IPPB Scale I & IBPS SO 2017 - Practice Quantitative Aptitude Questions (Quadratic Equation & Application Problems) Set-96:
Dear Readers, Important Practice Aptitude Questions for IPPB & IBPS SO Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.
Direction (Q. 1-5):In each of these questions, two equations (I) and (II) with variable 'x' and 'y' are given. You have to solve both the equations to find the value of 'x' and 'y' and mark answer if-
a) If x > y
b) If x ≥ y
c) If x < y
d) If x ≤ y
e) If x = y or relationship between 'x' and 'y' can't be established
1). I.20x2 – 59x + 42 = 0
II. 2y2 – 7y + 6 = 0
2). I.8x2 + 3x – 5 = 0
II.6 y2 + 17 y + 12 = 0
3). I.2 x2 + x –1 = 0
II. 12 y2 – 25 y + 12 = 0
4). I.x = 3 √4913
II. y2 – 28 y + 187 = 0
5). I.7 x + 4 y + 5 = 0
II. 3 x – 7 y + 37 = 0
6). The perimeter of a sector of a circle of radius 6.5 cm. is 23.4 cm. What is the area of sector ?
a) 67.6 cm2
b) 48.4 cm2
c) 33.8 cm2
d) 24.2 cm2
e) None of these
7). A sum amounts to Rs. 882 in 2 years at 5% compound interest. The sum is
a) Rs. 800
b) Rs. 822
c) Rs. 840
d) Rs. 816
e) None of these
8). A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. B alone can complete the work in --- hours.
a) 12 hours
b) 6 hours
c) 8 hours
d) 10 hours
e) None of these
9). If selling price of an article is Rs. 250, profit percentage is 25%. Find the ratio of the cost price and the selling price
a) 5: 3
b) 3 : 5
c) 4 : 5
d) 5 : 4
e) None of these
10). A boatman can row 3 km against the stream in 20 minutes and return in 18 minutes. Find the rate of current
a) 1⁄2 km/hr
b) 1 km/hr
c) 1⁄3 km/hr
d) 2⁄3 km/hr
e) None of these
1) e 2) a 3) c 4) b 5) c 6) c 7) a 8) a 9) c 10) a
1). 20x2 - 35x - 24 x + 42 = 0
= 5x (4x - 7) - 6 ( 4x - 7) = 0
= (4x - 7) (5x - 6) = 0
x = 7/4, 6/5
= 2y2 - 4y - 3y + 6 = 0
= 2y (y - 2) - 3 (y - 2) = 0
= (2y - 3) ( y - 2) = 0
y = 2, 3/2
i.e. No relationship between ‘x’ and ‘y’
Answer: e)
2). 8x2 + 8x – 5x – 5 = 0
8x (x+1) – 5(x+1) = 0
(8x-5) (x+1) = 0
x = -1, (5/8)
6y2 + 9y + 8y + 12 = 0
3y (2y + 3) + 4 (2y + 3) = 0
(3y + 4) (2y + 3) = 0
y = (-4/3), (-3/2)
Answer: a)
3). 2x2 + 2x - x - 1 = 0 12y2 - 16y - 9y + 12 = 0
2x ( x + 1) - 1 ( x + 1) = 0 4y ( 3y - 4) - 3 (3y - 4) = 0
(2x - 1) ( x + 1) = 0 (3y - 4) ( 4y - 3) = 0
x = -1, (1/2) y = (3/4), (4/3)
Answer: c)
4). x = 3 √ 4913
y2 - 11y - 17 y + 187 = 0
y (y - 11) - 17 (y - 11) = 0
(y - 11) ( y - 17) = 0
y = 11, 17
Answer: b)
5). equn (I) × 3 - equn (II) × 7
(21x + 12 y + 15 = 0) - (21x – 49y + 259 = 0) = 61y – 244 = 0
y = 244/61 = 4
From this x = -3,
Answer: c)
6). Length of arc = 23.4 - 6.5 - 6.5 = 10.4 cm.
Area of sector = 1/2 × 10.4 × 6.5 = 33.8cm2
Answer: c)
7). Let the sum be P
Amount After 2 years =P[1+(R/100)]T=P[1+(5/100)]2=P(105/100)2=P(21/20)2
Given that amount After 2 years = 882
=>P(21/20)2=882
=>P= (882×20×20)/ (21×21) =2×20×20=Rs. 800
Answer: a)
8). Work done by A in 1 hour = 1/4
Work done by B and C in 1 hour = 1/3
Work done by A and C in 1 hour = 1/2
Work done by A,B and C in 1 hour = 1/4+1/3 = 7/12
Work done by B in 1 hour = 7/12 – 1/2 = 1/12 => B alone can complete the work in 12 hours
Answer: a)
9). SP = 250
Profit = 25%
CP=100 / (100+Profit%)×SP = [100 / (100+25)] ×250=(100/125)×250=200
Required Ratio = 200 : 250 = 4:5
Answer: c)
10). Speed upstream = 3/(20/60) = 9km/hr
Speed downstream = 3 / (18/60) = 10 km/hr
Rate of current =(10-9) / 2 = 1/2km/hr
Answer: a)
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