Practice Quantitative Aptitude Questions For Upcoming SBI PO Prelims & TMB Exams 2017 (Quadratic Equation & Data Interpretation):
Dear Readers, Important Practice Aptitude Questions for Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this.
Direction (Q. 1-5):In each of the following questions two equations I and II are given. You have to solve both the equations and give answer:
a) x < y
b) x > y
c) x ≥ y
d) x ≤ y
e) x = y or the relationship cannot be established.
1). I. 20x2 - 31x + 12 = 0
II. 20y2 + y - 12 = 0
2). I.2x2 - 27x + 91 = 0
II.2y2 + y - 136 = 0
3). I. 2x -13 √x + 21 = 0
II. 2y -15 √y + 28 = 0
4). I. x2 = 3136
II.y2 = 1764
5). I. x2 - 20x + 91 = 0
II.y2 - 6y - 91 = 0
Directions (Q. 6-10): Study the following graph and table carefully and answer the questions given below it.
Percentage of Obese Men, Obese Women and Obese Children in a State in Various Years
Total Number of Men, Women and Children in the state Over the Years
6).What was the approximate average of obese men, obese women and obese children in the year 2007?
a) 12683
b) 12795
c) 12867
d) 12843
e) 12787
7).The number of obese men in the year 2009 were what per cent of the men not suffering from obesity in the same year?
a) 55%
b) 60%
c) 50.5%
d) 65.5%
e) None of these
8).What was the respective ratio of the obese women in the year 2006 to the obese men in the year 2008?
a) 6 : 7
b) 21 : 65
c) 15 : 73
d) 48 : 77
e) None of these
9).What is the difference between the number of obese women and obese children together in the year 2006 and the number of obese men in the same year?
a) 5475
b) 5745
c) 5045
d) 5075
e) None of these
10).What, was the total number of children not suffering from obesity in the year 2004 and 2005 together?
a) 4350
b) 31560
c) 4530
d) 31650
e) Not le of these
Answer Key:
1).c 2).e 3).d 4).e 5).e 6).c 7).b 8).d 9).a 10).d
1). I.20 x2 – 31 x + 12 = 0
(4 x - 3) (5x – 4) = 0
x = (3/4), (4/5)
II.20y2 + y – 12 = 0
or, (4y – 3) (5y + 4) = 0
y = (3/4), (-4/5)
Hence x ≥ y
Answer: c)
2). I. 2x2 –27x + 91 = 0
or, (x– 7) (2x – 13) = 0
x = 7, (13/2)
II.2y2 + y –136 = 0
or, (y –8) (2y + 17)= 0
y = 8, (-17/2)
Answer: e)
3). I.2x – 13√x + 21 =0
= (√x – 3) ( 2√x –7) = 0
x = 9, 49/4
II.2y – 15√y + 28 =0
or, (2 √y –7) ( √y –4) = 0
y = 49/4 y = 16, Hence, x ≤ y
Answer: d)
4). I. x2 = 3136
II.y2 = 1764
Answer: e)
5). I. x2 – 20 x + 91 = (x –7) (x – 13)= 0 x = 7, 13
II.y2 – 6 y – 91 = (y –13) (y + 7) = 0 y = 13, –7
Answer: e)
6).Average or obese men, obese women and obese children in the year 2007.
= { [ 66000 x (35/100)] + [54000 × (25/100)] + [16000 x (125/100)] } / 3
= (23100 + 13500 + 2000) / 3 = 12866.7 ≈ 12867
Answer: c)
7). Obese men in the year 2009
= 78000 x (375/100) = 29250
In 2009, those are not obese = 78000 – 29250 = 48750
According to the question, 48750 x x% = 29250
48750 x (x/100) = 29250
x = (29250 × 100)/48750 = 60%
Answer: b)
8). Obese women in the year 2006
= 60000 x (20/100) = 12000
Obese men in the year 2008
= 70000 x (275/100) = 19250
Required ratio = 12000 / 19250 = 48/77 = 48 : 77
Answer: d)
9). Number of obese women and obese children in the year 2006 = [60000 x (20/100)] + [12000 x(25/ 100)]
= 12000 + 3000 = 15000
Obese men in the year 2006 = 63000 x (325/100) = 20475
Difference = 20475 – 15000 = 5475
Answer: a)
10). In 2004 and 2005, those who are obese
= [15000 x (15/100)] + [21000 x (10/100)] = 2250 + 2100 = 4350
Those who are not obese are given by
= Total children in (2004 and 2005) — Number of children
Who are obese in the same year = 36000 — 4350 = 31,650
Answer: d)
More Practice Aptitude Questions for Upcoming Exams 2017 -Click Here
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