Real Numbers-HCF Calculator

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Real Numbers - HCF Calculator

Find the Highest Common Factor (HCF) of two numbers using Euclid's algorithm.

First Number:

Second Number:

Formula Sheet

Interactive formula bank for mathematics and science

Algebra Formulas

Quadratic Formula

x = [-b ± √(b² - 4ac)] / 2a

Solution for equations of the form ax² + bx + c = 0

Arithmetic Sequence

aₙ = a₁ + (n - 1)d

nth term of an arithmetic sequence

Geometric Sequence

aₙ = a₁ × rⁿ⁻¹

nth term of a geometric sequence

Sum of Arithmetic Series

Sₙ = n/2 × (a₁ + aₙ)

Sum of first n terms of arithmetic sequence

Sum of Geometric Series

Sₙ = a₁(1 - rⁿ)/(1 - r)

Sum of first n terms of geometric sequence (r ≠ 1)

Exponent Rules

aᵐ × aⁿ = aᵐ⁺ⁿ

Product of powers with same base

Logarithm Properties

logₐ(mn) = logₐ m + logₐ n

Product rule for logarithms

Distance Formula

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

Distance between two points in coordinate plane

Geometry Formulas

Pythagorean Theorem

a² + b² = c²

For right triangles, where c is the hypotenuse

Area of Circle

A = πr²

Area of a circle with radius r

Circumference of Circle

C = 2πr

Circumference of a circle with radius r

Area of Triangle

A = (1/2)bh

Area of a triangle with base b and height h

Area of Rectangle

A = l × w

Area of a rectangle with length l and width w

Volume of Cube

V = s³

Volume of a cube with side length s

Volume of Cylinder

V = πr²h

Volume of a cylinder with radius r and height h

Surface Area of Sphere

SA = 4πr²

Surface area of a sphere with radius r

Trigonometry Formulas

Pythagorean Identity

sin²θ + cos²θ = 1

Fundamental trigonometric identity

Sine Rule

a/sinA = b/sinB = c/sinC

For any triangle with sides a, b, c and angles A, B, C

Cosine Rule

a² = b² + c² - 2bc cosA

For any triangle with sides a, b, c and angle A

Tangent Identity

tanθ = sinθ / cosθ

Definition of tangent function

Reciprocal Identities

cscθ = 1/sinθ, secθ = 1/cosθ, cotθ = 1/tanθ

Reciprocal trigonometric functions

Area of Triangle (Trig)

A = (1/2)ab sinC

Area using two sides and included angle

Angle Sum Identity

sin(A+B) = sinA cosB + cosA sinB

Sine of sum of two angles

Double Angle Formula

sin(2θ) = 2 sinθ cosθ

Double angle formula for sine

Calculus Formulas

Power Rule

d/dx(xⁿ) = nxⁿ⁻¹

Derivative of x raised to a power

Product Rule

d/dx(uv) = u'v + uv'

Derivative of product of two functions

Quotient Rule

d/dx(u/v) = (u'v - uv')/v²

Derivative of quotient of two functions

Chain Rule

d/dx[f(g(x))] = f'(g(x)) × g'(x)

Derivative of composite functions

Derivative of sin x

d/dx(sin x) = cos x

Derivative of sine function

Derivative of cos x

d/dx(cos x) = -sin x

Derivative of cosine function

Power Rule for Integration

∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ -1)

Integration of power functions

Fundamental Theorem

∫ₐᵇ f(x) dx = F(b) - F(a)

Where F'(x) = f(x)

Physics Formulas

Newton's Second Law

F = ma

Force equals mass times acceleration

Kinetic Energy

KE = (1/2)mv²

Energy of motion

Potential Energy

PE = mgh

Gravitational potential energy

Ohm's Law

V = IR

Voltage equals current times resistance

Density Formula

ρ = m/V

Density equals mass divided by volume

Work Formula

W = Fd cosθ

Work equals force times displacement times cosine of angle

Power Formula

P = W/t

Power equals work divided by time

Speed Formula

v = d/t

Speed equals distance divided by time

Similar Triangle Construction

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Constructing Similar Triangles with Given Ratio

A geometric construction using triangle sides 5cm, 5.4cm, 6cm with 3/4 scale factor

Problem Statement

Construct a triangle with sides 5 cm, 5.4 cm, and 6 cm. Then construct a similar triangle whose sides are \(\frac{3}{4}\) times the sides of the original triangle.

Mathematical Foundation

For similar triangles, the ratio of corresponding sides is constant:

\[ \frac{A'B'}{AB} = \frac{B'C'}{BC} = \frac{A'C'}{AC} = k \] where \(k = \frac{3}{4}\) in this case.

The sides of the new triangle will be:

\[ \begin{align*} 5 \times \frac{3}{4} &= 3.75 \text{ cm} \\ 5.4 \times \frac{3}{4} &= 4.05 \text{ cm} \\ 6 \times \frac{3}{4} &= 4.5 \text{ cm} \end{align*} \]

Step-by-Step Construction Process

Part 1: Constructing the Original Triangle

Draw the base: Draw segment AB = 6 cm

Construct point C using compass: With A as center, radius 5 cm, draw an arc. With B as center, radius 5.4 cm, draw another arc. The intersection point is C.

Complete the triangle: Join AC and BC to form triangle ABC.

Part 2: Constructing the Similar Triangle

Method: Dilation from a Point

Choose a center of dilation: Select point A as the center.

Draw rays: Draw ray from A through B and ray from A through C.

Mark points for \(\frac{3}{4}\) ratio: On ray AB, mark point B' such that AB' = \(\frac{3}{4}\) × AB = 4.5 cm. On ray AC, mark point C' such that AC' = \(\frac{3}{4}\) × AC = 3.75 cm.

Complete the similar triangle: Join B'C' to form triangle AB'C'.

Verification

Original Triangle Sides

Side	Length
AB	6.00 cm
BC	5.40 cm
AC	5.00 cm

Similar Triangle Sides

Side	Length
A'B'	4.50 cm (6 × 3/4)
B'C'	4.05 cm (5.4 × 3/4)
A'C'	3.75 cm (5 × 3/4)

Geometric Proof of Similarity

The triangles are similar by Side-Side-Side (SSS) Similarity:

\[ \frac{A'B'}{AB} = \frac{B'C'}{BC} = \frac{A'C'}{AC} = \frac{3}{4} \]

Interactive Features

In the Geogebra applet, you can:

Adjust the scale factor using the slider

Move vertices of the original triangle

Observe how the similar triangle maintains the ratio

Verify measurements in real-time

Live Geogebra Construction

Interact with the construction below:

Click here to open in a new window

Construction Notes

The Geogebra construction demonstrates:

Original triangle ABC with sides 5cm, 5.4cm, 6cm
Similar triangle AB'C' with sides 3.75cm, 4.05cm, 4.5cm
Dynamic adjustment of the scale factor
Real-time measurement display

Multiple Choice Questions (MCQ) Practice

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Identifying 2D/3D shapes, completing number patterns, place value, mental addition & subtraction.

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Comparing large numbers, measurement (length, mass, capacity), multiplication, division intro.

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Understanding fractions, multi-digit multiplication, time & calendar, perimeter & area.

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Area & perimeter, equivalent fractions, angles, decimals, data handling, volume basics.

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Algebra Calculator

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Algebra Calculator

Solve equations, simplify expressions, and perform algebraic operations

Enter Expression or Equation

Quadratic Formula

x = [-b ± √(b² - 4ac)] / 2a

For equations of the form: ax² + bx + c = 0

Result

Quick Examples (Click to try):

2x + 5 = 13

x² - 5x + 6 = 0

2x² + 3x - 2 = 0

x² - 4x + 4 = 0

x² + 2x + 5 = 0

3x + 2x - 5 + 3

2(3x + 4)

Geometry Calculator

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Geometry Calculator

Calculate area, perimeter, volume, and other properties of geometric shapes

Select a Shape:

Square

Rectangle

Triangle

Circle

Sphere

Cylinder

Cone

Cube

◼️

Side Length (a):

Square Formulas

Area = a²

Perimeter = 4a

Diagonal = a√2

▭

Length (l):

Width (w):

Rectangle Formulas

Area = l × w

Perimeter = 2(l + w)

Diagonal = √(l² + w²)

▲

Base (b):

Height (h):

Side 1 (a):

Side 2 (c):

Triangle Formulas

Area = ½ × b × h

Perimeter = a + b + c

●

Radius (r):

Circle Formulas

Area = πr²

Circumference = 2πr

Diameter = 2r

🔴

Radius (r):

Sphere Formulas

Volume = (4/3)πr³

Surface Area = 4πr²

⛽

Radius (r):

Height (h):

Cylinder Formulas

Volume = πr²h

Surface Area = 2πr(h + r)

🔺

Radius (r):

Height (h):

Cone Formulas

Volume = (1/3)πr²h

Slant Height = √(r² + h²)

Surface Area = πr(r + √(r² + h²))

⬛

Side Length (a):

Cube Formulas

Volume = a³

Surface Area = 6a²

Space Diagonal = a√3

Calculation Results

10th Maths Public Exam Valuation Key 2025

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SSC-2025 Principles of Valuation

How to Construct an Equilateral Triangle

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An equilateral triangle is a triangle with all three sides equal in length and all three angles measuring 60°. Here’s a step-by-step guide to constructing one using a compass and straightedge (ruler).

Materials Needed:

A ruler (straightedge)
A compass
A pencil
A sheet of paper

Steps:

Draw the Base Segment (AB):

Use a ruler to draw a straight line segment of your desired length. Label the endpoints as A and B.

Set the Compass:

Adjust the compass to the length of AB.

Draw an Arc from Point A:

Place the compass at point A and draw an arc above the line segment.

Draw an Arc from Point B:

Without changing the compass width, place it at point B and draw another arc that intersects the first arc.

Mark the Third Vertex (C):

The intersection point of the two arcs is the third vertex of the triangle, C.

Complete the Triangle:

Use the ruler to draw lines from A to C and from B to C.

Now, you have a perfect equilateral triangle (△ABC) with all sides and angles equal!

Verification:

Measure all three sides—they should be equal.
Use a protractor to confirm that each angle is 60°.

CLICK on Play button

10th Maths Statistics Exercise 14.4 Solutions

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Exercise 14.4 Solutions – Class X Mathematics

Exercise 14.4 Solutions

From Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad

Question 1

The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rupees)	250-300	300-350	350-400	400-450	450-500
Number of workers	12	14	8	6	10

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Solution:

Step 1: Convert to less than type cumulative frequency distribution:

Daily income less than (in Rupees)	300	350	400	450	500
Cumulative frequency	12	12+14=26	26+8=34	34+6=40	40+10=50

Question 2

During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg)	Less than 38	Less than 40	Less than 42	Less than 44	Less than 46	Less than 48	Less than 50	Less than 52
Number of students	0	3	5	9	14	28	32	35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution:

Step 1: The data is already in less than type cumulative frequency form.

Step 2: Drawing the ogive:

Ogive (Less than type) for weights of students

Step 3: Finding median from the graph:

Total number of students (n) = 35

Median position = n/2 = 17.5

From the graph, the x-coordinate corresponding to y=17.5 is approximately 46.5 kg.

Step 4: Verifying using the formula:

The median class is 46-48 (since 17.5 falls in the cumulative frequency of 28)

Using the formula:

\[ \text{Median} = L + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h \]

Where:
L = 46 (lower limit of median class)
cf = 14 (cumulative frequency before median class)
f = 14 (frequency of median class)
h = 2 (class width)

\[ \text{Median} = 46 + \left(\frac{17.5 – 14}{14}\right) \times 2 = 46 + \left(\frac{3.5}{14}\right) \times 2 = 46 + 0.5 = 46.5 \text{ kg} \]

This matches our graphical estimate.

Question 3

The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield (Quintal/Hectare)	50-55	55-60	60-65	65-70	70-75	75-80
Number of farmers	2	8	12	24	38	16

Change the distribution to a more than type distribution, and draw its ogive.

Solution:

Step 1: Convert to more than type cumulative frequency distribution:

Production yield more than (Quintal/Hectare)	50	55	60	65	70	75
Cumulative frequency	100	100-2=98	98-8=90	90-12=78	78-24=54	54-38=16

Step 2: Drawing the ogive (more than type):

Ogive (More than type) for production yield

10th Maths Statistics Exercise 14.3 Solutions

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Exercise 14.3 Solutions – Class X Mathematics

Exercise 14.3 Solutions

Class X Mathematics – State Council of Educational Research and Training, Telangana, Hyderabad

1. Electricity Consumption Problem

Monthly consumption	65-85	85-105	105-125	125-145	145-165	165-185	185-205
Number of consumers	4	5	13	20	14	8	4

Solution:

Median:

First, we calculate cumulative frequencies:

Class	Frequency	Cumulative Frequency
65-85	4	4
85-105	5	9
105-125	13	22
125-145	20	42
145-165	14	56
165-185	8	64
185-205	4	68

Median class is where cumulative frequency ≥ n/2 = 34 → 125-145

Using median formula:

Median = L + [(n/2 – CF)/f] × h

Where L = 125, n = 68, CF = 22, f = 20, h = 20

Median = 125 + [(34 – 22)/20] × 20 = 125 + 12 = 137

Mean:

Using assumed mean method with A = 135:

Class	Midpoint (x)	Frequency (f)	d = (x – A)/h	f × d
65-85	75	4	-3	-12
85-105	95	5	-2	-10
105-125	115	13	-1	-13
125-145	135	20	0	0
145-165	155	14	1	14
165-185	175	8	2	16
185-205	195	4	3	12
Total				7

Mean = A + (Σfd/Σf) × h = 135 + (7/68) × 20 ≈ 137.06

Mode:

Modal class is 125-145 (highest frequency = 20)

Using mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where L = 125, f₁ = 20, f₀ = 13, f₂ = 14, h = 20

Mode = 125 + [(20 – 13)/(40 – 13 – 14)] × 20 ≈ 125 + (7/13) × 20 ≈ 135.77

Comparison: Mean (137.06) > Median (137) > Mode (135.77)

2. Median Problem with Missing Frequencies

Class interval	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	5	x	20	15	y	5

Given median = 28.5, n = 60

Solution:

Total observations: 5 + x + 20 + 15 + y + 5 = 60 ⇒ x + y = 15

Median class is where cumulative frequency ≥ n/2 = 30 → 20-30

Using median formula:

28.5 = 20 + [(30 – (5 + x))/20] × 10

8.5 = (25 – x)/2 ⇒ 17 = 25 – x ⇒ x = 8

Since x + y = 15 ⇒ y = 7

Solution: x = 8, y = 7

3. Insurance Policy Holders Age Distribution

Age (in years)	Below 20	Below 25	Below 30	Below 35	Below 40	Below 45	Below 50	Below 55	Below 60
Number of policy holders	2	6	24	45	78	89	92	98	100

Solution:

First convert to frequency distribution:

Class	Frequency	Cumulative Frequency
18-20	2	2
20-25	4	6
25-30	18	24
30-35	21	45
35-40	33	78
40-45	11	89
45-50	3	92
50-55	6	98
55-60	2	100

Median class is where cumulative frequency ≥ n/2 = 50 → 35-40

Using median formula:

Median = 35 + [(50 – 45)/33] × 5 ≈ 35 + 0.76 ≈ 35.76 years

4. Leaves Length Measurement

Length (in mm)	118-126	127-135	136-144	145-153	154-162	163-171	172-180
Number of leaves	3	5	9	12	5	4	2

Solution:

Convert to continuous classes as suggested:

Class	Frequency	Cumulative Frequency
117.5-126.5	3	3
126.5-135.5	5	8
135.5-144.5	9	17
144.5-153.5	12	29
153.5-162.5	5	34
162.5-171.5	4	38
171.5-180.5	2	40

Median class is where cumulative frequency ≥ n/2 = 20 → 144.5-153.5

Using median formula:

Median = 144.5 + [(20 – 17)/12] × 9 = 144.5 + 2.25 = 146.75 mm

5. Neon Lamps Life Time

Life time (in hours)	1500-2000	2000-2500	2500-3000	3000-3500	3500-4000	4000-4500	4500-5000
Number of lamps	14	56	60	86	74	62	48

Solution:

Calculate cumulative frequencies:

Class	Frequency	Cumulative Frequency
1500-2000	14	14
2000-2500	56	70
2500-3000	60	130
3000-3500	86	216
3500-4000	74	290
4000-4500	62	352
4500-5000	48	400

Median class is where cumulative frequency ≥ n/2 = 200 → 3000-3500

Using median formula:

Median = 3000 + [(200 – 130)/86] × 500 ≈ 3000 + 406.98 ≈ 3406.98 hours

6. Surnames Letters Distribution

Number of letters	1-4	4-7	7-10	10-13	13-16	16-19
Number of surnames	6	30	40	16	4	4

Solution:

Median:

First make classes continuous and calculate cumulative frequencies:

Class	Frequency	Cumulative Frequency
0.5-4.5	6	6
4.5-7.5	30	36
7.5-10.5	40	76
10.5-13.5	16	92
13.5-16.5	4	96
16.5-19.5	4	100

Median class is where cumulative frequency ≥ n/2 = 50 → 7.5-10.5

Median = 7.5 + [(50 – 36)/40] × 3 = 7.5 + 1.05 = 8.55

Mean:

Using midpoint method:

Class	Midpoint (x)	Frequency (f)	f × x
1-4	2.5	6	15
4-7	5.5	30	165
7-10	8.5	40	340
10-13	11.5	16	184
13-16	14.5	4	58
16-19	17.5	4	70
Total		100	832

Mean = Σfx/Σf = 832/100 = 8.32

Mode:

Modal class is 7-10 (highest frequency = 40)

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where L = 7, f₁ = 40, f₀ = 30, f₂ = 16, h = 3

Mode = 7 + [(40 – 30)/(80 – 30 – 16)] × 3 ≈ 7 + (10/34) × 3 ≈ 7.88

7. Students Weight Distribution

Weight (in kg)	40-45	45-50	50-55	55-60	60-65	65-70	70-75
Number of students	2	3	8	6	6	3	2

Solution:

Calculate cumulative frequencies:

Class	Frequency	Cumulative Frequency
40-45	2	2
45-50	3	5
50-55	8	13
55-60	6	19
60-65	6	25
65-70	3	28
70-75	2	30

Median class is where cumulative frequency ≥ n/2 = 15 → 55-60

Using median formula:

Median = 55 + [(15 – 13)/6] × 5 ≈ 55 + 1.67 ≈ 56.67 kg

10th Maths Statistics Exercise 14.2 Solutions

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Exercise 14.2 Solutions – Class X Mathematics

Exercise 14.2 Solutions

Class X Mathematics
State Council of Educational Research and Training, Telangana, Hyderabad

Problem 1

The following table shows the ages of the patients admitted in a hospital on a particular day:

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
Number of patients	6	11	21	23	14	5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:

Mode Calculation:

The modal class is the class with the highest frequency. Here, the highest frequency is 23, which corresponds to the class 35-45.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 35
f₁ = frequency of modal class = 23
f₀ = frequency of class preceding modal class = 21
f₂ = frequency of class succeeding modal class = 14
h = class width = 10

Mode = 35 + [(23 – 21)/(2×23 – 21 – 14)] × 10
= 35 + [2/(46 – 35)] × 10
= 35 + (2/11) × 10
= 35 + 1.818 ≈ 36.82

Mean Calculation:

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
5-15	10	6	60
15-25	20	11	220
25-35	30	21	630
35-45	40	23	920
45-55	50	14	700
55-65	60	5	300
Total		80	2830

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2830}{80} = 35.375\)

Comparison and Interpretation:

Mode = 36.82 years, Mean = 35.38 years

The mode (36.82) is slightly higher than the mean (35.38), indicating that the most common age group of patients is slightly older than the average age of all patients. Both values suggest that most patients admitted are in their mid-30s to mid-40s.

Problem 2

The following data gives the information on the observed life times (in hours) of 225 electrical components:

Lifetimes (in hours)	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	10	35	52	61	38	29

Determine the modal lifetimes of the components.

Solution:

The modal class is the class with the highest frequency. Here, the highest frequency is 61, which corresponds to the class 60-80.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 60
f₁ = frequency of modal class = 61
f₀ = frequency of class preceding modal class = 52
f₂ = frequency of class succeeding modal class = 38
h = class width = 20

Mode = 60 + [(61 – 52)/(2×61 – 52 – 38)] × 20
= 60 + [9/(122 – 90)] × 20
= 60 + (9/32) × 20
= 60 + 5.625 = 65.625

Modal lifetime of components = 65.63 hours

Problem 3

The following data gives the distribution of total monthly household expenditure of 200 families of Gummadidala village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure (in rupees)	1000-1500	1500-2000	2000-2500	2500-3000	3000-3500	3500-4000	4000-4500	4500-5000
Number of families	24	40	33	28	30	22	16	7

Solution:

Mode Calculation:

The modal class is the class with the highest frequency. Here, the highest frequency is 40, which corresponds to the class 1500-2000.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 1500
f₁ = frequency of modal class = 40
f₀ = frequency of class preceding modal class = 24
f₂ = frequency of class succeeding modal class = 33
h = class width = 500

Mode = 1500 + [(40 – 24)/(2×40 – 24 – 33)] × 500
= 1500 + [16/(80 – 57)] × 500
= 1500 + (16/23) × 500
= 1500 + 347.83 ≈ 1847.83

Mean Calculation:

We’ll use the step-deviation method with assumed mean a = 2750 and class width h = 500

Class Interval	Midpoint (x_i)	Frequency (f_i)	d_i = x_i – a	u_i = d_i/h	f_iu_i
1000-1500	1250	24	-1500	-3	-72
1500-2000	1750	40	-1000	-2	-80
2000-2500	2250	33	-500	-1	-33
2500-3000	2750 (a)	28	0	0	0
3000-3500	3250	30	500	1	30
3500-4000	3750	22	1000	2	44
4000-4500	4250	16	1500	3	48
4500-5000	4750	7	2000	4	28
Total					-35

Mean = \(a + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times h = 2750 + \left(\frac{-35}{200}\right) \times 500 = 2750 – 87.5 = 2662.5\)

Modal monthly expenditure = ₹1847.83
Mean monthly expenditure = ₹2662.50

Problem 4

The following distribution gives the state-wise, teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of students per teacher	15-20	20-25	25-30	30-35	35-40	40-45	45-50	50-55
Number of States	3	8	9	10	3	0	0	2

Solution:

Mode Calculation:

The modal class is the class with the highest frequency. Here, the highest frequency is 10, which corresponds to the class 30-35.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 30
f₁ = frequency of modal class = 10
f₀ = frequency of class preceding modal class = 9
f₂ = frequency of class succeeding modal class = 3
h = class width = 5

Mode = 30 + [(10 – 9)/(2×10 – 9 – 3)] × 5
= 30 + [1/(20 – 12)] × 5
= 30 + (1/8) × 5
= 30 + 0.625 = 30.625

Mean Calculation:

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
15-20	17.5	3	52.5
20-25	22.5	8	180
25-30	27.5	9	247.5
30-35	32.5	10	325
35-40	37.5	3	112.5
40-45	42.5	0	0
45-50	47.5	0	0
50-55	52.5	2	105
Total		35	1022.5

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{1022.5}{35} \approx 29.21\)

Interpretation:

Mode = 30.62 students per teacher
Mean = 29.21 students per teacher

The mode (30.62) is slightly higher than the mean (29.21), indicating that the most common student-teacher ratio is slightly higher than the average ratio across all states. This suggests that while most states have about 30-31 students per teacher, some states with lower ratios bring the average down slightly.

Problem 5

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs	3000-4000	4000-5000	5000-6000	6000-7000	7000-8000	8000-9000	9000-10000	10000-11000
Number of batsmen	4	18	9	7	6	3	1	1

Find the mode of the data.

Solution:

The modal class is the class with the highest frequency. Here, the highest frequency is 18, which corresponds to the class 4000-5000.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 4000
f₁ = frequency of modal class = 18
f₀ = frequency of class preceding modal class = 4
f₂ = frequency of class succeeding modal class = 9
h = class width = 1000

Mode = 4000 + [(18 – 4)/(2×18 – 4 – 9)] × 1000
= 4000 + [14/(36 – 13)] × 1000
= 4000 + (14/23) × 1000
= 4000 + 608.70 ≈ 4608.70

Modal runs scored by batsmen = 4608.70 runs

Problem 6

A student noted the number of cars passing through a spot on a road for 100 periods, each of 3 minutes, and summarised this in the table given below.

Number of cars	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	7	14	13	12	20	11	15	8

Find the mode of the data.

Solution:

The modal class is the class with the highest frequency. Here, the highest frequency is 20, which corresponds to the class 40-50.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 40
f₁ = frequency of modal class = 20
f₀ = frequency of class preceding modal class = 12
f₂ = frequency of class succeeding modal class = 11
h = class width = 10

Mode = 40 + [(20 – 12)/(2×20 – 12 – 11)] × 10
= 40 + [8/(40 – 23)] × 10
= 40 + (8/17) × 10
= 40 + 4.71 ≈ 44.71

Mode number of cars passing = 44.71 cars per 3-minute period

10th Maths Statistics Exercise 14.1 Solutions

zaheerpashamd

Exercise 14.1 Solutions – Class X Mathematics

Exercise 14.1 Solutions

Class X Mathematics
State Council of Educational Research and Training, Telangana, Hyderabad

Problem 1

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants	0 – 2	2 – 4	4 – 6	6 – 8	8 – 10	10 – 12	12 – 14
Number of houses	1	2	1	5	6	2	3

Solution:

To find the mean, we’ll use the midpoint of each class interval and multiply by frequency.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
0-2	1	1	1
2-4	3	2	6
4-6	5	1	5
6-8	7	5	35
8-10	9	6	54
10-12	11	2	22
12-14	13	3	39
Total		20	162

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{162}{20} = 8.1\)

Mean number of plants per house = 8.1

Problem 2

Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages in Rupees	200 – 250	250 – 300	300 – 350	350 – 400	400 – 450
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

We’ll use the step-deviation method with assumed mean a = 325 and class width h = 50

Class Interval	Midpoint (x_i)	Frequency (f_i)	d_i = x_i – a	u_i = d_i/h	f_iu_i
200-250	225	12	-100	-2	-24
250-300	275	14	-50	-1	-14
300-350	325 (a)	8	0	0	0
350-400	375	6	50	1	6
400-450	425	10	100	2	20
Total					-12

Mean = \(a + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times h = 325 + \left(\frac{-12}{50}\right) \times 50 = 325 – 12 = 313\)

Mean daily wages = ₹313

Problem 3

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency \( f \).

Daily pocket allowance (in Rupees)	11 – 13	13 – 15	15 – 17	17 – 19	19 – 21	21 – 23	23 – 25
Number of children	7	6	9	13	\( f \)	5	4

Solution:

Let’s calculate the sum of frequencies and products of midpoints and frequencies.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
11-13	12	7	84
13-15	14	6	84
15-17	16	9	144
17-19	18	13	234
19-21	20	f	20f
21-23	22	5	110
23-25	24	4	96
Total		44 + f	752 + 20f

Given mean = 18

\(\frac{752 + 20f}{44 + f} = 18\)

752 + 20f = 792 + 18f

2f = 40

f = 20

The missing frequency \( f = 20 \)

Problem 4

Thirty women were examined in a hospital by a doctor and their heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats/minute	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of women	2	4	3	8	7	4	2

Solution:

We’ll use the direct method to find the mean.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
65-68	66.5	2	133
68-71	69.5	4	278
71-74	72.5	3	217.5
74-77	75.5	8	604
77-80	78.5	7	549.5
80-83	81.5	4	326
83-86	84.5	2	169
Total		30	2277

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2277}{30} = 75.9\)

Mean heart beats per minute = 75.9

Problem 5

In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges.

Number of oranges	10-14	15-19	20-24	25-29	30-34
Number of baskets	15	110	135	115	25

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose?

Solution:

We’ll use the step-deviation method with assumed mean a = 22 and class width h = 5

Class Interval	Midpoint (x_i)	Frequency (f_i)	d_i = x_i – a	u_i = d_i/h	f_iu_i
10-14	12	15	-10	-2	-30
15-19	17	110	-5	-1	-110
20-24	22 (a)	135	0	0	0
25-29	27	115	5	1	115
30-34	32	25	10	2	50
Total					25

Mean = \(a + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times h = 22 + \left(\frac{25}{400}\right) \times 5 = 22 + \frac{125}{400} = 22 + 0.3125 = 22.3125\)

Mean number of oranges per basket = 22.31 (approx)

We chose the step-deviation method because the class intervals are equal and the frequencies are large.

Problem 6

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rupees)	100-150	150-200	200-250	250-300	300-350
Number of households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.

Solution:

We’ll use the direct method to find the mean.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
100-150	125	4	500
150-200	175	5	875
200-250	225	12	2700
250-300	275	2	550
300-350	325	2	650
Total		25	5275

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{5275}{25} = 211\)

Mean daily expenditure on food = ₹211

Problem 7

To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO₂ in ppm	0.00-0.04	0.04-0.08	0.08-0.12	0.12-0.16	0.16-0.20	0.20-0.24
Frequency	4	9	9	2	4	2

Find the mean concentration of SO₂ in the air.

Solution:

We’ll use the direct method to find the mean.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
0.00-0.04	0.02	4	0.08
0.04-0.08	0.06	9	0.54
0.08-0.12	0.10	9	0.90
0.12-0.16	0.14	2	0.28
0.16-0.20	0.18	4	0.72
0.20-0.24	0.22	2	0.44
Total		30	2.96

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2.96}{30} \approx 0.0987\)

Mean concentration of SO₂ = 0.099 ppm (approx)

Problem 8

A class teacher has the following attendance record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term.

Number of days	35-38	38-41	41-44	44-47	47-50	50-53	53-56
Number of students	1	3	4	4	7	10	11

Solution:

We’ll use the direct method to find the mean.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
35-38	36.5	1	36.5
38-41	39.5	3	118.5
41-44	42.5	4	170
44-47	45.5	4	182
47-50	48.5	7	339.5
50-53	51.5	10	515
53-56	54.5	11	599.5
Total		40	1961

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{1961}{40} = 49.025\)

Mean number of days present = 49.03 days (approx)

Problem 9

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate in %	45-55	55-65	65-75	75-85	85-95
Number of cities	3	10	11	8	3

Solution:

We’ll use the direct method to find the mean.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
45-55	50	3	150
55-65	60	10	600
65-75	70	11	770
75-85	80	8	640
85-95	90	3	270
Total		35	2430

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2430}{35} \approx 69.43\)

Mean literacy rate = 69.43%

10th Maths Probability Exercise 13.2 Solutions

zaheerpashamd

Exercise 13.2 Solutions – Class X Mathematics

Exercise 13.2 Solutions

Probability – Class X Mathematics

Question 1

A bag contains 3 red balls and 5 black balls. A ball is selected at random from the bag. What is the probability that the ball selected is:

(i) red? (ii) not red?

Solution:

Total number of balls = 3 (red) + 5 (black) = 8

(i) Probability of selecting a red ball = \(\frac{\text{Number of red balls}}{\text{Total balls}} = \frac{3}{8}\)

(ii) Probability of not selecting a red ball = 1 – Probability of selecting red ball = \(1 – \frac{3}{8} = \frac{5}{8}\)

Question 2

A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be:

(i) red? (ii) white? (iii) not green?

Solution:

Total marbles = 5 (red) + 8 (white) + 4 (green) = 17

(i) Probability of red marble = \(\frac{5}{17}\)

(ii) Probability of white marble = \(\frac{8}{17}\)

(iii) Probability of not green marble = 1 – Probability of green marble = \(1 – \frac{4}{17} = \frac{13}{17}\)

Question 3

A Kiddy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin:

(i) will be a 50p coin? (ii) will not be a ₹5 coin?

Solution:

Total coins = 100 (50p) + 50 (₹1) + 20 (₹2) + 10 (₹5) = 180

(i) Probability of 50p coin = \(\frac{100}{180} = \frac{5}{9}\)

(ii) Probability of not ₹5 coin = 1 – Probability of ₹5 coin = \(1 – \frac{10}{180} = \frac{170}{180} = \frac{17}{18}\)

Question 4

Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Diagram Description: A rectangular fish tank containing 5 male fish (shown as smaller, colorful fish) and 8 female fish (shown as larger, less colorful fish). The shopkeeper is using a net to catch one fish at random.

Solution:

Total fish = 5 (male) + 8 (female) = 13

Probability of male fish = \(\frac{5}{13}\)

Question 5

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (See figure), and these are equally likely outcomes. What is the probability that it will point at:

(i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?

Diagram Description: A circular spinner divided into 8 equal sectors numbered 1 through 8 clockwise. An arrow is fixed at the center that can spin freely.

Solution:

Total possible outcomes = 8

(i) Probability of pointing at 8 = \(\frac{1}{8}\)

(ii) Odd numbers = {1, 3, 5, 7} → 4 outcomes. Probability = \(\frac{4}{8} = \frac{1}{2}\)

(iii) Numbers > 2 = {3, 4, 5, 6, 7, 8} → 6 outcomes. Probability = \(\frac{6}{8} = \frac{3}{4}\)

(iv) Numbers < 9 = {1, 2, 3, 4, 5, 6, 7, 8} → 8 outcomes. Probability = \(\frac{8}{8} = 1\)

Question 6

One card is selected from a well-shuffled deck of 52 cards. Find the probability of getting:

(i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds

Solution:

Total cards = 52

(i) King of red colour: There are 2 (King of Hearts and King of Diamonds). Probability = \(\frac{2}{52} = \frac{1}{26}\)

(ii) Face cards: Jack, Queen, King in each suit → 3 × 4 = 12. Probability = \(\frac{12}{52} = \frac{3}{13}\)

(iii) Red face cards: Face cards in hearts and diamonds → 3 × 2 = 6. Probability = \(\frac{6}{52} = \frac{3}{26}\)

(iv) Jack of hearts: Only 1 card. Probability = \(\frac{1}{52}\)

(v) Spade cards: 13. Probability = \(\frac{13}{52} = \frac{1}{4}\)

(vi) Queen of diamonds: Only 1 card. Probability = \(\frac{1}{52}\)

Question 7

Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is selected at random.

(i) What is the probability that the card is the queen?

(ii) If the queen is selected and put aside (without replacement), what is the probability that the second card selected is (a) an ace? (b) a queen?

Solution:

Total cards initially = 5

(i) Probability of queen = \(\frac{1}{5}\)

(ii) After removing queen, remaining cards = 4

(a) Probability of ace = \(\frac{1}{4}\) (only ace of diamonds left)

(b) Probability of queen = \(\frac{0}{4} = 0\) (queen has been removed)

Question 8

12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Solution:

Total pens = 12 (defective) + 132 (good) = 144

Probability of good pen = \(\frac{132}{144} = \frac{11}{12}\)

Question 9

A lot of 20 bulbs contain 4 defective ones. One bulb is selected at random from the lot. What is the probability that this bulb is defective? Suppose the bulb selected in previous case is not defective and is not replaced. Now one bulb is selected at random from the rest. What is the probability that this bulb is not defective?

Solution:

First selection:

Total bulbs = 20, Defective = 4

Probability of defective bulb = \(\frac{4}{20} = \frac{1}{5}\)

Second selection (after removing one good bulb):

Remaining bulbs = 19, Non-defective = 16 – 1 = 15

Probability of not defective bulb = \(\frac{15}{19}\)

Question 10

A box contains 90 discs which are numbered from 1 to 90. If one disc is selected at random from the box, find the probability that it bears:

(i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.

Solution:

Total discs = 90

(i) Two-digit numbers: 10 to 90 → 81 numbers. Probability = \(\frac{81}{90} = \frac{9}{10}\)

(ii) Perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 → 9 numbers. Probability = \(\frac{9}{90} = \frac{1}{10}\)

(iii) Divisible by 5: 5, 10, 15, …, 90 → 18 numbers. Probability = \(\frac{18}{90} = \frac{1}{5}\)

Question 11

Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1m?

Diagram Description: A rectangular region with length 3m and width 2m. Inside it, there’s a circle with diameter 1m (radius 0.5m) centered in the rectangle.

Solution:

Area of rectangle = length × width = 3 × 2 = 6 m²

Area of circle = πr² = π(0.5)² = 0.25π m²

Probability = \(\frac{\text{Area of circle}}{\text{Area of rectangle}} = \frac{0.25π}{6} = \frac{π}{24}\)

Question 12

A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that:

(i) She will buy it? (ii) She will not buy it?

Solution:

Assuming Sudha will buy if the pen is good:

Good pens = 144 – 20 = 124

(i) Probability she will buy = Probability of good pen = \(\frac{124}{144} = \frac{31}{36}\)

(ii) Probability she will not buy = Probability of defective pen = \(\frac{20}{144} = \frac{5}{36}\)

Question 13

Two dice are rolled simultaneously and counts are added:

(i) complete the table given below:

Event: ‘Sum on 2 dice’	2	3	4	5	6	7	8	9	10	11	12
Probability	\(\frac{1}{36}\)	\(\frac{2}{36}\)	\(\frac{3}{36}\)	\(\frac{4}{36}\)	\(\frac{5}{36}\)	\(\frac{6}{36}\)	\(\frac{5}{36}\)	\(\frac{4}{36}\)	\(\frac{3}{36}\)	\(\frac{2}{36}\)	\(\frac{1}{36}\)

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac{1}{11}\). Do you agree with this argument? Justify your answer.

Solution:

(i) Table completed above.

(ii) No, I don’t agree. While there are 11 possible sums, they are not equally likely. For example, there’s only one way to get a sum of 2 (1+1) but six ways to get a sum of 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1). Therefore, the probability of each sum is different.

Question 14

A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Deskhtha wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that she will lose the game.

Solution:

Total possible outcomes when tossing a coin 3 times = 2³ = 8

Favorable outcomes for winning: HHH, TTT → 2 outcomes

Probability of winning = \(\frac{2}{8} = \frac{1}{4}\)

Probability of losing = 1 – Probability of winning = \(1 – \frac{1}{4} = \frac{3}{4}\)

Question 15

A dice is thrown twice. What is the probability that:

(i) 5 will not come up either time? (ii) 5 will come up at least once?

Solution:

Total possible outcomes when throwing a die twice = 6 × 6 = 36

(i) Outcomes where 5 doesn’t appear either time: Each die has 5 options (1,2,3,4,6). Total = 5 × 5 = 25

Probability = \(\frac{25}{36}\)

(ii) Probability that 5 comes at least once = 1 – Probability that 5 doesn’t appear at all = \(1 – \frac{25}{36} = \frac{11}{36}\)

10th Maths Probability Exercise 13.1 Solutions

zaheerpashamd

Exercise 13.1 Solutions – Class X Mathematics

Exercise 13.1 Solutions

Probability – Class X Mathematics

State Council of Educational Research and Training, Telangana, Hyderabad

Question 1: Complete the following statements

(i) Probability of an event \( E + \text{Probability of the event ‘not } E’ = \) ______.

Solution: The sum of the probability of an event and its complement is always 1.

\[ P(E) + P(\text{not } E) = 1 \]

(ii) The probability of an event that cannot happen is ______. Such an event is called ______.

Solution: The probability of an impossible event is 0. Such an event is called an impossible event.

\[ P(\text{Impossible event}) = 0 \]

(iii) The probability of an event that is certain to happen is ______. Such an event is called ______.

Solution: The probability of a certain event is 1. Such an event is called a sure event.

\[ P(\text{Certain event}) = 1 \]

(iv) The sum of the probabilities of all the elementary events of an experiment is ______.

Solution: The sum of probabilities of all elementary events of an experiment is 1.

\[ \sum P(E_i) = 1 \]

(v) The probability of an event is greater than or equal to ______ and less than or equal to ______.

Solution: The probability of any event \( E \) satisfies:

\[ 0 \leq P(E) \leq 1 \]

Question 2: Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

Solution: Not equally likely. The probability depends on the car’s condition, fuel, etc.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

Solution: Not equally likely. The outcome depends on the player’s skill.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

Solution: Equally likely. There’s a 50% chance of guessing correctly.

\[ P(\text{Right}) = P(\text{Wrong}) = 0.5 \]

(iv) A baby is born. It is a boy or a girl.

Solution: Equally likely (assuming equal probability for biological sexes).

\[ P(\text{Boy}) = P(\text{Girl}) ≈ 0.5 \]

Question 3: If \( P(E) = 0.05 \), what is the probability of ‘not E’?

Solution:

We know that \( P(E) + P(\text{not } E) = 1 \)

\[ P(\text{not } E) = 1 – P(E) = 1 – 0.05 = 0.95 \]

Question 4: A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag.

(i) an orange flavoured candy?

Solution: Since the bag contains only lemon flavored candies:

\[ P(\text{Orange}) = 0 \] (Impossible event)

(ii) a lemon flavoured candy?

Solution: Since all candies are lemon flavored:

\[ P(\text{Lemon}) = 1 \] (Certain event)

Diagram: A bag containing multiple lemon candies (all yellow) with one being drawn out.

[Illustration of a bag with yellow candies and one hand taking a candy out]

Question 5: Rahim removes all the hearts from the cards. What is the probability of:

i. Getting an ace from the remaining pack.

Solution: Original deck has 52 cards. After removing 13 hearts, 39 cards remain.

Number of aces in remaining cards: 3 (since Ace of Hearts was removed)

\[ P(\text{Ace}) = \frac{3}{39} = \frac{1}{13} \]

ii. Getting a diamonds.

Solution: All 13 diamonds remain in the deck of 39 cards.

\[ P(\text{Diamond}) = \frac{13}{39} = \frac{1}{3} \]

iii. Getting a card that is not a heart.

Solution: Since all hearts have been removed, all remaining 39 cards are not hearts.

\[ P(\text{Not heart}) = \frac{39}{39} = 1 \]

iv. Getting the Ace of hearts.

Solution: Since all hearts have been removed:

\[ P(\text{Ace of Hearts}) = 0 \] (Impossible event)

Diagram: A standard deck of cards with all heart cards (red) removed, showing only clubs, diamonds, and spades remaining.

[Illustration of a partial deck with hearts missing]

Question 6: In a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Solution:

Let \( P(\text{Same}) \) be the probability that two students share a birthday.

Given \( P(\text{Not same}) = 0.992 \)

\[ P(\text{Same}) = 1 – P(\text{Not same}) = 1 – 0.992 = 0.008 \]

Question 7: A die is rolled once. Find the probability of getting:

(i) a prime number

Solution: Prime numbers on a die: 2, 3, 5 (3 outcomes)

Total possible outcomes: 6

\[ P(\text{Prime}) = \frac{3}{6} = \frac{1}{2} \]

(ii) a number lying between 2 and 6

Solution: Numbers between 2 and 6: 3, 4, 5 (3 outcomes)

\[ P(\text{Between 2 and 6}) = \frac{3}{6} = \frac{1}{2} \]

(iii) an odd number

Solution: Odd numbers on a die: 1, 3, 5 (3 outcomes)

\[ P(\text{Odd}) = \frac{3}{6} = \frac{1}{2} \]

Diagram: A standard six-faced die showing numbers 1 through 6 with prime numbers (2,3,5) highlighted.

[Illustration of a die with some numbers highlighted]

Question 8: What is the probability of selecting a red king from a deck of cards?

Solution:

Total cards in deck: 52

Red kings: King of Hearts and King of Diamonds (2 cards)

\[ P(\text{Red King}) = \frac{2}{52} = \frac{1}{26} \]

Diagram: A deck of cards showing the two red kings (King of Hearts and King of Diamonds).

[Illustration of two red king cards]

Question 9: Make 5 more problems of this kind using dice, cards or birthdays

Sample Problems:

What is the probability of rolling a number greater than 4 on a standard die?
If two cards are drawn from a standard deck without replacement, what is the probability both are spades?
In a group of 30 people, what is the probability that at least two people share the same birthday?
What is the probability of drawing a face card (Jack, Queen, King) from a standard deck?
If you roll two dice, what is the probability that the sum is 7?

These problems can be discussed with friends and teachers to understand probability concepts better.

10th Maths Applications of Trigonometry Exercise 12.2 Solutions

zaheerpashamd

Exercise 11.3 Solutions - Class X Mathematics

Problem 1

A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is \( 60^\circ \). From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is \( 30^\circ \). Find the height of the tower and the width of the road.

Diagram Description: A vertical TV tower on one side of a road. Point A is directly opposite the tower on the other side of the road, with angle of elevation 60°. Point B is 10m from A towards the tower, with angle of elevation 30°.

Solution:

Let the height of the tower be \( h \) meters and the width of the road be \( x \) meters.

From point A (directly opposite):

\[ \tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3} \]

From point B (10m from A towards the tower):

\[ \tan 30^\circ = \frac{h}{x + 10} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 10} \]

Substitute \( h = x\sqrt{3} \):

\[ \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x + 10} \]

\[ x + 10 = 3x \]

\[ 2x = 10 \Rightarrow x = 5 \text{ meters} \]

Then \( h = 5\sqrt{3} \) meters.

The height of the tower is \( 5\sqrt{3} \) meters and the width of the road is 5 meters.

Problem 2

A 1.5 m tall boy is looking at the top of a temple which is 30 meter in height from a point at certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30° to 60° as he walks towards the temple. Find the distance he walked towards the temple.

Diagram Description: A temple of height 30m. A boy of height 1.5m stands at point A (initial position) with angle of elevation 30°, then moves to point B (closer position) with angle of elevation 60°.

Solution:

Effective height of temple above boy's eye level = 30 - 1.5 = 28.5 meters

Let initial distance be \( x \) meters and distance walked be \( d \) meters.

From initial position:

\[ \tan 30^\circ = \frac{28.5}{x} \Rightarrow x = 28.5\sqrt{3} \]

From final position:

\[ \tan 60^\circ = \frac{28.5}{x - d} \Rightarrow x - d = \frac{28.5}{\sqrt{3}} \]

Substitute \( x = 28.5\sqrt{3} \):

\[ 28.5\sqrt{3} - d = \frac{28.5}{\sqrt{3}} \]

\[ d = 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}} \]

\[ d = 28.5\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = 28.5\left(\frac{3 - 1}{\sqrt{3}}\right) = \frac{57}{\sqrt{3}} = 19\sqrt{3} \text{ meters} \]

The boy walked \( 19\sqrt{3} \) meters towards the temple.

Problem 3

A statue stands on the top of a 2m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the statue.

Diagram Description: A pedestal of height 2m with a statue on top. From a point on the ground, angle to top of pedestal is 45° and angle to top of statue is 60°.

Solution:

Let the height of the statue be \( h \) meters and distance from point to pedestal be \( x \) meters.

For pedestal (2m height):

\[ \tan 45^\circ = \frac{2}{x} \Rightarrow x = 2 \text{ meters} \]

For statue (2 + h meters height):

\[ \tan 60^\circ = \frac{2 + h}{2} \Rightarrow \sqrt{3} = \frac{2 + h}{2} \]

\[ 2 + h = 2\sqrt{3} \Rightarrow h = 2\sqrt{3} - 2 = 2(\sqrt{3} - 1) \text{ meters} \]

The height of the statue is \( 2(\sqrt{3} - 1) \) meters.

Problem 4

From the top of a building, the angle of elevation of the top of a cell tower is 60° and the angle of depression to its foot is 45°. If distance of the building from the tower is 7m, then find the height of the tower.

Diagram Description: A building and a cell tower 7m apart. From the top of the building, angle of elevation to tower top is 60° and angle of depression to tower base is 45°.

Solution:

Let height of building be \( h \) meters and height of tower be \( H \) meters.

From angle of depression (45°):

\[ \tan 45^\circ = \frac{h}{7} \Rightarrow h = 7 \text{ meters} \]

From angle of elevation (60°):

\[ \tan 60^\circ = \frac{H - h}{7} \Rightarrow \sqrt{3} = \frac{H - 7}{7} \]

\[ H - 7 = 7\sqrt{3} \Rightarrow H = 7 + 7\sqrt{3} = 7(1 + \sqrt{3}) \text{ meters} \]

The height of the tower is \( 7(1 + \sqrt{3}) \) meters.

Problem 5

A wire of length 18 m had been tied with electric pole at an angle of elevation 30° with the ground. Because it was covering a long distance, it was cut and tied at an angle of elevation 60° with the ground. How much length of the wire was cut?

Diagram Description: An electric pole with two positions of a wire - original at 30° (18m) and shortened at 60°.

Solution:

Let height of pole be \( h \) meters.

Original wire (18m at 30°):

\[ \sin 30^\circ = \frac{h}{18} \Rightarrow h = 18 \times \frac{1}{2} = 9 \text{ meters} \]

New wire (at 60°):

\[ \sin 60^\circ = \frac{9}{L} \Rightarrow L = \frac{9}{\sin 60^\circ} = \frac{9}{\sqrt{3}/2} = \frac{18}{\sqrt{3}} = 6\sqrt{3} \text{ meters} \]

Length cut = Original - New = \( 18 - 6\sqrt{3} = 6(3 - \sqrt{3}) \) meters

\( 6(3 - \sqrt{3}) \) meters of wire was cut.

Problem 6

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 30 m high, find the height of the building.

Diagram Description: A tower and building separated by some distance. From tower base to building top is 30°, from building base to tower top is 60°.

Solution:

Let height of building be \( h \) meters and distance between them be \( d \) meters.

From building's foot to tower's top (30m):

\[ \tan 60^\circ = \frac{30}{d} \Rightarrow d = \frac{30}{\sqrt{3}} = 10\sqrt{3} \text{ meters} \]

From tower's foot to building's top:

\[ \tan 30^\circ = \frac{h}{d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{10\sqrt{3}} \]

\[ h = \frac{10\sqrt{3}}{\sqrt{3}} = 10 \text{ meters} \]

The height of the building is 10 meters.

Problem 7

Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.

Diagram Description: Two poles of equal height on opposite sides of a 120ft road. A point between them has angles of elevation 60° to one pole and 30° to the other.

Solution:

Let height of poles be \( h \) feet.

Let distances from point to poles be \( x \) and \( 120 - x \) feet.

For first pole (60°):

\[ \tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3} \]

For second pole (30°):

\[ \tan 30^\circ = \frac{h}{120 - x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{120 - x} \]

Substitute \( h = x\sqrt{3} \):

\[ \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{120 - x} \]

\[ 120 - x = 3x \]

\[ 4x = 120 \Rightarrow x = 30 \text{ feet} \]

Then \( h = 30\sqrt{3} \) feet

Distances: 30 feet and 90 feet

The poles are \( 30\sqrt{3} \) feet high. The point is 30 feet from one pole and 90 feet from the other.

Problem 8

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Find the height of the tower.

Diagram Description: A tower with two observation points at 4m and 9m from base. Their angles of elevation are complementary (sum to 90°).

Solution:

Let height of tower be \( h \) meters.

Let angles be \( \theta \) and \( 90^\circ - \theta \).

From first point (4m):

\[ \tan \theta = \frac{h}{4} \]

From second point (9m):

\[ \tan(90^\circ - \theta) = \frac{h}{9} \Rightarrow \cot \theta = \frac{h}{9} \]

Since \( \tan \theta \times \cot \theta = 1 \):

\[ \frac{h}{4} \times \frac{h}{9} = 1 \]

\[ h^2 = 36 \Rightarrow h = 6 \text{ meters} \]

The height of the tower is 6 meters.

Problem 9

The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of \( 1500\sqrt{3} \) meter, find the speed of the jet plane.

Diagram Description: A jet flying at constant height. From point A, initial angle is 60°, after 15 seconds angle is 30°.

Solution:

Let initial distance be \( x \) meters and height \( h = 1500\sqrt{3} \) meters.

Initial position (60°):

\[ \tan 60^\circ = \frac{h}{x} \Rightarrow x = \frac{1500\sqrt{3}}{\sqrt{3}} = 1500 \text{ meters} \]

After 15 seconds (30°):

\[ \tan 30^\circ = \frac{h}{x + d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{1500\sqrt{3}}{1500 + d} \]

\[ 1500 + d = 1500 \times 3 = 4500 \]

\[ d = 3000 \text{ meters} \]

Speed = Distance/Time = \( \frac{3000}{15} = 200 \) m/s

Convert to km/h: \( 200 \times \frac{18}{5} = 720 \) km/h

The speed of the jet plane is 720 km/h.

Problem 10

The angle of elevation of the top of a tower from the foot of the building is 30° and the angle of elevation of the top of the building from the foot of the tower is 60°. What is the ratio of heights of tower and building.

Diagram Description: A tower and building separated by some distance. From building base to tower top is 30°, from tower base to building top is 60°.

Solution:

Let height of tower be \( h_t \) and building be \( h_b \), distance between them be \( d \).

From building's foot to tower's top:

\[ \tan 30^\circ = \frac{h_t}{d} \Rightarrow d = h_t\sqrt{3} \]

From tower's foot to building's top:

\[ \tan 60^\circ = \frac{h_b}{d} \Rightarrow \sqrt{3} = \frac{h_b}{h_t\sqrt{3}} \]

\[ h_b = 3h_t \]

Ratio \( \frac{h_t}{h_b} = \frac{h_t}{3h_t} = \frac{1}{3} \)

The ratio of heights of tower to building is 1:3.