Crack IBPS Clerk Prelims 2018 – Sectional Full Test-13 | Numerical Ability

Dear Readers, Take Free Numerical Ability Sectional Test of 35 Questions as like in the real exam to analyze your preparation level. Our Sectional Test Questions are taken as per the latest exam pattern and so it will be really useful for you to crack the prelims exam lucratively. Students who are weak in Reasoning Ability should utilize this chance constructively to accomplish a successful profession in Banking Field.

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Directions (Q. 1-5): What will come in the place of question mark (?) in the following series?

1) 97, 99, 105, 117, ?

a) 143

b) 151

c) 137

d) 147

e) 153

2) 10, 5, 5, 7.5, 15, ?

a) 37.5

b) 24.25

c) 40.75

d) 33.5

e) 29

3) 11, 16, 26, 41, 61, ?

a) 72

b) 94

c) 108

d) 86

e) 78

4) 9, 21, 51, 135, 381, ?

a) 753

b) 841

c) 907

d) 1055

e) 1113

5) 3, 4, 10,33, 136, ?

a) 565

b) 685

c) 840

d) 720

e) 460

Directions (Q. 6 – 10): In each of these questions two equations (I) and (II) are given. You have to solve both the equations and give answer as,

A) If a > b

B) If a < b

C) If a ≥ b

D) If a ≤ b

E) If a = b or no relation can be established between x and y

6)

I) a² – 32a + 252 = 0

II)b² – 28b + 192 = 0

7)

I) a² – 32a + 247 = 0

II)b² – 22b + 117 = 0

8)

I) a² + 29a + 208 = 0

II)b² + 19b + 78 = 0

9)

I) a² – 22a + 105 = 0

II)b² – 27b + 162 = 0

10)

I) (a – 18)² = 0

II) b² = 324

11) If the ratio of investment of Abhishek and Abhijit is 3:4, then the total profit at the end of the year is Rs. 50000 and Abhishek’s share in total profit is Rs. 10000. What is the ratio of their time period of investment?

a) 1:2

b) 1:3

c) 1:4

d) 2:3

e) 3:5

12) By giving a discount of 20%, a person is making a profit of Rs. 50. If the markup percentage is 30, then find the cost price of the item?

a) Rs 1300

b) Rs 1500

c) Rs 1250

d) Rs 1400

e) None of These

13) The simple interest received on a certain sum is Rs. 35000 in five years at the rate of 10%. What would be the compound interest received on a same sum at a rate of 6% in 2 years?

a) Rs. 7924

b) Rs. 8236

c) Rs. 8418

d) Rs. 8652

e) None of these

14) Rishi and Kunal entered into a partnership with investment in the ratio 6: 7 respectively. After one year, Nimi joined them with double the initial investment of Rishi. At the end of two years, Rishi doubled his investment. At the end of three years, they earned a total profit of Rs.69000. Find the difference between the shares of Kunal and Nimi in the profit.

a) Rs.4000

b) Rs.3000

c) Rs.5000

d) Rs.6000

e) None of these

15) The difference between the two numbers is 540. 32 % of one number is equal to 48 % of another number. Find the two numbers?

a) 1620 and 1080

b) 1510 and 970

c) 1480 and 940

d) 1350 and 810

e) 1730 and 1190

16) Priyanka is 8 years older than Renuka. The ratio of the present ages of Priyanka to Myna is 4 : 5. Renuka is 16 years younger than Myna. What is Renuka’s present age?

a) 28 years

b) 24 years

c) 20 years

d) 32 years

e) None of these

17) If downstream speed of a boat is 200% more than the upstream speed of the boat and an object flowing on river covers 100 m in 40 seconds, then how much distance boat can cover in still water in 1 hour?

a) 15 km

b) 16 km

c) 27 km

d) 18 km

e) None of these

18) Pipe A and pipe B can fill the tank while working alone in 12 hour and 10 hours respectively but there is an another pipe C which can empty the fully filled tank in 15 hours. If there is 200 liters of water available in the tank and pipe A, B and C were opened then tank will be fully filled in 8 hours. What is the capacity of the tank?

a) 3200 liter

b) 2500 liter

c) 3000 liter

d) 3600 liter

e) 4000 liter

19) The marked price of Sofa is Rs. 1500 more than the cost price. When the discount of Rs. 250 is allowed, the profit of 20 % is earned. At what price should Sofa be sold to earn the profit of 30 %?

a) Rs. 8125

b) Rs. 7340

c) Rs. 8845

d) Rs. 7920

e) None of these

20) A box contains 7 red balls, 5 black balls, 4 orange balls. If three balls are drawn out randomly, then the probability of getting a ball of different colours?

a) 5/8

b) 3/7

c) ¼

d) 4/5

e) None of these

Directions (Q. 21 – 25): Study the following information carefully and answer the given questions.

The following table shows the total number of students in different branches from 2001 to 2005 in a certain college.

21) Find the ratio of total students in Aeronautical and chemical to that of total students in civil and electronics in all the given years together?

a) 35:27

b) 119:270

c) 191:230

d) 357:811

e) None of these

22) What is the average number of students in the year 2004 in all the given branches together?

a) 84

b) 92

c) 80

d) 76

e) None of these

23) What was the percentage increase of students in 2003 in all the given branches together as compared to the previous year?

a) 23.56%

b) 27.98%

c) 37.88%

d) 26.89%

e) 32.26%

24) Find the difference between the total Electronic, Computer Science and Aeronautical students to that of total Mechanical, Chemical and Civil students in all the given years together?

a) 562

b) 304

c) 458

d) 420

e) None of these

25) In which year, was the growth being maximum in Mechanical while comparing with the previous year?

a) 2005

b) 2004

c) 2003

d) 2002

e) Either b or d

Directions (Q. 26-30): What value will come in place of question mark (?) in the following questions?

26) ∛157464 + √8836 – 45 % of 800 =? + (1188 ÷ 9) – 362

a) 952

b) 816

c) 878

d) 924

e) None of these

27) 160% of 150 + ?% of 900 = 1770

a) 165

b) 170

c) 190

d) 180

e) 175

28) √1225 + ? = 4 2/7 of 140

a) 720

b) 565

c) 345

d) 630

e) 440

29) 15 1/3 – 4 1/6 + 8 5/12 – 7 2/3 = ? + 2 7/12

a) 11 2/5

b) 13 ¼

c) 8 3/7

d) 9 1/3

e) None of these

30) 11 * ? *9 = 542 + 54

a) 28

b) 33

c) 30

d) 27

e) 22

Directions (Q. 31 – 35) what approximate value should come in the place of question mark (?) in the following questions?

31) 26 % of 397 + 48 % of 703 – 25 % of 451 = ?

a) 370

b) 250

c) 330

d) 410

e) 290

32) (8/36) ÷ (216/63) × 756 – 341.67 + 67 % of 599 = ?

a) 110

b) 170

c) 230

d) 250

e) 190

33) 41 % of 697 + 69 % of 804 – 35 % of 751 = ?

a) 523

b) 577

c) 642

d) 675

e) 440

34) 33450 ÷ 11 + (5/7) of 12542 + 52 % of 2499 = ?2 – 387.78

a) 86

b) 105

c) 78

d) 92

e) 117

35)120 % of 1799 + (3/7) of 6889 = (?) + 218.81

a) 5160

b) 4350

c) 5740

d) 6320

e) 4890

Answers :

1) Answer : C)

+ (12+1), + (22+2), + (32+3), + (42+4),…

2) Answer : A)

*0.5, *1, *1.5, *2, *2.5,…

3) Answers : D)

Difference of difference: 5, 5, 5, 5…

4) Answer: E)

*3-6, *3-12, *3-18, *3-24, *3-30,…

5) Answer : B)

*1+1, *2+2, *3+3, *4+4,…

6) Answer : E)

I) a² – 32a + 252 = 0

(a – 18) (a – 14) = 0

a = 18, 14

II) b² – 28b + 192 = 0

(b – 12) (b – 16) = 0

b = 12, 16

Can’t be determined

7) Answer : C)

I) a² – 32a + 247 = 0

(a – 13) (a – 19) = 0

a = 13, 19

II) b² – 22b + 117 = 0

(b – 13) (b – 9) = 0

b = 13, 9

a ≥ b

8) Answer : D)

I) a² + 29a + 208 = 0

(a + 13) (a + 16) = 0

a = -13, -16

II) b² + 19b + 78 = 0

(b + 13) (b + 6) = 0

b = -13, -6

a ≤ b

9) Answer : E)

I) a² – 22a + 105 = 0

(a – 15) (a – 7) = 0

a = 15, 7

II) b² – 27b + 162 = 0

(b – 18) (b – 9) = 0

b = 18, 9

Can’t be determined

10) Answer : C)

I) (a – 18)² = 0

(a – 18) (a – 18) = 0

a = 18, 18

II) b² = 324

b = ±18

a ≥ b

11) Answers : B)

The ratio of investment of Abhishek and Abhijit = 3 : 4

The ratio of share of Abhishek and Abhijit = 10000 : 40000 = 1: 4

Investment*Period = Ratio of profit

The ratio of time period of Abhishek and Abhijit = (1/3) : (4/4) = 1 : 3

12) Answer : C)

Let CP = 100n

Hence MP = 130n

And SP = 130n x 0.8 = 104n

Profit = 4n = 50

=> n = 12.5

Hence cost price = Rs. 1250

13) Answers : D)

S.I = (P*n*r)/100

35000 = (P*5*10)/100

P = (35000*100)/50

P = Rs. 70000

Compound Interest:

70000*(6/100) = 4200

74200*(6/100) = 4452

C.I = 4200 + 4452 = Rs. 8652

14) Answer : B)

The share of Rishi, Kunal and Nimi,

= > [6*2 + 12*1] : [7*3] : [12*2]

= >24 : 21 : 24

= >8 : 7 : 8

23’s = 69000

1’s = 3000

The difference between the shares of Kunal and Nimi in the total profit

= > (8’s – 7’s) = Rs. 3000

15) Answers : A)

Let the two numbers be x and y,

X – Y = 540

(32/100)*X = (48/100)*Y

(X/Y) = 3/2

X : y = 3 : 2

1’s = 540

The two numbers be,

= > 3’s = (540*3) = 1620

= > 2’s = (540*2) = 1080

16) Answer : B)

Priyanka = Renuka + 8

The ratio of the present ages of Priyanka to Myna = 4 : 5 (4x, 5x)

Renuka = Myna – 16

Renuka’s present age = 5x – 16

Priyanka = Renuka + 8

4x = 5x – 16 + 8

8 = 5x – 4x

X = 8

Present age of Renuka = 5x – 16 = 40 – 16 = 24 years

17) Answer : D)

The ratio of speed of downstream to that of upstream = 3 : 1

Speed of stream = 100/40 = 5/2 m/s = (5/2)*(18/5) = 9 Km/hr

(1/2)*(3x – x) = 9

X = 9

Speed of boat in still water = (1/2)*(4x) = 2x = 18 Km/hr

Distance covered in 1 hour = 18 km

18) Answer : C)

Let total capacity of the tank be T litres

Part of tank filled when all the pipes are opened = T*(1/12+1/10-1/15) = 7T/60

Total tank filled in 8 hours = (7T/60)*8 = 56T/60 =14T/15

According to the question,

= > (14T/15) + 200 = T

= > T/15 = 200

= > T = 3000 litres

19) Answer : A)

Let the C.P be x,

M.P = 1500 + C.P = 1500 + x

  1. P = (1500 + x) – 250 = 1250 + x

S.P = (120/100)*x =(6/5)x

1250 + x = (6/5)x

6250 + 5x = 6x

C.P(X) = 6250

To earn a profit of 30 %,

  1. P = (130/100)*6250 = Rs. 8125

20) Answer : C)

Total probability n(S) = 16C3

Required probability n(E) = 7C1 and 5C1 and 4C1

P(E) = n(E)/n(S)

P(E) = (7C1 and 5C1 and 4C1)/ 16C3

P(E) = ¼

21) Answer : B)

Total students in Aeronautical and Chemical = 216 + 141 = 357

Total students in Civil and electronics = 500 + 310 = 810

Required ratio = 357 : 810 = 119 : 270

22) Answer : A)

Total students in 2004 in all the given branches together

= > 109 + 110 + 115 + 80 + 60 + 30 = 504

Required average = 504/6 = 84

23) Answer : E)

Total number of students in 2002

= > 75 + 80 + 65 + 46 + 25 + 19 = 310

Total number of students in 2003

= > 85 + 95 + 108 + 60 + 35 + 27 = 410

Required percentage = [(410 – 310)/310] * 100 = 32.26%

24) Answer : B)

Total Students in Mechanical = 50 + 75 + 85 + 109 + 120 = 439

Total Students in Chemical = 15 + 19 + 27 + 30 + 50 = 141

Total Students in Civil = 24 + 46 + 60 + 80 + 100 = 310

Total Electronic, Computer Science and Aeronautical students in all the given years together

= > 439 + 141 + 310 = 890

Total Students in Electronic = 65 + 80 + 95 + 110 + 150 = 500

Total Students in Computer Science = 45 + 65 + 108 + 115 + 145 = 478

Total Students in Aeronautical = 11 + 25 + 35 + 60 + 85 = 216

Total Mechanical, Chemical and Civil students in all the given years together

= > 500 + 478 + 216 = 1194

Required difference = 1194 – 890 = 304

25) Answer : D)

From the given figure,

Growth in mechanical students:

In 2002: [(75 – 50)/150] * 100 = 50%

In 2003: [(85 – 75)/175] * 100 = 13.33%

In 2004: [009 – 85)/85] * 100 = 28.23%

In 2005: [020 – 109)/109] * 100 = 10.09%

26) Answer : A)

54 + 94 – (45/100)*800 = x + (1188/9) – 1296

54 + 94 – 360 + 1296 = x + 132

54 + 94 – 360 + 1296 – 132 = x

X = 952

27) Answer : B)

X % of 900 = 1770 – 160 % of 150

(x/100)*900 = 1770 – (160 * 150)/100

9x = 1770 – 240

x = 1530/9 = 170

28) Answer : B)

x = 140 * (30/7) – √1225

x = 20 * 30 – 35

x = 600 – 35 = 565

29) Answer : D)

15 1/3 – 4 1/6 + 8 5/12 – 7 2/3 – 2 7/12 = x

(15 – 4 + 8 – 7 – 2) (1/3 – 1/6 + 5/12 – 2/3 – 7/12) = x

10 [(4 – 2 + 5 – 8 – 7)/12] = x

X = 10 (-8/12) = 10 (-2/3)

X = (30 – 2)/3 = 28/3 = 9 1/3

30) Answer : C)

11 *x* 9 = 542 + 54

99 x = 2916 + 54

x = 2970/99 = 30

31) Answer : C)

(26/100)*400 + (48/100)*700 – (25/100)*450 = x

104 + 336 – 112.5 = x

X = 327.5 = 330

32)  Answer : A)

(8/36)*(63/216)*756 – 342 + (67/100)*600 = x

X = 49 – 342 + 402

X = 109 = 110

33) Answer : B)

41 % of 700 + 69 % of 800 – 35 % of 750 = x

X = (41/100)*700 + (69/100)*800 – (35/100)*750

X = 287 + 552 – 262.5

X = 576.5 = 577

34) Answer : E)

(33451/11) + (5/7)*12544 + (52/100)*2500 = x2 – 388

3041 + 8960 + 1300 + 388 = x2

X2 = 13689

X = 117

35) Answer : E)

120 % of 1800 + (3/7)*6888 = x + 219

(120/100)*1800 + (3/7)*6888 = x + 219

2160 + 2952 – 219 = x

X = 4893 = 4890

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