BANK & INSURANCE (NEW PATTERN SERIES)

Total Questions: 50

21. Directions (21-22): Study the following information carefully and answer the questions given beside.

Both the series given below follow the exact same pattern. In Series I there is a wrong term and the first term of Series II is given.

Series I: 4, 8, 18, 38, 74, 150, 298
Series II: 5, j, k, l, m, n, o

21. What is the value of (o – n) ÷ (k – l)?

Correct Answer: (b) -7.50
Solution:

From the explanation, we have

(o − n) ÷ (k − l) = (362 − 182) ÷ (22 − 46)
= 180 ÷ (−24)
= −7.50

22. What is the difference between the wrong term of Series I and the same numbered term of Series II as the wrong term of Series I?

Correct Answer: (b) 4
Solution:

From the explanation, we have

Since, the second term is the wrong term.
Wrong term of Series I = 8
The second term of Series II = 12

So the answer = 12 − 8 = 4

23. Directions (23-24): Solve the given series and answer the following question.

Series I: 8, 18, 50, 98, P, 338, Q
Series II: R, 192, 576, 144, 720, 120, S

23. If a series is in Arithmetic progression and its 2nd and 6th terms are R and Q respectively,
what is the 5th term of that series?

Correct Answer: (d) 529.50
Solution:

Series I:

Series Pattern             Given Series
2² × 2 = 8                        8
3² × 2 = 18                     18
5² × 2 = 50                     50
7² × 2 = 98                      98
11² × 2 = 242                 242           P
13² × 2 = 338                 338
17² × 2 = 578                 578           Q

Series II:

Series Pattern                     Given Series
192                                           192
192 × 3 = 576                           576
576 ÷ 4 = 144                           144
144 × 5 = 720                            720
720 ÷ 6 = 120                            120
120 × 7 = 840                             840           S

So R = 192 × 2 = 384

From common explanation, we have

R = 384
Q = 578
6 − 2 = 4 terms

Common Difference = (578 − 384) / 4
= 48.50

So the answer = 578 − 48.50 = 529.50

24. What is the difference between the values of P and S?

Correct Answer: (a) 598
Solution:

Series I:

Series Pattern             Given Series
2² × 2 = 8                        8
3² × 2 = 18                     18
5² × 2 = 50                     50
7² × 2 = 98                      98
11² × 2 = 242                 242           P
13² × 2 = 338                 338
17² × 2 = 578                 578           Q

Series II:

Series Pattern                     Given Series
192                                           192
192 × 3 = 576                           576
576 ÷ 4 = 144                           144
144 × 5 = 720                           720
720 ÷ 6 = 120                           120
120 × 7 = 840                           840           S

So R = 192 × 2 = 384

From common explanation, we have

Reqd. difference = 840 − 242
= 598

25. Direction (25): The two number series given below follow the same pattern.

Series I: 28, 10, 6, 5, 6, 11
Series II: 48, 20, A, B, C, D

What is the value of B + C?

Correct Answer: (b) 56
Solution:

Series I:

Series pattern                 Given series
28                        28
28 × 0.5 − 4 = 10                        10
10 × 1 − 4 = 6                         6
6 × 1.5 − 4 = 5                         5
5 × 2 − 4 = 6                         6
6 × 2.5 − 4 = 11                        11

Series II:

Series pattern                        Given series
48                              48
48 × 0.5 − 4 = 20                              20
20 × 1 − 4 = 16                              16
16 × 1.5 − 4 = 20                              20
20 × 2 − 4 = 36                              36
36 × 2.5 − 4 = 86                              86

B + C = 20 + 36 = 56

26. Direction (26): The two number series given below follow the same pattern.

Series I: 15, 28, 44, 65, 93, 130
Series II: 60, 73, A, B, C, D

What is the value of 2D - C + B?

Correct Answer: (a) 322
Solution:

27. Directions (27-28): Study the data carefully and answer the following questions:

Three series I, II and III are given below and P, Q and R are missing terms of these series respectively.

Series I: 63, 67, 75, P, 123, 187
Series II: 1156, 1108, 1036, 940, 820, Q
Series III: R, 1395, 3122, 5318, 8061, 11435, 15530

What is the HCF of P, Q and R?

Correct Answer: (d) 13
Solution:

Series I: 63, 67, 75, P, 123, 187

63 + 4 = 67
67 + 8 = 75
75 + 16 = 91 = P
91 + 32 = 123
123 + 64 = 187

Series II: 1156, 1108, 1036, 940, 820, Q

1156 − 24 × 2 = 1108
1108 − 24 × 3 = 1036
1036 − 24 × 4 = 940
940 − 24 × 5 = 820
820 − 24 × 6 = 676 = Q

Series III: R, 1395, 3122, 5318, 8061, 11435, 15530

(R = 65) + 11³ − 1 = 1395
1395 + 12³ − 1 = 3122
3122 + 13³ − 1 = 5318
5318 + 14³ − 1 = 8061
8061 + 15³ − 1 = 11435
11435 + 16³ − 1 = 15530

HCF of P, Q and R = HCF of 91, 676, 65 = 13

28. If series IV follows the same logic as series III and it starts from 9, and ‘K’ and ‘L’ being the 3rd and 6th term respectively or respectively of series IV then find (L - K).

Correct Answer: (a) 8313
Solution:

Series I: 63, 67, 75, P, 123, 187

63 + 4 = 67
67 + 8 = 75
75 + 16 = 91 = P
91 + 32 = 123
123 + 64 = 187

Series II: 1156, 1108, 1036, 940, 820, Q

1156 − 24 × 2 = 1108
1108 − 24 × 3 = 1036
1036 − 24 × 4 = 940
940 − 24 × 5 = 820
820 − 24 × 6 = 676 = Q

Series III: R, 1395, 3122, 5318, 8061, 11435, 15530

(R = 65) + 11³ − 1 = 1395
1395 + 12³ − 1 = 3122
3122 + 13³ − 1 = 5318
5318 + 14³ − 1 = 8061
8061 + 15³ − 1 = 11435
11435 + 16³ − 1 = 15530

9 + 1331 − 1 = 1339
1339 + 1728 − 1 = 3066 (3rd) = K
3066 + 2197 − 1 = 5262
5262 + 2744 − 1 = 8005
8005 + 3375 − 1 = 11379 (6th) = L

Hence, L − K = 11379 − 3066 = 8313

29. Three series I, II and III are given below and in each of them, one term is wrong.

P, Q and R are wrong terms in series I, II and III respectively while S, T and U are correct terms that will replace P, Q and R respectively.

Series I: 4, 22, 58, 112, 144, 274
Series II: 390, 270, 226, 190, 174, 170
Series III: 500, 284, 159, 83, 68, 60

Following table represents the data regarding the number of girls and boys participating in painting competitions
from schools A, B and C.

Total number of students participating in the painting competition from school A and C together is ___ more than
the total number of students participating in the painting competition from school B.

Correct Answer: (c) 140
Solution:

Logic for Series I:
4 + 18 × 1 = 22
22 + 18 × 2 = 58
58 + 18 × 3 = 112
112 + 18 × 4 = 184
184 + 18 × 5 = 274

So, wrong term = P = 144
Correct term = S = 184

Logic for Series II:
390 − 10² = 290
290 − 8² = 226
226 − 6² = 190
190 − 4² = 174
174 − 2² = 170

So, wrong term = Q = 270
Correct term = T = 290

Logic for Series III:
500 − 6³ = 284
284 − 5³ = 159
159 − 4³ = 95
95 − 3³ = 68
68 − 2³ = 60

So, wrong term = R = 83
Correct term = U = 95

Now, U − 16 = 95 − 16 = 79
P − R = 144 − 83 = 61
P + 20 = 144 + 20 = 164

T = 290
S = 184
Q = 270

Then, the given table becomes:

SchoolNumber of girls participated in painting competitionNumber of boys participated in painting competition
A7961
B164290
C184270

Then, total number of students participated in painting competition from school A
= 79 + 61 = 140

Total number of students participated in painting competition from school C
= 184 + 270 = 454

Now, total number of students participated in painting competition from schools A and C together
= 140 + 454

And the total number of students participated in painting competition from school B
= 164 + 290 = 454

Therefore, difference = (140 + 454) − 454 = 140

30. Three series I, II and III are given below. Study the information carefully and answer the question below.

Series I: 22, 28, 40, 58, 80, 112
Series II: 156, 145, 130, 115, 96, 73
Series III: 13, 17, 33, 69, 135, 233

P, Q and R are wrong terms in series I, II and III respectively while S, T and U are correct terms that will replace P, Q and R respectively.

A shopkeeper bought two items A and B at Rs. 4000 together. He marked item A at (P/2)% above its cost price
and item B at (R - P)% above its cost price. After allowing [(Q - P)/2]% discount on both the items together,
he sold them together at Rs. 4537.5. If the cost price of item A is Rs. a, then which of the following statements is correct?

(i) a > 8U     (ii) 20S < a < 15T     (iii) a < 4(T + U)

Correct Answer: (b) Only (iii)
Solution:

Logic for series I:
22 + 6 × 1 = 28
28 + 6 × 2 = 40
40 + 6 × 3 = 58
58 + 6 × 4 = 82
82 + 6 × 5 = 112

So, wrong term = P = 80, S = 82

Logic for series II:
156 − 11 = 145
145 − 13 = 132
132 − 17 = 115
115 − 19 = 96
96 − 23 = 73

So, wrong term = Q = 130, T = 132

Logic for series III:
13 + 2² = 17
17 + 4² = 33
33 + 6² = 69
69 + 8² = 133
133 + 10² = 233

So, wrong term = R = 135, U = 133

A shopkeeper bought two items A and B at Rs.4000 together.
Cost price of item A = Rs. a

Then, cost price of item B = Rs. (4000 − a)

He marked item A at (P/2)% above its cost price and item B at (R − P)% above its cost price.

P/2 = 80/2 = 40
R − P = 135 − 80 = 55

Now, marked price of item A = (100 + 40)% of a
= 1.4a

Marked price of items B = (100 + 55)% of (4000 − a)
= 6200 − 1.55a

Now, total marked price = 1.4a + 6200 − 1.55a
= 6200 − 0.15a

After allowing [(Q − P)/2]% discount on both the items together, he sold them together at Rs.4537.5.

(Q − P)/2 = (130 − 80)/2 = 25

Then, total selling price = (100 − 25)% of (6200 − 0.15a)
= (3/4) × (6200 − 0.15a)

Now, 4537.5 = (3/4) × (6200 − 0.15a)

6050 = 6200 − 0.15a
0.15a = 150
a = 1000

From (i): Now, 8U = 8 × 133 = 1064 > a
So, statement (i) is not correct.

From (ii): 20S = 20 × 82 = 1640
15T = 15 × 132 = 1980
20S > a, 15T > a
So, statement (ii) is not correct.

From (iii): 4(T + U) = 4(132 + 133) = 1060
Then, a < 1060
So, statement (iii) is correct.