IBPS Clerk Prelims 2019 question paper with answers and detailed solutions: IBPS Clerk exam was conducted on December 07, 2019. IBPS exam was successfully organised by the Institute of Banking and Personnel Selection. IBPS Clerk Prelims is assumed as easy to moderate level in terms of difficulty.
In this article, we will cover Previous Year 2019 Question papers with solutions.
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Direction (1-10): Find the value of (?) in the following questions. 3,8,10,16,26
IBPS Clerk Prelims question paper comprised of 100 questions from 3 sections- English Language, Numerical Ability and Reasoning Ability. The duration of the exam was 60 minutes. The candidates had to choose the 1 correct option out of the 5 given there. Each correct answer was awarded 1 mark. The exam also had a negative marking of 0.25 for every wrong answer with no penalty for unanswered questions.
1. Question
√(124+? +169) = 18
(a) 34
(b) 31
(c) 33
(d) 35
(e) 32
Answer :- B
Explanation :-
√(124+?+169)=18
=> 124+?+169 = 324 ( squaring of the both side)
=> ? = 324 – 293
:. ? = 31
2.Question
136 ÷ 2² ×?= 17% of 500 ÷ 10
(a) 0.75
(b) 1.33
(c) 0.25
(d) 1.66
(e) 0.5
Answer:-C
Explanation:-
136 ÷ 2² × ? = 17% of 500 ÷ 10
=> 34 ×? = 85 ÷ 10
=> ? = 8.5/34
:. ? = .25
3.Question
65 × 3 ÷ 13 + 67 −?² = 81 ÷ 9 × 2
(a) 9
(b) 5
(c) 12
(d) 8
(e) 14
Answer- D
Explanation:-
65 × 3 ÷ 13 + 67 – ?² = 81 ÷ 9 × 2
=> 65×3×1/13 + 67 – ?² = 9 × 2
=> 15 + 67 – ?² = 18
=> ?² = 64
:. ? = 8
4.Question
(2744)â…“ + (18)² – 121 = ? −69 × 5
(a) 658
(b) 568
(c) 666
(d) 656
(e) 562
Answer :-E
Explanation:-
(2744)â…“ + (18)² – 121 = ? – 69 × 5
=> 14 + 324 – 121 = ? – 345
:. ? = 562
5.Question
115 ÷ 5 + 12 × 6 =? +64 ÷ 4 − 35
(a) 114
(b) 118
(c) 108
(d) 116
(e) 111
Answer:-A
Explanation:-
115 ÷ 5 + 12 × 6 =? +64 ÷ 4 − 35
=> 23 + 72 =? + 16 – 35
:. ? = 114
6. Question
41% of 600 − 250 =? −77% of 900
(a) 693
(b) 675
(c) 684
(d) 679
(e) 689
Answer :-E
Explanation:-
41% of 600 − 250 =? −77% of 900
=> 246 – 250 = ? – 693
:. ? = 689
7.Question
1111 ÷ 11 + 2002 ÷ 26 + 750 ÷ 25 = ?
(a) 204
(b) 212
(c) 208
(d) 206
(e) 210
Answer:-C
Explanation:-
1111 ÷ 11 + 2002 ÷ 26 + 750 ÷ 25 =?
=> 101 + 77 + 30 =?
:. ? = 208
8.Question
360/? = 73 + 3³
(a) 3.4
(b) 4.3
(c) 3.1
(d) 3.6
(e) 3.9
Answer- D
Explanation:-
360/? = 73 + 3³
=> 360/? = 100
:. ? = 3.6
9.Question
1/2 – 3/5 +4 â…” = ? + 5/6
(a) 56/15
(b) 55/16
(c) 33/4
(d) 24/7
(e) 60/11
Answer :-A
Explanation:-
1/2 – 3/5 + 4 â…” = ? + 5/6
=> 1/2 – 3/5 + 14/3 = ? + 5/6
=> ? = 1/2 +14/3 – 3/5 – 5/6
=> ? = 112/30
:. ? = 56/15
10.Question
5/11× 121 + 13/9× 288 = 141+ ?
(a) 333
(b) 327
(c) 335
(d) 330
(e) 329
Answer- D
Explanation:-
5/11× 121 + 13/9× 288 = 141+?
=> 55 + 416 – 141 =?
:. ? = 330
Directions (11 – 15): Given bar graph shows the number of residents residing in 4 societies in years 2008, 2018. Read the data carefully and answer the questions carefully.
11. Question
What is the average of residents residing in society A in 2008, B in 2018, C in 2018 & D in 2008?
(a) 355
(b) 360
(c) 365
(d) 370
(e) 350
Answer :-B
Explanation:-
Required average = (250+370+420+400)/4 = 1440/4 = 360
12. Question
Residents residing in society B in 2008 are what per cent more/less than average of residents residing in society D in 2008 & 2018?
(a) 61/11%
(b) 39/11%
(c) 72/11%
(d) 83/11%
(e) 50/11%
Answer :-E
Explanation:-
B in 2008= 420
Average of D in 2008 and 2018=(400+480)/2 = 440
Required percentage = 20/440 × 100 % = 50/11 %
13.Question
Which society shows the maximum percentage increase in the no. of residents from the year 2008 to 2018?
(a) Both A & C
(b) Both A & D
(c) Both C & D
(d) Both A & B
(e) None of these
Answer :-A
Explanation:-
A = (350-250)/250 × 100 = 40 %
B = Decrease
C = (420-300)/300 × 100 = 40%
D = ( 480-400)/400 ×100 = 20%
Thus, A and C both show a maximum percentage increase.
14. Question
What is the ratio of all residents in all societies in 2008 to that of in 2018?
(a) 142 : 157
(b) 157 : 142
(c) 162 : 137
(d) 137 : 162
(e) 97 : 114
Answer- D
Explanation:-
Required ratio
= (250+420+300+400) : (350+370+420+480)
= 1370 : 1620
= 137 : 162
15. Question
What is the difference between the number of residents residing in societies A & B in 2018 together and that in societies C & D together in 2008?
(a) 30
(b) 24
(c) 20
(d) 28
(e) 26
Answer :-C
Explanation:-
(A+B) in 2018 =( 350 + 370 )= 720
(C+D) in 2008 = (300+400) = 700
Required difference = (720-700) = 20
16. Question
A shopkeeper sells two pens, he sold 1 pen at a profit and another pen at loss. SP of each of the two pens is Rs.300 and profit percentage on 1 pen is equal to loss percentage on other. If the overall loss of shopkeepers is 6.25%, then find the difference between the cost price of both pens.
(a) Rs.350
(b) Rs.100
(c) Rs.240
(d) Rs.160
(e) Rs.300
Answer- D
Explanation:-
Let, profit percentage = Loss percentage = x %
Now,
x²/100 = 6.25
x² = 625
x = 25 %
Cost price of 1st pen = 300 × 100/(100+25)
= 240
Cost price of 2nd pen = 300 × 100/(100-25)
= 400
Thus, the difference of cost price between two pen
=(400-240) = 160
17. Question
A man received Rs.3456 when he invested Rs. P at 12% p.a. at SI for 3 years. If he invested Rs. (P +4400) at 15% p.a. at CI compounding annually for 2 years, then find the interest received by him.
(a) Rs.4515
(b) Rs.4960
(c) Rs.4725
(d) Rs.4185
(e) Rs.4345
Answer :-A
Explanation:-
We know,
S.I = (P × R × T)/100
=> 3456 × 100 = P × 12 × 3
=> P = 9600
Now,
Principal for compound interest = (9600+4400) = 14000
C.I = P [ (1 + r/100)^t – 1]
=> C.I = 14000 [ ( 1 + 15/100)² – 1]
=> C.I = 14000 [ (23/20)² – 1]
=> C.I = 14000 ( 529/400 – 1)
=> C.I = 14000 × 129/400
:. C.I = 4515
Thus, the compound interest received by him = 4515
18. Question
Time taken by a boat to cover 162 km each in downstream and in upstream is 14 hours and 24 minutes. If the speed of the stream is 6 km/hr., then find the time taken by boat to cover 240 km in upstream.
(a) 7â…“hours
(b) 18 â…” hours
(c) 9â…“ hours
(d) 16 â…” hours
(e) 13 â…“ hours
Answer :-E
Explanation:-
Let the speed of boat = x km/hr
Time = Distance/speed
14 hours & 24 mins = (14+24/60) hrs = 72/5 hours
Now,according to question
162/(x+6) + 162/(x-6) = 72/5
Solving the equation we get the value of x
x = 24 km/ hrs
Now, upstream speed =( x -6) = (24-6) = 18 km/hrs
:. Required time taken = 240/18 = 13 â…“ hours
19. Question
C is 100% more efficient than B. A alone can complete a piece of work in 9 days and B & C together can complete the same work in 2â…”days. Find what portion of work will be completed, if A & B work together for 4 days.
(a) 13/18
(b) 8/9
(c) 5/6
(d) 2/3
(e) 17/18
Answer:- E
Explanation:-
Since C is 100% more efficient than B,
So, Efficiency of B = 1
Efficiency of C = 2
We know, Total Work = efficiency × time
Total work = 3×8/3 = 8
A’s efficiency = total work/time = 8/9
(A+B)’s 4 days work = 4×(8/9+1) = 4×17/9 = 68/9
:. Required portion of work completed
= 68/9 × 1/8 = 68/72 = 17/18
20. Question
The ratio of age of P 2 years ago to the age of R 2 years hence is 1: 2 and Q’s present age is 25% more than P’s present age. If the average of the present age of P & R is 39 years, then find the difference between P’s age of 5 years hence and R’s present age.
(a) 12 years
(b) 17 years
(c) 21 years
(d) 15 years
(e) 14 years
Answer :-B
Explanation:-
(P – 2)/(R + 2) = 1/2
=> 2P – 4 = R + 2
=> 2P – 6 = R —————– (1)
Average age of P and R = 39
:. Total age of P and R= 39×2 = 78
P + R = 78 ——————–(2)
Solving both the equations (1) and (2),
P = 28
R = 50
P’s age after 5 years = (28+5) = 33 years
Required age difference between P and R = (50-33) = 17 years
Directions ( 21-25): Solve the following quadratic equation and mark the answer as per instructions.
21.Question
I. x² – 2x – 143 = 0
II. y² – 169 = 0
(a) x > y
(b) x < y
(c) x ≤ y
(d) x ≥ y
(e) x = y or no relation can be established
Answer :E
Explanation:-
x² – 2x – 143 = 0
=> x² – 13x + 11x – 143 = 0
=> (x – 13) ( x + 11) = 0
:. x = 13, -11
y2 – 169 = 0
=> y² = 169
:. y = 13, -13
x = y or no relation can be established.
22.Question
I. x² – 7x – 18 = 0
II. y² − 19y + 90 = 0
(a) x ≤ y
(b) x = y or no relation can be established
(c) x > y
(d) x ≥ y
(e) x < y
Answer:-A
Explanation:-
x² – 7x – 18 = 0
=> x² – 9x + 2x – 18 = 0
=> ( x – 9) ( x + 2) = 0
:. x = 9, – 2
y² – 19y + 90 = 0
=> y² – 10y – 9y + 90 = 0
=> ( y -10) ( y -9 ) = 0
:. y = 10, 9
x ≤ y
23.Question
I. 2x² + 5x + 3 = 0
II. y² + 4y − 12 = 0
(a) x ≤ y
(b) x > y
(c) x = y or no relation can be established
(d) x < y
(e) x ≥ y
Answer:-C
Explanation:-
2x² + 5x + 3 = 0
=> 2x² + 2x + 3x + 3 = 0
=> (2x + 3) ( x + 1) = 0
:. x = -1, -3/2
y² + 4y − 12 = 0
=> y² +6y -2y -12 = 0
=> ( y +6) ( y -2) = 0
:. y = 2, -6
no relation can be established.
24.Question
I. 9x + 3y = 15
II. 4x + 5y = 14
(a) x = y or no relation can be established
(b) x > y
(c) x ≤ y
(d) x < y
(e) x ≥ y
Answer- D
Explanation:-
9x + 3y = 15
=> 3x + y = 5
=> y = 5 – 3x ———-(1)
4x + 5y = 14
=> 5y = 14 – 4x
=> y = (14-4x)/5 ——–(2)
Equating equation (1) & (2),
5 – 3x = (14 – 4x)/5
=> 25 – 15x = 14 – 4x
=> 25 – 14 = 15x – 4x
=> 11x = 11
:. X = 1
Putting the value of x in equation (1), we get
y = 5 – 3×1
:. y = 2
x<y
25.Question
I. 2x² − x − 1 = 0
II. 3y² − 5y + 2 = 0
(a) x ≤ y
(b) x < y
(c) x = y or no relation can be established
(d) x ≥ y
(e) x > y
Answer :-C
Explanation:-
2x² – x – 1 = 0
=> 2x² – 2x + x – 1 = 0
=> ( 2x +1) ( x – 1) = 0
:. x = 1, – 1/2
3y² – 5y + 2 = 0
=> 3y² – 3y – 2y + 2 = 0
=> (3y – 2) ( y -1) = 0
:. y = 1, 2/3
x = y or no relation can be established.
Directions (26 – 30): Find the wrong number in the following number series.
26.Question
26.Question
2, 3, 6, 15, 45, 156.5, 630
(a) 2
(b) 15
(c) 3
(d) 156.5
(e) 630
Answer- D
Explanation:-
2 × .5 + 2 = 3
3 ×1 + 3 = 6
6 × 1.5 + 6 = 15
15 × 2 + 15 = 45
45 × 2.5 + 45 = 157.5
157.5 × 3 + 157.5 = 630
27.Question
36, 20, 12, 8, 6, 5.5, 4.5
(a) 8
(b) 36
(c) 5.5
(d) 4.5
(e) 6
Answer :-C
Explanation:-
36 – 16 = 20
20 – 8 = 12
12 – 4 = 8
8 – 2 = 6
6 – 1 = 5
5 – .5 = 4.5
28.Question
1, 3, 9, 31, 128, 651, 3913
(a) 31
(b) 3
(c) 1
(d) 3913
(e) 128
Answer:-E
Explanation:-
1 × 1 + 2 = 3
3 × 2 + 3 = 9
9 × 3 + 4 = 31
31 × 4 + 5 = 129
129 × 5 + 6 = 651
651 × 6 + 7 = 3913
29.Question
2, 3, 10, 40, 172, 885, 5346
(a) 40
(b) 885
(c) 172
(d) 3
(e) 10
Answer:-A
Explanation:-
(2+1) × 1 = 3
(3 + 2)×2 = 10
(10+3)×3 = 39
(39+4)×4 = 172
(172+5)×5 = 885
(885+6)×6 = 5346
30.Question
5, 8, 16, 26, 50, 98, 194
(a) 5
(b) 194
(c) 8
(d) 16
(e) 98
Answer- D
Explanation:-
5 + 3 = 8
8 + 6 = 14
14 + 12 = 26
26 + 24 = 50
50 + 48 = 98
98 + 96 = 194
31.Question
A rectangular path of width 3m is surrounding the garden whose length is 3m more than its width. If cost of painting the path at rate of 0.5Rs/m² is Rs 273 then find the area of garden .
(a) 1525m²
(b) 1804 m²
(c) 1776 m²
(d) 1906 m²
(e) 1664 m²
Answer:-B
Explanation:-
Let, the width of the rectangular garden = ‘x’ m
Length of the garden = ( x + 3) m
Width of the garden along with path = ( x + 3×2) = (x+6) m
Length of the garden along with path = ( x + 3 + 3×2) = ( x + 9)
According to the question,
( x +9)(x +6) – (x+3) * x = 273/0.5
=> x² + 9x + 6x + 54 – x² – 3x = 546
=> 12 x = 546 – 54
:. x = 492/12 = 41
Width = 41 m
Length = (41+3) = 44 m
Area of the garden = length × width = 44 × 41 = 1804 m²
32.Question
In a class percentage of students who passed the exam is 60% and number of boys & girls who passed the exam is same. If boys who failed the exam are 200% more than girls who failed in exam then find the percentage of girls who failed out of total students .
(a) 9%
(b) 13%
(c) 10%
(d) 12%
(e) 15%
Answer:-C
Explanation:-
Let, total student = 100
Passed students number = 60
Failed students number = 40
Ratio of boys and girls in failed student= 300 : 100 = 3:1
So, number of failed boys = 40×3/4 = 30
Number of failed girls = 40×1/4 = 10
Required percentage = 10/100 ×100 % = 10%
33.Question
A man invested Rs.X at 15% p.a. at SI for 4 years and Rs.(1.35X) at 18% p.a. at SI for 3 years. If total interest received by man is Rs.15948, then find value of Rs. (3.12X).
(a) Rs.50544
(b) Rs.42764
(c) Rs.32580
(d) Rs.47372
(e) Rs.37440
Answer :-E
Explanation:
1st case,
S.I = (P×R×T)/100
=> S.I = (X * 15 * 4)/100 = 3X/5
2nd Case,
S.I = (P*R*T)/100
=> S.I = (1.35X * 18* 3)/100 = 729X/1000
Now, according to the question,
3X/5 + 729X/1000 =15948
=> (600X + 729X)/1000 = 15948
=> 1329 X = 15948000
=> X = 15948000/1329
:. X = 12000
Required value = 3.12 X = 3.12 * 12000 = 37440
34.Question
A man covers 6 ¼ % distance via bus at 80 km/hr, 25% of the distance via car at 120 km/hr., 30% distance via bicycle at 32 km/hr. and remaining distance via train at 62 km/hr. If total distance covered by man is 640km, then find the total time taken man during the entire journey.
(a) 65/6 hours
(b) 13 hours
(c) 44/3 hours
(d) 31/2 hours
(e) 71/6 hours
Answer:-E
Explanation:-
Total distance = 640 km
Distance Travel by bus = 640 × 6 ¼% = 640× 25/400 = 40 km
Distance travel by car = 640 × 25% = 640 × 1/4 = 160 km
Distance travel by bicycle = 640 × 30/100 = 192 km
Distance travel by train = 640 – (40+160+192) = 248 km
We know, time = distance/speed
required time = (40/80 + 160/120 + 192/32 +248/62)
= ( 1/2 + 4/3 + 6 + 4) = 71/6 hours
35.Question
Average weight of a class is 60kg and average weight of boys in the class is 80kg. Ratio of boys to girls in the class is 5 : 4. If there are 72 students in the class, then find the average weight of girls in the class.
(a) 54 kg
(b) 42 kg
(c) 35 kg
(d) 45 kg
(e) 38 kg
Answer:-C
Explanation:-
Number of boys = 72 × 5/9 = 40
Number of girls = 72 × 4/9 = 32
Total weight of the students = 60 × 72 = 4320
Total weight of boys = 80 × 40 = 3200
:. Total weight of girls = (4320 – 3200) = 1120
Thus, the average weight of girls = 1120/32 = 35 kg
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