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# IBPS Clerk Prelims Quantitative Aptitude Question Paper 2019

• Last Updated : 01 Jun, 2022

Direction (1-10): Find the value of (?) in the following questions. 3,8,10,16,26

1.Question
√(124+? +169) = 18
(a) 34
(b) 31
(c) 33
(d) 35
(e) 32

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

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

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)⅓ + (18)² – 121 = ? −69 × 5
(a) 658
(b) 568
(c) 666
(d) 656
(e) 562

Explanation:-
(2744)⅓ + (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

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

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

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

Explanation:-
360/? = 73 + 3³
=> 360/? = 100
:. ? = 3.6

9.Question
1/2 – 3/5 +4 ⅔ = ? + 5/6
(a) 56/15
(b) 55/16
(c) 33/4
(e) 60/11

Explanation:-
1/2 – 3/5 + 4 ⅔ = ? + 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

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

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%

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

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

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

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

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

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) 7⅓hours
(b) 18 ⅔ hours
(c) 9⅓ hours
(d) 16 ⅔ hours
(e) 13 ⅓ hours

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 ⅓ 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 2⅔days. 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

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

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

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

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

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

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

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

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

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

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

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

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

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%

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

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

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