ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

This is a G.P.(general process) in which a = 2, r = 2^2/2 = 2 and n = 9
S_n = a(rⁿ - 1)/(r - 1)
= 2 x (2⁹ - 1)/(2 - 1)
= 2 x (512 - 1)
= 2 x 511
= 1022.

Greatest number of five digits = 99999.
Required number must be divisible by L.C.M. of 15, 21 and 36, i.e 1260
On dividing 99999 by 1260, we get 459 as a reminder.
∴ Required number = (99999 - 459) = 99540
Answer : 99540

Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Lets, ten's and unit's be 2x and x respectively
Then, (10 times; 2x + x) - (10x + 2x) = 36
⇒ 9x = 36
⇒ x = 4

∴ Required difference (2x + x) - (2x - x ) = 2x = 4 × 2 = 8
Answer : 8

(n + 2)(n + 4)
= (2m + 2)(2m + 4)
= 2(m + 1)2(m + 2)
= 4(m +1)(m + 2)
= 4 × (product of two consecutive positive integers, one which must be even)
= 4 × (an even number), and this equals a number that is atleast a multiple of 8.
Hence, the answer is 8.

Clearly, 16² > 241.
∴ 16 > √241
Now,
Prime numbers less than 16 are 2, 3, 5, 7, 11, 13.
241 is not divisible by any of them. 
∴ 241 is a prime number
In that same fashion,
337 and 571 are also prime numbers.

Now, 391
clearly, 20² > 391
∴ 20 > √391
Now, Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19
Here, 391 is divisible by 17.
∴ 391 is not a prime number.

L.C.M of 3, 5, 6, 8, 10 and 12 = 120
So, the required number is of the form 120k + 2.
Least value of k for which (120k + 2) is divisible by 13 is k = 8.
∴ Required number = (120 × 8 + 2) = 962.
Answer : 962

⇒ x + 1/x = 13/6
⇒ (x² + 1)/x = 13/6
⇒ 6x² - 13x + 6 = 0
⇒ 6x² - 9x - 4x + 6 = 0
⇒ 3x(2x - 3) -2(2x - 3) = 0
⇒ (3x - 2)(2x - 3) = 0
⇒ x = 2/3 or 3/2

It is a G.P (general process) with r i.e;
2¹, 2², 2³,...
If number of term is n.Then,
2× 2^n-1 = 1024
2^n-1 = 512
2^n-1 = 2⁹
n -1 = 9
n = 10
Answer : 10

Choose n to be 0.
Then (n -2)/2
= (0 -2)/2
= -1 which is an integer.
So, eliminate
next, √n = √0 = 0.
Eliminate.
Next, 2/(n +1) = 2/1 = 2
eliminate, 
Next, √1/(n² + 2)
= √1/2
= 1/√2 which is not an integer
So, the Answer is: √1/(n² + 2)

(550 + 445) × 2 × 110 + 30 = 220030

Let the required number numbers be 33a and 33b
Then, 33a + 33b = 528
⇒a + b = 16
Now, co - primes with sum 16 are (1,15), (3,13), (5,11) and (7,9)
∴ Required numbers are (33 × 1, 33 × 15), (33 × 3, 33 × 13), ( 33 × 5, 33 × 11), (33 × 7, 33 ×9)
The numbers of such pairs are 4

The pages of the book may be divided into 10 groups;
(1 -100), (101 - 200), (201 -300),......, (901 - 1000).
Clearly, for the first group, one needs 11 zeros,
For second to ninth groups, one needs 20 zeros each.
So, total number of zeros required = 11 + 8 × 20 + 21 = 192
Answer : 192

Since it is a non - terminating and non - repeating decimal,
So it is an irrational number

n < 0 ⇒ 2n < 0,
again, - n > 0 and n² = (-n)2 > 0.
Thus, out of the numbers 0, -n, 2n and n²
We find that 2n is the least number here. 
Answer: 2n

On dividing 427398 by 15 we get the remainder 3, so 3 should be subtracted
Answer : 3

=> 3(2x+9) = 75
=> 2x + 9 = 25
=> x = 8
Answer : 8

We are given that the numbers m and n, when divided by 6, leave remainders of 2 and 3, respectively,
Hence, we can represent the numbers m and as 6p +2 and 6q + 3, respectively, where p and q are suitable integers.
Now, m - n = (6p + 2) - (6q + 3) = 6p - 6q - 1 = 6(p-q) - 1.
A remainder must be positive, so let's add 6 to this expression and compensate by subtracting 6 :
6(p - q) - 1 =
6 (p - q) - 6 + 6 - 1 =
6 (p - q) - 6 = 5 =
6 (p - q - 1) + 5
Thus, the remainder is 5,
and the answer is 5

H.C.F of two prime numbers is 1.
Product of numbers = (1 × 161) = 161.
Let the numbers be a and b.
Then, ab = 161.
Now, co - primes with product 161 are (1, 161) and (7, 23).
Since x and y are prime numbers and x > y, we have x = 23 and y = 7
∴ 3y -x = (3 × 7) - 23 = -2
Answer is : -2

Let the number be x and y, such that x > y.
then, 3x - 4y = 5 ..........(i)
And (x + y) - 6 (x - y) = 6 ⇔ -5x + 7y = 6 ........(ii)
Solving (i) and (ii), we get : x = 59 and y = 43
Answer : 59, 43.

The numbers 102 and 210 are themselves multiples of 3.
also a multiple of 3 exists once in every three consecutive integers.
Counting the multiples of 3 starting with 1 for 102,
2 {= 1 + (105 - 102)/3 = 1+ 1 = 2 } for 105,
3 { = 1 +(108 - 102)/3 = 1 + 2 = 3} for 108,
and so on, the count we get for 210 equals 1+(210 - 102)/3 = 1 + 36 = 37,
hence the answer is 37

L.C.M of A and B is ''B'';
L.C.M of B and C is ''B';
⇒ L.C.M of A, B, C is ''B''
Answer : B

Multiple of 7 between 11 and 200 are 14, 21, 28, 35, 42,....., 189, 196
T_m = 196 ⇒ 14 + (m - 1) × 7 = 196
⇒ (m - 1) × 7 = 182
⇒ (m - 1) = 26
⇒ m = 27.
Multiples of 7 and 3 both i.e. that of 21 are 21, 42, 63,......., 189
T_n = 189
⇒ 21 + (n - 1) × 21 = 189
⇒ (n - 1) × 21 = 168
⇒ (n - 1) = 8
⇒ n = 9.
Required number of terms = (27 - 9) = 18
Answer : 18

p < 1
⇒ 1/p > 1
⇒ 2/p > 2
2/p - p > 2 - p > 0
[∵ p < 1]
Hence, (2/p - p) is a positive number.

Number = (12 x 35)
Correct Quotient = 420 /21 = 20
Answer : 20

Let the numbers be 2x, 3x, 5x and 7x respectively.
Then, their L.C.M = (2 × 3 × 5 × 7)x = 210x
[∵ 2, 3, 5, 7 are prime numbers ]
So, 20x = 630
or x = 3
∵ The numbers are 6, 9, 15 and 21.
Required difference = 21 - 6 = 15.
Answer : 15

13p + 11 and x = 17q + 9
∵ 13p + 11 = 17q + 9
17q - 13p = 2
q = (2 + 13p)/17
∵ The least value of p for which q = (2 + 13p)/17 is a whole number p = 26
x = (13 x 26 + 11)
= (338 + 11)
= 349

Required L.C.M = (L.C.M of 3, 6, 9)/(H.C.F of 4, 7, 8)
=18/1 = 18
Answer : 18