Dear Readers, Important Practice Aptitude Questions for IBPS PO and Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.
1).In racing over a given distance, A can beat B by 20m, B can beat C by 10m, A can beat C by 28m. The racing distance is?
a) 58 m
b) 116 m
c) 100 m
d) 120 m
e) cannot be determined
2). A wire can be bent to form a circle of radius 56 cm. if it is instead bent in the form of a square, then its area will be:
a) 6400 sq.cm
b) 2025 sq.cm
c) 7744 sq.cm
d) 6561 sq.cm
e) none of these
3). A man drives from his house to the station. If he drives at the rate of 10 kms per hour, he reaches the station at 6 pm. if he drives at 15 kms per hour, he would reach the station at 4 pm. At what speed, in kms per hour, should he drives so as to reach the station at 5 pm?
a) 12
b) 12.33
c) 5√5
d) 12.5
e) 13
4). The sum of 5 natural numbers A, B, C, D, E is 126. It is known that A:B = 7:8, B:C = 2:3, C:D = 4:5, D:E = 5:7.What is the difference between A and E?
a) 14
b) 22
c) 24
d) 28
e) 32
5). The difference between the amount of compound interest and simple interest accrued on an amount of 26000 at the end of 3 years is 28994. 134. What is the rate of interest p.c.p.a.?
a) 22
b) 17
c) 19
d) cannot be determined
e) none of these
Directions (Q. 6-10): In the following two equations numbered I and II are given. you have to solve both the equations and choose the correct option.
a) x < y
b) x ≥ y
c) x > y
d) x ≤ y
e) x = y or the relationship between x and y cannot be determined
6). I. x2 - 19x + 72 = 0
II. y2 + 23y + 126 = 0
7). I. 7x2 + 63x + 126 = 0
II. 9y2 + 36y + 27 = 0
8). I. x7 = 128
II. y10 = 1024
9). I. x3/2 = 252 + 104
II. y1/4 = 42 – 13
10). I. x2 – 5.8x + 8.41 = 0
II. y2 - 3.8y + 3.12 = 0
1)c 2)c 3)a 4)d 5)c 6)c 7)d 8)e 9)e 10)c
1).Let the uniform speeds of A,B,C be vA,vB,vC, respectively.
Let t1 be the time taken for the race between A and B and t2 be the time taken for race between B and C
Since A covers the distance in time t1, race between A and C also takes time t1.Therefore,
VAt1-VBt1 =20
VBt2 –Vct2=10
VAt1 - Vct1=28
Also, VAt1=VBt2 = d. Thus VBt1 =d-20, Vct2=d – 10 and VCt2= d – 10 and VCt1=d-28.
(d-20)/(d-28) = VB/ VC = d/(d-10)
Thus (d-20)(d-10)=d(d-28).Hence, d=100.
Answer: c)
2). Length of the wire = circumference of the circle
2 × (22/7) × 56 = 2 × 22 × 8 = 352 cm
side of square = 352/4 = 88cm
area of square = 88 × 88 = 7744 sq.cm
Answer: c)
3). Let ‘x’ be the distance between house and station.
Time taken to travel a distance of x at the speed of 10km/hr, T1 = x/10 hours
Time taken to travel a distance of x at the speed of 15km/hr, T2 = x/15 hours
we know that T1 – T2 = 2 hours
therefore, we get x = 60 km
also, we can conclude that starting time is 12 pm [60/15 = 4, since he reach station at 4 pm travelling at 15 km/hr]
hence, to reach the station at 5 pm, speed = 60/5 = 12 km/hr [12 pm – 5pm, time = 5 hours]
Answer: a)
4). A:B = 7:8
B:C = 2:3 = 8:12
SO, A:B:C = 7:8:12
C:D = 4:5 = 12:15
A:B:C:D = 7:8:12:15
D:E = 5:7 = 15:21
SO, A:B:C:D:E = 7:8:12:15:21
given x(7+8+12+15+21) = 126
or 63x = 126 or x =2
E – A = 21X – 7X = 14X = 28
Answer: d)
5). let the annual rate of interest be R%
26000[1+ (R/100)]3 – (26000×3×R) / 100 = 28994.134
Substituting the options in the above equation, rate of interest is 19%
Answer: c)
6). from eqn I. x2 - 19x + 72 = 0
(x - 24) (x + 3) = 0
x = 24, -3
from eqn II. y2 + 23y + 126 = 0
(x + 14) (x + 9) = 0
x = -14, -9
Hence, x > y
Answer: c)
7). from eqn I. 7(x2 + 9x + 18) = 0
(x+6) (x+3) = 0
x = -6, -3
from eqn. II. 9(y2 + 4y + 3) = 0
(y+3) (y+1) = 0
y = -3, -1.
Hence, x ≤ y.
Answer: d)
8). From I, x7 = 27
From II, y10 = 210
Hence x = y
Answer: e)
9). From eqn I. x3/2 = 625 + 104 = 729
x = (729)2/3 = 81
from II, y1/4 = 16 – 13 = 3
y = 34 = 81
hence x = y
Answer: e)
10). From l,(x-2.9)2=0
From II,(y-1.2)(y-2.6)=0
=>y=1.2,2.6
Hence x > y
Answer: c)
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