# IBPS Clerk Mains Quantitative Aptitude Questions 2019 (Day-9) High Level New Pattern

Dear Aspirants, Quantitative Aptitude plays a crucial role in Banking and all other competitive exams. To enrich your preparation, here we have provided New Pattern Aptitude Questions for IBPS Clerk Mains. Candidates those who are going to appear in IBPS Clerk Mains can practice these questions daily and make your preparation effective.

[WpProQuiz 4708]

Direction (1 – 5): Read the following information carefully and answer the given questions:

In one of the village have five different schools. The number of boys in school A is one-eighth of the total students. The number of boys in school B is 10% of the total students in all the five schools together. The ratio between the numbers of girls to boys in school E is 1:1. The number of boys in school A is equal to the sum of number of boys in school B and the number of boys in school E. The number of girls in school C is 20% of the total number of girls. The number of boys in school C is three-two of the number of boys in school B. The number of students in school A is 50% more than the number of students in school D. The number of students in school B is 15% of the total students. The total boy in all the five schools together is 200 less than the total number of girls in all the five schools together. The total number of students in all the five schools together is 1200.

1) What is the difference between the total number students in which school has the highest number of girls and the total number of students in which school has the highest number of boys?

a) 64

b) 72

c) 68

d) 78

e) None of these

2) From a group of number of boys in school D and the number of girls in school E, two students are selected at random. Find the probability that at least one girl is selected?

a) 207/245

b) 183/245

c) 121/2450

d) 99/245

e) None of these

3) What is the ratio of girls to boys in School F?

Statement I: Total number of boys in School F is 200% of the number of students in School E.

Statement II: Total number of girls in School F is 130 less than the total number of students in School F.

a) Only I

b) Only II

c) Either I or II sufficient

d) All I and II necessary to the answer the question

e) The question can’t be answered even with all I and II

4) Quantity I: What is the average number of boys in school A, E, C together?

Quantity II: What is the average number of girls in School B and D together?

a) Quantity I > Quantity II

b) Quantity I ≥ Quantity II

c) Quantity II > Quantity I

d) Quantity II ≥ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

5) Quantity I: What is the difference between the total number of students in School A and B?

Quantity II: What is the difference between the total number of students in School D and E?

a) Quantity I > Quantity II

b) Quantity I ≥ Quantity II

c) Quantity II > Quantity I

d) Quantity II ≥ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

Directions (6 – 10): The questions below are based on the given Series-I. The series-I satisfy a certain pattern, follow the same pattern in Series-II and answer the questions given below.

6)

I. 8, 4, 6, 15, 52.5

II. 12 …. 78.75. If 78.75 is the nth term, then find the value of n?

a) 6

b) 4

c) 5

d) 7

e) 8

7)

I. 29, 59, 178, 891, 6238

II. 46 ….. 9808. If 9808 is the nth term, then find the value of n?

a) 6

b) 5

c) 7

d) 8

e) 9

8)

I. 45, 92, 372, 2984, 47760

II. 2.5 ….. 135712. If 135712 is the nth term, then find the value of n?

a) 9

b) 8

c) 7

d) 6

e) 11

9)

I. 11, 12, 26, 108, 872, 13968

II. 4 …… 217632. If 217632 is the nth term, then what value should come in the place of (n-1)th term?

a) 5016

b) 4820

c) 2560

d) 4520

e) 6800

10)

I. 1024, 1267, 1348, 1510, 1564, 1672, 1708

II. 961 ….. 1717. If 1717 is nth term, then what value should come in the place of (n – 2)th term?

a) 1339

b) 1559

c) 1609

d) 1753

e) 1497

Directions (1 – 5):

Total number of students=1200

Girls + Boys = 1200 —> (1)

Boys is 200 less than girls,

Girls – Boys = 200 —> (2)

So the number of boys =500

The number of girls=700

Number of Boys in school A = (1/8)*1200 = 150

Number of boys in school B = (10/100)*1200 = 120

Number of boys in school E = 150 – 120 = 30

Number of girls in school E = (1/1)*30 = 30

Total number of students in school E = 30+30 = 60

Number of boys in school C = (3/2)*120 = 180

Number of boys in school D = 500 – 150 – 120 – 30 – 180 = 20

Number of girls in school C = (20/100)*700 = 140

Total number of students in School C = 140 + 180 = 320

Total number of students in School B = (15/100)*1200 = 180

Number of girls in school B = 180 – 120 = 60

Number of students in A and D = 1200 – (180 + 320 + 60) = 640

Total number of students in School A=150/100 school D

D + (150/100)*D=640

D + (3/2)*D=640

5D = 640*2

Number of students in school D=256

Number of students in school A = 640 – 256 = 384

Number of girls in school A = 384 – 150 = 234

Number of girls in School D = 256 – 20 = 236 Highest number of girls in school D, the total number students in school D=256

Highest number of boys in school C, the total number of students in school C=320

Required difference = 320 – 256 = 64

Number of boys in school D=20

Number of girls in school E=30

50C2=50*49/1*2=1225

n(E)=20C1*30C1 + 30C2

=20*30+(30*29)/(1*2)

=600 + 435 =1035

Required probability = 1035/1225 = 207/245

From statement I,

Number of boys in school F=60*(200/100)=120

So, Statement I alone is not sufficient to the answer the question.

From Statement II,

Total number of students in school F = x

Number of girls in School F = x – 130

So, statement II alone is not sufficient to the answer the question.

From I, II

X=120+x-130

So, both statement I and II are also not sufficient to the answer the question.

From quantity I,

Average= (150+30+180)/3 = 120

From quantity II,

Average= (60+236)/2 = 148

Quantity II > Quantity I

From quantity I,

Difference = 384 – 180 = 204

From quantity II,

Difference = 256 – 60 = 196

Quantity I > Quantity II

Directions (6-10):

Series – I pattern:

8 is the 1st term

8 * 0.5 = 4

4 * 1.5 = 6

6 * 2.5 = 15

15 * 3.5 = 52.5

52.5 is the 5th term

Series – II pattern:

12 is the 1st term

12 * 0.5 = 6

6 * 1.5 = 9

9 * 2.5 = 22.5

22.5 * 3.5 = 78.75 = 5th term

Series –I pattern:

29 is the 1st term

29 * 2 + 1 = 59

59 * 3 + 1 = 178

178 * 5 + 1 = 891

891 * 7 + 1 = 6238 = 5th term

Multiply with prime numbers and add 1 after multiplication

Series –II pattern:

46 is the 1st term

46 * 2 +1 = 93

93 * 3 + 1 = 280

280 * 5 + 1 = 1401

1401 * 7 + 1 = 9808 = 5th term

Multiply with prime numbers and add 1 after multiplication

Series –I pattern:

45 is the 1st term

45 * 2 + 2 = 92

92 * 4 + 4 = 372

372 * 8 + 8 = 2984

2984 * 16 + 16 = 47760 = 5th term

Series – II pattern:

2.5 is the 1st term

2.5 * 2 + 2 = 7

7 * 4 + 4 = 32

32 * 8 + 8 = 264

264 * 16 + 16 = 4240

4240 * 32 + 32 = 135712 = 6th term

Series – I pattern:

11 is the 1st term

11 * 20 + 20 = 12

12 * 21 + 21 = 26

26 * 22 + 22 = 108

108 * 23 + 23 = 872

872 * 24 + 24 = 13968 = 6th term

Series – II pattern:

4 is the 1st term

4 * 20 + 20 = 5

5 * 21 + 21 = 12

12 * 22 + 22 = 52

52 * 23 + 23 = 424

424 * 24 + 24 = 6800 = (7 – 1)th term

6800 * 25 +25 = 217632 = 7th term

Series – I pattern:

1024 is the 1st term

1024 + 243 = 1267

1267 + (243/3) = 1348

1348 + (81*2) = 1510

1510 + (162/3) = 1564

1564 + (54*2) = 1672

1672 + (108/3) = 1708 = 7th term

Series – II pattern:

961 is the 1st term

961 + 243 = 1204

1204 + (243/3) = 1285

1285 + (81*2) = 1447

1447 + (162/3) = 1501

1501 + (54*2) = 1609 = (8-2)th term

1609 + (108/3) = 1645 = (8-1)th term

1645 + (36*2) = 1717 = 8th term 