Dear Aspirants, Quantitative Aptitude plays a crucial role in Banking and all other competitive exams. To enrich your preparation, here we have provided New Pattern Aptitude Questions for IBPS Clerk Mains. Candidates those who are going to appear in IBPS Clerk Mains can practice these questions daily and make your preparation effective.
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[WpProQuiz 4608]
Directions (Q. 1 – 5): The questions below are based on the given Series-I. The series-I satisfy a certain pattern, follow the same pattern in Series-II and answer the questions given below.
1) I) 52, 54, 60, 72, 92, 122
II) 36……148. If 148 is nth term, then find the value of n" layout="responsive" width="581" height="341">
Answers :
Direction (1-5) :
1) Answer: d)
Series I Pattern:
52 is the first term
52 + (12 + 1) = 54
54 + (22 + 2) = 60
60 + (32 + 3) = 72
72 + (42 + 4) = 92
92 + (52 + 5) = 122
122 is 6th term
Series II Pattern:
36 is the first term
36 + (12 + 1) = 38
38 + (22 + 2) = 44
44 + (32 + 3) = 56
56 + (42 + 4) = 76
76 + (52 + 5) = 106
106 + (62 + 6) = 148
148 is 7th term. So, n = 7
2) Answer: d)
Series I Pattern:
72 is the first term
72*0.5 + 1 = 37
37*1 + 1 = 38
38*1.5 + 1 = 58
58*2 + 1 = 117
117*2.5 + 1 = 293.5
293.5 is 6th term
Series II Pattern:
16 is the first term
16*0.5 + 1 = 9
9*1 + 1 = 10
10*1.5 + 1 = 16
16*2 + 1 = 33 = (n-3)th term
33*2.5 + 1 = 83.5 = (n-2)th term
83.5*3 + 1 = 251.5 = (n-1)th term
251.5*3.5 + 1 = 881.25 = nth term
881.25 is nth term
The answer is, 33
3) Answer: b)
Series I Pattern:
117 is the first term
117 – (13 + 1) = 115
115 + (23 + 1) = 124
124 – (33 + 1) = 96
96 + (43 + 1) = 161
161 – (53 + 1) = 35
35 is 6th term
Series II Pattern:
238 is the first term
238 – (13 + 1) = 236
236 + (23 + 1) = 245
245 – (33 + 1) = 217
217 + (43 + 1) = 282
282 – (53 + 1) = 156
156 + (63 + 1) = 373
373 is 7th term. So, n = 7
4) Answer: c)
Series I Pattern:
534 is the first term
534 ÷ 2 – 1 = 266
266 ÷ 2 – 1 = 132
132 ÷ 2 – 1 = 65
65 ÷ 2 – 1 = 31.5
31.5 ÷ 2 – 1 = 14.75
14.75 is 6th term
Series II Pattern:
890 is the first term
890 ÷ 2 – 1 = 444
444 ÷ 2 – 1 = 221
221 ÷ 2 – 1 = 109.5
109.5 ÷ 2 – 1 = 53.75
53.75 ÷ 2 – 1 = 25.875
25.875 is 6th term. So, n = 6
5) Answer: e)
Series I Pattern:
Series II Pattern:
Direction (6-10) :
6) Answer: c)
From Statement I,
Area of the rectangle= (11/100)*2200= 242
CSA of cone = (14/100)*2200 = 308
(22/7)*r*l = 308
So, Statement I alone is not sufficient to the answer the question.
From Statement II,
Area of the circle = (28/100)*2200= 616
(22/7)*r*r = 616
Radius of circle r = 14
Length of the cone = 14 cm
So, Statement II alone is not sufficient to the answer the question.
From statement I and II,
(22/7)*r*14=308
Radius of the cone=7
Breadth of the rectangle=7
Area of the rectangle=l*7=242
l=242/7
Perimeter of the rectangle=2*[(242/7) + 7) = 582/7 cm
Statement I and II are sufficient to answer the question.
From Statement III,
Area of the square = (22/100)*2200= 484
Length of the rectangle=22
Area of the rectangle = (11/100)*2200=242
22*b =242
b=11
Perimeter of the rectangle = 2*(11+22) = 66 cm
Statement III are sufficient to answer the question.
7) Answer: c)
From quantity I,
Volume of the cone = (1/3)*(22/7)*r*r*h
CSA = (14/100)*2200 = 308
(22/7)*r*l = 308
rl = 98
Area of square = (22/100)*2200= 441
Side of the square= radius of the cone=22
l=98/22=49/11
Height of cone = (5*49/11)/7=35/11
Volume of the cone = (1/3)*(22/7)*22*22*(35/11) = 4840/3
Volume of the sphere = 4840/3
From quantity II,
Area of the circle = (28/100)*2200 = 616
(22/7)*r*r = 616
R2 = 196
r=14
Volume of sphere = (4/3)*(22/7)*r*r*r
= > (4/3)*(22/7)*14*14*14
= > 34496/3
Quantity II > Quantity I
8) Answer: a)
From Statement I,
Volume of the cuboid=12*9*20=2160
CSA of cylinder = (15/100)*2200=330
330=2*(22/7)*r*(120/100)*r
12r*r=525
r*r=175/4
r = (5/2)*√7
Volume of cylinder=(22/7)*r*r*h
= > (22/7)*(175/4)*(120/100)*(5/2)*√7
= > 825 (√7/2)
Required ratio= 825 (√7/2) : 2160
So, Statement I alone is sufficient to the answer the question.
From Statement II,
Side of the square=22
Base of the cuboid=22
Radius of circle=14
Radius of cylinder= (2/7)*14= 4
So, Statement II alone is not sufficient to the answer the question.
9) Answer: c)
From Statement I,
Radius of the circle=14
Radius of the cone = (50/100)*14=7cm
CSA of cone = (14/100)*2200=308 cm2
(22/7)*7*l=308
L=14 cm
14*14=h*h+7*7
196-49=h*h
147=h*h
h=√147 = 7√3 cm
So, Statement I alone is sufficient to the answer the question.
From statement II,
Radius of the circle=14
Length of the cone=14*(150/100)= 21
CSA of cone=308
(22/7)*21*r=308
r=14/3
441=h*h + (196/9)
h=√3773/3
So, Statement II alone is sufficient to the answer the question.
10) Answer: a)
Area of rectangle and cuboid = [(11+10)/100]*2200=462 cm2
Area of square and cylinder= [(22+15)/100]*2200=814 cm2
Difference=814-462=352 cm2
This post was last modified on December 25, 2018 5:34 pm