Quantitative Aptitude quiz is basically to test your mathematical calculation and approach. It is to find out how fast you can solve a given question with the right methodology of solving the problem. If you know the formulas and short tricks of some important topic in Maths, you will definitely score good marks. So, it is important to know the basic concepts of all the topics so that you can apply the short tricks and solve the question with a new concept in lesser time while giving the quiz. Quantitative Aptitude Quiz helps to evaluate your preparation for banking exam so you can improve your preparation level. Mahendra Guru provides you Quantitative Aptitude Quiz for Bank examination based on the latest pattern so that you can practice on regular basis. It will definitely help you to score good marks in the exam. It is the most important section for all the govt exam like IBPS PO/ Clerk/SO/RRB, RBI, SBI, Insurance, SSC-MTS, CGL, CHSL, State Level and other Competitive exams.
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Q-(1-3) In each of the following question two equations are given you have to solve them and give answer.
निम्नलिखित प्रश्नों में दो समीकरण दिये गये हैं, आपको उन्हें हल करना है तथा उत्तर दीजिये –
Q-1 (I) x2
– x – 6 = 0
(II) y2
+ 7y + 10 = 0
(1) If/यदि x < y
(2) If/यदि x > y
(3) If/यदि x < y
(4) If/यदि x > y
(5) If x = y or no relation between x and y can be established/यदि x = y या x और y के मध्य कोई सम्बन्ध स्थापित नहीं किया जा सकता है |
Q-2 (I) 4x – 3y = 12
(II) 2x + 3y – 30 = 0
(1) If/यदि x < y
(2) If/यदि x > y
(3) If/यदि x < y
(4) If/यदि x > y
(5) If x = y or no relation between x and y can be established/यदि x = y या x और y के मध्य कोई सम्बन्ध स्थापित नहीं किया जा सकता है |
Q-3 (I) 2x2 + 5x + 2 = 0
(II) 2y2 + 11y + 14 = 0
(1) If/यदि x < y
(2) If/यदि x > y
(3) If/यदि x < y
(4) If/यदि x > y
(5) If x = y or no relation between x and y can be established/यदि x = y या x और y के मध्य कोई सम्बन्ध स्थापित नहीं किया जा सकता है |
Q-4 (I) x2 – 19x = – 88
(II) y2 – 48y + 576 = 0
(1) If/यदि x < y
(2) If/यदि x > y
(3) If/यदि x < y
(4) If/यदि x > y
(5) If x = y or no relation between x and y can be established/यदि x = y या x और y के मध्य कोई सम्बन्ध स्थापित नहीं किया जा सकता है |
Q-5 (I) x2 – 9x + 14 = 0
(II) y2 – y – 2 = 0
(1) If/यदि x < y
(2) If/यदि x > y
(3) If/यदि x < y
(4) If/यदि x > y
(5) If x = y or no relation between x and y can be established/यदि x = y या x और y के मध्य कोई सम्बन्ध स्थापित नहीं किया जा सकता है |
Q-(6-10) – What approximate value will come in place of question mark (?) in the following question?
निम्नलिखित प्रश्न में प्रश्नवाचक चिन्ह (?) के स्थान पर लगà¤à¤— क्या मान आयेगा?
Q-6 8724.997 × 3.987 ÷ 16.009 = ? + 821.44
(1) 1250
(2) 1190
(3) 1450
(4) 1200
(5) 1360
Q-7 
(1) 95
(2) 108
(3) 130
(4) 100
(5) 125
Q-8 
(1) 120
(2) 100
(3) 165
(4) 140
(5) 180
Q-9 (32.12)2 + (4.92)3
– (14.04)2 = ?
(1) 780
(2) 1150
(3) 950
(4) 800
(5) 1200
Q-10 
(1) 1800
(2) 2200
(3) 1750
(4) 1920
(5) 2300
ANSWER KEY
Q1-(4)
(I) x2 – x
– 6 = 0
x2 – 3x + 2x – 6 = 0
x(x – 3) +2(x – 3) = 0
(x – 3) (x+2) = 0
x = 3, x = –2
(II) y2 + 7n + 10 = 0
y2 + 5n + 2y + 10 = 0
y(y + 5) + 2(y + 5) = 0
(y + 5) (y + 2) = 0
y = – 5, y = – 2
x > y
x2 – 3x + 2x – 6 = 0
x(x – 3) +2(x – 3) = 0
(x – 3) (x+2) = 0
x = 3, x = –2
(II) y2 + 7n + 10 = 0
y2 + 5n + 2y + 10 = 0
y(y + 5) + 2(y + 5) = 0
(y + 5) (y + 2) = 0
y = – 5, y = – 2
x > y
Q2-(2)
4x – 3y = 12
2x + 3y = 30
Solving (I) and (II)/ (I) and (II) को हल करने पर,
x = 7
y =
x > y
Q3-(4)
(I) 2x2 + 5x + 2 = 0
2x2 + 4x + x + 2 = 0
2x (x + 2) + 1(x + 2) = 0
(x + 2) (2x + 1) = 0
2x2 + 4x + x + 2 = 0
2x (x + 2) + 1(x + 2) = 0
(x + 2) (2x + 1) = 0
x = – 2, x = 
(II)2y2 + 11y +
14 = 0
2y2 + 7y + 4y + 14 = 0y (2y + 7) + 2 (2y + 7) = 0
(2y + 7) (y + 2) = 0
y =
, y = –2
, y = –2
Q4-(1) (I) x2 – 19x + 88 = 0
x2 – x (11 + 8) + 88 = 0
x2 – 11x – 8x + 88 = 0
x(x – 11) – 8(x – 11) = 0
(x – 11) (x – 8) = 0
x = 11, x = 8
x2 – 11x – 8x + 88 = 0
x(x – 11) – 8(x – 11) = 0
(x – 11) (x – 8) = 0
x = 11, x = 8
(II) y2 – 48y +
576 = 0
y2 – 24y – 24y + 576 = 0
(y – 24)2 = 0
y = 24, 24
x < y
y2 – 24y – 24y + 576 = 0
(y – 24)2 = 0
y = 24, 24
x < y
Q5-(4) (I) x2 –
9x + 14 = 0
x2 – x (7 + 2) + 14
= 0
x2 – 7x – 2x + 14 = 0
x(x – 7) – 2(x – 7) = 0
(x – 7) (x – 2) = 0
x = 7, x = 2
(II) y2 – y – 2 = 0
y2 – 2y + y – 2 = 0
y(y – 2) +1(y – 2) = 0
(y – 2) (y +1) = 0
y = 2, y = –1
x > y
x2 – 7x – 2x + 14 = 0
x(x – 7) – 2(x – 7) = 0
(x – 7) (x – 2) = 0
x = 7, x = 2
(II) y2 – y – 2 = 0
y2 – 2y + y – 2 = 0
y(y – 2) +1(y – 2) = 0
(y – 2) (y +1) = 0
y = 2, y = –1
x > y
Q6-(5)
8725 × 4 ÷ 16 – 821 = ?
2181.25 – 821 = ?
? = 1360.25 ≈ 1360
Q7-(2)
4 × 9 × 3 ≈ ?
? ≈ 108
Q8-(4)
? = 10.3×11.3 ×1.2
? ≈ 140
Q9-(3)
1024 + 125 – 196 = ?
? = 1149 – 196
? = 953 ≈ 950
Q10-(4)
56.5 × 37.4 – 95 = ?
2113.1 – 95 = ?
? = 1923.1 ≈ 1920
