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ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

পরীক্ষাব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্সতারিখতারিখ অনির্ধারিতসময়45 minutes
মোট প্রশ্ন২৪
সিলেবাস
Math - 10 - Equation Solving, Geometry, Trigonometry (Area, Volume, Heights and Distances)
প্রশ্ন দেখুনঅনুশীলন
ঘনত্ব
উত্তর
উত্তরিতবর্তমানপুনরায় দেখুনঅসম্পূর্ণ

ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স · তারিখ অনির্ধারিত · ২৪ প্রশ্ন

১.
A line intersecting a circle in two points is called a _______.
  1. ক
    ক) Secant
  2. খ
    খ) Chord
  3. গ
    গ) Diameter
  4. ঘ
    ঘ) Tangent
  5. ঙ
    ঙ) None of these
ব্যাখ্যা
A line intersecting a circle in two points is called a Secant.
২.
The angle of elevation of the top of a building from a point on the ground, which is 30 m away from the foot of the building, is 30°. The height of the building is:
  1. ক
    ক) 10 m
  2. খ
    খ) 30/√3 m
  3. গ
    গ) √3/10 m
  4. ঘ
    ঘ) 30 m
  5. ঙ
    ঙ) 10√5 m
ব্যাখ্যা

Say x is the height of the building.
a is a point 30 m away from the foot of the building.
Here, height is the perpendicular and distance between point a and foot of building is the base.
The angle of elevation formed is 30°.

Hence,
tan 30° = perpendicular/base = x/30
1/√3 = x/30
x = 30/√3

৩.
The pairs of equations 9x + 3y + 12 = 0 and 18x + 6y + 26 = 0 have-
  1. ক
    ক) Unique solution
  2. খ
    খ) Exactly two solutions
  3. গ
    গ) Infinitely many solutions
  4. ঘ
    ঘ) No solution
  5. ঙ
    ঙ) Exactly three solutions
ব্যাখ্যা

Given,
9x + 3y + 12 = 0 and 18x + 6y + 26 = 0
a1/a2 = 9/18 = 1/2
b1/b2 = 3/6 = 1/2
c1/c2 = 12/26 = 6/13
Since, a1/a2 = b1/b2 ≠ c1/c2

So, the pairs of equations are parallel and the lines never intersect each other at any point, therefore there is no possible solution.

৪.
A cylindrical pencil sharpened at one edge is the combination of-
  1. ক
    ক) a cone and a cylinder
  2. খ
    খ) frustum of a cone and a cylinder
  3. গ
    গ) a hemisphere and a cylinder
  4. ঘ
    ঘ) two cylinders
  5. ঙ
    ঙ) None of these
ব্যাখ্যা
A cylindrical pencil sharpened at one edge is the combination of a cone and a cylinder.

৫.
If θ is an acute angle, and cosθ =15/17, then find the value of cot(90 - θ)?
  1. ক
    ক) 3/4
  2. খ
    খ) 7/19
  3. গ
    গ) 8/15
  4. ঘ
    ঘ) 11/17
  5. ঙ
    ঙ) 2/5
ব্যাখ্যা

Given that,
cos θ = 15/17
⇒ secθ = 17/15
⇒ sec2θ = 289/225
⇒ 1 + tan2θ = 289/225
⇒ tan2θ = 289/225 - 1
⇒ tan2θ = 64/225
⇒ tanθ = 8/15
⇒ cot(90 - θ) = 8/15

৬.
If 1/2 is a root of the quadratic equation x2 - mx - 5/4 = 0, then value of m is:
  1. ক
    ক) 2
  2. খ
    খ) -2
  3. গ
    গ) -3
  4. ঘ
    ঘ) 3
  5. ঙ
    ঙ) 1
ব্যাখ্যা

Given,
x = 1/2 as the root of equation x2 - mx - 5/4 = 0.
(1/2)2 – m(1/2) – 5/4 = 0
1/4 - m/2 - 5/4 = 0
m = -2

৭.
The height of an equilateral triangle with a side 2 cm is -
  1. ক
    ক) √3 cm
  2. খ
    খ) √2 cm
  3. গ
    গ) √5 cm
  4. ঘ
    ঘ) √7 cm
  5. ঙ
    ঙ) 4√5 cm
ব্যাখ্যা

আমরা জানি, সমবাহু ত্রিভুজের শীর্ষবিন্দু হতে ভূমির উপর অঙ্কিত লম্ব ভূমিকে সমদ্বিখণ্ডিত করে
তাই, BD = DC = 1 cm
পীথাগোরাসের সূত্র অনুসারে ΔABD হতে পাই,
⇒ AB2 = BD2 + AD2
⇒ AD2 = AB2 - BD2 = 22 - 12
∴ AD = √3

৮.
  1. ক
    ক) 5
  2. খ
    খ) 3
  3. গ
    গ) 2
  4. ঘ
    ঘ) 1
  5. ঙ
    ঙ) 7
ব্যাখ্যা


Hence,
we can write, √(6 + x) = x
⇒ 6 + x = x2
⇒ x2 - x - 6 = 0
⇒ x2 - 3x + 2x - 6 = 0
⇒ x(x - 3) + 2(x - 3) = 0
⇒ (x + 2) (x - 3) = 0
⇒ x = -2, 3

Since, we cannot be negative, therefore, x = 3.
৯.
The area of a triangle with sides 3 cm, 5 cm and 6 cm is -
  1. ক
    ক) 3√15 cm2
  2. খ
    খ) 2√14 cm2
  3. গ
    গ) 5√7 cm2
  4. ঘ
    ঘ) 7√11 cm2
  5. ঙ
    ঙ) 4√3 cm2
ব্যাখ্যা

অর্ধপরিসীমা, s = (3 + 5 + 6)/2
= 7 সে.মি
∴ ক্ষেত্রফল = √{s(s - a)(s - b)(s - c)} বর্গএকক
= √{7 (7 - 3) (7 - 5) (7 - 6)} বর্গসে.মি
= √(7 × 4 × 2 × 1)
= 2√14 বর্গসে.মি

১০.
A quadratic equation ax2 + bx + c = 0 has no real roots, if-
  1. ক
    ক) b2 – 4ac > 0
  2. খ
    খ) b2 – 4ac = 0
  3. গ
    গ) b2 – 4ac < 0
  4. ঘ
    ঘ) b2 – ac > 0
  5. ঙ
    ঙ) b2 – 2ac > 0
ব্যাখ্যা
A quadratic equation ax2 + bx + c = 0 has no real roots, if b2 – 4ac < 0.
That means, the quadratic equation contains imaginary roots.
১১.
The area of a circle inscribed in an equilateral triangle is 154 cm2. Find the perimeter of the triangle:
  1. ক
    ক) 81.9 cm
  2. খ
    খ) 25.8 cm
  3. গ
    গ) 33.5 cm
  4. ঘ
    ঘ) 72.7 cm
  5. ঙ
    ঙ) 85.3 cm
ব্যাখ্যা

Radius of incircle = a/2√3.
Area of incircle = (π × a2)/12 cm2
∴ πa2/12 = 154
⇒ a2 = (154 × 12 × 7)/22
⇒ a = 14√3
∴ Perimeter of the triangle = (3 × 14√3) cm
= (14 × 1.732) cm
= 72.7 cm (approx.)

১২.
A square and an equilateral triangle have equal perimeter. if the diagonal of the square is 12√2 cm then area of triangle is -
  1. ক
    ক) 56√3 cm2
  2. খ
    খ) 64√3 cm2
  3. গ
    গ) 47√5 cm2
  4. ঘ
    ঘ) 43√5 cm2
  5. ঙ
    ঙ) 37√7 cm2
ব্যাখ্যা

Let the side of the square be a cm.
Then, its diagonal = √2 a cm.
Now, √2 a = 12√2
⇒ a = 12 cm.
Perimeter of square = 4a = 48 cm.
Perimeter of equilateral triangle = 48 cm.
Each side of the triangle = 16 cm.
Area of the triangle = ((√3/4)×16×16) cm2
= 64√3 cm2

১৩.
The graph of linear equation x + 2y = 2, cuts the y-axis at:
  1. ক
    ক) (2, 0)
  2. খ
    খ) (0, 2)
  3. গ
    গ) (0, 1)
  4. ঘ
    ঘ) (1, 1)
  5. ঙ
    ঙ) (1, 2)
ব্যাখ্যা

x + 2y = 2
y = (2 - x)/2

If x = 0, then,
y = (2 - 0)/2
= 2/2
= 1

Hence, x + 2y = 2 cuts the y-axis at (0, 1).

১৪.
A ladder makes an angle of 60° with the ground, when placed along a wall. If the foot of ladder is 8 m away from the wall, the length of ladder is
  1. ক
    ক) 4 m
  2. খ
    খ) 8 m
  3. গ
    গ) 8√3 m
  4. ঘ
    ঘ) 16 m
  5. ঙ
    ঙ) 3√2 m
ব্যাখ্যা
Let AB be the wall, AC be the length of the ladder.



In right angled triangle ABC,
cos 60° = BC/AC
1/2 = 8/AC
AC = 8 × 2 = 16

Therefore, the length of the ladder is 16 m.

১৫.
The angle of depression of an object on the ground, from the top of a 25 m high tower is 30°. The distance of the object from the base of tower is
  1. ক
    ক) 25√3 m
  2. খ
    খ) 50√3 m
  3. গ
    গ) 75√3 m
  4. ঘ
    ঘ) 50 m
  5. ঙ
    ঙ) 53√5 m
ব্যাখ্যা
Let AB be the tower and BC be the distance of the object (at C) from the base of the tower.



In right triangle ABC,
tan 30° = AB/BC
1/√3 = 25/BC
BC = 25√3 m
১৬.
If the radius of a cylinder is 4cm and height is 10cm, then the total surface area of a cylinder is:
  1. ক
    ক) 440 sq.cm
  2. খ
    খ) 352 sq.cm.
  3. গ
    গ) 400 sq.cm
  4. ঘ
    ঘ) 412 sq.cm
  5. ঙ
    ঙ) 395 s.q. cm
ব্যাখ্যা

Total Surface Area of a Cylinder = 2πr(r + h)
TSA = 2 × 22/7 × 4(4 + 10)
= (2 × 22 × 4 × 14)/7
= (2 × 22 × 4 × 2)
= 352 sq.cm

১৭.
The total surface area of a cube is 96 cm2. The volume of the cube is:
  1. ক
    ক) 8 cm3
  2. খ
    খ) 512 cm3
  3. গ
    গ) 64 cm3
  4. ঘ
    ঘ) 27 cm3
  5. ঙ
    ঙ) 32 cm3
ব্যাখ্যা

We know that the Total Surface Area of the cone = 6a2.
6a2 = 96 cm2
a2 = 96/6 = 16
a = 4 cm

The volume of cone = a3 cubic units
V = 43 = 64cm3.

১৮.
In a cylinder, the radius is doubled and height is halved, the curved surface area will be-
  1. ক
    ক) Halved
  2. খ
    খ) Doubled
  3. গ
    গ) Same
  4. ঘ
    ঘ) Four times
  5. ঙ
    ঙ) None of these
ব্যাখ্যা

We know that the curved surface area of a cylinder is 2πrh
Given that, r = 2R, h= H/2
Hence, the CSA of new cylinder = 2π(2R)(H/2) = 2πRH
Therefore, the answer is “Same”.

১৯.
The surface area of cuboid-shaped box having length = 80 cm, breadth = 40cm and height = 20cm is:
  1. ক
    ক) 11200 sq.cm
  2. খ
    খ) 13000 sq.cm
  3. গ
    গ) 13400 sq.cm
  4. ঘ
    ঘ) 12000 sq.cm
  5. ঙ
    ঙ) 13467 sq.cm
ব্যাখ্যা

Surface area of the box = 2(lb + bh + hl)
S.A. = 2[(80 × 40) + (40 × 20) + (20 × 80)]
= 2(3200 + 800 + 1600)
= 2 × 5600
= 11200 sq.cm.

২০.
To construct a rectangle, we need to know:
  1. ক
    ক) All the interior angles
  2. খ
    খ) All the Sides
  3. গ
    গ) Only Length and breadth
  4. ঘ
    ঘ) Only one angle measure
  5. ঙ
    ঙ) None of these
ব্যাখ্যা
A rectangle has its parallel sides equal and all the interior angles measure 90 degrees. Hence, if the length and breadth of the rectangle is known, then we can construct it easily.
২১.
The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:
  1. ক
    ক) 75°
  2. খ
    খ) 45°
  3. গ
    গ) 35°
  4. ঘ
    ঘ) 60°
  5. ঙ
    ঙ) 30°
ব্যাখ্যা


Consider the diagram is shown above where QR represents the tree and PQ represents its shadow.
We have, QR = PQ
Let, ∠QPR = θ
tanθ = QR/PQ
= QR/QR [since QR = PQ]
= 1
= tan 45°
⇒ θ = 45°
i.e., the required angle of elevation = 45°

২২.
The sum of all the interior angles of a hexagon is:
  1. ক
    ক) 540°
  2. খ
    খ) 360°
  3. গ
    গ) 180°
  4. ঘ
    ঘ) 270°
  5. ঙ
    ঙ) 720°
ব্যাখ্যা

Number of sides in hexagon, n = 6
Sum of interior angles = (n - 2) x 180°
= (6 – 2) x 180°
= 720°

২৩.
The area of the base of a cylinder is 100π m2. The volume of the cylinder is 900π m3. What is the height of the cylinder?
  1. ক
    ক) 5 m
  2. খ
    খ) 9 m
  3. গ
    গ) 7 m
  4. ঘ
    ঘ) 4 m
  5. ঙ
    ঙ) 6 m
ব্যাখ্যা

Area of the base of a cylinder, πr2 = 100π
The volume of the cylinder, πr2h = 900π
∴ h = πr2h/πr2
= 900/100
= 9 m

২৪.
An observer 2m tall is 10√3 m away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is:
  1. ক
    ক) 10 m
  2. খ
    খ) 7 m
  3. গ
    গ) 15 m
  4. ঘ
    ঘ) 25 m
  5. ঙ
    ঙ) 12 m
ব্যাখ্যা



SR = PQ = 2 m
PS = QR = 10√3
tan 30° = TS/PS
1/√3 = TS/10√3
TS = 10√3/√3
= 10
TR = TS + SR
= 10 + 2
= 12 m.