ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

AC = Ladder
BC = 6.5 metres
In△ABC
Cos60^∘ = BC/AC
1/2 = 7/AC
AC = 14m

We can multiply equation 2 by -1 ad equation 4 by -1, and we have:
x + u= 6
-x + v =- 5
x + w = 4
-x + z = -3

Adding these equations together we have:
u + v + w + z = 2

We know surface area of hemisphere = 3πr²
ATQ, 3πr² = 48π
⇒ r² = 16 
⇒ r = 4

sec (x − 30°) = 2
⇒  sec (x - 30°) = sec 60°
⇒  x - 30° = 60°
⇒  x = 90°
∴ cot 90° = 0

We know that, Sum of 3 angles of a triangle is 180°
Here, Sum of the ratios are = 3+4+5 = 12
The larger angle = 180° × 5/12 = 75°
And, the smaller angle = 180° × 3/12 = 45°

Difference between the biggest and smallest angel = 75° - 45° = 30°

2x + 9y = 3
⇒ 7× 2x + 7 × 9y = 7× 3
⇒ 14x + 63y = 21
The given equation 14x + my = 23
The pair of equations 2x + 9y = 3 and 14x + my = 23 does not have a unique solution if the value of m is 63.

The area of the triangle
= (16/4)√(4 × 10² - 16²)
= 4√(400 - 256)
= 4 × √144
= 4 × 12
= 48 sq cm

Consider an equilateral triangle ABC having sides a and a median AD of length x unit'.
In an equilateral triangle, the median is always the perpendicular bisector of the triangle. 
So, BD=a/2

In triangle ABD, by Pythagoras theorem, we have
AB²=AD²+BD²
⟹a²=x²+(a​/2)²
⟹a²=x²+ a²/4​
⟹3a²​/4=x²
or,a²=4x²/3​

Now, area of equilateral triangle =√3​a²​/4
                                                     =​​(√3​​/4) × (4x²/3​)
                                                   =√3x²/3

 length of a median of an equilateral triangle = 3 cm

so,  area of equilateral triangle = √3x²/3 = √3×3²/3 = 3√3 sq cm

Given,
x = 1/2 as the root of equation x² - mx - 3/4 = 0.
(1/2)² – m(1/2) – 3/4 = 0
=> 1/4 - m/2 - 3/4 = 0
=>  -2/4 = m/2
=> 4m = -4
=> m = -1

Area of Wall = ( 5 + 4 + 5 + 4 ) m. wall length × 3 m height
                     = 18 × 3  sq.m. 
                     = 54 sq.m.

Area of Ceiling =  5 m × 4m
                        = 20 sq.m.

hence total painting area of walls and ceiling = 54 sq m + 20 sq m = 74 sq m
74 square meter area of walls and ceiling to be painted.

area of a pizza (circle) = πr² = 49π
Or, r² = 49π/π = 49
Or, r = 7
So, the size of the pizza box would be =  2r = 2 × 7 = 14 in

ATQ, x + x + 3x + 15 = 180°
Or, 5x = 165°
Or, x = 33°

Area of the floor = 4 × 8 = 32 m²
Area of each carpet = 2 × 2 = 4 m²
Required number of carpet = 32/4 = 8
one carpet costs Tk. 14
8 carpet costs Tk. 14 × 8 = Tk. 112

Adding the given equations:
12a + 3b + 7b - 2a = 2 + 8
Or, 10a + 10b = 10
Or, 10(a + b) = 10
Or, a + b = 1
So, average of a and b is 1/2

Given, radius = r and
Height = 5r

ATQ, πr² × 5r = 625π
⇒ r³ = 125
⇒ r = 5

We need to divide each large field into smaller flower beds such that the area of each bed is same.

So, we find the HCF of the larger fields that gives us the area of the smaller field.

HCF (60, 84, 108) = 12

Now, this HCF is the area (in m²) of each flower bed.

Also, the area of a rectangular field = Length x Breadth

=> 12 = 4 x Breadth

=> Breadth = 3m

Hence, each flower bed would be 3m wide.

Let x and (x + 1) be the two consecutive integers.

According to the given,
x(x + 1) = 20
x² + x = 20
x² + x – 20= 0

The area of the garden, excluding the walkway, is
= {(50 - 2×2) × (40 - 2×2)}
= 46×36
= 1656 m²