Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for SBI Clerk Mains 2020 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.
Directions (01 – 05): Study the following information carefully and answer the questions given below.
The following pie chart1 shows the percentage distribution of total number of students appeared for the entrance examination from 6 different cities and the pie chart2 shows the percentage distribution of total number of girls appeared for the entrance examination from those cities. The ratio between the total number of boys to that of girls is 7: 6.
1) Find the difference between the total numbers of boys appeared for the entrance examination from city A and E together to that from city B and D together?
A) 2000
B) 2400
C) 2600
D) 2800
E) None of these
2) If 60 % of boys from city C and 65 % girls from city F cleared their entrance examination, then find the sum of the total number of boys cleared their entrance examination from city C and total number of girls cleared their entrance examination from city F?
A) 4920
B) 4650
C) 4840
D) 4530
E) None of these
3) Find the difference between the average number of boys appeared for the entrance examination from city A, D and F and together the average number of girls appeared for the entrance examination from city A, C and E together?
A) 150
B) 100
C) 200
D) 250
E) None of these
4) If the ratio between the total number of girls cleared the entrance examination and went to the abroad for higher education to that of not cleared the entrance examination from city B and city D is 2 : 5 and 4 : 7 respectively, then find the total number of girls cleared the entrance examination and went to the abroad for higher education from city B and city D together?
A) 3200
B) 3000
C) 2800
D) 4000
E) None of these
5) If the percentage of total number of boys cleared the entrance examination and studied in IITs from city B and city E is 40 % and 35 % respectively, then find the ratio between the total number of boys cleared the entrance examination and studied in IITs from city B to that of total number of boys not cleared the entrance examination from city E?
A) 42 : 75
B) 28 : 47
C) 31 : 65
D) 55 : 79
E) None of these
Directions (06 – 10): 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.
6) I)Â 4, 20, 60, 300, 900, 4500, 13500
II) 11 …… 2475. If 2475 is nthterm, then what value should come in place of (n + 2)thterm?
A) 34250
B) 18975
C) 12375
D) 37125
E) 25650
7) I)Â 924, 924, 462, 154, 38.5, 7.7
II) 534 ……. 4.45. If 4.45 is nth term, then what value should come in place of (n – 2)th term?
A) 89
B) 22.25
C) 72
D) 64.75
E) 45
8) I) 27, 1357, 2356, 3084, 3595, 3937
II) 55 ….. 4304. If 4304 is nth term, the find the value of n?
A) 7
B) 6
C) 8
D) 9
E) 5
9) I) 57, 66, 39, 84, 21, 102, 3
II) 72 ……. 36. If 36 is nthterm, then what value should come in place of (n+3)rdterm?
A) 117
B) 135
C) 18
D) 92
E) 53
10) I)Â 12, 24, 96, 576, 4608, 46080
II) 4 …… 184320. If 184320 is nthterm, then find the value of n?
A) 5
B) 6
C) 9
D) 7
E) 8
Answers :
Directions (1-5) :
1) Answer: A
The total number of boys appeared for the entrance examination from city A and E together
= > [65000 * (18/100) – 30000 * (17/100)] + [65000 * (22/100) – 30000 * (21/100)]
= > [11700 – 5100] + [14300 – 6300]
= > 6600 + 8000 = 14600
The total number of boys appeared for the entrance examination from city B and D together
= > [65000 * (16/100) – 30000 * (14/100)] + [65000 * (20/100) – 30000 * (22/100)]
= > [10400 – 4200] + [13000 – 6600]
= > 6200 + 6400 = 12600
Required difference = 14600 – 12600 = 2000
2) Answer: D
The total number of boys cleared their entrance examination from city C
= > [65000 * (14/100) – 30000 * (16/100)] * (60/100)
= > [9100 – 4800] * (3/5)
= > 2580
The total number of girls cleared their entrance examination from city F
= > 30000 * (10/100) * (65/100)
= > 1950
Required Sum = 2580 + 1950 = 4530
3) Answer: B
The average number of boys appeared for the entrance examination from city A, D and F together
= > {[65000 * (18/100) – 30000 * (17/100)] + [65000 * (20/100) – 30000 * (22/100)] + [65000 * (10/100) – 30000 * (10/100)]} / 3
= > {[11700 – 5100] + [13000 – 6600] + [6500 – 3000]} / 3
= > (6600 + 6400 + 3500) / 3
= > 5500
The average number of girls appeared for the entrance examination from city A, C and E together
= > 30000 * [(17 + 16 + 21) / 100] * (1/3)
= > 5400
Required difference = 5500 – 5400 = 100
4) Answer: E
The total number of girls cleared the entrance examination and went to the abroad for higher education from city B and city D together
= > 30000 * (14/100) * (2/7) + 30000 * (22/100) * (4/11)
= > 1200 + 2400 = 3600
5) Answer: C
The total number of boys cleared the entrance examination and studied in IITs from city B
= > [65000 * (16/100) – 30000 * (14/100)] * (40/100)
= > (10400 – 4200) * (2/5)
= > 2480
The total number of boys not cleared the entrance examination from city E
= > [65000 * (22/100) – 30000 * (21/100)] * (65/100)
= > (14300 – 6300) * (13/20)
= > 5200
Required ratio = 2480 : 5200 = 31 : 65
Directions (6-10) :
6) Answer: D
Series I pattern:
4*5 = 20
20*3 = 60
60*5 = 300
300*3 = 900
900*5 = 4500
4500*3 = 13500
Series II pattern:
11*5 = 55
55*3 = 165
165*5 = 825
825*3 = 2475 (nth) term
2475*5 = 12375 (n + 1)th term
12375*3 = 37125 (n + 2)th term
7) Answer: A
Series I pattern:
924 ÷ 1 = 924
924 ÷ 2 = 462
462 ÷ 3 = 154
154 ÷ 4 = 38.5
38.5 ÷ 5 = 7.7
Series II pattern:
534 ÷ 1 = 534
534 ÷ 2 = 267
267 ÷ 3 = 89 (n – 2)th term
89 ÷ 4 = 22.25 (n – 1)th term
22.25 ÷ 5 = 4.45 (nth) term
8) Answer: C
Series I pattern:
27 + (113 – 1) = 1357
1357 + (103 – 1) = 2356
2356 + (93 – 1) = 3084
3084 + (83 – 1) = 3595
3595 + (73 – 1) = 3937
Series II pattern:
55 is first term.
55 + (113 – 1) = 1385
1385 + (103 – 1) = 2384
2384 + (93 – 1) = 3112
3112 + (83 – 1) = 3623
3623 + (73 – 1) = 3965
3965 + (63 – 1) = 4180
4180 + (53 – 1) = 4304
4304 is 8th term.
So, n = 8
9) Answer: B
Series I pattern:
57 + (9*1) = 66
66 – (9*3) = 39
39 + (9*5) = 84
84 – (9*7) = 21
21 + (9*9) = 102
102 – (9*11) = 3
Series II pattern:
72 + (9*1) = 81
81 – (9*3) = 54
54 + (9*5) = 99
99 – (9*7) = 36 (nth term)
36 + (9*9) = 117 (n + 1)th term
117 – (9*11) = 18 (n + 2)th term
18 + (9*13) = 135 (n + 3)rd term
10) Answer: D
Series I pattern:
12*2 = 24
24*4 = 96
96*6 = 576
576*8 = 4608
4608*10 = 46080
Series II pattern:
4 is the first term
4*2 = 8
8*4 = 32
32*6 = 192
192*8 = 1536
1536*10 = 15360
15360*12 = 184320
184320 is 7th term. So, n = 7
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