Progressions – Solutions

zaheerpashamd

Progressions - 1 Mark Problems

Important Formulas:

Arithmetic Progression (AP):
• nth term: aₙ = a + (n-1)d
• Sum of first n terms: Sₙ = n/2 [2a + (n-1)d]
• Common difference: d = aₙ - aₙ₋₁

Geometric Progression (GP):
• nth term: aₙ = a · rⁿ⁻¹
• Common ratio: r = aₙ / aₙ₋₁
• Sum of first n terms: Sₙ = a(1-rⁿ)/(1-r) for r≠1

The hand-bore well dealer charges Rs 200/- for the first one meter only and raises drilling charges at the rate of rupees 30/- for every subsequent meter. Write a progression for the above data. (M'15)

Step 1: Identify the pattern

First meter: Rs 200

Second meter: Rs 200 + 30 = Rs 230

Third meter: Rs 230 + 30 = Rs 260

And so on...

∴ The progression is: 200, 230, 260, 290, ...

In a flower garden, there are 23 plants in the first row, 21 plants in the second row, 19 plants in the third row and so on. If there are 10 rows in that flower garden, then find the total number of plants in the last row with the help of the formula tₙ = a + (n – 1)d. (M'15)

Step 1: Identify values

First term (a) = 23

Common difference (d) = 21 - 23 = -2

Number of terms (n) = 10

Step 2: Apply the formula

tₙ = a + (n - 1)d

t₁₀ = 23 + (10 - 1)(-2)

t₁₀ = 23 + 9(-2)

t₁₀ = 23 - 18 = 5

∴ There are 5 plants in the last row.

Write the common difference of an Arithmetic Progression, whose nth term is given by tₙ = 3n + 7. (J'15)

Step 1: Find first term

t₁ = 3(1) + 7 = 3 + 7 = 10

Step 2: Find second term

t₂ = 3(2) + 7 = 6 + 7 = 13

Step 3: Calculate common difference

d = t₂ - t₁ = 13 - 10 = 3

∴ The common difference is 3.

Find the sum of first 200 natural numbers. (M'16)

Step 1: Use the formula for sum of first n natural numbers

Sₙ = n(n+1)/2

Step 2: Substitute n = 200

S₂₀₀ = 200(200+1)/2

S₂₀₀ = 200 × 201 / 2

S₂₀₀ = 100 × 201

S₂₀₀ = 20,100

∴ The sum of first 200 natural numbers is 20,100.

Is 'zero' a term of the Arithmetic Progression 31, 28, 25, ...? Justify your answer. (J'16)

Step 1: Identify values

First term (a) = 31

Common difference (d) = 28 - 31 = -3

Step 2: Check if 0 is a term

tₙ = a + (n-1)d

0 = 31 + (n-1)(-3)

0 = 31 - 3(n-1)

3(n-1) = 31

n-1 = 31/3 ≈ 10.33

n = 11.33

Step 3: Interpret the result

Since n is not a natural number (n = 11.33), 0 is not a term of this AP.

∴ No, zero is not a term of this AP since n is not a natural number.

In a G.P. tₙ = (-1)ⁿ·2017. Find the common ratio. (M'17)

Step 1: Find first term

t₁ = (-1)¹·2017 = -2017

Step 2: Find second term

t₂ = (-1)²·2017 = 2017

Step 3: Calculate common ratio

r = t₂ / t₁ = 2017 / (-2017) = -1

∴ The common ratio is -1.

The nth term of AP is 6n + 2. Find the common difference (n∈ N). (J'17)

Step 1: Find first term

t₁ = 6(1) + 2 = 6 + 2 = 8

Step 2: Find second term

t₂ = 6(2) + 2 = 12 + 2 = 14

Step 3: Calculate common difference

d = t₂ - t₁ = 14 - 8 = 6

∴ The common difference is 6.

The sequence √3, √6, √9, √12, ... form an Arithmetic Progression? Give reason. (M'18)

Step 1: Check for common difference

First term = √3

Second term = √6

Third term = √9 = 3

Fourth term = √12 = 2√3

Step 2: Calculate differences

d₁ = √6 - √3

d₂ = 3 - √6

d₃ = 2√3 - 3

Step 3: Compare differences

√6 - √3 ≈ 2.449 - 1.732 = 0.717

3 - √6 ≈ 3 - 2.449 = 0.551

Since 0.717 ≠ 0.551, the differences are not equal.

∴ No, it does not form an AP because the common difference is not constant.

For the A.P.: -3, -7, -11, ... ; can we find directly a₃₀ - a₂₀ without actually finding a₃₀ and a₂₀? (J'18)

Step 1: Use the formula for nth term

aₙ = a + (n-1)d

a₃₀ = a + 29d

a₂₀ = a + 19d

Step 2: Find the difference

a₃₀ - a₂₀ = (a + 29d) - (a + 19d) = 10d

Step 3: Find common difference

d = -7 - (-3) = -4

Step 4: Calculate a₃₀ - a₂₀

a₃₀ - a₂₀ = 10d = 10(-4) = -40

∴ Yes, we can find directly: a₃₀ - a₂₀ = 10d = -40.

Radha says "1, 1, 1, ... are in A.P. and also in G.P". Do you agree with Radha? Give reason. (M'19)

Step 1: Check for AP

Common difference d = 1 - 1 = 0

Since d is constant, it is an AP.

Step 2: Check for GP

Common ratio r = 1/1 = 1

Since r is constant, it is a GP.

∴ Yes, I agree with Radha. The sequence 1, 1, 1, ... is both an AP (with d=0) and a GP (with r=1).

Find the 11th term of the A.P.: 15, 12, 9, ... (J'19)

Step 1: Identify values

First term (a) = 15

Common difference (d) = 12 - 15 = -3

n = 11

Step 2: Apply the formula

tₙ = a + (n-1)d

t₁₁ = 15 + (11-1)(-3)

t₁₁ = 15 + 10(-3)

t₁₁ = 15 - 30 = -15

∴ The 11th term is -15.

Find the 10th term of the arithmetic progression 3, 5, 7, ... (May 2022)

Step 1: Identify values

First term (a) = 3

Common difference (d) = 5 - 3 = 2

n = 10

Step 2: Apply the formula

tₙ = a + (n-1)d

t₁₀ = 3 + (10-1)(2)

t₁₀ = 3 + 9(2)

t₁₀ = 3 + 18 = 21

∴ The 10th term is 21.

Find the 10th term of an arithmetic progression 5, 9, 13, ... (Additional Problem)

Step 1: Identify values

First term (a) = 5

Common difference (d) = 9 - 5 = 4

n = 10

Step 2: Apply the formula

tₙ = a + (n-1)d

t₁₀ = 5 + (10-1)(4)

t₁₀ = 5 + 9(4)

t₁₀ = 5 + 36 = 41

∴ The 10th term is 41.

Progressions - 2 Mark Problems

Important Formulas:

If 7 times of 7th term of an Arithmetic Progression is equal to the 11 times of 11th term of it, then find the 18th term of that Arithmetic Progression. (J'15)

Step 1: Write the given condition

7 × a₇ = 11 × a₁₁

Step 2: Express terms using AP formula

a₇ = a + 6d

a₁₁ = a + 10d

7(a + 6d) = 11(a + 10d)

Step 3: Solve for a in terms of d

7a + 42d = 11a + 110d

7a - 11a = 110d - 42d

-4a = 68d

a = -17d

Step 4: Find the 18th term

a₁₈ = a + 17d = -17d + 17d = 0

∴ The 18th term of the AP is 0.

Measures of sides of a triangle are in Arithmetic Progression. Its perimeter is 30 cm, the difference between the longest and shortest side is 4 cm; then find the measures of the sides. (M'16)

Step 1: Let the sides be in AP

Let the sides be: a-d, a, a+d

(Where a is the middle term and d is the common difference)

Step 2: Apply perimeter condition

Perimeter = (a-d) + a + (a+d) = 30

3a = 30

a = 10

Step 3: Apply difference condition

Longest side - Shortest side = 4

(a+d) - (a-d) = 4

2d = 4

d = 2

Step 4: Find the sides

First side = a-d = 10-2 = 8 cm

Second side = a = 10 cm

Third side = a+d = 10+2 = 12 cm

∴ The sides of the triangle are 8 cm, 10 cm, and 12 cm.

Explain the terms in the formula Sₙ = n/2 [2a + (n − 1)d]. (J'16)

Step 1: Write the formula

Sₙ = n/2 [2a + (n − 1)d]

Step 2: Explain each term

Sₙ = Sum of first n terms of the AP

n = Number of terms

a = First term of the AP

d = Common difference of the AP

Step 3: Explain the expression inside brackets

2a + (n-1)d = Sum of first and last terms

(Since last term = a + (n-1)d)

∴ Sₙ represents the sum of first n terms, n is number of terms, a is first term, and d is common difference.

Find the sum of the first 10 terms of an A.P. 3, 15, 27, 39, .............. (M'17)

Step 1: Identify values

First term (a) = 3

Common difference (d) = 15 - 3 = 12

Number of terms (n) = 10

Step 2: Apply sum formula

Sₙ = n/2 [2a + (n-1)d]

S₁₀ = 10/2 [2(3) + (10-1)(12)]

S₁₀ = 5 [6 + 9(12)]

S₁₀ = 5 [6 + 108]

S₁₀ = 5 × 114 = 570

∴ The sum of first 10 terms is 570.

Find the value of 'k' so that k + 2, 4k – 6 and 3k – 2 are the three consecutive terms of an A.P. (J'17)

Step 1: Condition for three consecutive terms in AP

If a, b, c are in AP, then 2b = a + c

Step 2: Apply the condition

2(4k - 6) = (k + 2) + (3k - 2)

8k - 12 = k + 2 + 3k - 2

8k - 12 = 4k

Step 3: Solve for k

8k - 4k = 12

4k = 12

k = 3

∴ The value of k is 3.

Find the 7th term from the end of the arithmetic progression 7, 10, 13, ..... 184. (M'18)

Step 1: Find the total number of terms

First term (a) = 7

Common difference (d) = 10 - 7 = 3

Last term (l) = 184

l = a + (n-1)d

184 = 7 + (n-1)(3)

184 - 7 = 3(n-1)

177 = 3(n-1)

n-1 = 59

n = 60

Step 2: Find the 7th term from the end

7th term from end = (n-7+1)th term from beginning

= (60-7+1)th term = 54th term

Step 3: Calculate the 54th term

a₅₄ = a + (54-1)d

a₅₄ = 7 + 53(3)

a₅₄ = 7 + 159 = 166

∴ The 7th term from the end is 166.

In a rangoli design of 13 rows, every row increases its previous row by two dots and first row contains 5 dots, then how many total dots are in the design? (J'18)

Step 1: Identify the AP

First term (a) = 5

Common difference (d) = 2

Number of terms (n) = 13

Step 2: Apply sum formula

Sₙ = n/2 [2a + (n-1)d]

S₁₃ = 13/2 [2(5) + (13-1)(2)]

S₁₃ = 13/2 [10 + 12(2)]

S₁₃ = 13/2 [10 + 24]

S₁₃ = 13/2 × 34

S₁₃ = 13 × 17 = 221

∴ The total number of dots in the design is 221.

Write the formula of nth term of G.P. and explain the terms in it. (M'19)

Step 1: Write the formula

aₙ = a · rⁿ⁻¹

Step 2: Explain each term

aₙ = nth term of the GP

a = First term of the GP

r = Common ratio of the GP

n = Position of the term

Step 3: Additional explanation

The formula shows that each term is obtained by multiplying the previous term by the common ratio r.

∴ aₙ = a·rⁿ⁻¹, where a is first term, r is common ratio, and n is term position.

Which term of the G.P.: √2, 2, 2√2, 4, ... is 32? (J'19)

Step 1: Identify values

First term (a) = √2

Common ratio (r) = 2/√2 = √2

nth term (aₙ) = 32

Step 2: Apply GP formula

aₙ = a · rⁿ⁻¹

32 = √2 · (√2)ⁿ⁻¹

32 = (√2)ⁿ

Step 3: Solve for n

32 = 2⁵

(√2)ⁿ = 2⁵

(2¹ᐟ²)ⁿ = 2⁵

2ⁿᐟ² = 2⁵

n/2 = 5

n = 10

∴ 32 is the 10th term of the GP.

5, 8, 11, 14, ........... is an arithmetic progression. Find the sum of first 20 terms of it. (May 22)

Step 1: Identify values

First term (a) = 5

Common difference (d) = 8 - 5 = 3

Number of terms (n) = 20

Step 2: Apply sum formula

Sₙ = n/2 [2a + (n-1)d]

S₂₀ = 20/2 [2(5) + (20-1)(3)]

S₂₀ = 10 [10 + 19(3)]

S₂₀ = 10 [10 + 57]

S₂₀ = 10 × 67 = 670

∴ The sum of first 20 terms is 670.

3, 6, 9, 12, ............ is an arithmetic progression. Find the sum of first 20 terms of the progression. (Aug 22)

Step 1: Identify values

First term (a) = 3

Common difference (d) = 6 - 3 = 3

Number of terms (n) = 20

Step 2: Apply sum formula

Sₙ = n/2 [2a + (n-1)d]

S₂₀ = 20/2 [2(3) + (20-1)(3)]

S₂₀ = 10 [6 + 19(3)]

S₂₀ = 10 [6 + 57]

S₂₀ = 10 × 63 = 630

∴ The sum of first 20 terms is 630.

Which term of the A.P. 21, 18, 15, ... is –81? Also find the term which becomes zero. (Apr'23)

Step 1: Identify values

First term (a) = 21

Common difference (d) = 18 - 21 = -3

Step 2: Find n when aₙ = -81

aₙ = a + (n-1)d

-81 = 21 + (n-1)(-3)

-81 - 21 = (n-1)(-3)

-102 = (n-1)(-3)

n-1 = 102/3 = 34

n = 35

Step 3: Find n when aₙ = 0

0 = 21 + (n-1)(-3)

-21 = (n-1)(-3)

n-1 = 21/3 = 7

n = 8

∴ -81 is the 35th term and 0 is the 8th term.

If 6 times of 6th term of an arithmetic progression is equal to 9 times of 9th term of it, then show that 15th term of that A.P. is zero. (Jun'23)

Step 1: Write the given condition

6 × a₆ = 9 × a₉

Step 2: Express terms using AP formula

a₆ = a + 5d

a₉ = a + 8d

6(a + 5d) = 9(a + 8d)

Step 3: Solve for a in terms of d

6a + 30d = 9a + 72d

6a - 9a = 72d - 30d

-3a = 42d

a = -14d

Step 4: Find the 15th term

a₁₅ = a + 14d = -14d + 14d = 0

∴ The 15th term is 0, as required to show.

Progressions - 4 Mark Problems

Important Formulas:

If the sum of the first 7 terms of an Arithmetic Progression is 49 and that of first 17 terms is 289, then find the sum of first 'n' terms. (M'15)

Step 1: Write the given conditions

S₇ = 7/2 [2a + (7-1)d] = 49

7/2 [2a + 6d] = 49

7(a + 3d) = 49

a + 3d = 7 ...(1)

Step 2: Use second condition

S₁₇ = 17/2 [2a + (17-1)d] = 289

17/2 [2a + 16d] = 289

17(a + 8d) = 289

a + 8d = 17 ...(2)

Step 3: Solve equations (1) and (2)

Subtract (1) from (2):

(a + 8d) - (a + 3d) = 17 - 7

5d = 10

d = 2

Substitute in (1): a + 3(2) = 7

a + 6 = 7

a = 1

Step 4: Find sum of first n terms

Sₙ = n/2 [2a + (n-1)d]

Sₙ = n/2 [2(1) + (n-1)(2)]

Sₙ = n/2 [2 + 2n - 2]

Sₙ = n/2 [2n]

Sₙ = n²

∴ The sum of first n terms is n².

A manufacturer of TV sets produced 500 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the production of TV sets in the 15th year and the total production of TV sets in the first 10 years. (J'15)

Step 1: Set up the AP

Let production in 1st year = a

Common difference = d

Production in 3rd year: a + 2d = 500 ...(1)

Production in 7th year: a + 6d = 700 ...(2)

Step 2: Solve for a and d

Subtract (1) from (2):

(a + 6d) - (a + 2d) = 700 - 500

4d = 200

d = 50

Substitute in (1): a + 2(50) = 500

a + 100 = 500

a = 400

Step 3: Find production in 15th year

a₁₅ = a + 14d = 400 + 14(50)

a₁₅ = 400 + 700 = 1100

Step 4: Find total production in first 10 years

S₁₀ = 10/2 [2a + (10-1)d]

S₁₀ = 5 [2(400) + 9(50)]

S₁₀ = 5 [800 + 450]

S₁₀ = 5 × 1250 = 6250

∴ Production in 15th year = 1100 sets, Total production in first 10 years = 6250 sets.

Find the sum of all the three digit numbers which are divisible by 4. (M'16)

Step 1: Identify the sequence

Three-digit numbers divisible by 4: 100, 104, 108, ..., 996

This is an AP with a = 100, d = 4

Step 2: Find number of terms

aₙ = a + (n-1)d

996 = 100 + (n-1)(4)

996 - 100 = 4(n-1)

896 = 4(n-1)

n-1 = 224

n = 225

Step 3: Find the sum

Sₙ = n/2 [a + l]

S₂₂₅ = 225/2 [100 + 996]

S₂₂₅ = 225/2 × 1096

S₂₂₅ = 225 × 548

S₂₂₅ = 123,300

∴ The sum of all three-digit numbers divisible by 4 is 123,300.

The sum of the three terms which are in an arithmetic progression is 33. If the product of the first and the third terms exceeds the second term by 29, find the Arithmetic Progression. (J'16)

Step 1: Let the three terms be

a-d, a, a+d

Step 2: Apply sum condition

(a-d) + a + (a+d) = 33

3a = 33

a = 11

Step 3: Apply product condition

(a-d)(a+d) = a + 29

a² - d² = a + 29

11² - d² = 11 + 29

121 - d² = 40

d² = 121 - 40 = 81

d = ±9

Step 4: Find the AP

For d = 9: Terms are 11-9, 11, 11+9 = 2, 11, 20

For d = -9: Terms are 11-(-9), 11, 11+(-9) = 20, 11, 2

Both give the same AP: 2, 11, 20

∴ The Arithmetic Progression is 2, 11, 20, ...

Find the sum of all three-digit natural numbers, which are divisible by 3 and not divisible by 6. (M'17)

Step 1: Numbers divisible by 3 but not by 6

These are numbers divisible by 3 but not by 2 (odd multiples of 3)

Step 2: Find the sequence

Three-digit numbers divisible by 3: 102, 105, 108, ..., 999

Odd multiples of 3: 105, 111, 117, ..., 999

This is an AP with a = 105, d = 6

Step 3: Find number of terms

aₙ = a + (n-1)d

999 = 105 + (n-1)(6)

999 - 105 = 6(n-1)

894 = 6(n-1)

n-1 = 149

n = 150

Step 4: Find the sum

Sₙ = n/2 [a + l]

S₁₅₀ = 150/2 [105 + 999]

S₁₅₀ = 75 × 1104

S₁₅₀ = 82,800

∴ The sum of all three-digit numbers divisible by 3 but not by 6 is 82,800.

The sum of 5th and 9th terms of A.P. is 72 and the sum of 7th and 12th terms is 97. Find the A.P. (J'17)

Step 1: Write the given conditions

a₅ + a₉ = 72

(a + 4d) + (a + 8d) = 72

2a + 12d = 72

a + 6d = 36 ...(1)

Step 2: Second condition

a₇ + a₁₂ = 97

(a + 6d) + (a + 11d) = 97

2a + 17d = 97 ...(2)

Step 3: Solve equations (1) and (2)

From (1): a = 36 - 6d

Substitute in (2): 2(36 - 6d) + 17d = 97

72 - 12d + 17d = 97

72 + 5d = 97

5d = 25

d = 5

a = 36 - 6(5) = 36 - 30 = 6

Step 4: Write the AP

AP: a, a+d, a+2d, ...

AP: 6, 11, 16, 21, ...

∴ The Arithmetic Progression is 6, 11, 16, 21, ...

Which term of G.P.: 3, 9, 27, ... is 2187? (J'17)

Step 1: Identify values

First term (a) = 3

Common ratio (r) = 9/3 = 3

nth term (aₙ) = 2187

Step 2: Apply GP formula

aₙ = a · rⁿ⁻¹

2187 = 3 · 3ⁿ⁻¹

2187 = 3ⁿ

Step 3: Solve for n

3ⁿ = 2187

3ⁿ = 3⁷ (since 3⁷ = 2187)

n = 7

∴ 2187 is the 7th term of the GP.

Find the sum of all two-digit odd positive integers which are divisible by 3 but not by 2. (M'18)

Step 1: Understand the condition

Numbers divisible by 3 but not by 2 are odd multiples of 3

Step 2: Find the sequence

Two-digit odd multiples of 3: 15, 21, 27, ..., 99

This is an AP with a = 15, d = 6

Step 3: Find number of terms

aₙ = a + (n-1)d

99 = 15 + (n-1)(6)

99 - 15 = 6(n-1)

84 = 6(n-1)

n-1 = 14

n = 15

Step 4: Find the sum

Sₙ = n/2 [a + l]

S₁₅ = 15/2 [15 + 99]

S₁₅ = 15/2 × 114

S₁₅ = 15 × 57 = 855

∴ The sum of all two-digit odd positive integers divisible by 3 is 855.

Find the sum of the integers between 100 and 500 that are divisible by 9. (J'18)

Step 1: Identify the sequence

Numbers between 100 and 500 divisible by 9: 108, 117, 126, ..., 495

This is an AP with a = 108, d = 9

Step 2: Find number of terms

aₙ = a + (n-1)d

495 = 108 + (n-1)(9)

495 - 108 = 9(n-1)

387 = 9(n-1)

n-1 = 43

n = 44

Step 3: Find the sum

Sₙ = n/2 [a + l]

S₄₄ = 44/2 [108 + 495]

S₄₄ = 22 × 603

S₄₄ = 13,266

∴ The sum of integers between 100 and 500 divisible by 9 is 13,266.

Find the sum of all two-digit odd multiples of 3. (M'19)

Step 1: Find the sequence

Two-digit odd multiples of 3: 15, 21, 27, ..., 99

This is an AP with a = 15, d = 6

Step 2: Find number of terms

aₙ = a + (n-1)d

99 = 15 + (n-1)(6)

99 - 15 = 6(n-1)

84 = 6(n-1)

n-1 = 14

n = 15

Step 3: Find the sum

Sₙ = n/2 [a + l]

S₁₅ = 15/2 [15 + 99]

S₁₅ = 15/2 × 114

S₁₅ = 15 × 57 = 855

∴ The sum of all two-digit odd multiples of 3 is 855.

Find the sum of all integers between 1 to 50 which are not divisible by 3. (J'19)

Step 1: Find sum of all integers from 1 to 50

S = 50/2 [1 + 50] = 25 × 51 = 1275

Step 2: Find sum of integers divisible by 3 from 1 to 50

Numbers: 3, 6, 9, ..., 48

This is an AP with a = 3, d = 3

aₙ = a + (n-1)d

48 = 3 + (n-1)(3)

48 - 3 = 3(n-1)

45 = 3(n-1)

n-1 = 15

n = 16

Sum = 16/2 [3 + 48] = 8 × 51 = 408

Step 3: Find required sum

Sum of numbers not divisible by 3 = Total sum - Sum of numbers divisible by 3

= 1275 - 408 = 867

∴ The sum of all integers from 1 to 50 not divisible by 3 is 867.

Quadratic Equations -Solutions

zaheerpashamd

Quadratic Equations - 1 Mark Problems

Important Formulas:

• Standard form: ax² + bx + c = 0
• Discriminant (D) = b² - 4ac
• Nature of roots:
  - D > 0: Two distinct real roots
  - D = 0: Two equal real roots
  - D < 0: No real roots
• Sum of roots = -b/a
• Product of roots = c/a

Check whether 1 and 3/2 are the roots of the equation 2x² – 5x + 3 = 0. (J'15)

Step 1: Check x = 1

2(1)² - 5(1) + 3 = 2 - 5 + 3 = 0

Since the equation equals 0, x = 1 is a root.

Step 2: Check x = 3/2

2(3/2)² - 5(3/2) + 3 = 2(9/4) - 15/2 + 3

= 9/2 - 15/2 + 6/2 = (9 - 15 + 6)/2 = 0/2 = 0

Since the equation equals 0, x = 3/2 is also a root.

∴ Yes, both 1 and 3/2 are roots of the equation.

If b² – 4ac > 0 in ax² + bx + c = 0, (a ≠ 0); then what can you say about roots of the equation? (M'16)

Step 1: Recall discriminant concept

For a quadratic equation ax² + bx + c = 0, the discriminant D = b² - 4ac determines the nature of roots.

∴ If b² - 4ac > 0, the equation has two distinct real roots.

Find the value of k, if 2 is one of the roots of the quadratic equation x² – kx + 6 = 0. (J'16)

Step 1: Substitute x = 2 in the equation

(2)² - k(2) + 6 = 0

4 - 2k + 6 = 0

10 - 2k = 0

2k = 10

k = 5

∴ The value of k is 5.

Write the nature of roots of the quadratic equation 2x² – 5x + 6 = 0. (M'17)

Step 1: Calculate discriminant

For 2x² - 5x + 6 = 0, a = 2, b = -5, c = 6

D = b² - 4ac = (-5)² - 4(2)(6) = 25 - 48 = -23

Step 2: Determine nature of roots

Since D < 0, the equation has no real roots.

∴ The equation has no real roots (complex roots).

Write the nature of the roots of the quadratic equation x² – 8x + 16 = 0. (J'17)

Step 1: Calculate discriminant

For x² - 8x + 16 = 0, a = 1, b = -8, c = 16

D = b² - 4ac = (-8)² - 4(1)(16) = 64 - 64 = 0

Step 2: Determine nature of roots

Since D = 0, the equation has two equal real roots.

∴ The equation has two equal real roots.

Find sum and product of the roots of the quadratic equation x² – 4√3x + 9 = 0. (M'18)

Step 1: Identify coefficients

For x² - 4√3x + 9 = 0, a = 1, b = -4√3, c = 9

Step 2: Calculate sum of roots

Sum of roots = -b/a = -(-4√3)/1 = 4√3

Step 3: Calculate product of roots

Product of roots = c/a = 9/1 = 9

∴ Sum of roots = 4√3, Product of roots = 9

Find the values of k for which the quadratic equation 4x² + 5kx + 25 = 0 has equal roots. (J'18)

Step 1: Condition for equal roots

For equal roots, discriminant D = 0

Step 2: Calculate discriminant

For 4x² + 5kx + 25 = 0, a = 4, b = 5k, c = 25

D = b² - 4ac = (5k)² - 4(4)(25) = 25k² - 400

Step 3: Set D = 0 and solve for k

25k² - 400 = 0

25k² = 400

k² = 16

k = ±4

∴ The values of k are 4 and -4.

Find the roots of the quadratic equation x² + 2x – 3 = 0. (M'19)

Step 1: Factor the equation

x² + 2x - 3 = 0

(x + 3)(x - 1) = 0

Step 2: Solve for x

x + 3 = 0 or x - 1 = 0

x = -3 or x = 1

∴ The roots are x = -3 and x = 1.

Find the discriminant of the quadratic equation 3x² – 5x + 2 = 0 and hence write the nature of its roots. (J'19)

Step 1: Calculate discriminant

For 3x² - 5x + 2 = 0, a = 3, b = -5, c = 2

D = b² - 4ac = (-5)² - 4(3)(2) = 25 - 24 = 1

Step 2: Determine nature of roots

Since D > 0, the equation has two distinct real roots.

∴ Discriminant = 1, and the equation has two distinct real roots.

Is (x + 2)² = x² + 3 a Quadratic Equation? Justify. (May 2022)

Step 1: Expand and simplify

(x + 2)² = x² + 3

x² + 4x + 4 = x² + 3

4x + 4 = 3

4x + 1 = 0

Step 2: Check if it's quadratic

A quadratic equation must have degree 2 (highest power of x is 2).

Here, the highest power of x is 1, so it's a linear equation.

∴ No, it's not a quadratic equation. It simplifies to a linear equation.

Is x(2x + 3) = x² + 5 a Quadratic Equation? Justify. (Aug 2022)

Step 1: Expand and simplify

x(2x + 3) = x² + 5

2x² + 3x = x² + 5

2x² - x² + 3x - 5 = 0

x² + 3x - 5 = 0

Step 2: Check if it's quadratic

The simplified equation is x² + 3x - 5 = 0

This has degree 2 (highest power of x is 2), so it's a quadratic equation.

∴ Yes, it is a quadratic equation as it simplifies to x² + 3x - 5 = 0.

Solve the quadratic equation 2sin²θ – 3sinθ + 1 = 0 where 0° < θ ≤ 90°. (Apr'23)

Step 1: Let y = sinθ

The equation becomes: 2y² - 3y + 1 = 0

Step 2: Factor the quadratic

2y² - 3y + 1 = 0

(2y - 1)(y - 1) = 0

Step 3: Solve for y

2y - 1 = 0 or y - 1 = 0

y = 1/2 or y = 1

Step 4: Solve for θ

sinθ = 1/2 or sinθ = 1

For sinθ = 1/2: θ = 30° (since 0° < θ ≤ 90°)

For sinθ = 1: θ = 90°

∴ The solutions are θ = 30° and θ = 90°.

Quadratic Equations - 2 Mark Problems

Important Formulas:

• Standard form: ax² + bx + c = 0
• Discriminant (D) = b² - 4ac
• For equal roots: D = 0
• Sum of roots = -b/a
• Product of roots = c/a
• Quadratic with roots α and β: x² - (α+β)x + αβ = 0

If 9x² + kx + 1 = 0 has equal roots, then find the value of k. (M'16)

Step 1: Condition for equal roots

For equal roots, discriminant D = 0

Step 2: Identify coefficients

For 9x² + kx + 1 = 0, a = 9, b = k, c = 1

Step 3: Calculate discriminant

D = b² - 4ac = k² - 4(9)(1) = k² - 36

Step 4: Set D = 0 and solve for k

k² - 36 = 0

k² = 36

k = ±6

∴ The values of k are 6 and -6.

The sum of a number and its reciprocal is 10/3. Find the number. (M'17)

Step 1: Formulate the equation

Let the number be x

Then, x + 1/x = 10/3

Step 2: Multiply through by 3x to eliminate denominators

3x(x + 1/x) = 3x(10/3)

3x² + 3 = 10x

3x² - 10x + 3 = 0

Step 3: Solve the quadratic equation

3x² - 10x + 3 = 0

3x² - 9x - x + 3 = 0

3x(x - 3) - 1(x - 3) = 0

(3x - 1)(x - 3) = 0

x = 1/3 or x = 3

∴ The number is 3 or 1/3.

Is it possible to design a rectangular Garden, whose length is twice of its breadth and area is 200 m²? If so, find its length and breadth. (J'17)

Step 1: Set up variables

Let breadth = x meters

Then length = 2x meters

Step 2: Formulate the equation

Area = length × breadth = 2x × x = 2x²

Given area = 200 m²

So, 2x² = 200

Step 3: Solve for x

2x² = 200

x² = 100

x = 10 (taking positive value as length can't be negative)

Step 4: Find dimensions

Breadth = x = 10 m

Length = 2x = 20 m

∴ Yes, it is possible. The garden has length 20 m and breadth 10 m.

If the equation kx² – 2kx + 6 = 0 has equal roots, then find the value of k. (M'18)

Step 1: Condition for equal roots

For equal roots, discriminant D = 0

Step 2: Identify coefficients

For kx² - 2kx + 6 = 0, a = k, b = -2k, c = 6

Step 3: Calculate discriminant

D = b² - 4ac = (-2k)² - 4(k)(6) = 4k² - 24k

Step 4: Set D = 0 and solve for k

4k² - 24k = 0

4k(k - 6) = 0

k = 0 or k = 6

Step 5: Check validity

If k = 0, the equation becomes 6 = 0, which is not quadratic

So we discard k = 0

∴ The value of k is 6.

Without calculating the roots of x² - 5x + 6 = 0, explain the nature of roots. (J'18)

Step 1: Identify coefficients

For x² - 5x + 6 = 0, a = 1, b = -5, c = 6

Step 2: Calculate discriminant

D = b² - 4ac = (-5)² - 4(1)(6) = 25 - 24 = 1

Step 3: Analyze discriminant

Since D > 0 and is a perfect square, the roots are:

1. Real and distinct

2. Rational numbers

∴ The equation has two distinct rational roots.

Write the Quadratic equation, whose roots are 2 + √3 and 2 - √3. (M'19)

Step 1: Find sum of roots

Sum = (2 + √3) + (2 - √3) = 4

Step 2: Find product of roots

Product = (2 + √3)(2 - √3) = 4 - 3 = 1

(Using identity: (a+b)(a-b) = a² - b²)

Step 3: Form the quadratic equation

Quadratic equation with roots α and β is: x² - (α+β)x + αβ = 0

So, x² - (4)x + (1) = 0

x² - 4x + 1 = 0

∴ The required quadratic equation is x² - 4x + 1 = 0.

Find the roots of quadratic equation x² + 4x + 3 = 0 by "completing square method". (J'19)

Completing Square Method:

Step 1: Write the equation

x² + 4x + 3 = 0

Step 2: Move constant term to RHS

x² + 4x = -3

Step 3: Add square of half the coefficient of x to both sides

Coefficient of x = 4, half of it = 2, square = 4

x² + 4x + 4 = -3 + 4

x² + 4x + 4 = 1

Step 4: Write LHS as perfect square

(x + 2)² = 1

Step 5: Take square root on both sides

x + 2 = ±1

Step 6: Solve for x

x + 2 = 1 or x + 2 = -1

x = -1 or x = -3

∴ The roots are x = -1 and x = -3.

Shashanka said that (x + 1)² = 2(x – 3) is a quadratic equation. Do you agree? (J'19)

Step 1: Expand and simplify

(x + 1)² = 2(x - 3)

x² + 2x + 1 = 2x - 6

Step 2: Bring all terms to one side

x² + 2x + 1 - 2x + 6 = 0

x² + 7 = 0

Step 3: Check if it's quadratic

A quadratic equation must have degree 2 (highest power of x is 2).

Here, the equation is x² + 7 = 0, which has degree 2.

∴ Yes, I agree with Shashanka. It is a quadratic equation.

Write a Quadratic Equation, whose roots are 2 – √3 and 2 + √3. (Aug 22)

Step 1: Find sum of roots

Sum = (2 - √3) + (2 + √3) = 4

Step 2: Find product of roots

Product = (2 - √3)(2 + √3) = 4 - 3 = 1

(Using identity: (a+b)(a-b) = a² - b²)

Step 3: Form the quadratic equation

Quadratic equation with roots α and β is: x² - (α+β)x + αβ = 0

So, x² - (4)x + (1) = 0

x² - 4x + 1 = 0

∴ The required quadratic equation is x² - 4x + 1 = 0.

Quadratic Equations - 4 Mark Problems

Important Formulas:

• Standard form: ax² + bx + c = 0
• Discriminant (D) = b² - 4ac
• For equal roots: D = 0
• Sum of roots = -b/a
• Product of roots = c/a
• Quadratic with roots α and β: x² - (α+β)x + αβ = 0
• Area of square = side²
• Perimeter of square = 4 × side
• Distance = Speed × Time

If the sum of the areas of two squares is 468 m² and the difference of their perimeters is 24 m, then find the measurements of their sides. (J'15)

Step 1: Set up variables

Let side of first square = x meters

Let side of second square = y meters

Step 2: Form equations from given conditions

Area of first square = x²

Area of second square = y²

Sum of areas: x² + y² = 468 ...(1)

Perimeter of first square = 4x

Perimeter of second square = 4y

Difference of perimeters: 4x - 4y = 24 ...(2)

Step 3: Simplify equation (2)

4x - 4y = 24

4(x - y) = 24

x - y = 6 ...(3)

Step 4: Solve the system of equations

From (3): x = y + 6

Substitute in (1): (y + 6)² + y² = 468

y² + 12y + 36 + y² = 468

2y² + 12y + 36 - 468 = 0

2y² + 12y - 432 = 0

Divide by 2: y² + 6y - 216 = 0

Step 5: Solve the quadratic equation

y² + 6y - 216 = 0

y² + 18y - 12y - 216 = 0

y(y + 18) - 12(y + 18) = 0

(y + 18)(y - 12) = 0

y = -18 or y = 12

Since side cannot be negative, y = 12

Step 6: Find x

x = y + 6 = 12 + 6 = 18

∴ The sides of the squares are 18 m and 12 m.

Sum of the squares of two consecutive positive even integers is 100; find those numbers by using quadratic equations. (M'16)

Step 1: Set up variables

Let first even integer = x

Then next consecutive even integer = x + 2

Step 2: Form the equation

Sum of squares: x² + (x + 2)² = 100

Step 3: Expand and simplify

x² + (x² + 4x + 4) = 100

2x² + 4x + 4 = 100

2x² + 4x + 4 - 100 = 0

2x² + 4x - 96 = 0

Divide by 2: x² + 2x - 48 = 0

Step 4: Solve the quadratic equation

x² + 2x - 48 = 0

x² + 8x - 6x - 48 = 0

x(x + 8) - 6(x + 8) = 0

(x + 8)(x - 6) = 0

x = -8 or x = 6

Step 5: Check for positive integers

Since we need positive integers, x = 6

Then x + 2 = 8

∴ The two consecutive positive even integers are 6 and 8.

If -4 is a common root for the quadratic equations 2x² + px + 8 = 0 and p(x² + x) + k = 0, find the value of k. (J'17)

Step 1: Substitute x = -4 in first equation

2x² + px + 8 = 0

2(-4)² + p(-4) + 8 = 0

2(16) - 4p + 8 = 0

32 - 4p + 8 = 0

40 - 4p = 0

4p = 40

p = 10

Step 2: Expand second equation

p(x² + x) + k = 0

px² + px + k = 0

Step 3: Substitute x = -4 and p = 10 in second equation

10(-4)² + 10(-4) + k = 0

10(16) - 40 + k = 0

160 - 40 + k = 0

120 + k = 0

k = -120

∴ The value of k is -120.

Sum of squares of two consecutive even numbers is 580. Find the numbers by writing a suitable quadratic equation. (M'18)

Step 1: Set up variables

Let first even number = x

Then next consecutive even number = x + 2

Step 2: Form the equation

Sum of squares: x² + (x + 2)² = 580

Step 3: Expand and simplify

x² + (x² + 4x + 4) = 580

2x² + 4x + 4 = 580

2x² + 4x + 4 - 580 = 0

2x² + 4x - 576 = 0

Divide by 2: x² + 2x - 288 = 0

Step 4: Solve the quadratic equation

x² + 2x - 288 = 0

x² + 18x - 16x - 288 = 0

x(x + 18) - 16(x + 18) = 0

(x + 18)(x - 16) = 0

x = -18 or x = 16

Step 5: Find the numbers

If x = 16, then x + 2 = 18

If x = -18, then x + 2 = -16

∴ The numbers are 16 and 18, or -18 and -16.

If a number when increased by 12, equals 160 times of its reciprocal, then find the numbers. (J'18)

Step 1: Set up variables

Let the number be x

Its reciprocal is 1/x

Step 2: Form the equation

Number increased by 12: x + 12

160 times its reciprocal: 160 × (1/x) = 160/x

Equation: x + 12 = 160/x

Step 3: Multiply both sides by x

x(x + 12) = 160

x² + 12x = 160

x² + 12x - 160 = 0

Step 4: Solve the quadratic equation

x² + 12x - 160 = 0

x² + 20x - 8x - 160 = 0

x(x + 20) - 8(x + 20) = 0

(x + 20)(x - 8) = 0

x = -20 or x = 8

∴ The numbers are 8 and -20.

Sum of the areas of two squares is 850 m². If the difference of their perimeters is 40 m. Find the sides of the two squares. (M'19)

Step 1: Set up variables

Let side of first square = x meters

Let side of second square = y meters

Step 2: Form equations from given conditions

Area of first square = x²

Area of second square = y²

Sum of areas: x² + y² = 850 ...(1)

Perimeter of first square = 4x

Perimeter of second square = 4y

Difference of perimeters: 4x - 4y = 40 ...(2)

Step 3: Simplify equation (2)

4x - 4y = 40

4(x - y) = 40

x - y = 10 ...(3)

Step 4: Solve the system of equations

From (3): x = y + 10

Substitute in (1): (y + 10)² + y² = 850

y² + 20y + 100 + y² = 850

2y² + 20y + 100 - 850 = 0

2y² + 20y - 750 = 0

Divide by 2: y² + 10y - 375 = 0

Step 5: Solve the quadratic equation

y² + 10y - 375 = 0

y² + 25y - 15y - 375 = 0

y(y + 25) - 15(y + 25) = 0

(y + 25)(y - 15) = 0

y = -25 or y = 15

Since side cannot be negative, y = 15

Step 6: Find x

x = y + 10 = 15 + 10 = 25

∴ The sides of the squares are 25 m and 15 m.

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train. (J'19)

Step 1: Set up variables

Let actual speed = x km/h

Increased speed = (x + 5) km/h

Distance = 360 km

Step 2: Calculate times

Actual time = Distance/Speed = 360/x hours

Time at increased speed = 360/(x + 5) hours

Step 3: Form the equation

Actual time - New time = 1 hour

360/x - 360/(x + 5) = 1

Step 4: Multiply both sides by x(x + 5)

360(x + 5) - 360x = x(x + 5)

360x + 1800 - 360x = x² + 5x

1800 = x² + 5x

x² + 5x - 1800 = 0

Step 5: Solve the quadratic equation

x² + 5x - 1800 = 0

x² + 45x - 40x - 1800 = 0

x(x + 45) - 40(x + 45) = 0

(x + 45)(x - 40) = 0

x = -45 or x = 40

Since speed cannot be negative, x = 40

∴ The speed of the train is 40 km/h.

The numerator of a fraction is 3 less than its denominator. If 2 is added to both numerator and denominator, the sum of the new fraction formed and original fraction is 29/20 then find the original fraction. (Jun'23)

Step 1: Set up variables

Let denominator = x

Then numerator = x - 3

Original fraction = (x - 3)/x

Step 2: Form the new fraction

When 2 is added to both numerator and denominator:

New numerator = (x - 3) + 2 = x - 1

New denominator = x + 2

New fraction = (x - 1)/(x + 2)

Step 3: Form the equation

Sum of original and new fraction = 29/20

(x - 3)/x + (x - 1)/(x + 2) = 29/20

Step 4: Find common denominator and simplify

[(x - 3)(x + 2) + x(x - 1)] / [x(x + 2)] = 29/20

[(x² - x - 6) + (x² - x)] / [x(x + 2)] = 29/20

[2x² - 2x - 6] / [x(x + 2)] = 29/20

Divide numerator and denominator by 2:

[x² - x - 3] / [x(x + 2)] = 29/20

Step 5: Cross multiply

20(x² - x - 3) = 29x(x + 2)

20x² - 20x - 60 = 29x² + 58x

20x² - 29x² - 20x - 58x - 60 = 0

-9x² - 78x - 60 = 0

Multiply by -1: 9x² + 78x + 60 = 0

Divide by 3: 3x² + 26x + 20 = 0

Step 6: Solve the quadratic equation

3x² + 26x + 20 = 0

Using quadratic formula: x = [-b ± √(b² - 4ac)] / (2a)

a = 3, b = 26, c = 20

Discriminant = 26² - 4(3)(20) = 676 - 240 = 436

√436 = √(4 × 109) = 2√109

x = [-26 ± 2√109] / 6 = [-13 ± √109] / 3

Step 7: Find the fraction

x = [-13 + √109] / 3 (taking positive value as denominator can't be negative)

Numerator = x - 3 = [-13 + √109] / 3 - 3 = [-13 + √109 - 9] / 3 = [-22 + √109] / 3

Original fraction = ([-22 + √109] / 3) / ([-13 + √109] / 3) = (-22 + √109) / (-13 + √109)

∴ The original fraction is (-22 + √109) / (-13 + √109).

Pair of Linear Equations in Two Variables-Solutions

zaheerpashamd

Linear Equations Problems and Solutions

Pair of Linear Equations in Two Variables

1-Mark Questions

For what value of k, the following system of equations has a unique solution: x - ky = 2 and 3x + 2y = -5 (M'15)

Step 1: Condition for unique solution
For a system of linear equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ to have a unique solution:
a₁/a₂ ≠ b₁/b₂

Step 2: Apply the condition
For equations: x - ky = 2 and 3x + 2y = -5
a₁ = 1, b₁ = -k, a₂ = 3, b₂ = 2
So, 1/3 ≠ -k/2

Step 3: Solve for k
1/3 ≠ -k/2
Cross-multiplying: 2 ≠ -3k
k ≠ -2/3

∴ The system has a unique solution for all values of k except k = -2/3

For what values of m, the pair of equations 3x + my = 10 and 9x + 12y = 30 have a unique solution. (M'16)

Step 1: Condition for unique solution
For a unique solution: a₁/a₂ ≠ b₁/b₂

Step 2: Apply the condition
For equations: 3x + my = 10 and 9x + 12y = 30
a₁ = 3, b₁ = m, a₂ = 9, b₂ = 12
So, 3/9 ≠ m/12

Step 3: Solve for m
1/3 ≠ m/12
Cross-multiplying: 12 ≠ 3m
m ≠ 4

∴ The system has a unique solution for all values of m except m = 4

In a rectangle ABCD, AB = x + y, BC = x - y, CD = 9 and AD = 3. Find the values of x and y. (J'16)

Step 1: Properties of a rectangle
In a rectangle, opposite sides are equal.
So, AB = CD and BC = AD

Step 2: Set up equations
AB = CD ⇒ x + y = 9 ...(1)
BC = AD ⇒ x - y = 3 ...(2)

Step 3: Solve the equations
Adding (1) and (2):
(x + y) + (x - y) = 9 + 3
2x = 12 ⇒ x = 6

Step 4: Find y
Substitute x = 6 in equation (1):
6 + y = 9 ⇒ y = 3

∴ x = 6, y = 3

Show that the pair of Linear Equations 7x + y = 10 and x + 7y = 10 are consistent. (M'17)

Step 1: Check for consistency
For equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂:
If a₁/a₂ ≠ b₁/b₂, the equations are consistent with a unique solution.

Step 2: Apply the condition
For equations: 7x + y = 10 and x + 7y = 10
a₁ = 7, b₁ = 1, a₂ = 1, b₂ = 7
a₁/a₂ = 7/1 = 7
b₁/b₂ = 1/7 = 1/7

Step 3: Compare the ratios
Since 7 ≠ 1/7, the equations are consistent with a unique solution.

∴ The pair of equations is consistent as a₁/a₂ ≠ b₁/b₂

Write the Condition for the pair of linear equations in two variables to be parallel lines. (J'17)

Condition for parallel lines:
For equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
The lines are parallel if a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Explanation:
When a₁/a₂ = b₁/b₂, the lines have the same slope but different intercepts, making them parallel.

∴ The condition for parallel lines is: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

If x = a and y = b is solution for the pair of equations x - y = 2 and x + y = 4, then find the values of a and b. (M'18)

Step 1: Set up the equations
Since (a, b) is a solution to both equations:
a - b = 2 ...(1)
a + b = 4 ...(2)

Step 2: Solve the equations
Adding (1) and (2):
(a - b) + (a + b) = 2 + 4
2a = 6 ⇒ a = 3

Step 3: Find b
Substitute a = 3 in equation (2):
3 + b = 4 ⇒ b = 1

∴ a = 3, b = 1

Whether the following pair of Linear Equations are parallel? Justify. 6x - 4y + 10 = 0, 3x - 2y + 6 = 0. (J'18)

Step 1: Condition for parallel lines
For parallel lines: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Step 2: Identify coefficients
For 6x - 4y + 10 = 0: a₁ = 6, b₁ = -4, c₁ = 10
For 3x - 2y + 6 = 0: a₂ = 3, b₂ = -2, c₂ = 6

Step 3: Compare ratios
a₁/a₂ = 6/3 = 2
b₁/b₂ = -4/-2 = 2
c₁/c₂ = 10/6 = 5/3

Step 4: Check the condition
Since a₁/a₂ = b₁/b₂ = 2, but 2 ≠ 5/3

∴ Yes, the lines are parallel as a₁/a₂ = b₁/b₂ ≠ c₁/c₂

For what value of 't' the following pair of linear equations has no solution? 2x - ty = 5 and 3x + 2y = 11. (M'19)

Step 1: Condition for no solution
For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Step 2: Identify coefficients
For 2x - ty = 5: a₁ = 2, b₁ = -t, c₁ = 5
For 3x + 2y = 11: a₂ = 3, b₂ = 2, c₂ = 11

Step 3: Apply the condition
a₁/a₂ = b₁/b₂ ⇒ 2/3 = -t/2

Step 4: Solve for t
2/3 = -t/2
Cross-multiplying: 4 = -3t
t = -4/3

Step 5: Verify c₁/c₂
c₁/c₂ = 5/11
Since 2/3 ≠ 5/11, the condition is satisfied.

∴ t = -4/3

The solution of the linear equation x + y = 5 are (1, 4), (2, 3) and (3, 2). The solution of another linear equation x - y = 1 are (3, 2), (2, 1) and (5, 4). Plot these points on a graph sheet and draw lines. (May 2022)

Step 1: Identify the equations
Equation 1: x + y = 5
Equation 2: x - y = 1

Step 2: Find the intersection point
Solving the equations:
x + y = 5 ...(1)
x - y = 1 ...(2)
Adding (1) and (2): 2x = 6 ⇒ x = 3
Substituting in (1): 3 + y = 5 ⇒ y = 2

Step 3: Graph description

Step 4: Interpretation
The lines intersect at (3, 2), which is the common solution to both equations.

∴ The lines intersect at (3, 2)

The solutions of the linear equation x + y = 8 are (1, 8), (2, 6) and (3, 5). The solutions of another linear equation 3x + 3y = 12 are (1, 3), (3, 1) and (9, 4). Plot these points on a graph sheet and draw lines.

Step 1: Simplify the equations
Equation 1: x + y = 8
Equation 2: 3x + 3y = 12 ⇒ Divide by 3: x + y = 4

Step 2: Analyze the equations
Both equations have the same left-hand side (x + y) but different right-hand sides (8 and 4).
This means the lines are parallel.

Step 3: Graph description

Step 4: Interpretation
The lines are parallel and will never intersect, so the system has no solution.

∴ The lines are parallel and the system has no solution

If the pair of linear equations 6x - 4y + 10 = 0 and 3x + ky + 6 = 0 represents parallel lines graphically, then find the value of 'k'. (Jun'23)

Step 1: Condition for parallel lines
For parallel lines: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Step 2: Identify coefficients
For 6x - 4y + 10 = 0: a₁ = 6, b₁ = -4, c₁ = 10
For 3x + ky + 6 = 0: a₂ = 3, b₂ = k, c₂ = 6

Step 3: Apply the condition
a₁/a₂ = b₁/b₂ ⇒ 6/3 = -4/k

Step 4: Solve for k
2 = -4/k
k = -4/2 = -2

Step 5: Verify c₁/c₂
c₁/c₂ = 10/6 = 5/3
Since 2 ≠ 5/3, the condition is satisfied.

∴ k = -2

Linear Equations - 2 Mark Questions

If we multiply or divide both sides of a linear equation by a non-zero number, then the roots of that linear equation will remain the same. Is it true? If so, justify with an example. (M'15)

Step 1: Understanding the concept
Yes, this statement is true. When we multiply or divide both sides of a linear equation by the same non-zero number, we are performing an equivalent transformation that doesn't change the solution set.

Step 2: Example to justify
Consider the equation: 2x + 4 = 10
Solution: 2x = 6 ⇒ x = 3

Step 3: Multiply both sides by 2
(2x + 4) × 2 = 10 × 2 ⇒ 4x + 8 = 20
Solution: 4x = 12 ⇒ x = 3

Step 4: Divide both sides by 2
(2x + 4) ÷ 2 = 10 ÷ 2 ⇒ x + 2 = 5
Solution: x = 3

∴ Yes, the statement is true. Multiplying or dividing both sides by a non-zero number doesn't change the solution.

If the present ages of A and B are in ratio of 9 : 4 and after 7 years the ratio of the ages will be 5 : 3 then find their present ages. (J'15)

Step 1: Set up variables
Let present age of A = 9x years
Let present age of B = 4x years

Step 2: Set up equation for after 7 years
After 7 years:
Age of A = 9x + 7
Age of B = 4x + 7
Ratio = (9x + 7) : (4x + 7) = 5 : 3

Step 3: Form the equation
(9x + 7)/(4x + 7) = 5/3
Cross-multiplying: 3(9x + 7) = 5(4x + 7)

Step 4: Solve for x
27x + 21 = 20x + 35
27x - 20x = 35 - 21
7x = 14 ⇒ x = 2

Step 5: Find present ages
Age of A = 9 × 2 = 18 years
Age of B = 4 × 2 = 8 years

∴ Present ages: A = 18 years, B = 8 years

Solve the following pair of linear equations by substitution method: 2x - 3y = 19 and 3x - 2y = 21 (M'16)

Substitution Method:

Step 1: Express one variable in terms of the other
From first equation: 2x - 3y = 19 ⇒ 2x = 19 + 3y ⇒ x = (19 + 3y)/2

Step 2: Substitute in second equation
3x - 2y = 21
3[(19 + 3y)/2] - 2y = 21
(57 + 9y)/2 - 2y = 21

Step 3: Solve for y
Multiply throughout by 2: 57 + 9y - 4y = 42
57 + 5y = 42
5y = 42 - 57 = -15
y = -3

Step 4: Find x
x = (19 + 3y)/2 = (19 + 3(-3))/2 = (19 - 9)/2 = 10/2 = 5

∴ x = 5, y = -3

If the measure of angles of a triangle are x°, y° and 40°, and difference between the measures of angles x° and y° is 30°, then find values of x° and y°. (J'16)

Step 1: Use angle sum property of triangle
Sum of angles of a triangle = 180°
So, x + y + 40 = 180
x + y = 140 ...(1)

Step 2: Use the given difference
x - y = 30 ...(2) [Assuming x > y]

Step 3: Solve the equations
Adding (1) and (2):
(x + y) + (x - y) = 140 + 30
2x = 170 ⇒ x = 85

Step 4: Find y
From equation (1): 85 + y = 140 ⇒ y = 55

∴ x = 85°, y = 55°

Given the linear equation 3x + 4y = 11, write linear equations in two variables such that their geometrical representations form parallel lines and intersecting lines. (M'18)

Step 1: For parallel lines
For parallel lines, the ratios of coefficients of x and y should be equal but the constant term should be different.
Example: 3x + 4y = 15 (same coefficients for x and y, different constant)

Step 2: For intersecting lines
For intersecting lines, the ratios of coefficients of x and y should not be equal.
Example: 2x + 5y = 12 (different ratio of coefficients)

∴ Parallel lines: 3x + 4y = 15
Intersecting lines: 2x + 5y = 12

Solve the pair of linear equations 2x + 3y = 8 and x + 2y = 5 by Elimination method. (M'19)

Elimination Method:

Step 1: Make coefficients of one variable equal
Equations: 2x + 3y = 8 ...(1)
x + 2y = 5 ...(2)
Multiply equation (2) by 2: 2x + 4y = 10 ...(3)

Step 2: Eliminate x
Subtract equation (1) from equation (3):
(2x + 4y) - (2x + 3y) = 10 - 8
y = 2

Step 3: Find x
Substitute y = 2 in equation (2):
x + 2(2) = 5 ⇒ x + 4 = 5 ⇒ x = 1

∴ x = 1, y = 2

For what values of m the following system of equations will have no solution? Why?
mx + 4y = 10 and 9x + 12y = 30 (J'19)

Step 1: Condition for no solution
For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Step 2: Identify coefficients
For mx + 4y = 10: a₁ = m, b₁ = 4, c₁ = 10
For 9x + 12y = 30: a₂ = 9, b₂ = 12, c₂ = 30

Step 3: Apply the condition
a₁/a₂ = b₁/b₂ ⇒ m/9 = 4/12 = 1/3
m/9 = 1/3 ⇒ m = 3

Step 4: Check c₁/c₂
c₁/c₂ = 10/30 = 1/3
Since a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/3, the lines are coincident, not parallel.

Step 5: Conclusion
For no solution, we need a₁/a₂ = b₁/b₂ ≠ c₁/c₂
But here when m = 3, a₁/a₂ = b₁/b₂ = c₁/c₂
So there is no value of m for which the system has no solution.

∴ There is no value of m for which the system has no solution.

Solve 2x + y = 5 and 5x + 3y = 11. (May 2022)

Step 1: Use substitution method
From first equation: 2x + y = 5 ⇒ y = 5 - 2x

Step 2: Substitute in second equation
5x + 3y = 11
5x + 3(5 - 2x) = 11
5x + 15 - 6x = 11

Step 3: Solve for x
-x + 15 = 11
-x = 11 - 15 = -4
x = 4

Step 4: Find y
y = 5 - 2x = 5 - 2(4) = 5 - 8 = -3

∴ x = 4, y = -3

Solve 3x + 2y = 11 and 2x + 3y = 4. (Aug 2022)

Elimination Method:

Step 1: Make coefficients of y equal
Equations: 3x + 2y = 11 ...(1)
2x + 3y = 4 ...(2)
Multiply (1) by 3: 9x + 6y = 33 ...(3)
Multiply (2) by 2: 4x + 6y = 8 ...(4)

Step 2: Eliminate y
Subtract (4) from (3):
(9x + 6y) - (4x + 6y) = 33 - 8
5x = 25 ⇒ x = 5

Step 3: Find y
Substitute x = 5 in equation (1):
3(5) + 2y = 11 ⇒ 15 + 2y = 11 ⇒ 2y = -4 ⇒ y = -2

∴ x = 5, y = -2

Linear Equations -4 Marks Solutions

Linear Equations - 4 Marks Solutions

Draw a graph for the following pair of linear equations in two variables and find their solution from the graph: 2x + y = 5 and 3x – 2y = 4 (M'15)

Graphical Method:

Step 1: Find points for 2x + y = 5

When x = 0: 2(0) + y = 5 → 0 + y = 5 → y = 5 → Point: (0, 5)

When x = 1: 2(1) + y = 5 → 2 + y = 5 → y = 3 → Point: (1, 3)

When x = 2: 2(2) + y = 5 → 4 + y = 5 → y = 1 → Point: (2, 1)

Step 2: Find points for 3x - 2y = 4

When x = 0: 3(0) - 2y = 4 → 0 - 2y = 4 → -2y = 4 → y = -2 → Point: (0, -2)

When x = 2: 3(2) - 2y = 4 → 6 - 2y = 4 → -2y = -2 → y = 1 → Point: (2, 1)

When x = 4: 3(4) - 2y = 4 → 12 - 2y = 4 → -2y = -8 → y = 4 → Point: (4, 4)

∴ The lines intersect at point (2, 1), so the solution is x = 2, y = 1

Draw the graphs of the following equations 3x – y – 2 = 0 and 2x + y – 8 = 0 on the graph paper.
i) Write down the co-ordinates of the point of intersection of the equations.
ii) Find the area of the triangle formed by the lines and the X-axis. (J'15)

Step 1: Find points for 3x - y - 2 = 0

When x = 0: 3(0) - y - 2 = 0 → 0 - y - 2 = 0 → -y = 2 → y = -2 → Point: (0, -2)

When x = 1: 3(1) - y - 2 = 0 → 3 - y - 2 = 0 → 1 - y = 0 → y = 1 → Point: (1, 1)

When x = 2: 3(2) - y - 2 = 0 → 6 - y - 2 = 0 → 4 - y = 0 → y = 4 → Point: (2, 4)

Step 2: Find points for 2x + y - 8 = 0

When x = 0: 2(0) + y - 8 = 0 → 0 + y - 8 = 0 → y = 8 → Point: (0, 8)

When x = 2: 2(2) + y - 8 = 0 → 4 + y - 8 = 0 → y - 4 = 0 → y = 4 → Point: (2, 4)

When x = 4: 2(4) + y - 8 = 0 → 8 + y - 8 = 0 → y = 0 → Point: (4, 0)

Step 3: Find intersection point

The lines intersect at (2, 4)

Step 4: Find area of triangle with X-axis

For 3x - y - 2 = 0: When y = 0, 3x - 0 - 2 = 0 → 3x = 2 → x = 2/3 ≈ 0.67

For 2x + y - 8 = 0: When y = 0, 2x + 0 - 8 = 0 → 2x = 8 → x = 4

Base = 4 - 2/3 = 10/3

Height = y-coordinate of intersection point = 4

Area = 1/2 × base × height = 1/2 × (10/3) × 4 = 20/3 ≈ 6.67 sq units

∴ i) Intersection point: (2, 4)
ii) Area of triangle: 20/3 sq units ≈ 6.67 sq units

Draw the graph for the equations 2x – 3y = 5 and 4x – 6y = 15 on the graph paper and check whether they are consistent or not. (J'15)

Step 1: Find points for 2x - 3y = 5

When x = 1: 2(1) - 3y = 5 → 2 - 3y = 5 → -3y = 3 → y = -1 → Point: (1, -1)

When x = 4: 2(4) - 3y = 5 → 8 - 3y = 5 → -3y = -3 → y = 1 → Point: (4, 1)

When x = 7: 2(7) - 3y = 5 → 14 - 3y = 5 → -3y = -9 → y = 3 → Point: (7, 3)

Step 2: Find points for 4x - 6y = 15

When x = 3: 4(3) - 6y = 15 → 12 - 6y = 15 → -6y = 3 → y = -0.5 → Point: (3, -0.5)

When x = 6: 4(6) - 6y = 15 → 24 - 6y = 15 → -6y = -9 → y = 1.5 → Point: (6, 1.5)

When x = 9: 4(9) - 6y = 15 → 36 - 6y = 15 → -6y = -21 → y = 3.5 → Point: (9, 3.5)

Step 3: Check consistency

For equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂:

If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, then lines are parallel and inconsistent

Here: 2/4 = 1/2, -3/-6 = 1/2, 5/15 = 1/3

Since 1/2 = 1/2 ≠ 1/3, the lines are parallel

∴ The equations are inconsistent (no solution) as the lines are parallel

Draw the graph for the following pair of linear equations in two variables and find their solution from the graph. 3x– 2y = 2 and 2x + y = 6 (M'16)

Step 1: Find points for 3x - 2y = 2

When x = 0: 3(0) - 2y = 2 → 0 - 2y = 2 → -2y = 2 → y = -1 → Point: (0, -1)

When x = 2: 3(2) - 2y = 2 → 6 - 2y = 2 → -2y = -4 → y = 2 → Point: (2, 2)

When x = 4: 3(4) - 2y = 2 → 12 - 2y = 2 → -2y = -10 → y = 5 → Point: (4, 5)

Step 2: Find points for 2x + y = 6

When x = 0: 2(0) + y = 6 → 0 + y = 6 → y = 6 → Point: (0, 6)

When x = 2: 2(2) + y = 6 → 4 + y = 6 → y = 2 → Point: (2, 2)

When x = 3: 2(3) + y = 6 → 6 + y = 6 → y = 0 → Point: (3, 0)

∴ The lines intersect at point (2, 2), so the solution is x = 2, y = 2

Draw the graph for the equations 2x – y – 4 = 0 and x + y = 0 on the graph paper and check whether they are consistent or not. (J'16)

Step 1: Find points for 2x - y - 4 = 0

When x = 0: 2(0) - y - 4 = 0 → 0 - y - 4 = 0 → -y = 4 → y = -4 → Point: (0, -4)

When x = 2: 2(2) - y - 4 = 0 → 4 - y - 4 = 0 → -y = 0 → y = 0 → Point: (2, 0)

When x = 4: 2(4) - y - 4 = 0 → 8 - y - 4 = 0 → 4 - y = 0 → y = 4 → Point: (4, 4)

Step 2: Find points for x + y = 0

When x = -2: -2 + y = 0 → y = 2 → Point: (-2, 2)

When x = 0: 0 + y = 0 → y = 0 → Point: (0, 0)

When x = 2: 2 + y = 0 → y = -2 → Point: (2, -2)

Step 3: Find intersection point algebraically

From x + y = 0, we have y = -x

Substitute in 2x - y - 4 = 0: 2x - (-x) - 4 = 0 → 2x + x - 4 = 0 → 3x = 4 → x = 4/3

Then y = -4/3

∴ The equations are consistent (have a unique solution) and intersect at (4/3, -4/3)

Draw the graph of 2x + y = 6 and 2x – y + 2 = 0 and find the solution from the graph. (M'17)

Step 1: Find points for 2x + y = 6

When x = 0: 2(0) + y = 6 → 0 + y = 6 → y = 6 → Point: (0, 6)

When x = 1: 2(1) + y = 6 → 2 + y = 6 → y = 4 → Point: (1, 4)

When x = 3: 2(3) + y = 6 → 6 + y = 6 → y = 0 → Point: (3, 0)

Step 2: Find points for 2x - y + 2 = 0

When x = 0: 2(0) - y + 2 = 0 → 0 - y + 2 = 0 → -y = -2 → y = 2 → Point: (0, 2)

When x = 1: 2(1) - y + 2 = 0 → 2 - y + 2 = 0 → 4 - y = 0 → y = 4 → Point: (1, 4)

When x = 2: 2(2) - y + 2 = 0 → 4 - y + 2 = 0 → 6 - y = 0 → y = 6 → Point: (2, 6)

∴ The lines intersect at point (1, 4), so the solution is x = 1, y = 4

Show that the following pair of equations are consistent and show them graphically: x + 3y = 6 and 2x– 3y = 12 (J'17)

Step 1: Check consistency

For equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂:

If a₁/a₂ ≠ b₁/b₂, then lines intersect and are consistent

Here: 1/2 = 0.5, 3/-3 = -1

Since 0.5 ≠ -1, the lines intersect and are consistent

Step 2: Find points for x + 3y = 6

When x = 0: 0 + 3y = 6 → 3y = 6 → y = 2 → Point: (0, 2)

When x = 3: 3 + 3y = 6 → 3y = 3 → y = 1 → Point: (3, 1)

When x = 6: 6 + 3y = 6 → 3y = 0 → y = 0 → Point: (6, 0)

Step 3: Find points for 2x - 3y = 12

When x = 0: 2(0) - 3y = 12 → 0 - 3y = 12 → -3y = 12 → y = -4 → Point: (0, -4)

When x = 3: 2(3) - 3y = 12 → 6 - 3y = 12 → -3y = 6 → y = -2 → Point: (3, -2)

When x = 6: 2(6) - 3y = 12 → 12 - 3y = 12 → -3y = 0 → y = 0 → Point: (6, 0)

∴ The equations are consistent (have a unique solution) and intersect at (6, 0)

Solve the following pair of linear equations by graph method. 2x + y = 6 and 2x – y + 2 = 0. (J'18)

Step 1: Find points for 2x + y = 6

When x = 0: 2(0) + y = 6 → 0 + y = 6 → y = 6 → Point: (0, 6)

When x = 1: 2(1) + y = 6 → 2 + y = 6 → y = 4 → Point: (1, 4)

When x = 3: 2(3) + y = 6 → 6 + y = 6 → y = 0 → Point: (3, 0)

Step 2: Find points for 2x - y + 2 = 0

When x = 0: 2(0) - y + 2 = 0 → 0 - y + 2 = 0 → -y = -2 → y = 2 → Point: (0, 2)

When x = 1: 2(1) - y + 2 = 0 → 2 - y + 2 = 0 → 4 - y = 0 → y = 4 → Point: (1, 4)

When x = 2: 2(2) - y + 2 = 0 → 4 - y + 2 = 0 → 6 - y = 0 → y = 6 → Point: (2, 6)

∴ The lines intersect at point (1, 4), so the solution is x = 1, y = 4

Solve the equations by graphically 3x + 4y = 10 and 4x – 3y = 5. (M'19)

Step 1: Find points for 3x + 4y = 10

When x = 0: 3(0) + 4y = 10 → 0 + 4y = 10 → 4y = 10 → y = 2.5 → Point: (0, 2.5)

When x = 2: 3(2) + 4y = 10 → 6 + 4y = 10 → 4y = 4 → y = 1 → Point: (2, 1)

When x = 4: 3(4) + 4y = 10 → 12 + 4y = 10 → 4y = -2 → y = -0.5 → Point: (4, -0.5)

Step 2: Find points for 4x - 3y = 5

When x = 0: 4(0) - 3y = 5 → 0 - 3y = 5 → -3y = 5 → y = -5/3 ≈ -1.67 → Point: (0, -1.67)

When x = 2: 4(2) - 3y = 5 → 8 - 3y = 5 → -3y = -3 → y = 1 → Point: (2, 1)

When x = 4: 4(4) - 3y = 5 → 16 - 3y = 5 → -3y = -11 → y = 11/3 ≈ 3.67 → Point: (4, 3.67)

∴ The lines intersect at point (2, 1), so the solution is x = 2, y = 1

Sum of the present ages of two friends are 23 years, five years ago product of their ages was 42. Find their ages 5 years hence. (M'19)

Step 1: Set up equations

Let present ages be x and y years

x + y = 23 ...(1)

Five years ago: (x-5)(y-5) = 42 ...(2)

Step 2: Expand equation (2)

(x-5)(y-5) = 42

xy - 5x - 5y + 25 = 42

xy - 5(x+y) = 17

From (1): x+y = 23, so

xy - 5(23) = 17

xy - 115 = 17

xy = 132

Step 3: Solve the system

We have: x + y = 23 and xy = 132

This forms a quadratic: t² - 23t + 132 = 0

Discriminant = 23² - 4(1)(132) = 529 - 528 = 1

t = [23 ± √1]/2 = [23 ± 1]/2

t = 12 or t = 11

Step 4: Find ages 5 years hence

Present ages: 12 and 11 years

Ages 5 years hence: 17 and 16 years

∴ Their ages 5 years hence will be 17 years and 16 years

Draw the graph of x + y = 11 and x - y = 5. Find the solution of the pair of linear equations (J'19)

Step 1: Find points for x + y = 11

When x = 0: 0 + y = 11 → y = 11 → Point: (0, 11)

When x = 5: 5 + y = 11 → y = 6 → Point: (5, 6)

When x = 11: 11 + y = 11 → y = 0 → Point: (11, 0)

Step 2: Find points for x - y = 5

When x = 0: 0 - y = 5 → -y = 5 → y = -5 → Point: (0, -5)

When x = 5: 5 - y = 5 → -y = 0 → y = 0 → Point: (5, 0)

When x = 10: 10 - y = 5 → -y = -5 → y = 5 → Point: (10, 5)

∴ The lines intersect at point (8, 3), so the solution is x = 8, y = 3

Polynomial Questions -Solutions

zaheerpashamd

Polynomial Problems and Solutions

1-Mark Questions

If x ≠ –1, then find the quotient of (x⁵ + x⁴ + x³ + x²) / (x³ + x² + x + 1) (M'15)

Step 1: Factor numerator and denominator
Numerator: x⁵ + x⁴ + x³ + x² = x²(x³ + x² + x + 1)
Denominator: x³ + x² + x + 1

Step 2: Simplify the expression
(x⁵ + x⁴ + x³ + x²) / (x³ + x² + x + 1) = [x²(x³ + x² + x + 1)] / (x³ + x² + x + 1)

Step 3: Cancel common factors
Since x ≠ -1, (x³ + x² + x + 1) ≠ 0, so we can cancel:
= x²

∴ Quotient = x²

"We can write a trinomial having degree 7". Justify the above statement by giving one example. (M'15)

Step 1: Understand the terms
- A trinomial is a polynomial with exactly three terms.
- Degree of a polynomial is the highest power of the variable.

Step 2: Create an example
Example: 3x⁷ + 2x³ + 5
This has:
- Three terms: 3x⁷, 2x³, and 5
- Highest power: 7
- Therefore, it's a trinomial of degree 7.

∴ Example: 3x⁷ + 2x³ + 5
This is a trinomial with degree 7, which justifies the statement.

Write an example for a quadratic Polynomial that has no zeros. (M'16)

Step 1: Condition for no zeros
A quadratic polynomial has no real zeros when its discriminant (D = b² - 4ac) is negative.

Step 2: Create an example
Let's take p(x) = x² + 2x + 3
Discriminant D = b² - 4ac = (2)² - 4(1)(3) = 4 - 12 = -8 < 0
Since D < 0, this quadratic has no real zeros.

∴ Example: x² + 2x + 3
This quadratic polynomial has no real zeros as its discriminant is negative.

If p(x) = x³ – 3x² + 2x – 3 is a polynomial, then find the value of p(1). (J'16)

Step 1: Substitute x = 1
p(1) = (1)³ - 3(1)² + 2(1) - 3

Step 2: Calculate
p(1) = 1 - 3 + 2 - 3

Step 3: Simplify
p(1) = (1 + 2) + (-3 - 3) = 3 - 6 = -3

∴ p(1) = -3

Srikar says that the order of the polynomial (x² – 5)(x³ + 1) is 6. Do you agree with him? (J'17)

Step 1: Expand the polynomial
(x² - 5)(x³ + 1) = x²(x³ + 1) - 5(x³ + 1) = x⁵ + x² - 5x³ - 5

Step 2: Rearrange in standard form
= x⁵ - 5x³ + x² - 5

Step 3: Find the degree
The highest power of x is 5, so the degree (order) is 5.

∴ I don't agree with Srikar.
The order of the polynomial is 5, not 6.

Find zeros of the polynomial P(x) = x² – 4. (J'17)

Step 1: Set P(x) = 0
x² - 4 = 0

Step 2: Solve for x
x² = 4
x = ±√4 = ±2

∴ Zeros are x = 2 and x = -2

Verify the relation between zeros and coefficients of the quadratic polynomial x² – 4. (M'18)

Step 1: Identify coefficients
For x² - 4:
a = 1, b = 0, c = -4

Step 2: Find zeros
Zeros: α = 2, β = -2

Step 3: Verify sum of zeros
α + β = 2 + (-2) = 0
-b/a = -0/1 = 0
∴ α + β = -b/a

Step 4: Verify product of zeros
αβ = (2)(-2) = -4
c/a = -4/1 = -4
∴ αβ = c/a

∴ The relation between zeros and coefficients is verified:
Sum of zeros = -b/a = 0
Product of zeros = c/a = -4

Whether 1/2 and 1 are zeros of the polynomial p(x) = 2x² – 3x + 1 or not? Justify. (J'18)

Step 1: Check if 1/2 is a zero
p(1/2) = 2(1/2)² - 3(1/2) + 1 = 2(1/4) - 3/2 + 1 = 1/2 - 3/2 + 1 = -1 + 1 = 0
∴ 1/2 is a zero.

Step 2: Check if 1 is a zero
p(1) = 2(1)² - 3(1) + 1 = 2 - 3 + 1 = 0
∴ 1 is a zero.

∴ Both 1/2 and 1 are zeros of the polynomial p(x) = 2x² - 3x + 1.

If P(x) = x⁴ + 1, then find P(2) – P(-2). (M'19)

Step 1: Calculate P(2)
P(2) = (2)⁴ + 1 = 16 + 1 = 17

Step 2: Calculate P(-2)
P(-2) = (-2)⁴ + 1 = 16 + 1 = 17

Step 3: Find P(2) - P(-2)
P(2) - P(-2) = 17 - 17 = 0

∴ P(2) - P(-2) = 0

–3, 0 and 2 are the zeroes of the polynomial p(x) = x³ + (a – 1)x² + bx + c. Find a and c. (J'19)

Step 1: Use the zero x = 0
p(0) = (0)³ + (a - 1)(0)² + b(0) + c = c
Since 0 is a zero, p(0) = 0, so c = 0

Step 2: Use the zero x = 2
p(2) = (2)³ + (a - 1)(2)² + b(2) + 0 = 8 + 4(a - 1) + 2b = 0
8 + 4a - 4 + 2b = 0
4a + 2b + 4 = 0
2a + b + 2 = 0 ...(1)

Step 3: Use the zero x = -3
p(-3) = (-3)³ + (a - 1)(-3)² + b(-3) + 0 = -27 + 9(a - 1) - 3b = 0
-27 + 9a - 9 - 3b = 0
9a - 3b - 36 = 0
3a - b - 12 = 0 ...(2)

Step 4: Solve equations (1) and (2)
From (1): b = -2a - 2
Substitute in (2): 3a - (-2a - 2) - 12 = 0
3a + 2a + 2 - 12 = 0
5a - 10 = 0
a = 2

∴ a = 2, c = 0

Write any two linear polynomials having one term under three terms. (J'19)

Step 1: Understand linear polynomials
A linear polynomial has degree 1 and can have 1, 2, or 3 terms.

Step 2: Examples with one term
1. 5x (degree 1, one term)
2. -3x (degree 1, one term)

Step 3: Examples with three terms
1. 2x + 3y + 4 (degree 1 in x and y, three terms)
2. x + y - 2 (degree 1 in x and y, three terms)

∴ One-term examples: 5x, -3x
Three-term examples: 2x + 3y + 4, x + y - 2

If p(x) = x² + 3x + 4, then find the values of p(0) and p(1). (May 2022)

Step 1: Calculate p(0)
p(0) = (0)² + 3(0) + 4 = 0 + 0 + 4 = 4

Step 2: Calculate p(1)
p(1) = (1)² + 3(1) + 4 = 1 + 3 + 4 = 8

∴ p(0) = 4, p(1) = 8

If p(x) = 2x² + 5x – 7, then find the value of p(0) and p(1). (Aug 2022)

Step 1: Calculate p(0)
p(0) = 2(0)² + 5(0) - 7 = 0 + 0 - 7 = -7

Step 2: Calculate p(1)
p(1) = 2(1)² + 5(1) - 7 = 2 + 5 - 7 = 0

∴ p(0) = -7, p(1) = 0

Polynomial Problems - 2 Marks

Polynomial-2 Marks

For what value of k, –4 is a zero of the polynomial x² – x – (2k + 2). (J'15)

Step 1: Substitute x = -4 in the polynomial
p(x) = x² - x - (2k + 2)
p(-4) = (-4)² - (-4) - (2k + 2)

Step 2: Simplify the expression
p(-4) = 16 + 4 - 2k - 2 = 18 - 2k

Step 3: Set p(-4) = 0 (since -4 is a zero)
18 - 2k = 0

Step 4: Solve for k
2k = 18
k = 9

∴ k = 9

Use the table given below to draw the graph. Use the graph drawn to find the values of a and b. (J'15)

Given table:

x	-2	0	2	1	b
y	-3	1	a	3	-7

Step 1: Assume the relationship is linear (y = mx + c)
Using points (-2, -3) and (0, 1):
When x = 0, y = 1 ⇒ c = 1
Using (-2, -3): -3 = m(-2) + 1 ⇒ -3 = -2m + 1 ⇒ -2m = -4 ⇒ m = 2
∴ Equation: y = 2x + 1

Step 2: Find value of a
When x = 2: y = 2(2) + 1 = 4 + 1 = 5
∴ a = 5

Step 3: Find value of b
When y = -7: -7 = 2b + 1 ⇒ 2b = -8 ⇒ b = -4

∴ a = 5, b = -4

Length of a rectangle is 5 units more than its breadth. Express its perimeter in polynomial form.

Step 1: Define variables
Let breadth = x units
Then length = (x + 5) units

Step 2: Write perimeter formula
Perimeter = 2 × (length + breadth)
= 2 × [(x + 5) + x]
= 2 × (2x + 5)

Step 3: Simplify to polynomial form
= 4x + 10

∴ Perimeter = 4x + 10

Show that 2 and −1/3 are zeros of the polynomial 3x² – 5x – 2. (J'16)

Given polynomial: p(x) = 3x² - 5x - 2

Step 1: Check if x = 2 is a zero
p(2) = 3(2)² - 5(2) - 2 = 3(4) - 10 - 2 = 12 - 10 - 2 = 0
∴ 2 is a zero.

Step 2: Check if x = -1/3 is a zero
p(-1/3) = 3(-1/3)² - 5(-1/3) - 2 = 3(1/9) + 5/3 - 2 = 1/3 + 5/3 - 2 = 6/3 - 2 = 2 - 2 = 0
∴ -1/3 is a zero.

∴ Both 2 and -1/3 are zeros of the polynomial 3x² - 5x - 2.

Which of √2 and 2 is a zero of the polynomial p(x) = x³ – 2x? Why? (M'17)

Given polynomial: p(x) = x³ - 2x

Step 1: Check if x = √2 is a zero
p(√2) = (√2)³ - 2(√2) = 2√2 - 2√2 = 0
∴ √2 is a zero.

Step 2: Check if x = 2 is a zero
p(2) = (2)³ - 2(2) = 8 - 4 = 4 ≠ 0
∴ 2 is not a zero.

∴ √2 is a zero of the polynomial, but 2 is not.

Divide x³ – 3x² + 5x – 3 by x² – 2. And verify the division lemma. (J'17)

Division:

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

Step 1: Result of division
Quotient = x - 3
Remainder = 7x - 9

Step 2: Verify division lemma
Division lemma: Dividend = Divisor × Quotient + Remainder
(x² - 2)(x - 3) + (7x - 9) = x³ - 3x² - 2x + 6 + 7x - 9 = x³ - 3x² + 5x - 3
This matches the original dividend.

∴ Quotient = x - 3, Remainder = 7x - 9
The division lemma is verified.

Complete the following table for the polynomial y = p(x) = x³ – 2x + 3. (M'18)

Given polynomial: y = x³ - 2x + 3

x	-1	0	1	2
x³	(-1)³ = -1	0³ = 0	1³ = 1	2³ = 8
-2x	-2(-1) = 2	-2(0) = 0	-2(1) = -2	-2(2) = -4
3	3	3	3	3
y	-1 + 2 + 3 = 4	0 + 0 + 3 = 3	1 - 2 + 3 = 2	8 - 4 + 3 = 7
(x, y)	(-1, 4)	(0, 3)	(1, 2)	(2, 7)

∴ The completed table is shown above.

If one of the zeros of the cubic polynomial p(x) = ax³ + bx² + cx + d is zero, then find the product of other two zeros of p(x). (a ≠ 0) (J'18)

Step 1: Let the zeros be α, β, γ with γ = 0
So p(x) = a(x - α)(x - β)(x - 0) = a(x - α)(x - β)x

Step 2: Expand the polynomial
p(x) = a[x³ - (α + β)x² + αβx] = ax³ - a(α + β)x² + aαβx

Step 3: Compare with given form
p(x) = ax³ + bx² + cx + d
Comparing coefficients:
-a(α + β) = b ⇒ α + β = -b/a
aαβ = c ⇒ αβ = c/a
d = 0 (since constant term is 0 when one zero is 0)

Step 4: Product of other two zeros
The product of the other two zeros = αβ = c/a

∴ Product of other two zeros = c/a

Divide x³ – 4x² + 5x – 2 by x – 2. (M'19)

Division using synthetic division:

         x² - 2x + 1
        -----------
x - 2 | x³ - 4x² + 5x - 2
          x³ - 2x²
        -----------
             -2x² + 5x
             -2x² + 4x
             -----------
                   x - 2
                   x - 2
                   -----
                     0

Step 1: Result of division
Quotient = x² - 2x + 1
Remainder = 0

Step 2: Verify
(x - 2)(x² - 2x + 1) = x³ - 2x² - 2x² + 4x + x - 2 = x³ - 4x² + 5x - 2
This matches the original dividend.

∴ Quotient = x² - 2x + 1, Remainder = 0

Polynomial Problems - 4 Mark Questions

Lakshmi does not want to disclose the length, breadth and height of a cuboid of her project. She has constructed a polynomial x³ – 6x² + 11x – 6 by taking the values of length, breadth and height as its zeros. Can you open the secret [i.e., find the measures of length, breadth and height]? (M'15)

Step 1: Find the zeros of the polynomial
Given polynomial: p(x) = x³ - 6x² + 11x - 6

Step 2: Check for possible rational zeros
Possible zeros: ±1, ±2, ±3, ±6

Step 3: Check x = 1
p(1) = 1 - 6 + 11 - 6 = 0
∴ x = 1 is a zero.

Step 4: Factor out (x - 1)
Using synthetic division:
Coefficients: 1, -6, 11, -6
Dividing by (x - 1):
Bring down 1, multiply by 1 → 1, add to -6 → -5, multiply by 1 → -5, add to 11 → 6, multiply by 1 → 6, add to -6 → 0
Quotient: x² - 5x + 6

Step 5: Factor the quadratic
x² - 5x + 6 = (x - 2)(x - 3)

Step 6: Complete factorization
p(x) = (x - 1)(x - 2)(x - 3)

∴ The zeros are 1, 2, and 3
Therefore, the dimensions of the cuboid are:
Length = 3 units, Breadth = 2 units, Height = 1 unit (or any permutation of these values).

Draw the graph for the polynomial p(x) = x² + 3x - 4 and find its zeroes from the graph. (M'15, J'19)

Step 1: Create a table of values

x	-5	-4	-3	-2	-1	0	1	2
p(x)	6	0	-4	-6	-6	-4	0	6

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (-1.5, -6.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -4 and x = 1.

Step 4: Verify algebraically
x² + 3x - 4 = 0
(x + 4)(x - 1) = 0
x = -4 or x = 1

∴ The zeros are x = -4 and x = 1

Draw the graph of the polynomial p(x) = 3x² + 2x - 1 on the graph paper. Find its zeros from the graph. (J'15)

Step 1: Create a table of values

x	-2	-1	-0.5	0	0.5	1	2
p(x)	7	0	-1.25	-1	0.75	4	15

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (-1/3, -4/3).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -1 and x = 1/3.

Step 4: Verify algebraically
3x² + 2x - 1 = 0
(3x - 1)(x + 1) = 0
x = 1/3 or x = -1

∴ The zeros are x = -1 and x = 1/3

Draw the graph for the polynomial p(x) = x² - 3x + 2 and find the zeroes from the graph. (M'16)

Step 1: Create a table of values

x	-1	0	1	1.5	2	3	4
p(x)	6	2	0	-0.25	0	2	6

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (1.5, -0.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 1 and x = 2.

Step 4: Verify algebraically
x² - 3x + 2 = 0
(x - 1)(x - 2) = 0
x = 1 or x = 2

∴ The zeros are x = 1 and x = 2

Draw the graph of the polynomial p(x) = x² - 5x + 4 on the graph paper. Find its zeros from the graph. (J'16)

Step 1: Create a table of values

x	-1	0	1	2	2.5	3	4	5
p(x)	10	4	0	-2	-2.25	-2	0	4

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (2.5, -2.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 1 and x = 4.

Step 4: Verify algebraically
x² - 5x + 4 = 0
(x - 1)(x - 4) = 0
x = 1 or x = 4

∴ The zeros are x = 1 and x = 4

On dividing x³ - 3x² + 5x - 7 by x² - 2x + 4, if the remainder is in the form of Ax + B, find the values of A and B. (J'16)

Step 1: Perform polynomial division

         x - 1
        ---------------
x² - 2x + 4 | x³ - 3x² + 5x - 7
          x³ - 2x² + 4x
        ---------------
             -x² + x - 7
             -x² + 2x - 4
             -----------
                  -x - 3

Step 2: Identify remainder
Remainder = -x - 3

Step 3: Compare with Ax + B
-x - 3 = Ax + B
A = -1, B = -3

∴ A = -1, B = -3

Divide 3x⁴ - 5x³ + 4x² + 3x - 5 by x² - 3 and verify the division algorithm. (M'17)

Step 1: Perform polynomial division

         3x² - 5x + 13
        ---------------
x² - 3 | 3x⁴ - 5x³ + 4x² + 3x - 5
         3x⁴ + 0x³ - 9x²
        ---------------
             -5x³ + 13x² + 3x
             -5x³ + 0x² + 15x
             ---------------
                  13x² - 12x - 5
                  13x² + 0x - 39
                  ---------------
                       -12x + 34

Step 2: Identify quotient and remainder
Quotient = 3x² - 5x + 13
Remainder = -12x + 34

Step 3: Verify division algorithm
Division algorithm: Dividend = Divisor × Quotient + Remainder
(x² - 3)(3x² - 5x + 13) + (-12x + 34)
= 3x⁴ - 5x³ + 13x² - 9x² + 15x - 39 - 12x + 34
= 3x⁴ - 5x³ + 4x² + 3x - 5
This matches the original dividend.

∴ Quotient = 3x² - 5x + 13, Remainder = -12x + 34
The division algorithm is verified.

The perimeter of a right-angle triangle is 60 cm and its hypotenuse is 25 cm. Then find the remaining two sides. (M'17)

Step 1: Set up equations
Let the sides be a, b, and c where c = 25 cm (hypotenuse)
Perimeter: a + b + c = 60 ⇒ a + b = 35
Pythagorean theorem: a² + b² = c² = 625

Step 2: Solve the system
From a + b = 35, we get b = 35 - a
Substitute in a² + b² = 625:
a² + (35 - a)² = 625
a² + 1225 - 70a + a² = 625
2a² - 70a + 1225 - 625 = 0
2a² - 70a + 600 = 0
Divide by 2: a² - 35a + 300 = 0

Step 3: Solve the quadratic
a² - 35a + 300 = 0
(a - 15)(a - 20) = 0
a = 15 or a = 20

Step 4: Find the corresponding sides
If a = 15, then b = 35 - 15 = 20
If a = 20, then b = 35 - 20 = 15

∴ The remaining two sides are 15 cm and 20 cm

Draw the graph of the polynomial p(x) = x² - 5x + 6 and find the zeros from the graph. (M'17)

Step 1: Create a table of values

x	-1	0	1	2	2.5	3	4	5
p(x)	12	6	2	0	-0.25	0	2	6

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (2.5, -0.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 2 and x = 3.

Step 4: Verify algebraically
x² - 5x + 6 = 0
(x - 2)(x - 3) = 0
x = 2 or x = 3

∴ The zeros are x = 2 and x = 3

Draw the graph of p(x) = x² - 2x - 8 and find the zeros of the polynomial from it. (J'17)

Step 1: Create a table of values

x	-3	-2	-1	0	1	2	3	4	5
p(x)	7	0	-5	-8	-9	-8	-5	0	7

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (1, -9).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -2 and x = 4.

Step 4: Verify algebraically
x² - 2x - 8 = 0
(x - 4)(x + 2) = 0
x = 4 or x = -2

∴ The zeros are x = -2 and x = 4

Total number of pencils required are given by 4x⁴ + 2x³ - 2x² + 62x - 66. If each box contains x² + 2x - 3 pencils, then find the number of boxes to be purchased. (M'18)

Step 1: Perform polynomial division
We need to divide 4x⁴ + 2x³ - 2x² + 62x - 66 by x² + 2x - 3

Step 2: Division process

         4x² - 6x + 22
        ---------------
x² + 2x - 3 | 4x⁴ + 2x³ - 2x² + 62x - 66
         4x⁴ + 8x³ - 12x²
        ---------------
             -6x³ + 10x² + 62x
             -6x³ - 12x² + 18x
             ---------------
                  22x² + 44x - 66
                  22x² + 44x - 66
                  ---------------
                            0

Step 3: Interpret the result
The quotient is 4x² - 6x + 22 and the remainder is 0.

∴ Number of boxes to be purchased = 4x² - 6x + 22

Draw the graph of the polynomial p(x) = x² + x - 2 on the graph paper. Find its zeroes from the graph. (J'18)

Step 1: Create a table of values

x	-3	-2	-1	-0.5	0	1	2
p(x)	4	0	-2	-2.25	-2	0	4

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (-0.5, -2.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -2 and x = 1.

Step 4: Verify algebraically
x² + x - 2 = 0
(x + 2)(x - 1) = 0
x = -2 or x = 1

∴ The zeros are x = -2 and x = 1

Draw the graph of the polynomial p(x) = x² - 7x + 12, then find its zeroes from the graph. (M'19)

Step 1: Create a table of values

x	0	1	2	3	3.5	4	5	6
p(x)	12	6	2	0	-0.25	0	2	6

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (3.5, -0.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 3 and x = 4.

Step 4: Verify algebraically
x² - 7x + 12 = 0
(x - 3)(x - 4) = 0
x = 3 or x = 4

∴ The zeros are x = 3 and x = 4

Draw the graph of the polynomial p(x) = x² + 2x - 3 and find the zeroes of the polynomial from the graph. (May 2022, Jun'23)

Step 1: Create a table of values

x	-4	-3	-2	-1	0	1	2
p(x)	5	0	-3	-4	-3	0	5

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (-1, -4).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -3 and x = 1.

Step 4: Verify algebraically
x² + 2x - 3 = 0
(x + 3)(x - 1) = 0
x = -3 or x = 1

∴ The zeros are x = -3 and x = 1

Draw the graph of the quadratic polynomial p(x) = x² - 4x + 3 and find the zeroes of the polynomial from the graph. (Apr'23)

Step 1: Create a table of values

x	-1	0	1	2	3	4	5
p(x)	8	3	0	-1	0	3	8

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (2, -1).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 1 and x = 3.

Step 4: Verify algebraically
x² - 4x + 3 = 0
(x - 1)(x - 3) = 0
x = 1 or x = 3

∴ The zeros are x = 1 and x = 3

Draw the graph of the polynomial p(x) = x² - x - 2 and find the zeros of the polynomial from the graph. (Aug 2022)

Step 1: Create a table of values

x	-2	-1	0	0.5	1	2	3
p(x)	4	0	-2	-2.25	-2	0	4

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (0.5, -2.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -1 and x = 2.

Step 4: Verify algebraically
x² - x - 2 = 0
(x - 2)(x + 1) = 0
x = 2 or x = -1

∴ The zeros are x = -1 and x = 2

Class 10 Maths Formulas

zaheerpashamd

Class 10 Maths Formula Sheet - Ultimate Revision Guide

Class 10 Maths Formula Sheet

All Chapters. All Formulas. One Place. Use the search bar to find anything instantly!

Chapter 1: Real Numbers

Euclid's Division Lemma:

For $a$ and $b$, $a = bq + r$, where $0 \le r < b$.
Product of HCF and LCM (Two Numbers):

$HCF(a, b) \times LCM(a, b) = a \times b$
Condition for Terminating Decimal $\frac{p}{q}$:

$q$ must be of the form $2^m \times 5^n$.

Chapter 2: Polynomials

Sum of Zeroes ($\alpha + \beta$) for $ax^2 + bx + c$:

$\alpha + \beta = -\frac{b}{a}$
Product of Zeroes ($\alpha\beta$) for $ax^2 + bx + c$:

$\alpha\beta = \frac{c}{a}$
Division Algorithm:

$p(x) = g(x) \times q(x) + r(x)$

Chapter 3 & 4: Linear & Quadratic Equations

Quadratic Formula ($ax^2 + bx + c = 0$):

$x = \frac{-b \pm \sqrt{D}}{2a}$, where $D = b^2 - 4ac$ (Discriminant)
Nature of Roots:

D > 0: Real & Distinct
D = 0: Real & Equal
D < 0: No Real Roots
Cross Multiplication Method:

$x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}$ and $y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}$

Chapter 5: Arithmetic Progression (AP)

$n^{\text{th}}$ Term ($T_n$):

$T_n = a + (n-1)d$
Sum of $n$ Terms ($S_n$):

$S_n = \frac{n}{2}[2a + (n-1)d] \quad \text{or} \quad S_n = \frac{n}{2}[a + l]$
Arithmetic Mean ($a, b, c$ in AP):

$2b = a + c$

Chapter 6: Triangles

Basic Proportionality Theorem (BPT/Thales):

If $\text{DE} \parallel \text{BC}$, then $\frac{AD}{BD} = \frac{AE}{CE}$
Ratio of Areas (Similar $\triangle$s):

$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \frac{AB^2}{PQ^2}$
Pythagoras Theorem ($\angle B = 90^\circ$):

$AB^2 + BC^2 = AC^2$

Chapter 7: Coordinate Geometry

Distance Formula:

$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Section Formula ($m:n$):

$\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$
Area of Triangle:

$\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Chapter 8 & 9: Trigonometry

Trigonometric Identity 1:

$\sin^2 \theta + \cos^2 \theta = 1$
Trigonometric Identity 2:

$1 + \tan^2 \theta = \sec^2 \theta$
Trigonometric Identity 3:

$1 + \cot^2 \theta = \csc^2 \theta$
Complementary Angle:

$\sin(90^\circ - \theta) = \cos \theta$ and $\tan(90^\circ - \theta) = \cot \theta$

Chapter 10: Circles

Tangent-Radius Theorem:

Tangent at any point is $\perp$ to the radius through point of contact.
Tangents from External Point:

Lengths of tangents drawn from an external point are equal ($PQ=PR$).
Cyclic Quadrilateral:

Sum of opposite angles is $180^\circ$ ($\angle A + \angle C = 180^\circ$).

Chapter 12: Area Related to Circles

Area of Circle:

$\pi r^2$
Arc Length (Angle $\theta$):

Arc Length $= \frac{\theta}{360} \times 2\pi r$
Area of Sector (Angle $\theta$):

Area of Sector $= \frac{\theta}{360} \times \pi r^2$
Area of Minor Segment:

Area of Sector $-\frac{1}{2} r^2 \sin \theta$

Chapter 13: Surface Area & Volume

Volume of Cuboid:

$L \times B \times H$
Volume of Cylinder:

$\pi r^2 h$
CSA of Cone:

$\pi r l$, where $l = \sqrt{r^2+h^2}$
Volume of Sphere:

$\frac{4}{3} \pi r^3$

Chapter 14: Statistics

Mean (Direct Method):

$\overline{X} = \frac{\sum f_i x_i}{\sum f_i}$
Median (Grouped Data):

$L + \left(\frac{\frac{n}{2} - \text{pcf}}{f}\right) \times h$
Mode (Grouped Data):

$L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$
Empirical Relation:

Mode = 3 Median - 2 Mean

Chapter 15: Probability

Probability of an Event $E$ ($P(E)$):

$P(E) = \frac{\text{Number of Favourable Outcomes}}{\text{Total Number of Possible Outcomes}}$
Complementary Event ($\overline{E}$):

$P(E) + P(\overline{E}) = 1$
Range of Probability:

$0 \le P(E) \le 1$

Set Questions-Solutions

zaheerpashamd

Set Theory Problems - 1 Mark Questions

Set Operations and Representations

If A = {x : x ∈ N and x < 20} and B = {x : x ∈ N and x ≤ 5} then write the set A – B in the set builder form. (M'15)

Step 1: Find the elements of set A
A = {x : x ∈ N and x < 20}
∴ A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}

Step 2: Find the elements of set B
B = {x : x ∈ N and x ≤ 5}
∴ B = {1, 2, 3, 4, 5}

Step 3: Find A - B
A - B = {x : x ∈ A and x ∉ B}
∴ A - B = {6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}

Step 4: Write in set builder form
A - B = {x : x ∈ N, 5 < x < 20}

∴ A - B = {x : x ∈ N, 5 < x < 20}

"B is a set of all months in a year having 30 days". Write the above set in the roster form. (J'15)

Step 1: Identify months with 30 days
Months with 30 days: April, June, September, November

Step 2: Write in roster form
B = {April, June, September, November}

∴ B = {April, June, September, November}

If A – B = {3, 4, 5}, B – A = {1, 8, 9} and A ∩ B = {6, 7}, then find A ∪ B. (J'15)

Step 1: Understand the given information
A - B = {3, 4, 5} → Elements in A but not in B
B - A = {1, 8, 9} → Elements in B but not in A
A ∩ B = {6, 7} → Elements common to both A and B

Step 2: Find A ∪ B
A ∪ B = (A - B) ∪ (A ∩ B) ∪ (B - A)
A ∪ B = {3, 4, 5} ∪ {6, 7} ∪ {1, 8, 9}

Step 3: Combine all elements
A ∪ B = {1, 3, 4, 5, 6, 7, 8, 9}

∴ A ∪ B = {1, 3, 4, 5, 6, 7, 8, 9}

If A = {1, 1/4, 1/9, 1/16, 1/25}, then write A in set builder form. (M'16)

Step 1: Observe the pattern
Elements: 1, 1/4, 1/9, 1/16, 1/25
These can be written as: 1/1², 1/2², 1/3², 1/4², 1/5²

Step 2: Write in set builder form
A = {1/n² : n ∈ N, 1 ≤ n ≤ 5}

∴ A = {1/n² : n ∈ N, 1 ≤ n ≤ 5}

A = {x: x ∈ N, x is a composite number and x < 13}. Write set A in the roster form. (J'16)

Step 1: Identify composite numbers less than 13
Composite numbers are positive integers that have at least one divisor other than 1 and themselves.

Step 2: List composite numbers < 13
4, 6, 8, 9, 10, 12

Step 3: Write in roster form
A = {4, 6, 8, 9, 10, 12}

∴ A = {4, 6, 8, 9, 10, 12}

Represent A ∩ B through Venn diagram, where A = {1, 4, 6, 9, 10} and B = {x / x is a perfect square less than 25}. (M'17)

Step 1: Find elements of set B
B = {x / x is a perfect square less than 25}
Perfect squares less than 25: 1, 4, 9, 16
∴ B = {1, 4, 9, 16}

Step 2: Find A ∩ B
A = {1, 4, 6, 9, 10}
B = {1, 4, 9, 16}
A ∩ B = {1, 4, 9}

Step 3: Describe the Venn diagram
In a Venn diagram, A ∩ B would be represented by the overlapping region of two circles, one for set A and one for set B, containing the elements {1, 4, 9}.

∴ A ∩ B = {1, 4, 9}

If A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6}, Find A ∩ B. (J'17)

Step 1: Identify common elements
A = {1, 2, 3, 4, 5}
B = {3, 4, 5, 6}

Step 2: Find intersection
A ∩ B = {x : x ∈ A and x ∈ B}
A ∩ B = {3, 4, 5}

∴ A ∩ B = {3, 4, 5}

Give one example each for a finite set and an infinite set. (M'18)

Finite Set Example:
A set with a countable number of elements.
Example: A = {1, 2, 3, 4, 5} or B = {x : x is a day of the week}

Infinite Set Example:
A set with an unlimited number of elements.
Example: N = {1, 2, 3, 4, ...} or Z = {..., -3, -2, -1, 0, 1, 2, 3, ...}

∴ Finite Set: {1, 2, 3, 4, 5}
Infinite Set: {1, 2, 3, 4, ...}

List all the subsets of the set A = {x, y, z} (J'18)

Step 1: Subsets with 0 elements (Empty set)
∅ or {}

Step 2: Subsets with 1 element
{x}, {y}, {z}

Step 3: Subsets with 2 elements
{x, y}, {x, z}, {y, z}

Step 4: Subset with 3 elements (The set itself)
{x, y, z}

∴ All subsets: ∅, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}

If A = {x: x is a factor of 24}, then find n(A). (M'19)

Step 1: Find factors of 24
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Step 2: Write set A
A = {1, 2, 3, 4, 6, 8, 12, 24}

Step 3: Find n(A) - number of elements in A
n(A) = 8

∴ n(A) = 8

If A = {1, 2, 3}, B = {3, 4, 5} Then find A - B and B - A. (J'19)

Step 1: Find A - B
A - B = {x : x ∈ A and x ∉ B}
A = {1, 2, 3}, B = {3, 4, 5}
A - B = {1, 2}

Step 2: Find B - A
B - A = {x : x ∈ B and x ∉ A}
B - A = {4, 5}

∴ A - B = {1, 2}
B - A = {4, 5}

A = {x : x is a factor of 8}, B = {x : x is a factor of 36}. Is A ⊂ B? Justify. (Jun'23)

Step 1: Find elements of A
Factors of 8: 1, 2, 4, 8
∴ A = {1, 2, 4, 8}

Step 2: Find elements of B
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
∴ B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

Step 3: Check if A ⊂ B
A ⊂ B if every element of A is also an element of B.
Check each element of A:
1 ∈ B, 2 ∈ B, 4 ∈ B, 8 ∈ B?
8 is NOT an element of B

∴ A is NOT a subset of B because 8 ∈ A but 8 ∉ B

Set Theory Problems - 2 Mark Questions

If A = {x : x ∈ N and x < 6} and B = {x : x ∈ N and 3 < x < 8} then Show that A – B ≠ B – A with the help of Venn diagram. (M'15)

Step 1: Find the elements of set A
A = {x : x ∈ N and x < 6}
∴ A = {1, 2, 3, 4, 5}

Step 2: Find the elements of set B
B = {x : x ∈ N and 3 < x < 8}
∴ B = {4, 5, 6, 7}

Step 3: Find A - B
A - B = {x : x ∈ A and x ∉ B}
A - B = {1, 2, 3}

Step 4: Find B - A
B - A = {x : x ∈ B and x ∉ A}
B - A = {6, 7}

Step 5: Venn Diagram Description

In the Venn diagram:
- Circle A contains elements: {1, 2, 3, 4, 5}
- Circle B contains elements: {4, 5, 6, 7}
- The overlapping region (A ∩ B) contains: {4, 5}
- A - B is represented by the region in A but not in B: {1, 2, 3}
- B - A is represented by the region in B but not in A: {6, 7}
These two regions are clearly different.

∴ A - B = {1, 2, 3} and B - A = {6, 7}
Since {1, 2, 3} ≠ {6, 7}, we have shown that A - B ≠ B - A.

Answer the following questions and justify your answers. A = {x : x ∈ N, x < 2015}, is it a finite set or infinite set? B = {x : x + 5 = 5} is it a null set or a Universal set? (J'15)

For Set A:
A = {x : x ∈ N, x < 2015}
This means A contains all natural numbers less than 2015.
These are: 1, 2, 3, ..., 2014
The count is finite (2014 elements).

For Set B:
B = {x : x + 5 = 5}
Solving x + 5 = 5, we get x = 0
So B = {0}
This is not an empty set as it contains one element (0).
It's also not a universal set as it doesn't contain all possible elements.

∴ A is a finite set because it has a finite number of elements (2014).
B is neither a null set nor a universal set - it's a singleton set containing {0}.

A = {x : x ∈ N, and x is a factor of 30}; B = {x : x ∈ N, and x is a prime factor of 30} draw Venn diagram for A∪B (J'16)

Step 1: Find elements of set A
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30
∴ A = {1, 2, 3, 5, 6, 10, 15, 30}

Step 2: Find elements of set B
Prime factors of 30: 2, 3, 5
∴ B = {2, 3, 5}

Step 3: Find A ∪ B
A ∪ B = {1, 2, 3, 5, 6, 10, 15, 30}

Step 4: Venn Diagram Description

In the Venn diagram:
- Circle A contains: {1, 2, 3, 5, 6, 10, 15, 30}
- Circle B contains: {2, 3, 5}
- Since all elements of B are already in A, circle B is completely inside circle A
- The union A ∪ B is represented by the entire circle A
- The overlapping region (A ∩ B) is exactly equal to B: {2, 3, 5}

∴ A ∪ B = {1, 2, 3, 5, 6, 10, 15, 30}

If A = {x : x ∈ N, x < 10}, B = {x : x is a prime number and x < 10}, Then show that A – B ≠ B – A with the help of Venn diagram. (J'17)

Step 1: Find the elements of set A
A = {x : x ∈ N, x < 10}
∴ A = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Step 2: Find the elements of set B
B = {x : x is a prime number and x < 10}
Prime numbers less than 10: 2, 3, 5, 7
∴ B = {2, 3, 5, 7}

Step 3: Find A - B
A - B = {x : x ∈ A and x ∉ B}
A - B = {1, 4, 6, 8, 9}

Step 4: Find B - A
B - A = {x : x ∈ B and x ∉ A}
Since all elements of B are in A, B - A = ∅

Step 5: Venn Diagram Description

In the Venn diagram:
- Circle A contains: {1, 2, 3, 4, 5, 6, 7, 8, 9}
- Circle B contains: {2, 3, 5, 7}
- The overlapping region (A ∩ B) contains: {2, 3, 5, 7}
- A - B is represented by the region in A but not in B: {1, 4, 6, 8, 9}
- B - A is represented by the region in B but not in A: ∅ (empty)
These two regions are clearly different.

∴ A - B = {1, 4, 6, 8, 9} and B - A = ∅
Since {1, 4, 6, 8, 9} ≠ ∅, we have shown that A - B ≠ B - A.

If A = {1, 2, 3, 4}, B = {2, 4, 6, 8, 10}, then represent the Venn diagram of A - B. (J'18)

Step 1: Find A - B
A - B = {x : x ∈ A and x ∉ B}
A = {1, 2, 3, 4}, B = {2, 4, 6, 8, 10}
A - B = {1, 3}

Step 2: Venn Diagram Description

In the Venn diagram:
- Circle A contains: {1, 2, 3, 4}
- Circle B contains: {2, 4, 6, 8, 10}
- The overlapping region (A ∩ B) contains: {2, 4}
- A - B is represented by the region in A but not in B: {1, 3}
This region should be shaded or highlighted to represent A - B.

∴ A - B = {1, 3}

If μ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {2, 3, 5, 8} and B = {0, 3, 5, 7, 10}. Then represent A∩B in the Venn diagram. (M'19)

Step 1: Find A ∩ B
A ∩ B = {x : x ∈ A and x ∈ B}
A = {2, 3, 5, 8}, B = {0, 3, 5, 7, 10}
A ∩ B = {3, 5}

Step 2: Venn Diagram Description

In the Venn diagram with universal set μ:
- There is a rectangle representing the universal set μ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
- Circle A contains: {2, 3, 5, 8}
- Circle B contains: {0, 3, 5, 7, 10}
- The overlapping region (A ∩ B) contains: {3, 5}
This overlapping region should be shaded or highlighted to represent A ∩ B.

∴ A ∩ B = {3, 5}

If A = {x : x is a factor of 12} and B = {x: x is a factor of 6} then find A∪B and A∩B. (J'19)

Step 1: Find elements of set A
Factors of 12: 1, 2, 3, 4, 6, 12
∴ A = {1, 2, 3, 4, 6, 12}

Step 2: Find elements of set B
Factors of 6: 1, 2, 3, 6
∴ B = {1, 2, 3, 6}

Step 3: Find A ∪ B
A ∪ B = {1, 2, 3, 4, 6, 12}

Step 4: Find A ∩ B
A ∩ B = {1, 2, 3, 6}

∴ A ∪ B = {1, 2, 3, 4, 6, 12}
A ∩ B = {1, 2, 3, 6}

If A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8}. Then show that n(A ∪ B) = n(A) + n(B) – n(A ∩ B). (May 2022)

Step 1: Find n(A)
A = {1, 2, 3, 4, 5}
n(A) = 5

Step 2: Find n(B)
B = {2, 4, 6, 8}
n(B) = 4

Step 3: Find A ∩ B
A ∩ B = {2, 4}
n(A ∩ B) = 2

Step 4: Find A ∪ B
A ∪ B = {1, 2, 3, 4, 5, 6, 8}
n(A ∪ B) = 7

Step 5: Verify the formula
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
7 = 5 + 4 - 2
7 = 7

∴ The formula n(A ∪ B) = n(A) + n(B) – n(A ∩ B) is verified
as 7 = 5 + 4 - 2.

If A = {4, 8, 12, 16, 20}, B = {6, 12, 18, 24, 30}, then show that n(A∪B) = n(A) + n(B) – n(A∩B). (Aug 2022)

Step 1: Find n(A)
A = {4, 8, 12, 16, 20}
n(A) = 5

Step 2: Find n(B)
B = {6, 12, 18, 24, 30}
n(B) = 5

Step 3: Find A ∩ B
A ∩ B = {12}
n(A ∩ B) = 1

Step 4: Find A ∪ B
A ∪ B = {4, 6, 8, 12, 16, 18, 20, 24, 30}
n(A ∪ B) = 9

Step 5: Verify the formula
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
9 = 5 + 5 - 1
9 = 9

∴ The formula n(A ∪ B) = n(A) + n(B) – n(A ∩ B) is verified
as 9 = 5 + 5 - 1.

Set Theory Problems and Solutions

4-Mark Questions

X is a set of factors of 24 and Y is a set of factors of 36, then find sets X∪Y and X∩Y by using Venn diagram and comment on the answer. (M'16)

Step 1: Find elements of set X
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
∴ X = {1, 2, 3, 4, 6, 8, 12, 24}

Step 2: Find elements of set Y
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
∴ Y = {1, 2, 3, 4, 6, 9, 12, 18, 36}

Step 3: Find X ∪ Y
X ∪ Y = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}

Step 4: Find X ∩ Y
X ∩ Y = {1, 2, 3, 4, 6, 12}

Step 5: Venn Diagram Description

In the Venn diagram:
- Circle X contains: {1, 2, 3, 4, 6, 8, 12, 24}
- Circle Y contains: {1, 2, 3, 4, 6, 9, 12, 18, 36}
- The overlapping region (X ∩ Y) contains: {1, 2, 3, 4, 6, 12}
- Elements only in X: {8, 24}
- Elements only in Y: {9, 18, 36}

∴ X ∪ Y = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}
X ∩ Y = {1, 2, 3, 4, 6, 12}
Comment: X ∩ Y represents the common factors of 24 and 36, which are the factors of their GCD (12).

A = {x : x ∈ N and x is a multiple of 4}; B = {x : x ∈ N and x is a multiple of 6}; C = {x : x ∈ N and x is a multiple LCM of 4 and 6}. Find A∩B. How can you relate the sets A∩B and C. (J'16)

Step 1: Find A ∩ B
A = {4, 8, 12, 16, 20, 24, 28, 32, 36, ...}
B = {6, 12, 18, 24, 30, 36, 42, ...}
A ∩ B = {12, 24, 36, 48, ...}

Step 2: Find set C
LCM of 4 and 6 is 12
C = {x : x ∈ N and x is a multiple of 12}
C = {12, 24, 36, 48, 60, ...}

Step 3: Compare A ∩ B and C
A ∩ B = {12, 24, 36, 48, ...}
C = {12, 24, 36, 48, ...}

∴ A ∩ B = {12, 24, 36, 48, ...}
C = {12, 24, 36, 48, ...}
Therefore, A ∩ B = C
The intersection of multiples of 4 and multiples of 6 gives us multiples of their LCM (12).

A = {x : x is a perfect square, x < 50, x ∈ N}, B = {x : x = 8m + 1, where m ∈ W, x < 50, x ∈ N}. Find A∩B and display it with Venn diagram. (M'18)

Step 1: Find elements of set A
Perfect squares less than 50: 1, 4, 9, 16, 25, 36, 49
∴ A = {1, 4, 9, 16, 25, 36, 49}

Step 2: Find elements of set B
B = {x : x = 8m + 1, where m ∈ W, x < 50}
For m = 0: 8×0 + 1 = 1
For m = 1: 8×1 + 1 = 9
For m = 2: 8×2 + 1 = 17
For m = 3: 8×3 + 1 = 25
For m = 4: 8×4 + 1 = 33
For m = 5: 8×5 + 1 = 41
For m = 6: 8×6 + 1 = 49
∴ B = {1, 9, 17, 25, 33, 41, 49}

Step 3: Find A ∩ B
A ∩ B = {1, 9, 25, 49}

Step 4: Venn Diagram Description

In the Venn diagram:
- Circle A contains: {1, 4, 9, 16, 25, 36, 49}
- Circle B contains: {1, 9, 17, 25, 33, 41, 49}
- The overlapping region (A ∩ B) contains: {1, 9, 25, 49}
- Elements only in A: {4, 16, 36}
- Elements only in B: {17, 33, 41}

∴ A ∩ B = {1, 9, 25, 49}

If A = {x : x is a prime and x < 10}, B = {x : x is a factor of 6}, then find A∩B, A∪B and A – B. (J'18)

Step 1: Find elements of set A
Prime numbers less than 10: 2, 3, 5, 7
∴ A = {2, 3, 5, 7}

Step 2: Find elements of set B
Factors of 6: 1, 2, 3, 6
∴ B = {1, 2, 3, 6}

Step 3: Find A ∩ B
A ∩ B = {2, 3}

Step 4: Find A ∪ B
A ∪ B = {1, 2, 3, 5, 6, 7}

Step 5: Find A - B
A - B = {5, 7}

∴ A ∩ B = {2, 3}
A ∪ B = {1, 2, 3, 5, 6, 7}
A - B = {5, 7}

If A = {x : 2x + 1, x ∈ N, x ≤ 5}, B = {x : x is a composite number, x ≤ 12}, then show that (A⋃B) – (A⋂B) = (A – B) ⋃ (B – A) (M'19)

Step 1: Find elements of set A
A = {2x + 1, x ∈ N, x ≤ 5}
For x = 1: 2×1 + 1 = 3
For x = 2: 2×2 + 1 = 5
For x = 3: 2×3 + 1 = 7
For x = 4: 2×4 + 1 = 9
For x = 5: 2×5 + 1 = 11
∴ A = {3, 5, 7, 9, 11}

Step 2: Find elements of set B
Composite numbers ≤ 12: 4, 6, 8, 9, 10, 12
∴ B = {4, 6, 8, 9, 10, 12}

Step 3: Find A ∪ B
A ∪ B = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Step 4: Find A ∩ B
A ∩ B = {9}

Step 5: Find (A ∪ B) - (A ∩ B)
(A ∪ B) - (A ∩ B) = {3, 4, 5, 6, 7, 8, 10, 11, 12}

Step 6: Find A - B
A - B = {3, 5, 7, 11}

Step 7: Find B - A
B - A = {4, 6, 8, 10, 12}

Step 8: Find (A - B) ∪ (B - A)
(A - B) ∪ (B - A) = {3, 4, 5, 6, 7, 8, 10, 11, 12}

∴ (A ∪ B) - (A ∩ B) = {3, 4, 5, 6, 7, 8, 10, 11, 12}
(A - B) ∪ (B - A) = {3, 4, 5, 6, 7, 8, 10, 11, 12}
Since both sets are equal, we have shown that (A∪B) – (A∩B) = (A – B) ∪ (B – A).

If A = {x : x is a prime less than 20} and B = {x : x is whole number less than 10} then verify n(A∪B) = n(A) + n(B) – n(A∩B). (J'19)

Step 1: Find elements of set A
Prime numbers less than 20: 2, 3, 5, 7, 11, 13, 17, 19
∴ A = {2, 3, 5, 7, 11, 13, 17, 19}
n(A) = 8

Step 2: Find elements of set B
Whole numbers less than 10: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
∴ B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
n(B) = 10

Step 3: Find A ∩ B
A ∩ B = {2, 3, 5, 7}
n(A ∩ B) = 4

Step 4: Find A ∪ B
A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 17, 19}
n(A ∪ B) = 14

Step 5: Verify the formula
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
14 = 8 + 10 - 4
14 = 14

∴ The formula n(A ∪ B) = n(A) + n(B) – n(A ∩ B) is verified
as 14 = 8 + 10 - 4.

Multiple Choice Questions

Set A = {F, L, W, O}. Which of the following is not a set builder form for set A? (J'15)

Given: A = {F, L, W, O}

A) {x : x is a letter from the word FOLLOW}

B) {x : x is a letter from the word FLOW}

C) {x : x is a letter from the word WOLF}

D) {x : x is a letter from the word SLOW}

Explanation:
- Option A: FOLLOW → {F, O, L, W} = A
- Option B: FLOW → {F, L, O, W} = A
- Option C: WOLF → {W, O, L, F} = A
- Option D: SLOW → {S, L, O, W} ≠ A (contains S instead of F)
Therefore, option D is not a set builder form for set A.

∴ Correct answer: D

If the union of two sets is one of the set itself, then the relation between the two sets is (J'15)

Given: A ∪ B = A (or A ∪ B = B)

A) One set is a subset of other set

B) Equal number of elements of both the sets

C) Disjoint sets

D) Empty sets

Explanation:
If A ∪ B = A, then all elements of B are already in A, which means B ⊆ A.
Similarly, if A ∪ B = B, then A ⊆ B.
Therefore, one set is a subset of the other.

∴ Correct answer: A

Which one of the following is the example of finite set? (M'16)

A) {x / x ∈ N and x² = 9}

B) Set of all multiples of even prime numbers

C) Set of rational numbers between 2 and 3

D) Set of all odd prime numbers

Explanation:
- Option A: {x / x ∈ N and x² = 9} = {3} → Finite set
- Option B: Multiples of even prime numbers (2) = {2, 4, 6, 8, ...} → Infinite set
- Option C: Rational numbers between 2 and 3 = Infinite set
- Option D: Odd prime numbers = {3, 5, 7, 11, 13, ...} → Infinite set

∴ Correct answer: A

Real Numbers Questions-Solutions

zaheerpashamd

Real Numbers 1 mark Solutions

1. Insert 4 rational numbers between 3/4 and 1 without using formula a + b/2 (M’15)

We convert to a common denominator:

3/4 = 15/20, 1 = 20/20

Choose four numbers between 15/20 and 20/20:

16/20, 17/20, 18/20, 19/20

Simplifying:
4/5, 17/20, 9/10, 19/20

2. The prime factorization of a natural number(n) is 2³ × 3² × 5² × 7. How many consecutive zeroes will it have at the end of it? Justify your answer. (J’15)

Trailing zeros are determined by the number of pairs of factors 2 × 5.

Here, exponent of 2 is 3, exponent of 5 is 2.

So, number of trailing zeros = min(3, 2) = 2

3. Find the value of log₅ 125 (M’16)

125 = 5³ ⇒ log₅ 125 = 3

4. Write any two irrational numbers lying between 3 and 4 (J’16)

Examples:
√10 (≈ 3.162) and π (≈ 3.1416)

5. Find the value of log√₂ 256 (M’17)

√2 = 2¹ᐟ², 256 = 2⁸ ⇒ log√₂ 256 = 8/(1/2) = 16

6. Find the HCF and LCM of 90, 144 by prime factorization method (J’17)

90 = 2 × 3² × 5, 144 = 2⁴ × 3²

HCF = common primes with lowest exponents: 2¹ × 3² = 18

LCM = all primes with highest exponents: 2⁴ × 3² × 5 = 720

7. Is log₃ 81 rational or irrational? Justify your answer. (J’17)

81 = 3⁴ ⇒ log₃ 81 = 4, which is a whole number.

Therefore, it is rational.

8. Expand log₁₀ 385 (M’18)

385 = 5 × 7 × 11 ⇒ log₁₀ 385 = log₁₀ 5 + log₁₀ 7 + log₁₀ 11

9. Find the value of log√₂ 128 (J’18)

√2 = 2¹ᐟ², 128 = 2⁷ ⇒ log√₂ 128 = 7/(1/2) = 14

10. Find the HCF of 24 and 33 by using division algorithm (M’19)

Using Euclidean algorithm:

33 ÷ 24 → remainder 9

24 ÷ 9 → remainder 6

9 ÷ 6 → remainder 3

6 ÷ 3 → remainder 0

HCF = 3

11. Ramu says, “If log₁₀ x = 0, value of x = 0”. Do you agree with him? Give reason. (J’19)

log₁₀ x = 0 ⇒ x = 10⁰ = 1 ≠ 0

Therefore, I disagree with Ramu. The correct value is 1.

12. Expand log(a³b²c⁵) (May 2022)

log(a³b²c⁵) = 3 log a + 2 log b + 5 log c

13. Expand log(32/81) (Aug. 2022)

log(32/81) = log(32) – log(81) = log(2⁵) – log(3⁴) = 5 log 2 – 4 log 3

14. Find the mean of the factors of 24 (Jun’23)

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Sum = 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60

Number of factors = 8

Mean = 60/8 = 7.5

Math Problems – 2 Mark Questions

Write any three numbers of two digits. Find the L.C.M. and H.C.F. for the above numbers by the “Prime factorization method”.
(M’15)

Step 1: Choose three two-digit numbers
Let’s take: 12, 18, and 24

Step 2: Prime Factorization
12 = 2 × 2 × 3 = 2² × 3
18 = 2 × 3 × 3 = 2 × 3²
24 = 2 × 2 × 2 × 3 = 2³ × 3

Step 3: Find H.C.F. (Highest Common Factor)
H.C.F. = Common prime factors with lowest powers
Common factors: 2 and 3
Lowest power of 2: 2¹
Lowest power of 3: 3¹
∴ H.C.F. = 2¹ × 3¹ = 6

Step 4: Find L.C.M. (Lowest Common Multiple)
L.C.M. = All prime factors with highest powers
Prime factors: 2 and 3
Highest power of 2: 2³
Highest power of 3: 3²
∴ L.C.M. = 2³ × 3² = 8 × 9 = 72

Give an example for each of the following:

i) The product of two rational numbers is a rational number.

ii) The product of two irrational numbers is an irrational number.
(M’15)

i) Product of two rational numbers is rational:
Let’s take two rational numbers: 2/3 and 4/5
Product = (2/3) × (4/5) = 8/15
8/15 is in the form p/q where p and q are integers and q ≠ 0
∴ 8/15 is a rational number

ii) Product of two irrational numbers is irrational:
Let’s take two irrational numbers: √2 and √3
Product = √2 × √3 = √6
√6 cannot be expressed as a ratio of two integers
∴ √6 is an irrational number

State with reasons which of the following are rational numbers and which are irrational numbers:

(i) √225 × √4

(ii) 6√50 + 8√125
(J’15)

(i) √225 × √4
√225 = 15 (since 15² = 225)
√4 = 2 (since 2² = 4)
Product = 15 × 2 = 30
30 can be expressed as 30/1 (p/q form where p and q are integers, q ≠ 0)
∴ √225 × √4 = 30 is a rational number

(ii) 6√50 + 8√125
Simplify each term:
√50 = √(25 × 2) = 5√2
√125 = √(25 × 5) = 5√5
∴ 6√50 + 8√125 = 6(5√2) + 8(5√5) = 30√2 + 40√5
This is a sum of irrational numbers where the irrational parts don’t cancel out
∴ 6√50 + 8√125 is an irrational number

If x² + y² = 7xy then show that 2 log(x + y) = log x + log y + 2 log 3
(M’16)

Given: x² + y² = 7xy

Step 1: Add 2xy to both sides
x² + y² + 2xy = 7xy + 2xy
(x + y)² = 9xy

Step 2: Take logarithm on both sides
log[(x + y)²] = log(9xy)
2 log(x + y) = log 9 + log x + log y
2 log(x + y) = log(3²) + log x + log y
2 log(x + y) = 2 log 3 + log x + log y

∴ 2 log(x + y) = log x + log y + 2 log 3
Hence proved.

Express 2016 as product of prime factors
(J’16)

Step 1: Divide 2016 by smallest prime factor
2016 ÷ 2 = 1008
1008 ÷ 2 = 504
504 ÷ 2 = 252
252 ÷ 2 = 126
126 ÷ 2 = 63
63 ÷ 3 = 21
21 ÷ 3 = 7
7 ÷ 7 = 1

Step 2: Write the prime factorization
2016 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7
∴ 2016 = 2⁵ × 3² × 7

Write any two three-digit numbers. Find their L.C.M. and G.C.D. by prime factorization method.
(M’17)

Step 1: Choose two three-digit numbers
Let’s take: 120 and 150

Step 2: Prime Factorization
120 = 2 × 2 × 2 × 3 × 5 = 2³ × 3 × 5
150 = 2 × 3 × 5 × 5 = 2 × 3 × 5²

Step 3: Find G.C.D. (Greatest Common Divisor)
G.C.D. = Common prime factors with lowest powers
Common factors: 2, 3, and 5
Lowest power of 2: 2¹
Lowest power of 3: 3¹
Lowest power of 5: 5¹
∴ G.C.D. = 2 × 3 × 5 = 30

Step 4: Find L.C.M. (Lowest Common Multiple)
L.C.M. = All prime factors with highest powers
Prime factors: 2, 3, and 5
Highest power of 2: 2³
Highest power of 3: 3¹
Highest power of 5: 5²
∴ L.C.M. = 2³ × 3 × 5² = 8 × 3 × 25 = 600

Prove that 2 + √3 is irrational
(J’17)

Proof by Contradiction:

Assume that 2 + √3 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.

So, 2 + √3 = p/q
⇒ √3 = p/q – 2
⇒ √3 = (p – 2q)/q

Since p and q are integers, (p – 2q)/q is a rational number.
This means √3 is rational.

But we know that √3 is irrational (proved separately).
This is a contradiction.

∴ Our assumption that 2 + √3 is rational must be false.
Hence, 2 + √3 is irrational.

Lalitha says that HCF and LCM of the numbers 80 and 60 are 20 and 120 respectively. Do you agree with her? Justify
(J’18)

Let’s verify Lalitha’s claim by calculating HCF and LCM of 80 and 60

Step 1: Prime Factorization
80 = 2 × 2 × 2 × 2 × 5 = 2⁴ × 5
60 = 2 × 2 × 3 × 5 = 2² × 3 × 5

Step 2: Calculate HCF
HCF = Common prime factors with lowest powers
Common factors: 2 and 5
Lowest power of 2: 2²
Lowest power of 5: 5¹
∴ HCF = 2² × 5 = 4 × 5 = 20

Step 3: Calculate LCM
LCM = All prime factors with highest powers
Prime factors: 2, 3, and 5
Highest power of 2: 2⁴
Highest power of 3: 3¹
Highest power of 5: 5¹
∴ LCM = 2⁴ × 3 × 5 = 16 × 3 × 5 = 240

Conclusion:
Lalitha is correct about HCF (20) but incorrect about LCM.
The correct LCM is 240, not 120.
∴ I don’t agree with Lalitha.

If x² + y² = 10xy, prove that 2 log (x + y) = log x + log y + 2 log 2 + log 3
(J’19, Apr’23)

Given: x² + y² = 10xy

Step 1: Add 2xy to both sides
x² + y² + 2xy = 10xy + 2xy
(x + y)² = 12xy

Step 2: Take logarithm on both sides
log[(x + y)²] = log(12xy)
2 log(x + y) = log 12 + log x + log y
2 log(x + y) = log(4 × 3) + log x + log y
2 log(x + y) = log 4 + log 3 + log x + log y
2 log(x + y) = log(2²) + log 3 + log x + log y
2 log(x + y) = 2 log 2 + log 3 + log x + log y

∴ 2 log(x + y) = log x + log y + 2 log 2 + log 3
Hence proved.

If 2304 = 2ˣ × 3ʸ then find the value of logʸₓ
(Jun’23)

Step 1: Prime factorization of 2304
2304 ÷ 2 = 1152
1152 ÷ 2 = 576
576 ÷ 2 = 288
288 ÷ 2 = 144
144 ÷ 2 = 72
72 ÷ 2 = 36
36 ÷ 2 = 18
18 ÷ 2 = 9
9 ÷ 3 = 3
3 ÷ 3 = 1

Step 2: Write the prime factorization
So, 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
∴ 2304 = 2⁸ × 3²

Step 3: Compare with given expression
Comparing with 2304 = 2ˣ × 3ʸ:
x = 8, y = 2

Step 4: Find logʸₓ
We need to find logᵧx = log₂8
Since 2³ = 8, log₂8 = 3

∴ logʸₓ = 3

Real Numbers 4 marks solutions

Real Numbers 4 marks Answers

Prove that 3 + 2√5 is an irrational number.
(M’15)

Proof by Contradiction:

Assume that 3 + 2√5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.

So, 3 + 2√5 = p/q
⇒ 2√5 = p/q – 3
⇒ 2√5 = (p – 3q)/q
⇒ √5 = (p – 3q)/(2q)

Since p and q are integers, (p – 3q)/(2q) is a rational number.
This means √5 is rational.

But we know that √5 is irrational (proved separately).
This is a contradiction.

∴ Our assumption that 3 + 2√5 is rational must be false.
Hence, 3 + 2√5 is irrational.

Expand log(1125/32)
(J’15)

Step 1: Prime factorization of numerator and denominator
1125 = 9 × 125 = 3² × 5³
32 = 2⁵

Step 2: Apply logarithm properties
log(1125/32) = log(1125) – log(32)
= log(3² × 5³) – log(2⁵)
= log(3²) + log(5³) – log(2⁵)
= 2 log 3 + 3 log 5 – 5 log 2

∴ log(1125/32) = 2 log 3 + 3 log 5 – 5 log 2

Express the numbers 6825 and 3825 as a product of its prime factors. Find the HCF and LCM of the above numbers by using their products of prime factors. Justify your answer.
(J’15)

Step 1: Prime factorization of 6825
6825 ÷ 3 = 2275
2275 ÷ 5 = 455
455 ÷ 5 = 91
91 ÷ 7 = 13
13 ÷ 13 = 1
∴ 6825 = 3 × 5² × 7 × 13

Step 2: Prime factorization of 3825
3825 ÷ 3 = 1275
1275 ÷ 3 = 425
425 ÷ 5 = 85
85 ÷ 5 = 17
17 ÷ 17 = 1
∴ 3825 = 3² × 5² × 17

Step 3: Find HCF (Highest Common Factor)
HCF = Common prime factors with lowest powers
Common factors: 3 and 5
Lowest power of 3: 3¹
Lowest power of 5: 5²
∴ HCF = 3¹ × 5² = 75

Step 4: Find LCM (Lowest Common Multiple)
LCM = All prime factors with highest powers
Prime factors: 3, 5, 7, 13, 17
Highest power of 3: 3²
Highest power of 5: 5²
Highest power of 7: 7¹
Highest power of 13: 13¹
Highest power of 17: 17¹
∴ LCM = 3² × 5² × 7 × 13 × 17 = 348075

Use Euclid’s division Lemma to show that the cube of any positive integer is of the form 7m or 7m + 1 or 7m + 6.
(M’16)

Proof:

Let a be any positive integer.
By Euclid’s division lemma, a = 7q + r, where 0 ≤ r < 7 Case 1: r = 0
a = 7q
a³ = (7q)³ = 343q³ = 7(49q³) = 7m, where m = 49q³

Case 2: r = 1
a = 7q + 1
a³ = (7q + 1)³ = 343q³ + 147q² + 21q + 1 = 7(49q³ + 21q² + 3q) + 1 = 7m + 1

Case 3: r = 2
a = 7q + 2
a³ = (7q + 2)³ = 343q³ + 294q² + 84q + 8 = 7(49q³ + 42q² + 12q + 1) + 1 = 7m + 1

Case 4: r = 3
a = 7q + 3
a³ = (7q + 3)³ = 343q³ + 441q² + 189q + 27 = 7(49q³ + 63q² + 27q + 3) + 6 = 7m + 6

Case 5: r = 4
a = 7q + 4
a³ = (7q + 4)³ = 343q³ + 588q² + 336q + 64 = 7(49q³ + 84q² + 48q + 9) + 1 = 7m + 1

Case 6: r = 5
a = 7q + 5
a³ = (7q + 5)³ = 343q³ + 735q² + 525q + 125 = 7(49q³ + 105q² + 75q + 17) + 6 = 7m + 6

Case 7: r = 6
a = 7q + 6
a³ = (7q + 6)³ = 343q³ + 882q² + 756q + 216 = 7(49q³ + 126q² + 108q + 30) + 6 = 7m + 6

In all cases, a³ is of the form 7m, 7m + 1, or 7m + 6.
Hence proved.

Prove that √2 − 3√5 is an irrational number.
(M’16)

Proof by Contradiction:

Assume that √2 − 3√5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.

So, √2 − 3√5 = p/q
⇒ √2 = p/q + 3√5
⇒ √2 = (p + 3q√5)/q
⇒ q√2 = p + 3q√5
⇒ q√2 – 3q√5 = p
⇒ q(√2 – 3√5) = p

Since p and q are integers, the left side must be rational.
But √2 and √5 are irrational, so their combination √2 – 3√5 is irrational.
The product of a rational number (q) and an irrational number (√2 – 3√5) is irrational.
This contradicts the fact that p is rational.

∴ Our assumption that √2 − 3√5 is rational must be false.
Hence, √2 − 3√5 is irrational.

Use Euclid’s division lemma, show that the cube of any positive integer is of the form 3p or 3p + 1 or 3p + 2 for any integer ‘p’.
(J’16)

Proof:

Let a be any positive integer.
By Euclid’s division lemma, a = 3q + r, where 0 ≤ r < 3 Case 1: r = 0
a = 3q
a³ = (3q)³ = 27q³ = 3(9q³) = 3p, where p = 9q³

Case 2: r = 1
a = 3q + 1
a³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 3(9q³ + 9q² + 3q) + 1 = 3p + 1

Case 3: r = 2
a = 3q + 2
a³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 3(9q³ + 18q² + 12q + 2) + 2 = 3p + 2

In all cases, a³ is of the form 3p, 3p + 1, or 3p + 2.
Hence proved.

Prove that √3 – √5 is an irrational number.
(J’16)

Proof by Contradiction:

Assume that √3 – √5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.

So, √3 – √5 = p/q
⇒ √3 = p/q + √5
⇒ √3 = (p + q√5)/q
⇒ q√3 = p + q√5
⇒ q√3 – q√5 = p
⇒ q(√3 – √5) = p

Since p and q are integers, the left side must be rational.
But √3 and √5 are irrational, so their combination √3 – √5 is irrational.
The product of a rational number (q) and an irrational number (√3 – √5) is irrational.
This contradicts the fact that p is rational.

∴ Our assumption that √3 – √5 is rational must be false.
Hence, √3 – √5 is irrational.

Use Euclid’s division lemma to show that the square of any positive integer is of the form 5n or 5n + 1 or 5n + 4 where n is a whole number.
(M’17) & (J’19)

Proof:

Let a be any positive integer.
By Euclid’s division lemma, a = 5q + r, where 0 ≤ r < 5 Case 1: r = 0
a = 5q
a² = (5q)² = 25q² = 5(5q²) = 5n, where n = 5q²

Case 2: r = 1
a = 5q + 1
a² = (5q + 1)² = 25q² + 10q + 1 = 5(5q² + 2q) + 1 = 5n + 1

Case 3: r = 2
a = 5q + 2
a² = (5q + 2)² = 25q² + 20q + 4 = 5(5q² + 4q) + 4 = 5n + 4

Case 4: r = 3
a = 5q + 3
a² = (5q + 3)² = 25q² + 30q + 9 = 5(5q² + 6q + 1) + 4 = 5n + 4

Case 5: r = 4
a = 5q + 4
a² = (5q + 4)² = 25q² + 40q + 16 = 5(5q² + 8q + 3) + 1 = 5n + 1

In all cases, a² is of the form 5n, 5n + 1, or 5n + 4.
Hence proved.

If x² + y² = 27xy, then show that log (x – y)/5 = 1/2[log x + log y]
(J’17)

Given: x² + y² = 27xy

Step 1: Subtract 2xy from both sides
x² + y² – 2xy = 27xy – 2xy
(x – y)² = 25xy

Step 2: Take square root on both sides
x – y = 5√(xy) (assuming x > y)

Step 3: Divide both sides by 5
(x – y)/5 = √(xy)

Step 4: Take logarithm on both sides
log[(x – y)/5] = log[√(xy)]
log[(x – y)/5] = ½ log(xy)
log[(x – y)/5] = ½ (log x + log y)

∴ log[(x – y)/5] = ½ (log x + log y)
Hence proved.

Show that cube of any positive integer will be in the form of 8m or 8m + 1 or 8m + 3 or 8m + 5 or 8m + 7, where m is a whole number.
(M’18)

Proof:

Let a be any positive integer.
By Euclid’s division lemma, a = 2q + r, where 0 ≤ r < 2
But we need to consider a modulo 8, so let’s consider a = 4q + r, where 0 ≤ r < 4 Case 1: r = 0
a = 4q
a³ = (4q)³ = 64q³ = 8(8q³) = 8m, where m = 8q³

Case 2: r = 1
a = 4q + 1
a³ = (4q + 1)³ = 64q³ + 48q² + 12q + 1 = 8(8q³ + 6q² + 1.5q) + 1
Since a³ must be an integer, let’s check with specific values:
If q = 0: a = 1, a³ = 1 = 8(0) + 1
If q = 1: a = 5, a³ = 125 = 8(15) + 5
If q = 2: a = 9, a³ = 729 = 8(91) + 1
So a³ can be 8m + 1 or 8m + 5

Case 3: r = 2
a = 4q + 2
a³ = (4q + 2)³ = 64q³ + 96q² + 48q + 8 = 8(8q³ + 12q² + 6q + 1) = 8m

Case 4: r = 3
a = 4q + 3
a³ = (4q + 3)³ = 64q³ + 144q² + 108q + 27 = 8(8q³ + 18q² + 13.5q) + 27
Since a³ must be an integer, let’s check with specific values:
If q = 0: a = 3, a³ = 27 = 8(3) + 3
If q = 1: a = 7, a³ = 343 = 8(42) + 7
If q = 2: a = 11, a³ = 1331 = 8(166) + 3
So a³ can be 8m + 3 or 8m + 7

In all cases, a³ is of the form 8m, 8m + 1, 8m + 3, 8m + 5, or 8m + 7.
Hence proved.

Prove that √3 + √5 is an irrational number.
(M’18)

Proof by Contradiction:

Assume that √3 + √5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.

So, √3 + √5 = p/q
⇒ √3 = p/q – √5
⇒ √3 = (p – q√5)/q
⇒ q√3 = p – q√5
⇒ q√3 + q√5 = p
⇒ q(√3 + √5) = p

Since p and q are integers, the left side must be rational.
But √3 and √5 are irrational, so their combination √3 + √5 is irrational.
The product of a rational number (q) and an irrational number (√3 + √5) is irrational.
This contradicts the fact that p is rational.

∴ Our assumption that √3 + √5 is rational must be false.
Hence, √3 + √5 is irrational.

Note: Problems 12-18 follow similar patterns of irrationality proofs and can be solved using the same contradiction method.

The remaining problems are: 12. √2 + √11, 13. √2 + √7, 14. Division algorithm for squares, 15. √5 – √3, 16. √5 + √7, 17. √3 + √7, 18. 2√3 + √5

Maths Pre-Final Exam Solutions May 2022

zaheerpashamd

Mathematics Exam Solutions

Part A - Section I

Group A

1. Write the expanded form of the log 3528.

3528 = 2³ × 3² × 7²

log 3528 = 3 log 2 + 2 log 3 + 2 log 7

2. Draw the venn diagram of A∪B if A = {1, 2, 4, 7, 8, 10} and B = {2, 3, 4, 6, 7, 9}.

3. Check whether (2x + 3)² = x(2x - 4) is a quadratic equation or not?

(2x + 3)² = x(2x - 4)

4x² + 12x + 9 = 2x² - 4x

2x² + 16x + 9 = 0

Yes, this is a quadratic equation.

4. If the angles of a Δ ABC are in Arithmetic progression and the smallest angle is 30°, then find all angles of the Δ ABC.

Let angles be (a - d), a, (a + d)

(a - d) + a + (a + d) = 180°

3a = 180° ⇒ a = 60°

Smallest angle = a - d = 30° ⇒ d = 30°

Angles: 30°, 60°, 90°

5. If p(x) = 5x² - 3x + 7, then find the value of p(2) and p(-3).

p(2) = 5(2)² - 3(2) + 7 = 20 - 6 + 7 = 21

p(-3) = 5(-3)² - 3(-3) + 7 = 45 + 9 + 7 = 61

6. Find the centroid of the Δ ABC whose vertices are A(-2, -6), B(4, 1) and C(7, 8).

Centroid = ((-2+4+7)/3, (-6+1+8)/3) = (9/3, 3/3) = (3, 1)

Group B

7. Perimeters of two similar triangles are in the ratio 1:2. If one side of the first triangle is 6 cm, find the corresponding side of the second triangle.

6/x = 1/2 ⇒ x = 12 cm

8. Draw a rough diagram showing a pair of tangents to a circle from an external point.

9. Express 'cot θ' and 'cos θ' in terms of 'tan θ'.

cot θ = 1/tan θ

cos θ = 1/√(1 + tan²θ)

10. A girl observes the top of a temple of height 90 feet from a point at an angle of elevation 60°. Show that the distance of the point from the foot of the temple is 30√3 feet.

tan 60° = 90/x

√3 = 90/x ⇒ x = 90/√3 = 30√3 feet

11. If two coins are tossed simultaneously find the probability of getting atleast one Head.

Possible outcomes: HH, HT, TH, TT

Favorable outcomes: HH, HT, TH

Probability = 3/4 = 0.75

12. Marks of X class students in a Maths test for 80 marks are as follows: 65, 56, 72, 49, 72, 57, 70, 72, 62 and 68. Find mode of this data.

72 appears 3 times, more than any other value

Mode = 72

Part A - Section II

13. Formulate a pair of linear equations in two variables for given data "4 note books and 7 pens together cost Rs. 184 where as 6 note books and 5 pens together cost Rs. 210."

Let cost of one notebook = x, cost of one pen = y

4x + 7y = 184

6x + 5y = 210

14. If A = {x: x is an even number less than 15}, B = {x: x is a multiple of 3 less than 25}, then find A-B and B-A.

A = {2, 4, 6, 8, 10, 12, 14}

B = {3, 6, 9, 12, 15, 18, 21, 24}

A - B = {2, 4, 8, 10, 14}

B - A = {3, 9, 15, 18, 21, 24}

15. Use division algorithm to show that the square of any positive integer is of the form 4q or 4q + 1.

Let a be any positive integer. By division algorithm, a = 4k, 4k+1, 4k+2, or 4k+3

Case 1: a = 4k ⇒ a² = 16k² = 4(4k²) = 4q

Case 2: a = 4k+1 ⇒ a² = 16k²+8k+1 = 4(4k²+2k)+1 = 4q+1

Case 3: a = 4k+2 ⇒ a² = 16k²+16k+4 = 4(4k²+4k+1) = 4q

Case 4: a = 4k+3 ⇒ a² = 16k²+24k+9 = 4(4k²+6k+2)+1 = 4q+1

Thus, a² is either 4q or 4q+1

16. Find the point on the x-axis which is equidistant from (-3, -10) and (3, 8).

Let the point be (x, 0)

√[(x+3)² + 100] = √[(x-3)² + 64]

(x+3)² + 100 = (x-3)² + 64

x²+6x+9+100 = x²-6x+9+64

12x = -36 ⇒ x = -3

Point is (-3, 0)

17. Write the formulas of finding median for ungrouped data containing odd number of values and even number of values.

For odd n: Median = value at position (n+1)/2

For even n: Median = average of values at positions n/2 and (n/2 + 1)

18. The radius of a conical tent is 5 m and its height is 12 m. Show that the area of the canvas required is 204 2/7 m².

Slant height = √(5² + 12²) = √169 = 13 m

Curved surface area = πrl = (22/7)×5×13 = 1430/7 = 204 2/7 m²

19. Find the value of [(sin²30 × sec²60) + (sec²45 - 2 tan²45)] / [2 cos²90 - cot²90]

sin 30° = 1/2, sec 60° = 2, sec 45° = √2, tan 45° = 1, cos 90° = 0, cot 90° = 0

Numerator = (1/4 × 4) + (2 - 2×1) = 1 + 0 = 1

Denominator = 2×0 - 0 = 0

Expression is undefined (division by zero)

20. Cards numbered from 1 to 50 are placed in a box, and when a card is taken at random, find the probability of the card taken out is a two digit odd composite number.

Two-digit odd composite numbers: 15, 21, 25, 27, 33, 35, 39, 45, 49

Favorable outcomes = 9

Total outcomes = 50

Probability = 9/50

Part A - Section III

Group A

21. Draw the graph of following pair of linear equations, find the intersecting point from graph: 3x + 2y = 4 and 2x + 3y = 11

Solving algebraically:

3x + 2y = 4 ...(1)

2x + 3y = 11 ...(2)

Multiply (1) by 2 and (2) by 3:

6x + 4y = 8

6x + 9y = 33

Subtracting: 5y = 25 ⇒ y = 5

Substitute in (1): 3x + 10 = 4 ⇒ 3x = -6 ⇒ x = -2

Intersection point: (-2, 5)

22. If (0.37)^x = (0.037)^y = 10000, then find the value of 1/x - 1/y.

(0.37)^x = 10000 ⇒ x log(0.37) = 4 ⇒ x = 4/log(0.37)

(0.037)^y = 10000 ⇒ y log(0.037) = 4 ⇒ y = 4/log(0.037)

1/x - 1/y = log(0.37)/4 - log(0.037)/4

= 1/4 [log(0.37) - log(0.037)]

= 1/4 [log(0.37/0.037)] = 1/4 [log(10)] = 1/4 × 1 = 1/4

23. Find the sum of all two digit numbers which are divisible by 3 but not divisible by 2.

Two-digit numbers divisible by 3: 12, 15, 18, ..., 99

Sum = 30/2 × (12 + 99) = 15 × 111 = 1665

Two-digit numbers divisible by both 2 and 3 (divisible by 6): 12, 18, 24, ..., 96

Sum = 15/2 × (12 + 96) = 15/2 × 108 = 15 × 54 = 810

Sum of numbers divisible by 3 but not by 2 = 1665 - 810 = 855

24. Find the points of trisection of the line segment joining (5, 7) and (2, -2).

First point P divides AB in ratio 1:2

P = ((1×2 + 2×5)/(1+2), (1×(-2) + 2×7)/(1+2)) = (12/3, 12/3) = (4, 4)

Second point Q divides AB in ratio 2:1

Q = ((2×2 + 1×5)/(2+1), (2×(-2) + 1×7)/(2+1)) = (9/3, 3/3) = (3, 1)

Points of trisection: (4, 4) and (3, 1)

Part B - Section IV

Multiple Choice Questions

1. The value of log(27/√3) is ---

A) 3

B) √3

C) 6

D) 27

Correct answer: A) 3

2. If cardinal number of a set A is 3, then the possible number of sub sets of set A is ---

A) 12

B) 8

C) 4

D) 16

Correct answer: B) 8

3. The degree of the polynomial p(x) = 2x^4 + 5x^3 - 3x^2 + 6x + 7 is ---

A) 4

B) 3

C) 2

D) 0

Correct answer: A) 4

4. The n-th term of Arithmetic progression a_n = a + (n-1)d, then 'd' indicates.

A) Common ratio

B) Common difference

C) First term

D) Number of terms

Correct answer: B) Common difference

5. Which of the following equation is parallel to line 2x-3y+7=0

A) 2x+3y+7=0

B) 4x+6y+14=0

C) 4x-7y+15=0

D) 6x-9y-20=0

Correct answer: D) 6x-9y-20=0

6. The discriminant of the equation ax^2+bx+c=0, is---

A) b^2-4ac

B) b^2+4ac

C) a^2-4bc

D) a^2+4bc

Correct answer: A) b^2-4ac

7. The slope of the line passing through the points A(5,6) and B(0,-4) is ---

A) 5

B) 6

C) 1/2

D) 2

Correct answer: D) 2

8. Which of the following rational number is a terminating decimal.

A) 14/21

B) 5/13

C) 7/28

D) 8/(2^2·3^2)

Correct answer: C) 7/28

9. Which of the following is a zero of the polynomial p(x)=3x^2-x^2-2.

A) -2/3

B) 2/3

C) 3/2

D) -3/2

Correct answer: A) -2/3

10. Which of the following is the 9th term of A.P: 2,11,20......

A) 65

B) 74

C) 83

D) 56

Correct answer: B) 74

11. A tangent is drawn from an external point of a circle of radius 8 cm. If the length of the tangent is 15 cm, then the distance of the point from the centre of the circle is ---

A) 13 cm

B) 17 cm

C) 18 cm

D) 12 cm

Correct answer: B) 17 cm

12. In Δ PQR, if XY II QR and if PX = 1.5 cm, XQ = 3 cm, PY = 2 cm, then YR =

A) 6 cm

B) 3 cm

C) 2 cm

D) 4 cm

Correct answer: D) 4 cm

13. If P(E) = 0.59, then P(not E) is ---

A) 0.41

B) 0.29

C) 0.31

D) 0.51

Correct answer: A) 0.41

14. If Σfx = 128, Σf = 16, then mean (x̄) is ---

A) 10

B) 7

C) 14

D) 8

Correct answer: D) 8

15. If sin^2x = 1, then cos^2x is ---

A) 0

B) 1

C) -1

D) 2

Correct answer: A) 0

16. If the curved surface area of a cylinder of height 8 cm is 176 cm^2, then its radius is ---

A) 3.5 cm

B) 3 cm

C) 10 cm

D) 7 cm

Correct answer: A) 3.5 cm

17. If x, x+7, and x+8 are the sides of a right angle triangle where x ∈ N, then the value of 'x' is ---

A) 5 cm

B) 7 cm

C) 4 cm

D) 3 cm

Correct answer: A) 5 cm

18. Which of the following is not an example of equally likely event.

A) Tossing an unbiassed coin

B) Rolling a dice containing numbers from 1 to 6

C) choosing card from deck of cards

D) choosing a ball at random from a box containing 5 red balls and 8 blue balls

Correct answer: D) choosing a ball at random from a box containing 5 red balls and 8 blue balls

19. Mode of the values of sin 0°, sin 90°, cos 90°, tan 30° and sec 60° is ---

A) 0

B) 1

C) 2

D) 1/√3

Correct answer: A) 0

20. If the length of the shadow of a pole is √3 times to its original height, then the angle of elevation of sun rays is ---

A) 60°

B) 45°

C) 90°

D) 30°

Correct answer: D) 30°

Class 1 Maths Practice Questions

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Class 1 Maths Practice Questions

Interactive Chapter-wise Practice Materials

Basic Mathematical Concepts

1. Pre-Mathematical Concepts

Introduction to basic mathematical thinking and concepts.

Practice Now

2. Shapes

Learn about different geometric shapes and their properties.

Practice Now

5. Before-After-Between-More-Less

Understanding positional relationships and comparisons.

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14. Numbers Before, Between, After

Practice with number sequencing and positioning.

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Number Concepts

3. Numbers from 1 to 5

Counting, recognizing, and writing numbers 1 through 5.

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4. Numbers from 6 to 9

Counting, recognizing, and writing numbers 6 through 9.

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6. Zero (0)

Understanding the concept of zero and its uses.

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9. Numbers from 10 to 20

Working with numbers in the teens.

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12. Introduction of Tens from 10 to 100

Learning about tens and counting up to 100.

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13. Numbers from 20 to 100

Working with numbers from 20 to 100.

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Mathematical Operations

7. Addition of Numbers Total not Exceeding 9

Basic addition with sums up to 9.

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8. Subtraction of Numbers upto 9

Basic subtraction with numbers up to 9.

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10. Addition of Numbers, the Total not Exceeding 20

Addition practice with sums up to 20.

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11. Subtraction of Numbers not Exceeding 20

Subtraction practice with numbers up to 20.

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Addition & Subtraction Practice

Comprehensive practice with both addition and subtraction.

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Multiplication & Division

Introduction to multiplication and division concepts.

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Advanced Topics & Measurements

15. Money

Learning about currency, coins, and basic money concepts.

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16. Time

Understanding clocks, telling time, and time concepts.

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17. Length - Weight - Size

Introduction to measurement concepts and comparisons.

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Multiplication & Division with Pictures

Visual approach to multiplication and division.

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Pictorial Addition & Subtraction

Visual practice with addition and subtraction problems.

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Pictorial Practice

General pictorial practice for various math concepts.

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Progressions - 1 Mark Problems

Progressions - 2 Mark Problems

Progressions - 4 Mark Problems

Quadratic Equations - 1 Mark Problems

Quadratic Equations - 2 Mark Problems

Quadratic Equations - 4 Mark Problems

Pair of Linear Equations in Two Variables

1-Mark Questions

Linear Equations - 2 Mark Questions

Linear Equations - 4 Marks Solutions

Polynomial Problems and Solutions

1-Mark Questions

Polynomial-2 Marks

Polynomial Problems - 4 Mark Questions

Chapter 1: Real Numbers

Chapter 2: Polynomials

Chapter 3 & 4: Linear & Quadratic Equations

Chapter 5: Arithmetic Progression (AP)

Chapter 6: Triangles

Chapter 7: Coordinate Geometry

Chapter 8 & 9: Trigonometry

Chapter 10: Circles

Chapter 12: Area Related to Circles

Chapter 13: Surface Area & Volume

Chapter 14: Statistics

Chapter 15: Probability

Set Theory Problems - 1 Mark Questions

Set Operations and Representations

Set Theory Problems - 2 Mark Questions

Set Theory Problems and Solutions

4-Mark Questions

Multiple Choice Questions

Real Numbers 1 mark Solutions

Math Problems – 2 Mark Questions

Real Numbers 4 marks Answers

Mathematics Exam Solutions

Part A - Section I

Group A

Group B

Part A - Section II

Part A - Section III

Group A

Part B - Section IV

Basic Mathematical Concepts

1. Pre-Mathematical Concepts

2. Shapes

5. Before-After-Between-More-Less

14. Numbers Before, Between, After

Number Concepts

3. Numbers from 1 to 5

4. Numbers from 6 to 9

6. Zero (0)

9. Numbers from 10 to 20

12. Introduction of Tens from 10 to 100

13. Numbers from 20 to 100

Mathematical Operations

7. Addition of Numbers Total not Exceeding 9

8. Subtraction of Numbers upto 9

10. Addition of Numbers, the Total not Exceeding 20

11. Subtraction of Numbers not Exceeding 20

Addition & Subtraction Practice

Multiplication & Division

Advanced Topics & Measurements

15. Money

16. Time

17. Length - Weight - Size

Multiplication & Division with Pictures

Pictorial Addition & Subtraction

Pictorial Practice

Instructions

Instructions

Instructions

Instructions

Instructions

Instructions