ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

(2√2 + 5√3) + (√2 - 4√3)
= (2 + 1)√2 + (5 - 4)√3
= 3√2 + √3

Let
the ten's digit be x 
unit's digit be y.

Then,
(10x + y) - (10y + x) = 36
9(x - y) = 36
x - y = 4

Divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

Let
The three integers be x, x + 2 and x + 4.

Then,
3x = 2(x + 4) + 3      
3x = 2x + 8 + 3
3x - 2x = 11
x = 11.

Third integer = x + 4 = 11 + 4 = 15

On dividing 8567 by 4, the remainder is 3.

To make it divisible by 4, we must add (4 - 3) =1 

Let
the larger numbers be x and
the smaller numbers be y.

Then, xy = 9375 and
          x/y = 15
now 
xy/(x/y) = 9375/15
y² = 625
y² = 25²
y = 25 

x = 25 × 15 = 375

x + y = 375 + 25 = 400

Given that 
HCF = 24
LCM = 168
Ratio of numbers = 1 ∶ 7. 
Let numbers be x and 7x.

We know 
Product of numbers = LCM × HCF
 x × 7x = 24 × 168
⇒ x² = 24 × 24
⇒ x = 24
∴ Larger number = 7x
                            = 24 × 7
                             = 168

HCF of two number is 4
Let, those numbers be 4x and 4y where x and y are prime to each other Accordingly,
4x + 4y = 36
⇒ 4(x + y) = 36
⇒ (x + y) = 9
Now,
9 = 8 + 1
9 = 7 + 2
9 = 6 + 3
9 = 5 + 4
In all these cases only (8, 1); (7, 2); and (5, 4) are prime to each other.
So, such three pair is possible
∴ Such three pair of number is possible.

Let
the numbers be x and x + 2.

Then,
(x + 2)² - x² = 84
x² + 2.x.2 + 2² - x² =84
4x + 4 = 84
4x = 80
x = 20.

The required sum = x + (x + 2)
                             = 2x + 2
                             = (2 × 20) + 2
                             = 42

Here
20 − 14 = 6
25 − 19 = 6
35 − 29 = 6
40 − 34= 6

Required number (L.C.M of 20, 25, 35, and 40) - 6

L.C.M of 20, 25, 35, and 40 = 1400
Required number= 1400 - 6 = 1394

Let us assume the least perfect square be X
⇒ 7 = 7 × 1
⇒ 11 = 11 × 1
⇒ 12 = 2² × 3 ⇒

The LCM of (7, 11, 12) = 2² × 3 × 11 × 7
⇒ The least perfect square = 2² × 3² × 11² × 7² = 213444
∴ The required result will be 213444.

Given:
The sum of two numbers is 40.
The difference of two number is 4.
Let
the larger number be x.
 the smaller number be y.
According to the question,
(x + y)/(x - y) = 40/4 
(x + y)/(x - y) =10/1
10(x - y) = x + y
10x - 10y = x + y
10x - x = 10y + y
9x = 11y 
x/y = 11/9
x : y = 11 : 9 

∴ The ratio of the number is 11 : 9

On dividing 1111 by 99, the quotient is 11 and the remainder is 22.
Hence, the required number is 11

Divisibility of 6 = Number should be multiple of 3 and 2
6*2 it is definitely divisible by 2
To be divisible by 3 Sum of all digits of the number be divisible by 3
We just try to put value in place of *
632 = 6 + 3 + 2 = 11(not divisible by 3)
662 = 6 + 6 + 2 = 14(not divisible by 3)
672 = 6 + 7 + 2 = 15( divisible by 3)
622 = 6 + 2 + 2 = 10(not divisible by 3)
∴ * is replaced by 7

6x² - x - 1 = 6x² - 3x + 2x - 1
                = 3x(2x - 1) + 1(2x - 1)
                 = (2x - 1) (3x + 1)


(3x² + 7x + 2) = 3x² + 6x + x + 2
                      = 3x(x + 2) + 1(x + 2)
                       = (x + 2)(3x + 1)

(2x² + 3x - 2) = 2x² + 4x - x - 2
                      = 2x(x + 2) - 1(x + 2)
                      = (x + 2) (2x - 1)

∴ Required L.C.M. is = (2x - 1) (3x + 1) (x + 2)

Let us consider the number be 3x,4x and 5x
The L.C.M of the three numbers be =60x
60x = 2400
x = 2400/60
x = 40

Then the number be
3x = 3 × 40 = 120
4x = 4 × 40 = 160
5x = 5 × 40 = 200

∴ The HCF of the three numbers be 40

Let
the first three consecutive numbers be x, x + 1 and x + 2.
Then, 
x + x + 1 + x + 2 = 30
⇒ 3x + 3 = 30
⇒ 3x = 27
⇒ x = 9
∴ The first three numbers are  9, 10,11
⇒ The next three numbers are 12, 13, 14.
Hence, the required sum =12 + 13 + 14
                                        = 39

The descending order of the number from 1 to 24  which are divisible by 2 is:
24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2
Now count the 7th place number from the right.

The number will be at the 8th place from the bottom is 14.

Let 
the numbers be a and b.
According to the problem,
the first equation is:
a/5=5b/8
a/5=5b/8
25b = 8a
a = 25b/8

   a + 35=4b
⟹a = 4b − 35
(25b/8) - 4b = - 35 
(25b - 32b)/8 = - 35 
- 7b/8 = -35
b = (35 × 8)/7
b = 40

Let
the number be x
According to problems condition 
(x/5) + 4 = (x/4) - 10
(x/5) - (x/4) = - 10 - 4
(4x - 5x)/20 = - 14
-x/20 = - 14 
x = 280

Let,
the numbers are a, b and c
According to the question,
⇒ a + b = 10
⇒ b + c = 16
⇒ c + a = 24

From above question,
⇒ 2(a + b + c) = 50
⇒ (a + b + c) = 25
Value of a = 25 - (b + c) = 25 - 16 = 9
Value of c = 25 - (a + b) = 25 - 10 = 15

Required difference = (15 - 9) = 6
∴ Difference is 6