বিগত সালের প্রশ্ন ও প্রশ্ন ব্যাংক

Solution:

Given that,

Total surface area of a cube, S = 150 square units.

We know,

Surface area of a cube, S = 6a²

⇒ 6a² = 150

⇒ a² = 150/6

⇒ a² = 25

⇒ a² = 5²

∴ a = 5

And we know, Volume of a cube, V = a³

= 5³

= 125

So the volume of the cube is 125 cubic units.

Solution:

Given that, Diagonal of rectangle, d = 13

Width, w = 5

We know,

Pythagorean theorem, l² + w² = d²

⇒ 13² = l² + 5²

⇒ 169 = l² + 25

⇒ l² = 169 - 25

⇒ l² = 144 = 12²

∴ l = 12

∴ length In a rectangle is, l = 12

Perimeter of a rectangle, P = 2(l + w)

= 2(12 + 5) = 2 × 17 = 34

So the perimeter of the rectangle is 34.

Solution:

Given that,

Surface area of a cube = 96 square units

We know,

Surface area of a cube, S = 6a²

⇒ 6a² = 96

⇒ a² = 96/6

⇒ a² = 16 = 4²

∴ a = 4

The longest stick that can fit inside the cube runs along the space diagonal.

So the space diagonal of a cube, d = a√3 = 4√3 ; [a = 4]

So the length of the longest stick that can be placed inside the cube is 4√3 units.

Solution:

Given that,

Trapezoid with bases a = 10 and b = 16

Height, h = 4

We know,

Area of trapezoid = (1/2) × (sum of bases) × height

= (1/2) × (a + b) × h

= (1/2) × (10 + 16) × 4

= (1/2) × 104

= 52

So the area of the trapezoid is 52 square units.

Solution:

Given that,

Two similar triangles Areas ratio = 4 : 9

Perimeter of smaller triangle = 20

For similar triangles, the ratio of areas = square of the ratio of corresponding sides.

(Area of larger/Area of smaller) ​=(side of larger/side of smaller​)²

⇒ (9/4) = (k/1)² ;

[Let the ratio of sides = k : 1 (larger : smaller)]

⇒ k² = 9/4

∴ k = 3/2

∴ Perimeter of larger : Perimeter of smaller = 3 : 2

∴ Perimeter of larger = 20 × (3/2) = 30 ;

[Perimeter of smaller = 20]

So the perimeter of the larger triangle is 30.

Solution:

Given that,

log₂(x + 3) = 4

⇒ x + 3 = 2⁴

⇒ x + 3 = 16

⇒ x = 16 - 3

∴ x = 13

Solution:

Given inequality,

4(x - 6) + 4 < 8(x - 4)

⇒ 4x - 24 + 4 < 8x - 32

⇒ 4x - 20 < 8x - 32

⇒ 4x - 8x < - 32 + 20

⇒ - 4x < - 12

∴ x > 3

Solution:

We know that for any positive real number,

a⁰ = 1 and b⁰ = 1

So, (a⁰ - 3b⁰)⁵

= (1 - 3 × 1)⁵

= (1 - 3)⁵

= (- 2)⁵ = - 32

Solution:

Given that,

3^{x + y} = 9⁴

⇒ 3^{x + y} = (3²)⁴

⇒ 3^{x + y} = 3⁸

∴ x + y = 8 ........(1)

And, 2^{x - y} = 16

⇒ 2^{x - y} = 2⁴

∴ x - y = 4 ....... (2)

Now (1) + (2) than we get,

⇒ (x + y) + (x - y) = 8 + 4

⇒ 2x = 12

⇒ x = 12/2

∴ x = 6

From (1), 6 + y = 8

⇒ y = 8 - 6

∴ y = 2

∴ x² - y² = (6)² - (2)²

= 36 - 4 = 32

Solution: 
12 workers can complete work in 15 days
∴ 1 workers can complete work in  = 12 × 15 days 
∴ 9 workers can complete work in = (12 × 15)/9 days 
= 20 days 

It will take 9 workers 20 days to complete the same project. 

Solution: 
Given that, 
The sum doubles itself in 8 years.
Amount after 8 years = 2P
Simple Interest for 8 years = P

We know,
SI = (P × r × n)/100
⇒ P = (P × r × 8)/100
⇒ 8r = 100
⇒ r = 100/8
∴ r = 12.5% per year

Now, we want to find in how many years the sum becomes four times itself.
Amount = 4P
Interest needed = 4P - P = 3P

We know,
SI = (P × r × n)/100 
⇒ 3P = (P × 12.5 × n)/100.  ; [r = 12.5%]
⇒ 3 = (12.5 × n)/100
⇒ n = (3 × 100)/12.5
∴ n = 24 years

∴ The sum will become four times itself in 24 years.

Solution:

Let,
the cost price (CP) be Tk. 100

Marked Price = 40% more than cost price
= 100 + 40
= Tk. 140 

Discount = 15% of 140
= (15/100) ​× 140
= Tk. 21

Selling Price (SP)= 140 - 21 = Tk. 119 

∴ Profit = SP - CP = 119 - 100 = Tk. 19 

∴ overall percentage profit = (profit/cost price) × 100%
= (19/100) × 100%
= 19%

Solution:
Given that,
A : B = 2 : 3
B : C = 4 : 5
And A = 16

Now, 
A : B = 2 : 3 
⇒ A/B = 2/3
⇒ 2B = 3A
⇒ B = (16 × 3)/2  ; [A = 16]
∴ B = 24

And,
B : C = 4 : 5
B/C = 4/5
⇒ 24/C = 4/5
⇒ C = (24 × 5) / 4
∴ C = 30

So the value of C is 30.

Solution:
Given that,
Total students = 40
Study French = 18
Study Spanish = 15
Study both = 7

Students studying at least one language = F + S - Both
= 18 + 15 - 7
= 26

Therefore, Students who study neither language = Total students - at least one language
= 40 - 26 = 14

∴ 14 students study neither language.

Solution:
Let the beginning price of January = 100x
The price of a stock increased by 20% in January.
Price of stock become = 100x + 100x of 20%
= 100x + (100x × 20)/100
= 120x

And then the stock price decreased by 10% in February
Price of stock become = 120x - 120x of 10%
= 120x - (120x × 10)/100
= 108x

ATQ,
108x = 108
⇒ x = 108/108
∴ x = 1

Therefore the price at the beginning of January
= 100 × 1
= Tk. 100

The price at the beginning of January was Tk. 100.

Solution:

Given that,

Fixed service fee = Tk. 50

Charge per hour = Tk. 30

Total bill = Tk. 200

Let h = number of full hours the technician works.

Now, total cost equation,

50 + 30h ≤ 200

⇒ 30h ≤ 200 - 50

⇒ 30h ≤ 150

⇒ h ≤ 150 / 30

∴ h ≤ 5

So the technician can work a maximum of 5 full hours.