ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: A circle and a rectangle have the same perimeter. The sides of the rectangle are 9 cm and 13 cm. What is the area of the circle?

Solution:
The sides of the rectangle are 9 cm and 13 cm.
Perimeter of the rectangle =2(9 + 13) = 44 cm
Circumference of circle = 44 cm.

Here
2πr = 44
(22/7)r = 22
r/7 = 1
r = 7

Area of circle = πr²
= (22/7) × 7²
= (22/7) × 49
= (22 × 7)
= 154 cm²

Question: If 1 + sinθ = x cosθ, then cotθ is-

Solution:
Given,
1+ sinθ = xcos θ
⇒ (1 + sinθ)/cosθ = x
⇒ 1/cosθ + sinθ/cosθ  = x
⇒  secθ + tanθ = x...............(i)

We know,
 sec²θ - tan²θ = 1
⇒ (secθ + tanθ) (secθ - tanθ) = 1  
⇒ x(secθ - tanθ) = 1
⇒ secθ - tanθ = 1/x.................(ii)

(i) - (ii) ⇒
secθ  +  tanθ - (secθ - tanθ) = x - 1/x
⇒ secθ +  tanθ - secθ + tanθ = (x² - 1)/x
⇒ 2tanθ = (x² - 1)/x
⇒ tanθ = (x² - 1)/2x
∴ cotθ = 2x/(x² - 1)

Question: The difference between the circumference and the radius of a circle is 185 cm. Find the radius of a circle is -

Solution:
Let r be the radius of circle

Given that,
2πr - r = 185
⇒ r(2π - 1) = 185
⇒ r{(44/7) - 1} = 185
⇒ r (44 - 7)/7 }= 185
⇒ r(37/7) = 185
⇒ r = 185 (7/37)
∴ r = 35

∴ Radius of circle is 35 cm.

Question: sin(θ + 15°) = 3/√12 হলে cos²θ = কত?

Solution:
sin(θ + 15°) = 3/√12
⇒ sin(θ + 15°) = 3/(2√3)
⇒ sin(θ + 15°) = √3/2
⇒ sin(θ + 15°) = sin60°
⇒ θ + 15° = 60°
⇒ θ = 45°

Now,
cos²θ = (cos 45°)²
= (1/√2)²
= 1/2

Question: How many cubes of 3 cm edge can be cut out of a cube of 18 cm edge? 

Solution: 
Number of cubes 
= (18 × 18 × 18)/(3 × 3 × 3) = 216

Question: Given that 1 cubic cm of marble weighs 25 gms, the weight of a marble block 28 cm in width and 5 cm thick is 112 kg. What is the length of the block? 

Solution: 
Let the length of the block = x cm
Then (x × 28 × 5 × (25/10000)) = 112
∴ x = (112 × (1/28) × (1/5) × (1000/25)) = 32 cm 

Question: A swimming pool is 25 m long and 18 m broad. When a number of men dive into the pool, the height of the water rises by 1 cm. If the average amount of water displaced by one of the men be 0.1 cubic meter, how many men are there in  the pool? 

Solution: 
Volume of water displaced= 25 × 18 × (1/100) m³ = 9/2 m³ 
Volume of water displaced by 1 man = 0.1m³ 
∴ Number of men are in the pool = (9/2)/0.1 = (9 × 10)/2 = 45  [0.1 m = 10 cm]

Question: If each edge of a cube is increased by 25%, then the percentage increase in its surface area is? 

Solution: 
Let the original edge = a 
Then, the surafce area = 6a² 
New edge = 125a/100 = 5a/4 
New surafce area = 6 × (5a/4)² = 75a²/8
Increase in surface area = (75a²/8 - 6a²) = 27a²/8
The percentage increase in its surface area = (27a² × 1/6a² × 100)% = 56.25% 

Question: The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 4400 cm³. Then its radius will be- 

Solution: 
Let the radius and height of the cylinder be 5x and 7x cm respectively.
Then, volume = πr²h
= (22/7) × (5x)² × 7x
= 550x³

ATQ,
550x³ = 4400 
x³ = 4400/550 = 8
x = ∛8 = 2 

Hence, the radius = (5 × 2) = 10 cm

Question: A right-angled triangle, whose perpendicular sides measure 2.4 cm and 1.8 cm, is inscribed in a circle. What is the diameter of the circle (in cm)?

Solution:

Let,
ABC is a right triangle with perpendicular AB = 2.4 cm and base BC = 1.8 cm

We know,
AC² = AB² + BC²
⇒ AC = √(2.4² + 1.8²)
⇒ AC = √(5.76 + 3.24)
⇒ AC = √9
∴ AC = 3 

∴ The diameter of the circle, 2r = 3 cm

Question: The right circular cone of height 24 cm has a volume of 1232 cm³, then the area of its curved surface is? 

Solution: 
Volume of the cone = (1/3) × r² × h = 1232
⇒ (1/3) × (22/7) × r² × 24 = 1232 
⇒ r² = (1232 × 7 × 3)/(22 × 24)
⇒ r² = 49
∴ r = 7 

So, slant height l = √(7)² + (24)² = √625 = 25 
So, curved surface area = πrl = (22/7) × 7 × 25 = 550 cm³ 

Question: Two cans have the same height equal to 21 cm. One can is cylindrical, the distance of whose base is 10 cm. The other can has square base of side 10 cm. What is the difference in their capacities?

Solution:
Difference in capacities = Volume of cubodial can - Volume of cylindrical can 
= [ (10 × 10 × 21) - ((22/7) × 5 × 5 × 21))] cm³
= (2100 - 1650) cm³ 
= 450 cm³

Question: The value of {(tan 15° - tan 60°)/(cot 30° - cot 75°)} + 1 is -

Solution:
(tan 15° - tan 60°)/(cot 30° - cot 75°) + 1
= (tan 15° - tan 60°)/cot (90° - 60°) - cot (90° - 15°) + 1
= (tan 15°- tan 60°)/(tan 60° - tan 15°) + 1
= (tan 15°- tan 60°)/(- 1)(tan 15° - tan 60°) + 1
= - 1 + 1
= 0

Question: If the height of a cone is doubled and its base diameter is trebled, then the ratio of the volume of the resultant cone to that of the original cone is? 

Solution:  
Let the original radius and height of the cone be r and h respectively.
Then, new radius = 3r and new height = 2h 
∴ New volume/Original volume = ((1/3) × π × (3r)² × 2h) : ((1/3) × π × r² × h) = 18 : 1 

Quetion: A right cylindrical vessel is full of water. How many right cones having the same radius and height those of the right cylinder will be needed to store that water?  

Solution: 
Let the radius be r and the height be h.

Then, number of cones needed = (Volume of cylinder/Volume of 1 cone)
= πr²h/(1/3)πr²h
= 3

Question: Angles of a quadrilateral are in the ratio 3 : 4 : 5 : 8. The largest angle is -

Solution:
Let First angle = 3x
Second angle = 4x
Third angle = 5x
and fourth angle = 8x
We know 3x + 4x + 5x + 8x = 360°
⇒ 20x = 360°
⇒ x = 18°

∴ Measure of largest angle = 8x
= (8 × 18°)
= 144°

Question: If a 64 meter ladder is placed against a 32√2 metre wall such that it just reaches the top of the wall, the angle of elevation of the wall is

Solution:
Given that 
Ladder's length = 64 m
Wall's height = 32√2 m

Perpendicular = Wall's height = 32√2 m
Hypotenuse = Ladder's length = 64 m

We know 
Sinθ = Perpendicular/Hypotenuse
⇒ Sinθ = 32√2/64
⇒ Sinθ = 1/√2
⇒ Sinθ = sin 45º
∴ θ = 45º

Question: If a regular square pyramid has a base of size 8 cm and height 30 cm. What is its volume?  

Solution: 
Volume of the pyramid = (1/3) × 8² × 30 cm³
= 640 cm³ or 640 cc

Question: The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 3.5 m away from the wall. The length of the ladder is-

Solution: 

Let AB be the wall and BC be the ladder.

 Then, ∠ACB = 60°
⇒ AC = 3.5 m
⇒ AC/BC= cos60^∘
⇒ AC/BC= 1/2
⇒ BC = 2 × AC
⇒ BC = 2 × 3.5
⇒ BC = 7 m

Question: If a 26 m ladder is placed against a 13 m wall such that it just reaches the top of the wall. What will be the elevation of the wall?

Solution:


AC = 26 meters
AB = 13 meters
∠ACB = θ

∴ sinθ = AB/AC
⇒ sinθ =13/26
⇒ sinθ =1/2
⇒ sinθ = sin30^∘
⇒ θ = 30^∘

Question: The angle of elevation of the top of a tower at a point on the ground 37 m away from the foot of the tower is 45°. What is the height of the tower? 

Solution:

 
Let AB be tower and C is a point on the ground 37 m away
From the foot of tower B
The angle of elevation is 45°

Let h be the height of the tower.
∴ tanθ = AB/BC
⇒ tan45^∘ = AB/37
⇒ 1 = AB/37
⇒ AB = 37 m

Question: The area of a circle whose radius is the diagonal of a square whose area is 16 square cm is -

Solution:
Area of square = 16
Side of square = √16 = 4

Diagonal of square = 4√2

So, the radius of the circle is 4√2 cm

Area of circle = πr²
= π(4√2)²
= 32π

Question: A wire can be bent in the form of a circle of radius 42 cm. If it is bent in the form of a square, then its area will be -

Solution:
দেয়া আছে,
বৃত্তের ব্যাসার্ধ r = 42 cm

বৃত্তের পরিধি = 2πr 
 = 2 × (22/7) × 42 
= 2 × 22 × 6
= 264 cm 

বর্গের এক বাহুর দৈর্ঘ্য = 264/4 = 66 cm

বর্গের ক্ষেত্রফল = (66)² cm² = 4356 cm²

Question: The area of a circle is increased by 22 square cm if its radius is increased by 1 cm. The original radius of the circle is -

Solution:
Let the original radius of the circle be r cm.

ATQ,
π(r + 1)² - πr² = 22
⇒ π{(r + 1)² - r²} = 22
⇒ π(r² + 2r + 1 -r²) = 22
⇒ 2r + 1 = 22/π
⇒ 2r + 1 = (22 × 7)/22
⇒ 2r + 1 = 7
⇒ 2r = 6
⇒ r = 3 cm

Question: If the breadth of a rectangle is decreased by 20%, then to double the area, its length is required to be increased by -

Solution:
ধরি,
প্রস্থ = ১০০ একক এবং দৈর্ঘ্য ১০০ একক
ক্ষেত্রফল = ১০০ × ১০০ = ১০০০০ বর্গএকক

২০% কমানোর পর প্রস্থ = ১০০ - ২০ = ৮০ একক
এবং নতুন ক্ষেত্রফল = ১০০০০ × ২ = ২০০০০ বর্গএকক

সুতরাং নতুন দৈর্ঘ্য = ২০০০০/৮০ = ২৫০ একক

দৈর্ঘ্য বৃদ্ধি = ২৫০ - ১০০ = ১৫০ একক
শতকরা দৈর্ঘ্য বৃদ্ধি = (১০০ × ১৫০)/১০০ = ১৫০%

Question: The area of the four walls of a room is 120 square meters and the length is twice the breadth. If the height of the room is 4 m, then the area of the floor is -

Solution: 
Let the breadth = x meters and
length = (2x) metres

Area of 4 walls = 2(2x + x) × 4
= 24x

ATQ,
∴ 24x = 120
⇒ x = 5

So, length = 10 m, and breadth = 5 m

Area of the floor = 10 × 5 = 50 square meters