BANK & INSURANCE (NEW PATTERN SERIES)

Total Questions: 50

1. Directions (1-2): Find the pattern of the series carefully and answer the questions given below.

A   B   169   192   199   184   141   64

1. A new series starts with B. The 2nd, 3rd, 4th and 5th terms are (B – 420),     (B + 320), (B – 220) and (B + 120) respectively. What is the value of the 6th term?

Correct Answer: (c) 116
Solution:

Given middle part of series: 169, 192, 199, 184, 141, 64

Now check:

10² − 1³ = 100 − 1 = 99
12² − 2³ = 144 − 8 = 136
14² − 3³ = 196 − 27 = 169
16² − 4³ = 256 − 64 = 192
18² − 5³ = 324 − 125 = 199
20² − 6³ = 400 − 216 = 184
22² − 7³ = 484 − 343 = 141
24² − 8³ = 576 − 512 = 64

So full series is: 99, 136, 169, 192, 199, 184, 141, 64

Thus:

A = 99
B = 136

Now new series starts with B = 136

136 − 420 = −284
136 + 320 = 456
136 − 220 = −84
136 + 120 = 256

Pattern in constants: −420, +320, −220, +120 (Increase by +100 each time)

Next constant = −20

136 − 20 = 116

Final Answer: 116

2. What is the difference between the fifth term of the given series and 200% of A? (a) 0 (b) 1 (c) 2 (d) 4 (e) 7

Correct Answer: (b) 1
Solution:

Given pattern:

10² − 1³ = 99 → A
12² − 2³ = 136 → B
14² − 3³ = 169
16² − 4³ = 192
18² − 5³ = 199 ← 5th term
20² − 6³ = 184
22² − 7³ = 141
24² − 8³ = 64

So, A = 99

5th term = 199

200% of A = 2 × 99 = 198

Difference = 199 − 198 = 1

Final Answer: 1

3. Directions (3–4): Study the following information carefully and answer the questions given below:

Series I:

a, 156, 110, 72, 42, 20, b

Series II:

a + c, a, a − c, a − 2c, a − 3c, b, d

3. What is the value of a ÷ c?

Correct Answer: (c) 3.08
Solution:

Series I: a, 156, 110, 72, 42, 20, b

Find differences of known terms:

156 − 110 = 46
110 − 72 = 38
72 − 42 = 30
42 − 20 = 22

Differences: 46, 38, 30, 22

Observe pattern:

46 − 38 = 8
38 − 30 = 8
30 − 22 = 8

So differences decrease by 8 each time.

Therefore previous difference: 46 + 8 = 54

So:

a − 156 = 54
a = 210

Now continue pattern after 22: (22 − 8 = 14)

So:

20 − b = 14
b = 6

Thus:

a = 210
b = 6

Series II:

a + c
a
a − c
a − 2c
a − 3c
b
d

Substitute a = 210 and b = 6:

210 + c
210
210 − c
210 − 2c
210 − 3c
6
d

This is clearly an Arithmetic Progression with common difference = −c

So: 210 − 3c = 6

Solve:

210 − 3c = 6
3c = 204
c = 68

Next term:

d = b − c
d = 6 − 68
d = −62

Finding a ÷ c = 210 ÷ 68 = 3.088

4. What is the value of b − d?

Correct Answer: (d) 68
Solution:

Series I: a, 156, 110, 72, 42, 20, b

Find differences of known terms:

156 − 110 = 46
110 − 72 = 38
72 − 42 = 30
42 − 20 = 22

Differences: 46, 38, 30, 22

Observe pattern:

46 − 38 = 8
38 − 30 = 8
30 − 22 = 8

So differences decrease by 8 each time.

Therefore previous difference: 46 + 8 = 54

So:

a − 156 = 54
a = 210

Now continue pattern after 22 : (22 − 8 = 14)

So:

20 − b = 14
b = 6

Thus:

a = 210
b = 6

Series II:

a + c
a
a − c
a − 2c
a − 3c
b
d

Substitute a = 210 and b = 6:

210 + c
210
210 − c
210 − 2c
210 − 3c
6
d

This is clearly an Arithmetic Progression with common difference = −c

So: 210 − 3c = 6

Solve:

210 − 3c = 6
3c = 204
c = 68

Next term:

d = b − c
d = 6 − 68
d = −62

Finding (b − d) = 6 − (−62) = 6 + 62 = 68 

5. The series given below follows a pattern. ‘a’ and ‘b’ are positive integers.

24, a + b, a + 3b, 60, a + 10b, 114

If ‘24’ is the first term, what is the 11th term of the series?

Correct Answer: (a) 354
Solution:

If the series is → a, a + b, a + 3b, a + 6b, a + 10b, a + 15b

a = 24 ---- (1)
a + 6b = 60 ---- (2)

Eq2 − Eq1
6b = 36
b = 6

Using in the last term:
a + 15b = 24 + 15 × 6 = 114 (Follows)

So the series:

11th term → a + 15b + 6b + 7b + 8b + 9b + 10b
= a + 15b + 8b × 5 = a + 55b = 24 + 55 × 6
= 354

6. Directions (6-7): Study the following information carefully and answer the questions given beside.

Below given are three different series. In series 1 there are two missing terms (e + 53) and a.
In series 2 there are two missing terms b and c.
In series 3 there is a missing term (d + 1).

Series 1: 40     45         58      (e + 53)      124      185          a
Series 2: 8        b         16          52             c        1016       6064
Series 3: 2    (d + 1)     38        135          564       2845      17094

6.

I. The ratio of the length and the breadth of a rectangle field is 16 : 11 respectively
and the total cost of fencing the land is Rs. 4185 at the rate of Rs. (b + 6.5) per metre
and the area of the land is ‘M’ meters².

II. Item A and Item B are sold at the same price. Item A at a loss of ‘d’%
and item B at a profit of ‘e’%. The cost price of item B is Rs. ‘a’.
The sum of the CP of item A and the SP of item B is ‘N’.

Correct Answer: (b) M < (5N + 795)
Solution:

I: b = 9

Let, the length and the breadth of the rectangle are 16x and 11x respectively

Perimeter → 4185 / (9 + 6.5) = 270 metres

2 × (16x + 11x) = 270
27x = 135
x = 5

Length = 16 × 5 = 80 metres
Breadth = 11 × 5 = 55 metres

Area = 80 × 55 = 4400 meters²

M = 4400

II:

d + 1 = 11
d = 10

e + 53 = 83
e = 30

a = 270

The cost price of item B = Rs. 270

The SP of item B = (270 × 130) / 100 = Rs. 351

The SP of item A = Rs. 351

The CP of item A = (351 × 100) / 90 = Rs. 390

N = 390 + 351
N = 741

Option A → 3M < 6N (Not satisfied)

3M = 3 × 4400 = 13200
6N = 6 × 741 = 4446

Option B → M < (5N + 795) (satisfied)

M = 4400
5N + 795 = 4500

Option C → M > 8N (not satisfied)

M = 4400
8N = 5928

Option D → 2M = 7N (not satisfied)

2M = 8800
7N = 5187

7. I. In a furnace, (17a/54)% of the copper ore gets wasted and out of the remaining ore only (c - 120)% is pure copper. If the quantity of pure copper obtained is ‘6b’ kg, the initial quantity of ore was ‘S’ kg.

II.  The area of a circular field is ‘T’ meter². The ratio of the radius of the circular field to the side of a square field is 14 : 13.
The total cost of ploughing the square field at the rate of Rs. 8 per meter² is Rs. 1,35,200.

Correct Answer: (a) S < T/110
Solution:

Quantity I:

17a / 54 = 17 × (270) / 54 = 85%

Remaining ore = 100% − 85% = 15%

c − 120 = 200 − 120 = 80%

15% of 80% of the ore = 54

Initial ore = S = (54 × 100 × 100) / (15 × 80)

S = 450

Quantity II:

Area of the square field = 135200 / 8 = 16900 meters²

The side of the square field = √16900 = 130 meters

The radius of the circular field = (130 × 14) / 13 = 140 metres

The area of the circular field = T = π × 140 × 140

T = 61,600 meters²

T = 61,600

Option A → S < T/110 (satisfied)

S = 450
T/110 = 560

8. Directions (8-9): Study the following information carefully and answer the questions given beside.

Both the series given below follow a similar pattern. Solve both the series and answer the following questions.
I. a, 154, 1548, 12390, 74344, 297378
II. b, c, 490, 4908, d, 235624

8. What is the value of “d ÷ a”?

Correct Answer: (b) 3272.50
Solution:

Series Pattern                          Given Series
12                                                    12           a
12 × 12 + 10 = 154                         154
154 × 10 +8 = 1548                       1548

1548 × 8 + 6 = 12390                   12390
12390 × 6 + 4 = 74344                 74344
74344 × 4 + 2 = 297378               297378

Series 2:

Series Pattern                            Given Series
2                                                        2             b
2 × 14 + 12 = 40                              40            c
40 × 12 + 10 = 490                          490
490 × 10 + 8 = 4908                       4908
4908 × 8 + 6 = 39270                    39270        d
39270 × 6 + 4 = 235624                235624

From common explanation, we have d ÷ a = 39270 / 12 = 3272.50

9. The series given below follows a particular pattern. What is the value of ‘e’ in the following series?

 a/3, 4b, c/2.50, 4b³, e

Correct Answer: (b) 64
Solution:

Series Pattern                          Given Series
12                                                    12           a
12 × 12 + 10 = 154                        154
154 × 10 +8 = 1548                       1548

1548 × 8 + 6 = 12390                    12390
12390 × 6 + 4 = 74344                  74344
74344 × 4 + 2 = 297378                297378

Series 2:

Series Pattern                            Given Series
2                                                        2           b
2 × 14 + 12 = 40                             40           c
40 × 12 + 10 = 490                         490
490 × 10 + 8 = 4908                      4908
4908 × 8 + 6 = 39270                    39270      d
39270 × 6 + 4 = 235624                235624

From common explanation, we have

a/3, 4b, c/2.50, 4b³, e

12/3, 4 × 2, 40/2.50, 4 × 8, e

4, 8, 16, 32, e

4 × 2 = 8
8 × 2 = 16
16 × 2 = 32
32 × 2 = 64

So the answer = 64

10. Directions (10-11): Study the following information carefully and answer the questions given beside.

Three series are given below. One of the first 5 terms of each series is wrong.
The wrong term of Series I is identified as ‘I’, the wrong term of Series II is identified as ‘J’ and
the wrong term of Series III is identified as ‘K’.

Series I: 2544, 1264, 624, 302, 144, L, 24
Series II: 5, 8, 15, 23, 38, 61, 99, M
Series III: 173, 179, 181, 187, 193, N, 199

10. What is the value of “I - (M + N)”?

Correct Answer: (d) -55
Solution:

Series 1:

Series Pattern                                    Given Series
2544                                                         2544
2544 ÷ 2 − 8 = 1264                               1264
1264 ÷ 2 − 8 = 624                                 624
624 ÷ 2 − 8 = 304                                   304          I 
304 ÷ 2 − 8 = 144                                   144
144 ÷ 2 − 8 = 64                                      64            L
64 ÷ 2 − 8 = 24                                        24

Series 2:

Series Pattern                                    Given Series
7 + 8 = 15                                                15
8 + 15 = 23                                              23
23 + 38 = 61                                            61
38 + 61 = 99                                            99
61 + 99 = 160                                         160            M

There must be 7 in place of 5

So J = 5

Series III:

173, 179, 181, 187, 193, N, 199

The series consists of consecutive prime numbers. Consecutive prime numbers from 173

→ 173, 179, 181, 191, 193, 197, 199

191 should be in place of 187.

So K = 187
N = 197

I          J     K         L        M        N
302    5     187     64     160     197

From the common explanation, we have

Reqd. value = 302 − (160 + 197) = −55