১.
If (2a + b)/(a + 4b) = 3 then find the value of (a + b)/(a + 2b)?
ব্যাখ্যা
Given,
(2a + b)/(a + 4b) = 3
2a + b = 3a + 12b
-a = 11b
a = -11b
∴ (a + b)/(a + 2b)
= (-11b + b)/(-11b + 2b)
= -10b/-9b
= 10/9.
ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স
মোট প্রশ্ন২৫
সিলেবাস
Math - 10 - Equation Solving, Geometry, Trigonometry (Area, Volume, Heights and Distances)
ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স
ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স · তারিখ অনির্ধারিত · ২৫ প্রশ্ন
২.
If x = 1 + √2 + √3 and y = 1 + √2 - √3 then the value of x2 + 4xy + y2/(x + y)?
ব্যাখ্যা
x = 1 + √2 + √3 ...........(i)
y = 1 + √2 - √3 .............(ii)
x2 + 4xy + y2/(x + y)
= {(x + y)2 + 2xy}/(x + y)
From (i) + (ii)
x + y = 2 + 2√2
xy = (1 + √2)2 - (√3)2
= 3 + 2√2 - 3
= 2√2
{(x + y)2 + 2xy}/(x + y)
= {(2 + 2√2)2 + (2 × 2√2)}/(2 + 2√2)
= (12 + 12√2)/(2 + 2√2)
= 12(1 + √2)/2((1 + √2)
= 12/2
= 6.
৩.
If a2 + b2 = 5ab then find the value of (a2/b2 + b2/a2)
ব্যাখ্যা
a2 + b2 = 5ab
a2/ab + b2/ab = 5
a/b + b/a = 5
Squaring the both sides
(a/b)2 + (b/a)2 = (5)2
a2/b2 + b2/a2 + 2 × (a/b) × (b/a)= 25
a2/b2 + b2/a2 + 2 = 25
a2/b2 + b2/a2 = 25 -2
a2/b2 + b2/a2 = 23.
৪.
If x = a(b - c), y = b(c - a), z = c(a - b), then the value of (x/a)3 + (y/b)3 + (z/c)3 is -
ব্যাখ্যা
x = a(b - c),
y = b(c - a),
z = c(a - b)
Let,
x/a = b - c = A
y/b = c - a = B
z/c = a - b = C
∴ A + B + C = b - c + c - a + a - b
= 0
A3 + B3 + C3 = 3ABC
= (x/a)3 + (y/b)3 + (z/c)3
= 3 × (x/a) × (y/b) × (z/c)
= 3xyz/abc
৫.
If x : y = 5 : 3, then (8x - 5y) : (8x + 5y) = ?
ব্যাখ্যা
let x = 5 and y = 3
Then,
(8x - 5y) : (8x + 5y)
= {(8 × 5) - (5 ×3)} : {(8 × 5} + (5 × 3)}
= (40 - 15) : (40 + 15)
= 25 : 55
= 5 : 11.
৬.
Find the value of a and b if (x - 1) and (x + 1) are factors of x4 + ax3 - 3x2 + 2x + b = ?
ব্যাখ্যা
If (x - 1) and (x + 1) are the factors y equation then,
x - 1 = 0
x = 1
Put x = 1 we get,
1 + a - 3 + 2 + b = 0
a + b = 0 ........(i)
x + 1 = 0
x = -1
Put x = -1 we get,
1 - a - 3 - 2 + b = 0
b - a = 4 ........(ii)
(i) - (ii) we get,
a + b - b + a = 0 - 4
2a = -4
a = -2
∴ b = 2
a, b = -2, 2.
৭.
If x = 1 - q and y = 2q + 1, then for what value of q, x is equal to y?
ব্যাখ্যা
According to math,
If,
x = y
Then, 1 - q = 2q + 1
⇒ 2q + q = 1 - 1
⇒ 3q = 0
⇒ q = 0.
৮.
If x/2 = y/3 = z/4 = (2x - 3y - + 5z)/k, then the value of k is -
ব্যাখ্যা
Let,
x/2 = y/3 = z/4 = m
Then, x = 2m, y = 3m, z = 4m
∴ x/2 = (2x - 3y + 5z)/k = 2m/2
⇒ (2 × 2m - 3 × 3m + 5 × 4m)/k = m
⇒ k = 4 - 9 + 20
= 15.
৯.
Shefali has a rectangular piece of cloth with dimensions 20 m and 15 m. She wants to paint a border of breadth 4 m inside the four sides of the rectangle. The paint would cost her Tk. 6 per sq m. Find the cost of painting the complete border.
ব্যাখ্যা
Paint Area = Total area - Non-Paint area
Subtracting width of the border from all sides we get,
Length = 20 - 4 - 4 = 12
Breadth = 15 - 4 - 4 = 7
∴ Paint Area = (20 x 15) - (12 x 7) = 300 - 84 = 216
Total Cost of painting the border = Rate x Area
= 6 x 216
= Tk. 1296
১০.
If the diagonal of a rectangle is 17 cm long its perimeter is 46 cm, find the area of the rectangle.
ব্যাখ্যা
Let length = x and breadth = y. Then,
2(x + y)= 46 or
x + y = 23 and
x2 + y2 = (17)2
= 289.
Now, (x + y)2 = (23)2
⇒ (x2 + y2) + 2xy = 529
⇒ 289 + 2xy = 529
⇒ xy = 120.
∴ Area = xy = 120 cm2
১১.
If the diagonal and the area of a rectangle are 25 m2 and 168 m2, what is the length of the rectangle?
ব্যাখ্যা
Let the length of the rectangle be x metre.
Then, a breath of the rectangle = (168/x)
∴ √{(x)2 + (168/x)2} = 25
⇒ √{(x2 + (28224/x2)} = 25
⇒ {(x2 + (28224/x2)} = 625
⇒ x4 - 625x2 + 28224 = 0
⇒ x2(x2 - 576) - 49(x2 - 576) = 0
⇒ (x2 - 576)(x2 - 49) = 0
⇒ x2 = 576 or x2 = 49
⇒ x = 24 or x = 7
Hence, length = 24 cm. and breadth = 7 m.
১২.
A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and the rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?
ব্যাখ্যা
Area of the park = (60 x 40) m2 = 2400 m2
Area of the lawn = 2109 m2
∴ Area of the crossroads = (2400 - 2109) m2 = 291 m2
Let the width of the road be x metres. Then,
60x + 40x - x2 = 291
⇒ x2 - 100x + 291 = 0
⇒ (x - 97)(x - 3) = 0
⇒ x = 3 m
১৩.
A rectangular grassy plot 110 m by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paisa per sq. metre.
ব্যাখ্যা
Area of the plot = (110 × 65)m2
= 7150m2
Area of the plot excluding the path = [(110 - 5) × (65 - 5)]m2
= 6300 m2
∴ Area of the path = (7150 - 6300) m2
= 850 m2
Cost of graveling the path = Tk. {850 × (80/100)}
= Tk. 680
১৪.
The perimeter of a rectangle and a square are 160 m each. The area of the rectangle is less than that of the square by 100 sq m. The length of the rectangle is -
ব্যাখ্যা
The perimeter of the square = 160 m.
Side of square = (160/4) m
= 40 m.
Area of square = (40 × 40) m2 = 1600 m2
Area of rectangle = (1600 - 100) m2 = 1500 m2
Let the length and breadth of the rectangle be 'l' and 'b' respectively.
Then, 2(l + b) = 160
⇒ (l + b) = 80
⇒ b = 80 - l.
∴ lb = 1500
⇒ l(80 - l) = 1500
⇒ 80l - l2 = 1500
⇒ l2 - 80 l + 1500 = 0
⇒ (l - 50)(l - 30) = 0
⇒ l = 50. or l = 30
Hence, length = 50 m, breadth = 30 m.
১৫.
A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?
ব্যাখ্যা
Given that,
The area of the field = 680 sq. feet
⇒ lb = 680 sq. feet
Length(l) = 20 feet
⇒ 20 × b = 680
⇒ b = 680/20
= 34 feet
∴ Required length of the fencing = l + 2b
= 20 + (2 × 34)
= 88 feet
১৬.
The breadth of a room is twice its height and half its length. The volume of the room is 512 cu. M. The length of the room is -
ব্যাখ্যা
Let the breadth be x metre,
Then, length = 4x metre
∴ Volume of the room = (4x × 2x × x) m3
= (8x3) m3
8x3 = 512
x3 = 64
x = 4.
The length of the room is (4 × 4) = 16 m.
১৭.
অনির্ধারিতThe dimensions of a wooden plank are in the ratio 6 : 5 : 3. Its surface area is 504 sq. m. Find the breadth of the plank -
ব্যাখ্যা
Let the common factor be K
∴ Length = 6K;
Breadth = 5K and
Height = 3K
We know,
Total Surface area of cuboid = 2(LB + BH + LH)
L = Length; B = Breadth/width; H = Height
Whole surface area of the rectangular plank = 2{(6K x 5K) + (5K x 3K) + (6K x 3K)}
∴ 504 = 126K2
∴ K = 2
∴ Breadth = 5K = 10 m
১৮.
By what percent the volume of a cube increases if the length of each edge was increased by 50%?
ব্যাখ্যা
Let original edge = a,
Then, original volume = a3
New edge = (150/100)a
= 3a/2
New volume = (3a/2)3
= 27a3/8
Increase in volume = (27a3)/8 - (a3)
= 19a3/8
∴ Increase% = {(19a3/8) × (1/a3) × 100}%
= 237.5%
১৯.
A rectangular block 6 cm by 12 cm by 15 cm is cut up into an exact number of equal cubes. Find the least possible number of cubes.
ব্যাখ্যা
Volume of the block = (6 × 12 × 15) cm3
= 1080 cm3
Side of the largest cube = H.C.F of 6 cm, 12 cm, 15 cm
= 3 cm.
Volume of this cube = (3 × 3 × 3) cm3
= 27 cm3
Number of cubes = 1080/27
= 40.
২০.
The dimensions of an open box are 50 cm, 40 cm and 23 cm. Its thickness is 3 cm. If a cubic cm of metal used in the box weighs 0.5 gms, Find the weight of the box.
ব্যাখ্যা
External dimensions,
l = 50 cm,
b = 40 cm,
h = 23 cm
Internal dimension,
l' = 50 - (2 × 3) = 44 cm
b' = 40 - (2 × 3) = 34 cm
h' = 23 - 3 = 20 cm
The volume of the metal used in the box = External Volume - Internal Volume
= [( 50 × 40 × 23) - (44 × 34 × 20)] cm3
= 16080 cm3.
∴ Weight of the metal =
(16080 × 0.5)/1000 kg
= 8.04 kg
২১.
A solid body is made up of a cylinder of radius r and height r, a cone of base radius r and height r fixed to the cylinder's one base and a hemisphere of radius r to its other base. The total volume of the body (given r = 2) is :
ব্যাখ্যা
Total volume of the body :
= Volume of the cylinder + Volume of the cone + Volume of the hemisphere
= πr2.r + (1/3)πr2.r + (2/3)πr3
= 2πr3
= 2π × (2)3
= 16π.
২২.
A ladder 10 m long just reaches the top of a wall and makes an angle of 60° with the wall. Find the distance of the foot of the ladder from the wall.
ব্যাখ্যা
Let BA be the ladder and AC be the wall as shown above.
Then the distance of the foot of the ladder from the wall = BC
Given that BA = 10 m , BAC = 60°
sin 60° = BC/BA
√3/2 = BC/10
BC = 10√3/2
= 5√3.
= 5 × 1.71
= 8.65 m.
২৩.
The angle of depression of a point situated at a distance of 70m from the base of a tower is 60°. The height of the tower is -
ব্যাখ্যা
Length of the tower AB = h meter.
∠DAC = ∠ACB = 60°
BC = 70 metre
In ABC,
tan 60° = AB/BC
⇒ √3 = h/70
⇒ h = 70√3 meter.
২৪.
On the same side of the tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m, the distance between the objects is -
ব্যাখ্যা
Let AB be the tower and C and D be the objects.
Then, AB = 150 m,
∠ACB = 45° and
∠ADB = 60°
AB/AD = tan 60° = √3
AD = AB/√3
= 150/√3 m.
AB/AC = tan 45° = 1
AC = AB = 150 m.
∴CD = (AC - AD)
= {150 - (150/√3)} m
= [{150(√3 - 1)/√3} × {(√3)/(√3)}] m
= 50(3 - √3) m
= (50 × 1.27) m
= 63.5 m.
২৫.
A telegraph post gets broken at a point against a storm and its top touches the ground at a distance 20 m from the base of the post making an angle 30° with the ground. What is the height of the post?
ব্যাখ্যা
Given, BC = 20 m
∠ACB = 30°
The total height of the telegraph post is (AB + CA) = ?
In ABC, tan 30° = AB/BC
1/√3 = AB 20
∴ AB = 20/√3m
Now, cos 30° = BC/AC
√3/2 = 20/AC
∴ AC = 40/√3 m
So, AB + CA
= (20/√3) + (40/√3)
= (60/√3)
= 20√3 m