Vardaan Learning Institute

Explanation: We know the relationship: HCF × LCM = Product of two numbers.
Given: HCF = 27, LCM = 162, One number = 54.
Let the other number be x.
27 × 162 = 54 × x
x = (27 × 162) / 54
x = 4374 / 54 = 81.
Thus, the other number is 81.

Explanation: The given equations are:
1) x + 2y - 5 = 0
2) -3x - 6y + 15 = 0
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
a₁/a₂ = 1/(-3) = -1/3
b₁/b₂ = 2/(-6) = -1/3
c₁/c₂ = -5/15 = -1/3
Since a₁/a₂ = b₁/b₂ = c₁/c₂, the lines are coincident and have infinitely many solutions.

Explanation: Given AP: 2, 7, 12...
First term (a) = 2
Common difference (d) = 7 - 2 = 5
We need to find the 10th term (a₁₀).
Formula: aₙ = a + (n-1)d
a₁₀ = 2 + (10 - 1)(5) = 2 + 9(5) = 2 + 45 = 47.

Explanation: The perpendicular distance of a point P(x, y) from the x-axis is given by the absolute value of its y-coordinate (|y|).
Here, for P(2, 3), the y-coordinate is 3.
Therefore, the distance is 3 units.

Explanation: The length of the shadow of a vertical object is inversely proportional to the angle of elevation of the sun. As the sun moves from a higher position (noon) towards the horizon (evening), the angle of elevation decreases, causing the shadow length to increase. Conversely, if the shadow is increasing, the angle of elevation is decreasing.

Explanation: The standard formula for the area of a sector with angle θ (in degrees) is:
Area = (θ/360) × πR²
Here, θ = P.
So, Area = (P/360) × πR².
Multiplying numerator and denominator by 2 gives:
Area = (P / (360×2)) × 2πR² = (P/720) × 2πR².
This matches option (d).

Explanation:
1. Modal Class: The class with the highest frequency. Max frequency is 20, so Modal Class is 15-20. Lower limit = 15.
2. Median Class: Total frequency N = 10+15+12+20+9 = 66. N/2 = 33.
Cumulative frequencies: 10, 25 (10+15), 37 (25+12).
The value 33 lies in the group 10-15 (since cf 37 > 33). So Median Class is 10-15. Lower limit = 10.
Sum = 15 + 10 = 25.

Explanation:
(a) √4 = 2 (Rational)
(b) 3.1414... is a terminating decimal (Rational)
(c) √7 is a non-perfect square root, which is non-terminating and non-repeating. Thus, it is irrational.
(d) 22/7 is in the form p/q (Rational).

Explanation: For a system of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to have no solution (parallel lines), the condition is:
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Here: 1/5 = 2/k ≠ -3/7
Taking the first two parts: 1/5 = 2/k ⇒ k = 10.
Also checking the third part: 2/10 = 1/5, which is not equal to -3/7. Condition satisfied.
Therefore, k = 10.

Explanation: Common difference (d) = Second term - First term.
d = [(1-p)/p] - [1/p]
d = (1 - p - 1) / p
d = -p / p = -1.

Explanation: Using distance formula between (x₁, y₁) = (a, b) and (x₂, y₂) = (-a, -b):
D = √[(-a - a)² + (-b - b)²]
D = √[(-2a)² + (-2b)²]
D = √[4a² + 4b²] = √[4(a² + b²)]
D = 2√(a² + b²).

Explanation: Let θ be the angle of elevation.
tan θ = Height of pole / Length of shadow
tan θ = 6 / 2√3 = 3 / √3 = √3
We know that tan 60° = √3.
Therefore, θ = 60°.

Explanation: Given: Numerical value of Perimeter = Numerical value of Area.
2πr = πr²
Dividing both sides by πr (since r ≠ 0):
2 = r
Thus, Radius = 2 units.

Explanation: The empirical relationship between the three measures of central tendency is given by the formula:
3 Median = Mode + 2 Mean
This matches option (a).

Explanation: Performing prime factorization of 140:
140 = 2 × 70
70 = 2 × 35
35 = 5 × 7
So, 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7.

Explanation: Dependent linear equations represent coincident lines, meaning the second equation must be a scalar multiple of the first.
Given equation: -5x + 7y = 2 (or -5x + 7y - 2 = 0).
Let's check option (d): 10x - 14y = -4.
Dividing the entire equation by -2:
(10/-2)x - (14/-2)y = (-4/-2)
-5x + 7y = 2.
This is exactly the given equation. Thus, they are dependent.

Explanation: Let points be A(-1, 7) and B(4, -3). Ratio m₁:m₂ = 2:3.
Using Section Formula:
x = (m₁x₂ + m₂x₁) / (m₁ + m₂)
x = (2(4) + 3(-1)) / (2+3) = (8 - 3) / 5 = 5/5 = 1.
y = (m₁y₂ + m₂y₁) / (m₁ + m₂)
y = (2(-3) + 3(7)) / (2+3) = (-6 + 21) / 5 = 15/5 = 3.
Coordinates: (1, 3).

Explanation:
Angle swept in 60 mins = 360°.
Angle swept in 5 mins (θ) = (360/60) × 5 = 30°.
Area = (θ/360) × πr²
Area = (30/360) × (22/7) × 14 × 14
Area = (1/12) × 22 × 2 × 14
Area = 616 / 12 = 51.33 cm².

Explanation: Mean = Sum of observations / Number of observations.
Given: Mean = 9, n = 6.
Sum = 9 × 6 = 54.
Sum = 6 + 7 + x + 8 + y + 14 = 35 + x + y.
35 + x + y = 54
x + y = 54 - 35 = 19.

Explanation: Given AP: 21, 18, 15...
a = 21, d = 18 - 21 = -3, aₙ = -81.
-81 = 21 + (n-1)(-3)
-81 - 21 = -3(n-1)
-102 = -3(n-1)
n-1 = 34 ⇒ n = 35.

Solution:
We use the property: HCF(a, b) × LCM(a, b) = a × b
Given: HCF(306, 657) = 9
9 × LCM = 306 × 657
LCM = (306 × 657) / 9
LCM = 34 × 657
LCM = 22,338.

Solution:
1) 2x + 3y = 11
2) 2x - 4y = -24
Subtracting (2) from (1):
(2x + 3y) - (2x - 4y) = 11 - (-24)
7y = 35 ⇒ y = 5.
Substituting y = 5 in equation (1):
2x + 3(5) = 11 ⇒ 2x = 11 - 15 = -4 ⇒ x = -2.
Now, find 'm' in y = mx + 3:
5 = m(-2) + 3
2 = -2m ⇒ m = -1.

Solution:
Let the point on the x-axis be P(x, 0). Let A = (2, -5) and B = (-2, 9).
Since P is equidistant from A and B, PA = PB, so PA² = PB².
(x - 2)² + (0 - (-5))² = (x - (-2))² + (0 - 9)²
(x - 2)² + 25 = (x + 2)² + 81
x² - 4x + 4 + 25 = x² + 4x + 4 + 81
-4x + 29 = 4x + 85
-8x = 85 - 29 = 56
x = 56 / -8 = -7.
Therefore, the point is (-7, 0).

Question: A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.
Solution:
Let AB be the vertical pole and AC be the rope length (hypotenuse).
Given: AC = 20 m, ∠ACB = 30°.
In right-angled ΔABC, we use the sine ratio:
sin 30° = Opposite / Hypotenuse = AB / AC
1/2 = AB / 20
AB = 20 / 2 = 10 m.
The height of the pole is 10 m.

Solution:
Circumference = 2πr = 22 cm.
2 × (22/7) × r = 22
r = (22 × 7) / 44 = 3.5 cm.
Area of Quadrant = (1/4)πr²
= (1/4) × (22/7) × 3.5 × 3.5
= 0.5 × 11 × 0.5 × 3.5 / 2 (Simplifying)
= 77/8 cm² or 9.625 cm².

Solution:
Let us assume, to the contrary, that 3 + 2√5 is rational.
That is, we can find coprime integers a and b (b ≠ 0) such that 3 + 2√5 = a/b.
Rearranging the equation:
2√5 = a/b - 3
2√5 = (a - 3b) / b
√5 = (a - 3b) / 2b
Since a and b are integers, (a - 3b) / 2b is rational. Thus, √5 should be rational.
But this contradicts the fact that √5 is irrational.
So, our assumption is incorrect.
Therefore, 3 + 2√5 is irrational.

Solution:
Let the fixed charge be ₹ x and the charge per km be ₹ y.
According to the problem:
1) x + 10y = 105
2) x + 15y = 155
Subtracting equation (1) from (2):
(x + 15y) - (x + 10y) = 155 - 105
5y = 50 ⇒ y = 10.
Substituting y = 10 in equation (1):
x + 10(10) = 105
x + 100 = 105 ⇒ x = 5.
Answer: Fixed charges = ₹ 5 and Charge per km = ₹ 10.

Solution:
The two-digit numbers divisible by 3 are: 12, 15, 18, ..., 99.
This forms an AP where a = 12, d = 3, and aₙ = 99.
Using aₙ = a + (n-1)d:
99 = 12 + (n-1)3
87 = 3(n-1)
n-1 = 87/3 = 29
n = 30.
There are 30 two-digit numbers divisible by 3.

Solution:
Let the x-axis divide the line segment joining A(1, -5) and B(-4, 5) in the ratio k:1.
Since the point lies on the x-axis, its y-coordinate is 0.
Using Section Formula for y:
y = (m₁y₂ + m₂y₁) / (m₁ + m₂)
0 = (k(5) + 1(-5)) / (k + 1)
0 = 5k - 5
5k = 5 ⇒ k = 1.
So, the ratio is 1:1 (midpoint).
Now finding the x-coordinate:
x = (1(-4) + 1(1)) / (1+1) = -3 / 2 = -1.5.
Coordinates of the point of division: (-1.5, 0).

Solution:
We construct a table to find the Mean (x̄ = Σfx / Σf).
Classes: 45-55, 55-65, 65-75, 75-85, 85-95
Class Marks (x): 50, 60, 70, 80, 90
Frequency (f): 3, 10, 11, 8, 3
Product (fx):
3 × 50 = 150
10 × 60 = 600
11 × 70 = 770
8 × 80 = 640
3 × 90 = 270
Total Frequency Σf = 35
Total Sum Σfx = 150 + 600 + 770 + 640 + 270 = 2430
Mean = 2430 / 35 = 69.428...
Mean Literacy Rate = 69.43 %.

Solution:
The first 15 multiples of 8 are: 8, 16, 24, ...
This is an AP with a = 8, d = 8, n = 15.
Sum formula: Sₙ = (n/2)[2a + (n-1)d]
S₁₅ = (15/2)[2(8) + (15-1)8]
S₁₅ = (15/2)[16 + 14(8)]
S₁₅ = (15/2)[16 + 112]
S₁₅ = (15/2)[128] = 15 × 64 = 960.

Solution:
Let speed of boat in still water = u km/h and speed of stream = v km/h.
Upstream speed = (u - v); Downstream speed = (u + v).
Case 1: 30/(u-v) + 44/(u+v) = 10
Case 2: 40/(u-v) + 55/(u+v) = 13
Let 1/(u-v) = X and 1/(u+v) = Y.
30X + 44Y = 10 ---(i)
40X + 55Y = 13 ---(ii)
Multiplying (i) by 4 and (ii) by 3:
120X + 176Y = 40
120X + 165Y = 39
Subtracting: 11Y = 1 ⇒ Y = 1/11.
Substitute Y in (i): 30X + 4 = 10 ⇒ 30X = 6 ⇒ X = 1/5.
So, u - v = 5 and u + v = 11.
Adding these two: 2u = 16 ⇒ u = 8 km/h.
Subtracting: 2v = 6 ⇒ v = 3 km/h.

Solution:
The lengths of semicircles are πr₁, πr₂, πr₃...
Radii are 0.5, 1.0, 1.5, ... (AP with a=0.5, d=0.5).
Total length L = π(0.5) + π(1.0) + ... + π(6.5) (for 13 terms).
L = π [0.5 + 1.0 + ... + 13 terms]
Sum of terms inside bracket (AP):
S = (n/2)[2a + (n-1)d] = (13/2)[2(0.5) + 12(0.5)]
S = 6.5 [1 + 6] = 6.5 × 7 = 45.5
Total Length = π × 45.5 = (22/7) × 45.5
Length = 22 × 6.5 = 143 cm.

Solution:
Let AB be the building (7m) and CD be the tower.
From the top of building A, angle of depression of foot D is 45°.
Let horizontal distance BD = x. In ΔABD, tan 45° = AB/BD ⇒ 1 = 7/x ⇒ x = 7m.
From A, angle of elevation of top C is 60°.
Let h be the height of the tower above the building level. Horizontal distance is same x = 7m.
tan 60° = h / x ⇒ √3 = h / 7 ⇒ h = 7√3 m.
Total height of tower = Building height + h
H = 7 + 7√3 = 7(1 + √3) m.

Solution:
Total Frequency n = 60.
Sum of frequencies = 5 + x + 20 + 15 + y + 5 = 45 + x + y.
45 + x + y = 60 ⇒ x + y = 15. --- (1)
Median = 28.5. This lies in class interval 20-30.
Lower limit (l) = 20, frequency (f) = 20, cumulative frequency (cf) of previous class = 5 + x, class size (h) = 10.
Median = l + [(n/2 - cf) / f] × h
28.5 = 20 + [(30 - (5+x)) / 20] × 10
8.5 = (25 - x) / 2
17 = 25 - x ⇒ x = 8.
From (1): 8 + y = 15 ⇒ y = 7.

Solution:
(i) Niharika runs on the 2nd line, so x = 2. She covers 1/4th of AD (100m).
y = 1/4 × 100 = 25.
Coordinates: (2, 25).

(ii) Preet runs on the 8th line, so x = 8. She covers 1/5th of AD (100m).
y = 1/5 × 100 = 20.
Coordinates: (8, 20).

(iii) Distance between flags (2, 25) and (8, 20):
D = √[(8-2)² + (20-25)²]
D = √[6² + (-5)²] = √[36 + 25] = √61 m.

Solution:
(i) Area of circle = πr² = (22/7) × 28 × 28 = 22 × 4 × 28 = 2464 cm².

(ii) Area of one sector (6 designs ⇒ 60° angle):
Area = (60/360) × 2464 = 2464 / 6 = 410.67 cm².

(iii) Area of one design segment = Area of Sector - Area of Triangle.
Area of Equilateral Δ (side 28) = (√3/4) × 28² = 1.7/4 × 784 = 1.7 × 196 = 333.2 cm².
Area of 1 Design = 410.67 - 333.2 = 77.47 cm².
Total Area of 6 Designs = 6 × 77.47 = 464.82 cm².
Cost = 464.82 × 0.35 = ₹ 162.69 (approx).

Solution:
(i) Distance to pick 1st potato: 5m to go + 5m to return = 10 m.

(ii) Distance to pick 2nd potato: (5+3)m to go + (5+3)m to return = 8 + 8 = 16 m.

(iii) The distances form an AP: 10, 16, 22...
a = 10, d = 6, n = 10 potatoes.
Total Distance S₁₀ = (10/2)[2(10) + (10-1)6]
S₁₀ = 5[20 + 54]
S₁₀ = 5 × 74 = 370 m.