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[WpProQuiz 2566]
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Directions (1-5) Study the following information carefully and answer the given questions:
Following table shows the percentage distribution of total number of students who completed their graduation from different universities and the ratio of male to female among them.
Total number of students = 56000
University | Total number of students | Male : Female |
P | 23 % | 7 : 3 |
Q | 18 % | 5 : 4 |
R | 12 % | 2 : 1 |
S | 27 % | 3 : 5 |
T | 20 % | 2 : 3 |
6) In a class, 40 boys and their average age is five-sixth of the number of girls in that class and 30 girls and their average age is half of the number of boys in that class. Find the overall average in the class (Approximately)
Directions (11 – 15) Study the following information carefully and answer the given questions:
Following line graph shows the % profit of two companies A and B in different years.
Profit % = [(Income – Expenditure)/Expenditure]*100
Directions (21 – 25): What value should come in place of question mark (?) in the following number series?
Directions (26 – 30): What value should come in place of question mark (?) in the following questions?
Directions (31-35): In each of these questions two equations (I) and (II) are given. You have to solve both the equations and give answer
II. 15y2 -19y +6 =0
II. 5y2 + 16y +3 =0
II. 2y2 + 3y + 1=0
II. y3 – 878 = 453
II. y2 – 7y+12 =0
Answers:
Directions (1-5):
University | Total number of students | Male | Female |
P | 12880 | 9016 | 3864 |
Q | 10080 | 5600 | 4480 |
R | 6720 | 4480 | 2240 |
S | 15120 | 5670 | 9450 |
T | 11200 | 4480 | 6720 |
1). Answer a
The total number of male graduates from the given universities together
= > 9016 + 5600 + 4480 + 5670 + 4480
= > 29246
2). Answer d
Total number of female graduates from university Q and R
= > 4480 + 2240 = 6720
Total number of male graduates from university P and T
= > 9016 + 4480 = 13496
Required % = (6720/13496)*100 = 49.79 % = 50 %
3). Answer c
The total number of male graduates from university Q, R and T together
= > 5600 + 4480 + 4480 = 14560
The total number of female graduates from university P, R and S together
= > 3864 + 2240 + 9450 = 15554
Required ratio = 14560: 15554 = 7280: 7777
4). Answer b
Required % = {[(23+ 18) – (12 + 27)]/(12 + 27)}*100
= > {(41 – 39)/39}*100 = 5.128 % = 5 % more
5). Answer d
University S has highest number of female graduates.
6). Answer b
Required average = ((40*25) + (30*20))/70=1600/70=23 years
7). Answer b
Let us take speed of current and speed of boat be x and y
Given,
Upstream speed = 40/5=8 km/hr
Downstream speed = 40/2.5=16 km/hr
Speed of boat = (16+8)/2=12 km/hr
Speed of current = (16-8)/2=4 km/hr
8). Answer: c
No of days taken by B = 25 days
No of days taken by C = 20/1*2 = 40 days
=>1/20+1/40-1/25
=>(15-8)/200=200/7 = 28 4/7 days
9). Answer: a
Milk and water ratio = 4:1
Given,
(4x-16)/(x-4+20) =12/13
52x-208=12x+192
=>40x=400=>x=10
10). Answer: b
Shamili and Kamal’s profit ratio = (4000×12):((4000+x)×12)=2400:3600
=>4000/(4000+x)=2/3
=>x=2000
Profit ratio = (4000×12):(6000×12):(8000×8)
=>6:9:8
Total profit = 2400/6×23=Rs.9200
Direction (11-15)
11). Answer c
The expenditure of Company A in the year 2012 = 32 lakhs
The income of Company B in the year 2014 = 13 + 32 = 45 lakhs
The income of company A in 2012
= > 16 = [(I – 32)/32]*100
= > 512 + 3200 = 100I
= > I = 3712/100 = 37.12 lakhs
The expenditure of Company B in the year 2014
= > 20 = [45 – e)/e]*100
= > 20e = 4500 – 100e
= > 120e = 4500
= > e = (4500/120) = 37.5 lakhs
Required % = (37.12/37.5)*100 = 98.98 % = 100 %
12). Answer b
The income of company A in the year 2013 = 44.25 lakhs
The expenditure of company A in the year 2013
= > 18 = [(44.25 – e)/e]*100
= > 118e = 4425
= > e = (4425/118) = 37.5 lakhs
The expenditure of company A in the year 2013 = The expenditure of company B in the year 2015
The expenditure of company B in the year 2015 = 37.5 lakhs
The income of company B in the year 2015
= > 36 = [(I – 37.5)/37.5]*100
= > 1350 = 100I – 3750
5100 = 100I
I = 51 lakhs
13). Answer a
Let the income of company A and B in the year 2014 be x,
The expenditure of company A in the year 2014
= > 25 = [(x – E1)/E1]*100
= > 125E1 = 100x
= > E1 = (100x/125) = (4x/5)
The expenditure of company B in the year 2014
= > 20 = [(I – E2)/E2]*100
= > 120E2 = 100x
= > E2 = (100x/120) = (5x/6)
Required ratio
= > (4x/5): (5x/6)
= > 24: 25
14). Answer d
The income of company B in the year 2016 = 37.8 lakhs
The expenditure of company B in the year 2016
= > 26 = [(37.8 – e)/e]*100
= > 126e = 3780
= > e = (3780/126) = 30 lakhs
The expenditure of company A in the year 2015 = 27 lakhs
The income of company A in the year 2015
= > 28 = [(I – 27)/27]*100
= > 756 = 100I – 2700
= > 3456 = 100I
= > I = 34.56 lakhs
Required % = [(34.56- 30)/34.56]*100 = 13.19 % = 13 % less
15). Answer c
The expenditure of company A in the year 2015 = 27 lakhs
The income of company A in the year 2015
= > 28 = [(I – 27)/27]*100
= > 756 = 100I – 2700
= > 3456 = 100I
= > I = 34.56 lakhs
The expenditure of company B in the year 2015 = 33 lakhs
The income of company B in the year 2015
= > 36 = [(I – 33)/33]*100
= > 1188 = 100I – 3300
= > I = (4488/100) = 44.88 lakhs
Required difference = 44.88 – 34.56 = 10.32 lakhs
16). Answer: d
Given,
1/x+1/(x+10)+1/10=11/60
(2x+10)/(x2+10x)=5/60
X2-14x-120=0
X=20
Required ratio = 1/20:1/30=3:2
17). Answer: c
Given,
Total no. of probability = 9+x
Required probability = 5/(9+x)=1/3
=>9+x=15=>x=6
18). Answer: c
Let us take CP be Rs.100
SP= 110 =80% of marked price
MP = 110/80*100=137.5
Required percentage = 37.5/100*100=37.5%
19). Answer: c
Given,
44/100*x-735=x*20/100+(x-2500)*15/100
9x=36000=>x=4000
20). Answer: d
1/x+1/(x+5)=9/100
Simplify the above equation we get x=20 days
Direction (21-25)
21). Answer c
The pattern is, *0.5 + 1, *1 + 1, *2 + 2, *4 + 4…
The answer is, 156
22). Answer b
The difference of difference is, 82 + 1, 92 + 1, 102 + 1, 112 + 1
The answer is, 1022
23). Answer d
The pattern is, +(1*3), +(2*6), +(3*9), +(4*12),..
The answer is, 117
24). Answer a
The pattern is, +172, -192, +212, -232,..
The answer is, 569
25). Answer b
The pattern is, ÷ 2 + 1
The answer is, 26.25
Direction (26-30)
26). Answer c
23 × 2 × 30 + 42 – 225 = x
X = 1380 + 42 – 225
X = 1197
27). Answer a
(6246/18)*40 = (x/15)*60
X = (6246/18)*40 * (60/15)
X = 3470
28). Answer d
(20/100)*1540 + (14/100)*1050 = (20/100)* x + 361
308 + 147 – 361 = (x/5)
94 = (x/5)
X = 470
29). Answer a
(378/9) + (124/8) + 1460*(5/4) = x + (20/100)*500
42 + 15.5 + 1825 – 100 = x
X = 1782.5
30). Answer d
((156/13) + (17*42)) / (9 + 7) = x
X = (12 + 714) / 16
X = 726 /16
X = 45 3/8
Direction (31-35)
31). Answer a
I. 10x2 -17x +7 =0
10x2 -10x-7x +7 =0
10x(x-1) -7(x-1) =0
(10x-7) (x-1) =0
X= 7/10, 1
II. 15y2 -19y +6 =0
15y2 -10y-9y +6 =0
5y(3y-2)-3 (3y-2) =0
(5y-3) (3y-2) =0
Y= 3/5, 2/3
x>y
32). Answer e
I. 12x2 +19x +5 =0
12x2 +4x +15x +5 =0
4x(3x+1)+5 (3x+1) =0
(4x+5)(3x+1) =0
X=-5/4, -1/3
II. 5y2 + 16y +3 =0
5y2 + y+15y +3 =0
Y(5y+1)+3 (5y+1) =0
(y+3) (5y+1) =0
Y=-3, -1/5
Can’t be determined
33). Answer c
I. 6x2 + 31x + 35 =0
6x2 + 21x+10x + 35 =0
3x(2x+7)+5 (2x+7)=0
(3x+5)(2x+7) =0
X= -5/3, -7/2
II. 2y2 + 3y + 1=0
2y2 + 2y +y+ 1=0
2y(y+1) +1(y+1) =0
(2y+1) (y+1) =0
Y= -1/2, -1
X<y
34). Answer e
I. X2 – 264 = 361
X2 = 361+264
X2 = 625
X= 25, -25
II. y3 – 878 = 453
Y3 = 453 + 878
Y3 = 1331
Y =11
Can’t be determined
35). Answer e
I. 4x2– 30x +56 =0
4x2– 16x-14x +56 =0
4x(x-4) -14(x-4) =0
(4x-14) (x-4) =0
X= 14/4, 4= 7/2, 4
II. y2 – 7y+12 =0
y2 – 4y-3y+12 =0
y(y-4)-3 (y-4) =0
(y-3) (y-4) =0
Y =3, 4
Can’t be determined
This post was last modified on March 19, 2019 3:47 pm