the following number series a wrong number is given. Find out the wrong number.

निम्नलिखित संख्या श्रृंखला में एक संख्या गलत है। गलत संख्या ज्ञात कीजिये।

Q.6. 26    146    66    115    90    99

(1) 26
(2) 66
(3) 115
(4) 146
(5) 99

Q.7. 7    8    9    21    9.25    51.25

(1) 7
(2) 9
(3) 8
(4) 9.25
(5) 51.25

Q.8. 144   72    24    6    1.2    0.3

(1) 72

(2) 24
(3) 6
(4) 1.2
(5) 0.3

Q.9. 4    11    10    48    173    888

(1) 4
(2) 11
(3) 10
(4) 173
(5) 888

Q.10. – 10    – 9    – 10    – 4    52    385

(1) – 10
(2) – 9
(3) 4
(4) 52
(5) 385

1. Sol. (2) (1 + 7 + 4) + = ?
12 + = ?
12 +

2. Sol. (5)

= 35

3. Sol. (3)
? = 45

4. Sol. (3) (I) p2 + 7p + 12 = 0
p2 + 4p + 3p + 12 = 0
p (p + 4) + 3 (p + 4) = 0
p = – 4, – 3
(II) q2 – 2q – 8 = 0
q2 – 4q + 2q – 8 = 0
q (q – 4) + 2(q – 4) =0
q = 4, – 2
q > p

5. Sol. (3) (I) p2 – 3p + 2 = 0
p2 – 2p – p + 2 = 0
p (p – 2) – 1 (p – 2) = 0
p = 2, 1
(II) q2 – 7q + 12 = 0
q2 – 4q – 3q + 12 = 0
q (q – 4) – 3 (q – 4) = 0
q = 4, 3
q > p

6. Sol. (4) 26 + 112 = 147
147 – 92 = 66
66 + 72 = 115
115 – 52 = 90
90 + 32 = 99

7. Sol. (2) 7 × 1 + 1 = 8
8 ÷ 2 + 2 = 6
6 × 3 + 3 = 21
21 ÷ 4 + 4 = 9.25
9.25 × 5 + 5 = 51.25

8. Sol. (4)

9. Sol. (3) 4 × 1 + 7 = 11
11 × 2 – 11 = 11
11 × 3 + 15 = 48
48 × 4 – 19 = 173
173 × 5 + 23 = 888

10. Sol. (3) – 10 × 1 + 13 = – 9
– 9 × 2 + 23 = – 10
– 10 × 3 + 33 = – 3
– 3 × 4 + 43 = 52
52 × 5 + 53 = 385

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