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[WpProQuiz 2474]
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Directions (1-5): Study the following and answer the following questions.
The following line graph shows that the profit percent of two companies X and Y in 2006 to 2011
Note: profit percentage= [(income-expenditure)/expenditure]*100
Directions (16-20): What will come in place of question mark in the following questions?
Directions (21-25): In each of the following questions, two equations (i) and (ii) are given. You have to solve them and find the correct option.
21 a) 45x+24y-56=0
b) 10x+ 6y-13=0
22.a)21x= 517+470
b)y= (√(6181-6132))2
23.a)x2= 527+351+643
b) y2+15y-324=0
24).
a)5x2+11x+6=0
b) 6y2+11y+5=0
25).
a) x= √(192-37)
b) y=(√484+√1024)/(√3)2
Directions (26-30): Study the following and answer the following questions.
The following pie charts Show the distribution of students in Primary and Secondary levels in seven different schools.
Total number of students in Primary level= 32700
Total number of students studying in Secondary level= 42600
26.The number of students studying in the Primary level is approximately what percentage more than that of Secondary level from School D?
Directions (31-35): Find the missing number in the following number series?
32. 750,1125,1500,1875, 2250,?
Answers:
1). Answer: d
Let the income of the companies X and Y in 2007 be Rs x.
Also Let us take the expenditure of the company X be E1 and that of the company Y be E2.
Now, 20= [(x- E1)/ E1]*100
x=120 E1/100………….. (1)
Also for company Y,
10 = [(x- E2)/ E2]*100
x= 110 E2/100………….. (2)
From (1) and (2), we have
120 E1/100=110 E2/100
E1/ E2= 110/120
=11/12
Hence E1: E2= 11:12
2). Answer: e
Given that the income of company Y in 2006= Rs 63 lakhs and profit percentage= 50%
Let Rs x be the expenditure of the company Y in 2006.
Then, 50= [(63-x)/x]*100
x = (63-x)*2
3x= 126
x=Rs 42 lakhs
Then in 2008, the expenditure of the company Y= Rs 21 lakhs
Let y be the income of the company Y in 2008.
Then, 30 = [(y-21)/21]*100
63= 10y-210
y= Rs 27.3 lakhs
3). Answer: a
Let x be the expenditure of the company X in 2010.
Given that, the profit of company X in 2010= Rs 67.5 lakhs
(i.e) 22.5/100*x=67.5
x= Rs 300 lakhs
Then the expenditure of the company X in 2011= Rs 200 crores
Now, profit of the company X in 2011= 200*10/100
= Rs 20 lakhs
4). Answer: e
Since we cannot find the income of a company Y for all the years separately, the total expenditure of the same company will not be determined.
5). Answer: e
Given that, Income2008/ Income2011= 5/7
Since the profit percentage of a company X in 2008 is 35% and that of in 2011 is 10%, we have
135% of Expenditure2008/110 % of Expenditure 2011= 5/7
(i.e) Expenditure2008/ Expenditure 2011= 5/7*110/135
= 110/189
6). Answer: e
Let the ages of Manu and Salim 15 years ago be 2x and 5x respectively.
Then, the present age of Manu= (2x+15)
And the present age of Salim= (5x+15)
Now, (2x+15+5)/ (5x+15+5) =3/5
(2x+20)/(5x+20)=3/5
5(2x+20) =3(5x+20)
x=8
Thus present age of Manu= (2x+15) =31
Present age of Salim= (5x+15) =55
Required ratio= 31:55
7). Answer: d
We know that surface area of the sphere= 4πr2.
= 4*22/7*35*35
=15400 m2
Then the cost of covering the spherical bowl= (15400*1.5) = Rs 23100
8). Answer: c
Let the price of LG TV in the last year be 4x and the price of Samsung TV in the last year be 3x.
Then, (4x-10000)/(3x*120/100)=5/6
6(4x-10000) =5(3x*6/5)
6x= 60000
x= 10000
Thus the price of LG TV in the last year =4x= Rs 40000
9). Answer: c
Let the rate of upstream be x km/hr and then rate of downstream be 3x km/hr.
Then, 2x= 28
x= 14 km/hr
(i.e) Rate of upstream= 14 km/hr
And rate of downstream= 42 km/hr
Thus, Rate of stream= ½(42-14)=1/2 *(28)= 14 km/hr
10). Answer: a
Given that profit= Cost price of 50 pens
Now, Total profit= Selling price of 400 pens- Cost price of 400 pens
Cost price of 50 pens = 3825 – Cost price of 400 pens
Cost price of 50 pens+ Cost price of 400 pens=3825
Cost price of 450 pen =3825
Thus, Cost price of 1 pen= 3825/450= Rs 8.5
11). Answer: e
Let x be age of Nancy, y be the age of Sofi and z be the age of Mithun.
According to the question, x+y+2z=89 ……………. (1)
3x+y+z= 98 …………… (2)
x+3y+3z=154 …………. (3)
After solving the equations, we have x= 35/2 = 17.5 years.
12). Answer: d
Let r be the rate of interest.
For question, we have simple interest for 1 year= 5124-4270=Rs 854
Then, 854= (4270*r)/100
r= 854*100/4270
r= 20%
13). Answer: d
Let the speed of Fathima be x km/hr
Then, 180/x-180/2x =6
(360-180)/2x =6
180=12x
x=15km/hr
14). Answer: a
Given that y= 120x/100 ….. (1)
Also xy=2430 … (2)
Substituting (1) in (2), we have
120x2/100=2430
x2= 2430*100/120
=2025
x=45
Then, y= 54
Thus, x+y =99
15). Answer: c
Let x be the length of the train
(x+270)/18=3x/27=speed
(x+270)/18= x/9
x+270=2x
x= 270
Thus speed of the train=3*270/27 = 30 m/s
Directions (16-20):
16). Answer: c
(42% of 6100+84)÷ (30% of 1350-63) =?
(2562+84) ÷ (405-216) =?
?=14
17). Answer: a
153+ (52*17) =34 *?
153 + 884= 34* ?
1037/34=?
?= 31
18). Answer: d
4176*3/4*5/6*7=?+ 208
18270-208=?
18062=?
19). Answer: e
19*23-197+153=?
437-197+153=?
393=?
20). Answer: b
700/270*405+576/12=?
1050+48=?
1098=?
Directions (21-25):
21). Answer: b
45x+24y-56=0 …………… (1)
10x+ 6y-13=0 …………….. (2)
Multiplying equation (2) by 4, then we have
40x+24y-52=0 ………….. (3)
Solving (1) and (3), we have x= 4/5 and y= 5/6
Hence x<y
22). Answer: b
a)21x= 517+470
21x= 987
x=47
b)y= (√(6181-6132))2
y= (√(49))2
y= 49
Hence x<y
23). Answer: e
a)x2= 527+351+643
x2=1521
x=39 and -39
b) y2+15y-324=0
y2+27y-12y-324=0
(y+27)(y – 12)=0
y= -27 and 12
Hence the relationship cannot be determined
24). Answer: c
a) 5x2+11x+6=0
5x2+5x+6x+6=0
(x+1)(5x+6)=0
x= -1 and -6/5
b) 6y2+11y+5=0
6y2+6y+5y+5=0
(6y+5)(y+1)=0
y= -1 and -5/6
Hence x≤y
25). Answer: e
a) x= √(192-37)
x= √(361-37)
x=√324
x= 18
b) y=(√484+√1024)/(√3)2
y= (√484+√1024)/(√3)2
y= (22+32)/3
y=54/3
y= 18
Hence x=y
Directions (26-30):
26). Answer: d
Number of students studying in the Secondary level in school D=10/100*42600
= 4260
Similarly, number of students studying in the Primary level in school D=24/100*32700
= 7848
Required percentage=3588/4260*100
= 84%
27).Answer: e
Number of students studying in the Secondary level in school A=18/100*42600
= 7668
Number of students studying in the Secondary level in school C=20/100*42600
= 8520
Number of students studying in the Secondary level in school G=13/100*42600
= 5538
Total number of students studying at Secondary level from A, C and G=7668+8520+5538=21726
28). Answer: c
Number of students studying in the Primary level in school F=8/100*32700
= 2616
Number of students studying in the Primary level in school G=14/100*32700
= 4578
Required percentage= 2616/4578*100=57%
29). Answer: d
Number of students studying in the Secondary level in school C=20/100*42600
= 8520
Total number of students studying primary level from school E=16/100*32700
= 5232
Required ratio= 8520:5232
= 355:218
30). Answer: a
Number of students studying in the Secondary level in school A=18/100*42600
= 7668
Number of students studying in the Secondary level in school B=14/100*42600
= 5964
Number of students studying in the Secondary level in school C=20/100*42600
= 8520
Number of students studying in the Secondary level in school D=10/100*42600
= 4260
Number of students studying in the Secondary level in school E=15/100*42600
= 6390
Total number of students studying secondary level from those schools= 7668+5964+8520+4260+6390 = 32802
Thus Number of students not attending the camp= 32802/2=16401
Directions (31-35):
31). Answer: A
15.6+0.7 =16.3
16.3+1.4 =17.7
17.7+2.1=19.8
19.8+2.8= 22.6
22.6+3.5= 26.1
32). Answer: C
750/2 *3 =1125
1125/3 *4 = 1500
1500/4*5 = 1875
1875/5*6 = 2250
2250/6*7 =2625
(or)
The difference of the difference is 375,
33). Answer: D
Series follows,
*1.5 + 5
34). Answer: B
*0.5 + 0.5,*1.5+1.5,*2.5+2.5, *3.5+3.5,*4.5 +4.5
35). Answer: D
Difference is prime square = 22, 32, 52, 72, 112
This post was last modified on June 8, 2018 10:36 am