ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Let the number of hens be x and the number of cows be y.
Then, x + y = 48 .... (i)
and 2x + 4y = 140
x + 2y = 70 .... (ii)
Solving (i) and (ii) we get: x = 26, y = 22.
The required answer = 26.

Given,
5 - [4 - {3 - (3 - 3 - 6)}]
= 5 - [4 - {3 - (-6)}]
= 5 - [4 - {3 +6}]
= 5 - [4 - {9}]
= 5 - [4 - 9]
= 5 - [-5]
= 5 + 5
= 10

Here,
(28+14+7+4+2+1)/ 28 = 56/ 28 =2

Let the number be = x
According to question,
x²=(75.15)²−(60.12)²
⇒x²=(75.15+60.12)(75.15−60.12)
⇒x²=135.27×15.03
⇒x²=2033.1081
⇒x=45.09

Let the three consecutive numbers are x, x + 1, x + 2
According to question
x² + (x + 1)² + (x + 2)² = 110
⇒ x² + x² + 1 + 2x + x² + 4 + 4x = 110
⇒ 3x² + 6x + 5 = 110
⇒ 3x² + 6x - 105 = 0
⇒ x² + 2x - 35 = 0
⇒ x_² + 7x - 5x - 35 = 0
⇒ x(x + 7) -5(x + 7) = 0
⇒ (x + 7)(x - 5) = 0
⇒ x = 5 & -7
[∴ (-) value can not considered]
∴ Smallest number is = 5

According to question,
3a+4b /3a−4b = 3c+4d/ 3c−4d
⇒ 3a/4b = 3c/4d
⇒ad=bc

Let the numbers be x and y
According to question,(x>y)
x²−y²=45
(x+y)(x−y)=45
Make factor of 45
15×3
9×5
45×1
Total 3pairs
These pairs gives the value of x and y which satisfy the given condition.

200 ÷ 25 × 4 + 12 - 3
= 200/ 25 × 4 + 12 - 3
= 8 × 4 + 12 - 3
= 32 + 12 - 3
= 44 - 3
= 41

4 x 162 = 648. Sum of decimal places = 6.
So,
0.04 x 0.0162
= 0.000648
= 6.48 x 10^-4

0.125125... = 0.125 = 125/999

(489.1375 x 0.0483 x 1.956) / (0.0873 x 92.581 x 99.749)
≈ 489 x 0.05 x 2 / 0.09 x 93 x 100
= 489/ (9 x 93 x 10)
= (163/279)x (1/10)
= 0.58/ 10
= .058
≈ .06

Sum of decimal places = 7.
Since the last digit to the extreme right will be zero (since 5 x 4 = 20),
so there will be 6 significant digits to the right of the decimal point.

Given,
52416/312=168
⇔ 52416/ 168 =312
Now, 52.416/ 0.0168
= 524160/ 168
= (52416/ 168 )×10
=312×10
=3120

9 × 7 = 63
Sum of decimal places = 5
∴ 0.09 × 0.007 = 0.00063

L.C.M. of 21, 36, 66 = 2772.
Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11
To make it a perfect square, it must be multiplied by 7 x 11.
So, required number = 2² x 3² x 7² x 11² = 213444

Money collected = (59.29 x 100) paise = 5929 paise.
Number of members = √5929 = 77.

Let the missing number be x
Given, 48.2 × 2.5 × 2.2 + x = 270
⇒ x = 270 - 48.2 × 2.5 × 2.2
⇒ x = 270 - 265.1
⇒ x = 4.9
Hence, the number is 4.9

Quantity of blood donated in 2 years
= (350 × 3) ml
= 1050 ml
= 1.05 litres
∴ Quantity of blood donated in 6 years
=( 1.05/2 ×6)
= 3.15 litres

(0.11)³+(0.22)³+....+(0.99)³
=(0.11)³(1³+2³+....+9³)
=0.001331×2025
=2.695275
≈2.695

Given expression :
= (10.3)³+1³ / (10.3)²−(10.3×1)+1²
= (a³+b³ / a²−ab+b²)
=(a+b)
=(10.3+1)
=11.3

Let x /1776 = 111/ x
Then,
⇔x²=111×1776
⇔x²=111×111×16
⇔x= √ {(111)²×(4)²}
⇔x=111×4
⇔x=444

Let the number be x
Then,⇔3/5 x² = 126.15
⇔x²=(126.15× 5/3)
⇔x²=210.25
⇔x= √ 210.25
⇔x=14.5

= √(8/3)
= √(8×3)/(3×3)
= √(24/3)
=4.899/3
=1.633

L.C.M. of 3, 4, 5, 6, 8 is 120
Now 120 = 2 × 2 × 2 × 3 × 5
To make it a perfect square, it must be multiplied by 2 × 3 × 5
So, required number
=22×22×32×52
=3600

21600=2⁵×3³×5²
To make it a perfect cube, it must be multiplied by (2 × 5), i.e., 10