১.
The angle of elevation of ladder leaning against a house is 60° and the foot of the ladder is 7 metres from the house. The length of the ladder is-
ব্যাখ্যা
AC = Ladder
BC = 6.5 metres
In△ABC
Cos60∘ = BC/AC
1/2 = 7/AC
AC = 14m
ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স
মোট প্রশ্ন১৮
সিলেবাস
Math - 10: Equation Solving, Geometry, Trigonometry (Area, Volume, Heights and Distances)
ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স
ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স · তারিখ অনির্ধারিত · ১৮ প্রশ্ন
AC = Ladder
BC = 6.5 metres
In△ABC
Cos60∘ = BC/AC
1/2 = 7/AC
AC = 14m
২.
A system of equations is shown below:
X + u= 6
X - v=5
X + w = 4
X - z = 3
What is the value of u+v+w+z?
X + u= 6
X - v=5
X + w = 4
X - z = 3
What is the value of u+v+w+z?
ব্যাখ্যা
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
x + u= 6
-x + v =- 5
x + w = 4
-x + z = -3
Adding these equations together we have:
u + v + w + z = 2
৩.
If the total surface area of the hemisphere be 48π sq. cm, then its radius is -
ব্যাখ্যা
We know surface area of hemisphere = 3πr2
ATQ, 3πr2 = 48π
⇒ r2 = 16
⇒ r = 4
ATQ, 3πr2 = 48π
⇒ r2 = 16
⇒ r = 4
৪.
If sec(x − 30°) = 2, then cot x = ?
ব্যাখ্যা
sec (x − 30°) = 2
⇒ sec (x - 30°) = sec 60°
⇒ x - 30° = 60°
⇒ x = 90°
∴ cot 90° = 0
⇒ sec (x - 30°) = sec 60°
⇒ x - 30° = 60°
⇒ x = 90°
∴ cot 90° = 0
৫.
Three angles of a triangle are in proportion 3 : 4 : 5. Then what is the difference in degrees between the biggest and the smallest angles?
ব্যাখ্যা
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°
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°
৬.
For what value of 'm' will be the pair of equations 2x + 9y = 3 and 14x + my = 23 does not have a unique solution?
ব্যাখ্যা
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.
⇒ 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.
৭.
If the land of the isosceles triangle is 16cm and the other two sides are 10cm each, what is the area of the triangle?
ব্যাখ্যা
The area of the triangle
= (16/4)√(4 × 102 - 162)
= 4√(400 - 256)
= 4 × √144
= 4 × 12
= 48 sq cm
= (16/4)√(4 × 102 - 162)
= 4√(400 - 256)
= 4 × √144
= 4 × 12
= 48 sq cm
৮.
If 3 cm is the length of a median of an equilateral triangle, then the area is -
ব্যাখ্যা
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 =√3a2/4
=(√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
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 =√3a2/4
=(√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
৯.
If 1/2 is a root of the quadratic equation x2 - mx - 3/4 = 0, then value of m is:
ব্যাখ্যা
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
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
১০.
The floor of a room is of size 5 m × 4m and its height is 3m. The walls and ceiling of the room require painting. What is the area to be painted?
ব্যাখ্যা
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.
= 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.
১১.
If the area of a small pizza is 49π in2, what size pizza box would best fit the small pizza?
ব্যাখ্যা
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
Or, r2 = 49π/π = 49
Or, r = 7
So, the size of the pizza box would be = 2r = 2 × 7 = 14 in
১২.
The angle measure of base angles of an isosceles triangle are represented by x and the vertex angle is 3x+15. Find the measure of base angle.
ব্যাখ্যা
ATQ, x + x + 3x + 15 = 180°
Or, 5x = 165°
Or, x = 33°
Or, 5x = 165°
Or, x = 33°
১৩.
A rectangular floor that measures 4 meters by 8 meters is to be covered with carpet that each measures 2 meters by 2 meters. If the carpet cost Tk. 14 a piece, what is the total cost to cover the floor?
ব্যাখ্যা
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
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
১৪.
If 12a + 3b = 2 and 7b – 2a = 8 , what is the average of a and b ?
ব্যাখ্যা
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
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
১৫.
The height of a cylinder is five times the radius of the cylinder. If the volume of the cylinder is 625π sq cm, what is the radius of the cylinder?
ব্যাখ্যা
Given, radius = r and
Height = 5r
ATQ, πr2 × 5r = 625π
⇒ r3 = 125
⇒ r = 5
Height = 5r
ATQ, πr2 × 5r = 625π
⇒ r3 = 125
⇒ r = 5
১৬.
Three rectangular fields having area 60 sq m, 84 sq m and 108 sq m are to be divided into identical rectangular flower beds, each having a length 4m. Find the breadth of each flower bed.
ব্যাখ্যা
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.
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.
১৭.
The product of two consecutive positive integers is 20. To find the integers, this can be represented in the form of quadratic equation as-
ব্যাখ্যা
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
According to the given,
x(x + 1) = 20
x2 + x = 20
x2 + x – 20= 0
১৮.
A garden of 50 meter length and 40 meter width has a walkway of 2 meter width on every side. What is the area of the garden, in square meters, excluding the walkway?
ব্যাখ্যা
The area of the garden, excluding the walkway, is
= {(50 - 2×2) × (40 - 2×2)}
= 46×36
= 1656 m2
= {(50 - 2×2) × (40 - 2×2)}
= 46×36
= 1656 m2