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Mathematics Mastery - Exam Preparation Compilation

Real Numbers

Solved Examples

1 Prove that √5 is irrational.

Assume that √5 is rational. Then it can be expressed as a fraction a/b where a and b are coprime integers.

Then √5 = a/b ⇒ 5 = a²/b² ⇒ a² = 5b².

This means a² is divisible by 5, so a is also divisible by 5. Let a = 5c.

Then (5c)² = 5b² ⇒ 25c² = 5b² ⇒ 5c² = b².

This means b² is divisible by 5, so b is also divisible by 5.

But this contradicts our assumption that a and b are coprime. Therefore, √5 is irrational.

2 Find the HCF of 96 and 404 by prime factorization method.

Prime factors of 96: 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3

Prime factors of 404: 2 × 2 × 101 = 2² × 101

HCF is the product of common prime factors with lowest powers: 2² = 4

Therefore, HCF of 96 and 404 is 4.

3 Show that every positive even integer is of the form 2q and every positive odd integer is of the form 2q+1 for some integer q.

According to Euclid's division lemma, for any positive integer a and b=2, there exist unique integers q and r such that:

a = 2q + r, where 0 ≤ r < 2

So possible values of r are 0 or 1.

If r = 0, then a = 2q → even integer

If r = 1, then a = 2q + 1 → odd integer

Hence, every positive even integer is of the form 2q and every positive odd integer is of the form 2q+1.

4 Find the LCM and HCF of 6 and 20 by prime factorization method.

Prime factors of 6: 2 × 3

Prime factors of 20: 2 × 2 × 5 = 2² × 5

HCF = product of common prime factors with lowest powers = 2

LCM = product of all prime factors with highest powers = 2² × 3 × 5 = 4 × 3 × 5 = 60

Therefore, HCF = 2 and LCM = 60.

5 Prove that 3 + 2√5 is irrational.

Assume that 3 + 2√5 is rational. Then it can be expressed as a fraction a/b where a and b are coprime integers.

Then 3 + 2√5 = a/b ⇒ 2√5 = a/b - 3 = (a - 3b)/b

⇒ √5 = (a - 3b)/(2b)

Since a and b are integers, (a - 3b)/(2b) is rational, which would mean √5 is rational.

But we know that √5 is irrational. This is a contradiction.

Therefore, our assumption is wrong, and 3 + 2√5 is irrational.

6 Use Euclid's division algorithm to find the HCF of 867 and 255.

Using Euclid's division algorithm:

867 = 255 × 3 + 102

255 = 102 × 2 + 51

102 = 51 × 2 + 0

Since the remainder is 0, the HCF is the divisor at this stage, which is 51.

Therefore, HCF of 867 and 255 is 51.

Practice Questions

1. Prove that √3 is irrational.
2. Find the HCF of 240 and 6552 using Euclid's division algorithm.
3. Show that the square of any positive integer is of the form 3m or 3m+1 for some integer m.
4. Find the LCM and HCF of 510 and 92 and verify that LCM × HCF = product of the two numbers.
5. Prove that 5 - 2√3 is irrational.

Polynomials

Solved Examples

1 Find the zeroes of the quadratic polynomial x² + 7x + 10 and verify the relationship between the zeroes and the coefficients.

Given polynomial: x² + 7x + 10

To find zeroes: x² + 7x + 10 = 0

⇒ (x + 2)(x + 5) = 0

⇒ x = -2 or x = -5

Sum of zeroes = -2 + (-5) = -7 = -coefficient of x/coefficient of x² = -7/1

Product of zeroes = (-2) × (-5) = 10 = constant term/coefficient of x² = 10/1

Hence verified.

2 Find a quadratic polynomial whose zeroes are 5 and -3.

If α and β are zeroes of a quadratic polynomial, then the polynomial is:

k[x² - (α + β)x + αβ], where k is a constant

Here α = 5, β = -3

α + β = 5 + (-3) = 2

αβ = 5 × (-3) = -15

So the polynomial is: k[x² - 2x - 15]

For simplest form, take k = 1: x² - 2x - 15

3 Divide the polynomial p(x) = x³ - 3x² + 5x - 3 by g(x) = x² - 2 and find quotient and remainder.

Performing polynomial division:

        x - 3
x² - 2 | x³ - 3x² + 5x - 3
        -(x³ + 0x² - 2x)
        ---------------
             -3x² + 7x - 3
             -(-3x² + 0x + 6)
             ----------------
                  7x - 9
        

Quotient = x - 3

Remainder = 7x - 9

4 Check whether the polynomial x² + 4x + 5 has real zeroes.

For quadratic polynomial ax² + bx + c, discriminant D = b² - 4ac

Here a = 1, b = 4, c = 5

D = 4² - 4×1×5 = 16 - 20 = -4 < 0

Since D < 0, the polynomial has no real zeroes.

5 If α and β are the zeroes of the polynomial 2x² + 5x + k such that α² + β² + αβ = 21/4, find the value of k.

For polynomial 2x² + 5x + k:

α + β = -b/a = -5/2

αβ = c/a = k/2

Given: α² + β² + αβ = 21/4

We know that α² + β² = (α + β)² - 2αβ

So (α + β)² - 2αβ + αβ = 21/4

⇒ (α + β)² - αβ = 21/4

⇒ (-5/2)² - (k/2) = 21/4

⇒ 25/4 - k/2 = 21/4

⇒ -k/2 = 21/4 - 25/4 = -4/4 = -1

⇒ k/2 = 1 ⇒ k = 2

6 Find all zeroes of the polynomial x⁴ - 3x³ - x² + 9x - 6 if two of its zeroes are √3 and -√3.

Since √3 and -√3 are zeroes, (x - √3)(x + √3) = x² - 3 is a factor.

Divide the polynomial by x² - 3:

        x² - 3x + 2
x² - 3 | x⁴ - 3x³ - x² + 9x - 6
        -(x⁴ + 0x³ - 3x²)
        ---------------
             -3x³ + 2x² + 9x
             -(-3x³ + 0x² + 9x)
             ------------------
                  2x² + 0x - 6
                  -(2x² + 0x - 6)
                  ---------------
                        0
        

Now solve x² - 3x + 2 = 0

⇒ (x - 1)(x - 2) = 0 ⇒ x = 1 or x = 2

So all zeroes are: √3, -√3, 1, 2

Practice Questions

1. Find the zeroes of the polynomial x² - 3 and verify the relationship with coefficients.
2. Form a quadratic polynomial whose zeroes are 2 + √3 and 2 - √3.
3. Divide the polynomial 3x⁴ - 4x³ - 3x - 1 by x - 1 and find quotient and remainder.
4. If α and β are zeroes of x² - 4x + 1, find the value of α³ + β³.
5. Find all zeroes of the polynomial 2x⁴ - 9x³ + 5x² + 3x - 1 if two of its zeroes are 2 + √3 and 2 - √3.

Statistics

Solved Examples

1 Find the mean of the following distribution:

x:	5	15	25	35	45
f:	3	5	8	3	1

Creating the table:

x	f	f×x
5	3	15
15	5	75
25	8	200
35	3	105
45	1	45
Total	20	440

Mean = Σ(f×x)/Σf = 440/20 = 22

2 Find the mode of the following data: 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18

Arranging data in ascending order: 14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28

14 appears 4 times (maximum frequency)

So mode = 14

3 Find the median of the following data: 31, 38, 27, 28, 36, 25, 35, 40

Arranging in ascending order: 25, 27, 28, 31, 35, 36, 38, 40

Number of observations (n) = 8 (even)

Median = average of (n/2)th and (n/2 + 1)th observations

= average of 4th and 5th observations = (31 + 35)/2 = 66/2 = 33

4 The following table shows the ages of patients admitted to a hospital. Find the mode.

Age (years)	5-15	15-25	25-35	35-45	45-55	55-65
No. of patients	6	11	21	23	14	5

Modal class is the class with highest frequency: 35-45

l = 35, f₁ = 23, f₀ = 21, f₂ = 14, h = 10

Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h

= 35 + [(23 - 21)/(2×23 - 21 - 14)] × 10

= 35 + [2/(46 - 35)] × 10

= 35 + (2/11) × 10 ≈ 35 + 1.82 = 36.82

5 Find the mean of the following frequency distribution using step deviation method:

Class	0-10	10-20	20-30	30-40	40-50
Frequency	12	18	27	20	17

Let assumed mean A = 25, class width h = 10

Class	f	Midpoint (x)	d = (x - A)/h	f×d
0-10	12	5	-2	-24
10-20	18	15	-1	-18
20-30	27	25	0	0
30-40	20	35	1	20
40-50	17	45	2	34
Total	94			12

Mean = A + (Σfd/Σf) × h = 25 + (12/94) × 10 ≈ 25 + 1.28 = 26.28

6 The median of the following data is 28.5. Find the values of x and y if the total frequency is 60.

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

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

Median is 28.5, which lies in class 20-30

Median class: 20-30, l = 20, f = 20, cf = 5 + x, h = 10

Median = l + [(n/2 - cf)/f] × h

28.5 = 20 + [(30 - (5 + x))/20] × 10

8.5 = [(25 - x)/20] × 10

8.5 = (25 - x)/2

17 = 25 - x ⇒ x = 8

Then y = 15 - x = 15 - 8 = 7

Practice Questions

1. Find the mean, median and mode of: 5, 7, 10, 12, 7, 12, 8, 13, 7
2. Calculate the mean of the following distribution:

   CI	0-10	10-20	20-30	30-40	40-50
   f	5	8	15	16	6

3. Find the mode of:

   CI	0-20	20-40	40-60	60-80	80-100
   f	5	10	12	6	3

4. If the median of the following distribution is 32.5, find the missing frequencies:

   CI	0-10	10-20	20-30	30-40	40-50	50-60	60-70
   f	5	9	15	?	?	14	7

   Total frequency = 80.
5. The mean of 20 numbers is 18. If 2 is added to every number, what will be the new mean?

Triangles

Solved Examples

1 In ΔABC, if DE ∥ BC, AD = 3 cm, DB = 4 cm, and AE = 6 cm, find EC.

By Basic Proportionality Theorem (Thales theorem):

AD/DB = AE/EC

3/4 = 6/EC

EC = (6 × 4)/3 = 24/3 = 8 cm

2 In ΔABC, ∠B = 90°, BD ⊥ AC. If AD = 4 cm and CD = 9 cm, find BD.

In right triangle, by altitude rule:

BD² = AD × DC

BD² = 4 × 9 = 36

BD = √36 = 6 cm

3 Prove that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Let ΔABC ~ ΔPQR

Then AB/PQ = BC/QR = AC/PR = k (scale factor)

Draw altitudes AD and PS from A and P to BC and QR respectively

Since triangles are similar, ∠B = ∠Q and AD/PS = AB/PQ = k

Area(ΔABC) = 1/2 × BC × AD

Area(ΔPQR) = 1/2 × QR × PS

Area ratio = [1/2 × BC × AD] / [1/2 × QR × PS] = (BC/QR) × (AD/PS) = k × k = k²

But k = AB/PQ, so area ratio = (AB/PQ)²

Similarly, it equals (BC/QR)² and (AC/PR)²

4 In ΔABC, D and E are points on AB and AC such that DE ∥ BC. If AD = 4x - 3, AE = 8x - 7, BD = 3x - 1, and CE = 5x - 3, find the value of x.

By Basic Proportionality Theorem:

AD/BD = AE/CE

(4x - 3)/(3x - 1) = (8x - 7)/(5x - 3)

Cross multiply: (4x - 3)(5x - 3) = (3x - 1)(8x - 7)

20x² - 12x - 15x + 9 = 24x² - 21x - 8x + 7

20x² - 27x + 9 = 24x² - 29x + 7

0 = 4x² - 2x - 2

Divide by 2: 2x² - x - 1 = 0

(2x + 1)(x - 1) = 0

x = 1 or x = -1/2 (not possible as lengths can't be negative)

So x = 1

5 In ΔABC, ∠B = 90°, AB = 6 cm, BC = 8 cm. Find the length of AC.

By Pythagoras theorem:

AC² = AB² + BC²

AC² = 6² + 8² = 36 + 64 = 100

AC = √100 = 10 cm

6 In ΔPQR, S and T are points on sides PR and QR respectively such that ST ∥ PQ. If PT divides QR in the ratio 2:3 and QS = 6 cm, find SR.

Since ST ∥ PQ, by Basic Proportionality Theorem:

RS/SQ = RT/TQ

Given that PT divides QR in ratio 2:3, so RT:TQ = 2:3

Thus RS/SQ = 2/3

Given QS = 6 cm, so SQ = 6 cm

RS/6 = 2/3

RS = (2/3) × 6 = 4 cm

Practice Questions

1. In ΔABC, DE ∥ BC. If AD = 2 cm, DB = 3 cm, and AE = 3 cm, find AC.
2. In right ΔABC, ∠B = 90°, BD ⊥ AC. If AB = 6 cm, BC = 8 cm, find BD.
3. Prove that in a right triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.
4. D and E are points on sides AB and AC of ΔABC such that DE ∥ BC. If AD = 4 cm, AB = 12 cm, and AC = 18 cm, find AE.
5. The areas of two similar triangles are 81 cm² and 144 cm². If the largest side of the smaller triangle is 27 cm, find the largest side of the larger triangle.

Trigonometry

Solved Examples

1 Evaluate: sin 30° + cos 60° - tan 45°

sin 30° = 1/2

cos 60° = 1/2

tan 45° = 1

So, 1/2 + 1/2 - 1 = 1 - 1 = 0

2 If sin A = 3/5, find the value of cos A and tan A.

sin A = opposite/hypotenuse = 3/5

Let opposite = 3k, hypotenuse = 5k

Then adjacent = √[(hypotenuse)² - (opposite)²] = √(25k² - 9k²) = √(16k²) = 4k

cos A = adjacent/hypotenuse = 4k/5k = 4/5

tan A = opposite/adjacent = 3k/4k = 3/4

3 Prove that: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan² θ + cot² θ

LHS = (sin θ + cosec θ)² + (cos θ + sec θ)²

= sin²θ + 2sinθcosecθ + cosec²θ + cos²θ + 2cosθsecθ + sec²θ

= (sin²θ + cos²θ) + 2(1) + 2(1) + (cosec²θ + sec²θ) [since sinθcosecθ = 1, cosθsecθ = 1]

= 1 + 2 + 2 + (1 + cot²θ + 1 + tan²θ) [since cosec²θ = 1 + cot²θ, sec²θ = 1 + tan²θ]

= 5 + 2 + tan²θ + cot²θ

= 7 + tan²θ + cot²θ = RHS

4 If 7 sin²θ + 3 cos²θ = 4, find tan θ.

7 sin²θ + 3 cos²θ = 4

Divide both sides by cos²θ:

7 tan²θ + 3 = 4 sec²θ

But sec²θ = 1 + tan²θ

So 7 tan²θ + 3 = 4(1 + tan²θ)

7 tan²θ + 3 = 4 + 4 tan²θ

7 tan²θ - 4 tan²θ = 4 - 3

3 tan²θ = 1

tan²θ = 1/3

tan θ = 1/√3

5 A ladder 15 m long leans against a wall making an angle of 60° with the ground. How high up the wall does the ladder reach?

Let AB be the wall, AC be the ladder

AC = 15 m, ∠ACB = 60°

sin 60° = AB/AC

√3/2 = AB/15

AB = 15 × √3/2 = (15√3)/2 ≈ 12.99 m

6 Prove that: (1 + cot θ - cosec θ)(1 + tan θ + sec θ) = 2

LHS = (1 + cot θ - cosec θ)(1 + tan θ + sec θ)

= [1 + cosθ/sinθ - 1/sinθ] [1 + sinθ/cosθ + 1/cosθ]

= [(sinθ + cosθ - 1)/sinθ] [(cosθ + sinθ + 1)/cosθ]

= [(sinθ + cosθ)² - 1²] / (sinθ cosθ)

= [sin²θ + 2sinθcosθ + cos²θ - 1] / (sinθ cosθ)

= [1 + 2sinθcosθ - 1] / (sinθ cosθ)

= (2sinθcosθ)/(sinθcosθ) = 2 = RHS

Practice Questions

1. Evaluate: tan² 60° + 2 cos² 45° - sec² 30°
2. If 3 tan θ = 4, find the value of (5 sin θ - 3 cos θ)/(5 sin θ + 3 cos θ)
3. Prove that: (sin A + cos A)(1 - sin A cos A) = sin³ A + cos³ A
4. The angle of elevation of the top of a tower from a point 20 m away from its base is 45°. Find the height of the tower.
5. Prove that: (cosec θ - sin θ)(sec θ - cos θ) = 1/(tan θ + cot θ)

Area Related to Circles

Solved Examples

1 Find the area of a circle whose circumference is 44 cm.

Circumference = 2πr = 44 cm

2 × (22/7) × r = 44

r = (44 × 7)/(2 × 22) = (308)/(44) = 7 cm

Area = πr² = (22/7) × 7 × 7 = 22 × 7 = 154 cm²

2 The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Area of first circle = π(8)² = 64π cm²

Area of second circle = π(6)² = 36π cm²

Sum of areas = 64π + 36π = 100π cm²

Let radius of new circle be R

Then πR² = 100π

R² = 100 ⇒ R = 10 cm

3 Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Area of sector = (θ/360°) × πr²

= (60/360) × (22/7) × 6 × 6

= (1/6) × (22/7) × 36

= (22/7) × 6 = 132/7 ≈ 18.857 cm²

4 The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km per hour?

Diameter = 80 cm, so radius = 40 cm

Circumference = 2πr = 2 × (22/7) × 40 = (1760/7) cm

Speed = 66 km/h = 66 × 1000 × 100 cm/h = 6,600,000 cm/h

Distance in 10 minutes = (10/60) × 6,600,000 = 1,100,000 cm

Number of revolutions = Total distance / Circumference

= 1,100,000 ÷ (1760/7) = 1,100,000 × (7/1760)

= (1,100,000/1760) × 7 = 625 × 7 = 4375

5 Find the area of the shaded region where ABCD is a square of side 14 cm and four semicircles are drawn with each side of the square as diameter.

Area of square = 14 × 14 = 196 cm²

Area of four semicircles = area of two circles with diameter 14 cm each

Radius of each circle = 7 cm

Area of two circles = 2 × π × 7² = 2 × (22/7) × 49 = 2 × 22 × 7 = 308 cm²

Area of shaded region = Area of square + Area of two circles - Area of four semicircles

Wait, actually the four semicircles are drawn outward from the square.

The shaded region is the area inside the square but outside the semicircles.

Area of one semicircle = (1/2) × π × 7² = (1/2) × (22/7) × 49 = 77 cm²

Area of four semicircles = 4 × 77 = 308 cm²

But these semicircles overlap in the square, so the shaded area is:

Area of square - area of the four unshaded regions

This problem requires more detailed solution with diagram.

6 In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle. Find the area of the design.

Area of design = Area of circle - Area of equilateral triangle

Area of circle = πr² = (22/7) × 32 × 32 = (22528/7) cm²

For equilateral triangle inscribed in circle of radius R:

Side of triangle = √3 R = √3 × 32 cm

Area of equilateral triangle = (√3/4) × side² = (√3/4) × (32√3)²

= (√3/4) × 1024 × 3 = (√3/4) × 3072 = 768√3 cm²

Area of design = (22528/7) - 768√3 cm²

Practice Questions

1. The circumference of a circle is 88 cm. Find its area.
2. Find the area of a quadrant of a circle whose circumference is 44 cm.
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
4. Find the area of the shaded region where a square of side 4 cm is inscribed in a circle.
5. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find the area of that part of the field in which the horse can graze.

Arithmetic Progressions

Solved Examples

1 Find the 10th term of the AP: 2, 7, 12, 17, ...

First term a = 2

Common difference d = 7 - 2 = 5

nth term aₙ = a + (n - 1)d

10th term a₁₀ = 2 + (10 - 1)×5 = 2 + 9×5 = 2 + 45 = 47

2 How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?

First term a = 9

Common difference d = 17 - 9 = 8

Sum Sₙ = n/2 [2a + (n - 1)d] = 636

n/2 [2×9 + (n - 1)×8] = 636

n/2 [18 + 8n - 8] = 636

n/2 [8n + 10] = 636

n(4n + 5) = 636

4n² + 5n - 636 = 0

Using quadratic formula: n = [-5 ± √(25 + 4×4×636)]/(2×4)

= [-5 ± √(25 + 10176)]/8 = [-5 ± √10201]/8 = [-5 ± 101]/8

Taking positive value: n = (96)/8 = 12

So 12 terms must be taken.

3 If the sum of first 7 terms of an AP is 49 and that of first 17 terms is 289, find the sum of first n terms.

S₇ = 7/2 [2a + 6d] = 49

⇒ 7/2 × 2(a + 3d) = 49 ⇒ 7(a + 3d) = 49 ⇒ a + 3d = 7 ...(1)

S₁₇ = 17/2 [2a + 16d] = 289

⇒ 17/2 × 2(a + 8d) = 289 ⇒ 17(a + 8d) = 289 ⇒ a + 8d = 17 ...(2)

Subtracting (1) from (2): 5d = 10 ⇒ d = 2

From (1): a + 3×2 = 7 ⇒ a + 6 = 7 ⇒ a = 1

Sum of first n terms Sₙ = n/2 [2×1 + (n - 1)×2]

= n/2 [2 + 2n - 2] = n/2 × 2n = n²

4 Find the sum of first 24 terms of the list of numbers whose nth term is given by aₙ = 3 + 2n.

aₙ = 3 + 2n

First term a₁ = 3 + 2×1 = 5

Second term a₂ = 3 + 2×2 = 7

Common difference d = 7 - 5 = 2

Sum S₂₄ = 24/2 [2×5 + (24 - 1)×2]

= 12 [10 + 23×2] = 12 [10 + 46] = 12 × 56 = 672

5 The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

a = 5, l = 45, Sₙ = 400

Sₙ = n/2 (a + l) ⇒ 400 = n/2 (5 + 45)

400 = n/2 × 50 ⇒ 400 = 25n ⇒ n = 16

l = a + (n - 1)d ⇒ 45 = 5 + (16 - 1)d

45 = 5 + 15d ⇒ 40 = 15d ⇒ d = 40/15 = 8/3

6 If the pth term of an AP is q and the qth term is p, prove that its nth term is (p + q - n).

Given: aₚ = a + (p - 1)d = q ...(1)

a_q = a + (q - 1)d = p ...(2)

Subtracting (2) from (1):

[a + (p - 1)d] - [a + (q - 1)d] = q - p

(p - 1 - q + 1)d = q - p

(p - q)d = q - p

d(p - q) = -(p - q) ⇒ d = -1 (since p ≠ q)

From (1): a + (p - 1)(-1) = q ⇒ a - p + 1 = q ⇒ a = p + q - 1

nth term aₙ = a + (n - 1)d = (p + q - 1) + (n - 1)(-1)

= p + q - 1 - n + 1 = p + q - n

Practice Questions

1. Find the 15th term of the AP: 6, 9, 12, 15, ...
2. Find the sum of first 40 positive integers divisible by 6.
3. If the sum of first n terms of an AP is given by Sₙ = 3n² + 2n, find the AP and its 15th term.
4. How many terms of the AP: 24, 21, 18, ... must be taken so that their sum is 78?
5. The sum of first 6 terms of an AP is 42 and the ratio of 10th term to 30th term is 1:3. Find the first term and common difference.

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