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HCF of 24 and 40 is 8

∴ 24: 40 = 24/40

= (24 ÷ 8) / (40 ÷ 4)

= 3/5

= 3:5

Hence the simplest form of 24:40 is 3:5

(ii) The box will be 9.6

The given ratio can be written as (13.5/15) = (135/150)

HCF of 135 and 150 is 15

∴ 135: 150 = 135/150

= (135 ÷ 15)/ (150 ÷ 15)

= 9/10

= 9:10

Hence the simplest form of 135:150 is 9:10

According to question 

Product of extreme terms = 19 × 24 = 456 …………… (i) 

Product of middle terms= 16 × 21 = 336 …………… (ii)

Both (i) and (ii) are not equal. 

19, 16, 21 and 24 are not in proportion.

First convert the given mixed fraction into improper fraction. [6(2/3)]: [7(1/2)] = (20/3): (15/2)

By cross multiplication, we get,

= 40:45

HCF of 40 and 45 is 5

= (40 ÷ 5) / (45 ÷ 5)

= (8/9)

= 8:9

Hence the simplest form of 40:45 is 8:9

(i) Simplest form of Ratio 2 m 20 cm : 11 cm, is 20 : 1

⇒ By ratio of 15 cm to 2 m. 

= 15 : 2 × 100 = 15 : 200 (∵ 1 m = 100 cm) 

⇒ by ratio of 10 sec to 3 min. 

= 10 sec : 3 × 60 sec ( ∵ 1 min = 60 sec ) 

= 10 : 180 = 1 : 8 

3/40 ≠ 1/18

Thus, they are not in proportion

Fraction from of ratios = 2/3 and 3/5 

Fractions with like denominator = 10/15 and 9/15 

Thus, greater ratio will be 10/15, i.e. 2 : 3

By cross multiplication, we get,

= 9:6

HCF of 9 and 6 is 3

= (9 ÷ 3) / (6 ÷ 3)

= 3/2

= 3:2

Hence the simplest form of 9:6 is 3:2

Converting both the given quantities in the same units, we have:

We know that,

= 1 hour = 60 minutes

Then,

= (1 × 60) minutes + 5 minutes: 45 minutes

= 65 minutes: 45 minutes

HCF of 65 and 45 is 5

= (65 ÷ 5) / (45 ÷ 5)

= 13/9

= 13minutes: 9minutes

(i) 15 : 20 = 15/20 = 3/4 = 3 : 4 

(ii) 28 : 35 = 28/35 = 4/5 = 4 : 5

Converting both the given quantities in the same units, we have:

We know that,

= 1m = 100cm

Then,

= (1 × 100) cm + 5 cm: 63 cm

= 105cm: 63cm

HCF of 105 and 63 is 21

= (105 ÷ 21) / (63 ÷ 21)

= 5/3

= 5cm: 3cm

Converting both the given quantities in the same units, we have:

We know that,

= 1 rupee = 100paise

∴ 3 rupees = 300 paise

Then,

= 75:300

HCF of 75 and 300 is 75

= (75 ÷ 75) / (300 ÷ 75)

= 1/4

= 1 paise : 4 paise

दिया है, sinθ – cosθ = 0

दोनों और वर्ग करने पर,

(sinθ – cosθ)² = 0

sin²θ + cos²θ – 2sinθ cosθ = 0

1 – 2 sinθ cosθ = 0

1 = 2sinθ cosθ

sinθ·cosθ = 1/2

L. H. S. = sin⁴θ + cos⁴θ

= (sin²θ)² + (cos²θ)² + 2sin²θcos²θ – 2sin²θcos²θ

= (sin²θ + cos²θ)² – 2(sinθ·cosθ)²

= (1)² – 2(1/2)² 

= 1 – 2 x 1/4 = 1 – 1/2 = 1/2 = R.H.S

LHS = 4cot² 45° – sec²60° + sin²30° 

= 4 x (1)² – (2)² + (1/2)²

= 4 - 4 + 1/4 = 1/4 = RHS

यदि θ = 45° 

∴ sin2θ 

= sin2 x 45° = sin90° = 1

Let P(n): n³ – 7n + 3 is divisible by 3, for all natural numbers n.

Now P(1): (1) – 7(1) + 3 = -3, which is divisible by 3.

Hence, P(1) is true.

Let us assume that P(n) is true for some natural number n = k.

P(k) = K³ – 7k + 3 is divisible by 3

or, K³ – 7k + 3 = 3m, m∈ N ........(i)

P(k+ 1 ):(k + 1)³ – 7(k + 1) + 3

= k³ + 1 + 3k(k + 1) – 7k— 7 + 3 = k -7k + 3 + 3k(k + 1)-6

= 3m + 3[k(k + 1) - 2]     [Using (i)]

= 3[m + (k(k + 1) – 2)], which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for all natural numbers n.

According to the question,

P(n) = xⁿ – yⁿ is divisible by x – y, x integers with x ≠ y.

So, substituting different values for n, we get,

P(0) = x⁰ – y⁰ = 0 Which is divisible by x − y.

P(1) = x − y Which is divisible by x − y.

P(2) = x² – y²

= (x +y)(x−y) Which is divisible by x−y.

P(3) = x³ – y³

= (x−y)(x²+xy+y²) Which is divisible by x−y.

Let P(k) = x^k – y^k be divisible by x – y;

So, we get,

⇒ x^k – y^k = a(x−y).

Now, we also get that,

⇒  P(k+1) = x^k+1 – y^k+1

= x^k(x−y) + y(x^k−y^k)

= x^k(x−y) +y a(x−y) Which is divisible by x − y.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) xⁿ – yⁿ is divisible by x – y, where x integers with x ≠ y which is true for any natural number n.

According to the question,

P(n) = 7ⁿ – 2ⁿ is divisible by 5.

So, substituting different values for n, we get,

P(0) = 7⁰ – 2⁰ = 0 Which is divisible by 5.

P(1) = 7¹ – 2¹ = 5 Which is divisible by 5.

P(2) = 7² – 2² = 45 Which is divisible by 5.

P(3) = 7³ – 2³ = 335 Which is divisible by 5.

Let P(k) = 7^k – 2^k be divisible by 5

So, we get,

⇒ 7^k – 2^k = 5x.

Now, we also get that,

⇒  P(k+1)= 7^k+1 – 2^k+1

= (5 + 2)7^k – 2(2^k)

= 5(7^k) + 2 (7^k – 2^k)

= 5(7^k) + 2 (5x) Which is divisible by 5.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = 7ⁿ – 2ⁿ is divisible by 5 is true for each natural number n.

According to the question,

P(n) = 3²ⁿ – 1 is divisible by 8.

So, substituting different values for n, we get,

P(0) = 3⁰ – 1 = 0 which is divisible by 8.

P(1) = 3² – 1 = 8 which is divisible by 8.

P(2) = 3⁴ – 1 = 80 which is divisible by 8.

P(3) = 3⁶ – 1 = 728 which is divisible by 8.

Let P(k) = 3^2k – 1 be divisible by 8

So, we get,

⇒ 3^2k – 1 = 8x.

Now, we also get that,

⇒  P(k+1) = 3^2(k+1) – 1

= 3²(8x + 1) – 1

= 72x + 8 is divisible by 8.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = 3²ⁿ – 1 is divisible by 8, for all natural numbers n.

According to the question,

P(n) = n³ – 7n + 3 is divisible by 3.

So, substituting different values for n, we get,

P(0) = 0³ – 7×0 + 3 = 3 which is divisible by 3.

P(1) = 1³ – 7×1 + 3 = −3 which is divisible by 3.

P(2) = 2³ – 7×2 + 3 = −3 which is divisible by 3.

P(3) = 3³ – 7×3 + 3 = 9 which is divisible by 3.

Let P(k) = k³ – 7k + 3 be divisible by 3

So, we get,

⇒ k³ – 7k + 3 = 3x.

Now, we also get that,

⇒  P(k+1) = (k+1)³ – 7(k+1) + 3

= k³ + 3k² + 3k + 1 – 7k – 7 + 3

= 3x + 3(k² + k – 2) is divisible by 3.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = n³ – 7n + 3 is divisible by 3, for all natural numbers n.

According to the question,

P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true

Let P(n) be 2ⁿ < n!

So, the examples of the given statements are,

P(0) ⇒ 2⁰ < 0!

i.e 1 < 1 ⇒ not true

P(1) ⇒ 2¹ < 1!

i.e 2 < 1 ⇒ not true

P(2) ⇒ 2² < 2!

i.e 4 < 2 ⇒ not true

P(3) ⇒ 2³ < 3!

i.e 8 < 6 ⇒ not true

P(4) ⇒ 2⁴ < 4!

i.e 16 < 24 ⇒ true

P(5) ⇒ 2⁵ < 5!

i.e 32 < 60 ⇒ true, etc.

According to the question,

P(n) = 2³ⁿ – 1 is divisible by 7.

So, substituting different values for n, we get,

P(0) = 2⁰ – 1 = 0 which is divisible by 7.

P(1) = 2³ – 1 = 7 which is divisible by 7.

P(2) = 2⁶ – 1 = 63 which is divisible by 7.

P(3) = 2⁹ – 1 = 512 which is divisible by 7.

Let P(k) = 2^3k – 1 be divisible by 7

So, we get,

⇒ 2^3k – 1 = 7x.

Now, we also get that,

⇒  P(k+1) = 2^3(k+1) – 1

= 2³(7x + 1) – 1

= 56x + 7

= 7(8x + 1) is divisible by 7.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = 2³ⁿ – 1 is divisible by7, for all natural numbers n.

According to the question,

P(n) = 4ⁿ – 1 is divisible by 3.

So, substituting different values for n, we get,

P(0) = 4⁰ – 1 = 0 which is divisible by 3.

P(1) = 4¹ – 1 = 3 which is divisible by 3.

P(2) = 4² – 1 = 15 which is divisible by 3.

P(3) = 4³ – 1 = 63 which is divisible by 3.

Let P(k) = 4^k – 1 be divisible by 3,

So, we get,

⇒ 4^k – 1 = 3x.

Now, we also get that,

⇒  P(k+1) = 4^k+1 – 1

= 4(3x + 1) – 1

= 12x + 3 is divisible by 3.

⇒ P(k+1) is true when P(k) is true

Therefore, by Mathematical Induction,

P(n) = 4ⁿ – 1 is divisible by 3 is true for each natural number n.

According to the question,

P(n) is 1 + 2 + 2² + … 2ⁿ = 2ⁿ⁺¹ – 1.

So, substituting different values for n, we get,

P(0) = 1 = 2⁰⁺¹ − 1 Which is true.

P(1) = 1 + 2 = 3 = 2¹⁺¹ − 1 Which is true.

P(2) = 1 + 2 + 2² = 7 = 2²⁺¹ − 1 Which is true.

P(3) = 1 + 2 + 2² + 2³ = 15 = 2³⁺¹ − 1 Which is true.

Let P(k) = 1 + 2 + 2² + … 2^k = 2^k+1 – 1 be true;

So, we get,

⇒ P(k+1) is 1 + 2 + 2² + … 2^k + 2^k+1 = 2^k+1 – 1 + 2^k+1

= 2×2^k+1 – 1

= 2^(k+1)+1 – 1

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

1 + 2 + 2² + … 2ⁿ = 2ⁿ⁺¹ – 1 is true for all natural numbers n.

To prove; Number of subsets of a set containing n distinct elements is 2ⁿ.

For a null set there is only one element Φ and therefore only one subset.

⇒ at n = 0 number of subsets = 1 = 2⁰

By considering a set with 1 element there will be 2 subsets with 1 and Φ

⇒ at n = 0 number of subsets = 2 = 2²

By considering a set S_k with k element

Let at n = k number of subsets = 2^k be true.

For a set S_k+1 containing k+1 element

The extra one element in set S_k+1 when compared with S_k will form an extra collection of subsets by combing with the existing 2^k subsets and as a result an extra 2^k subsets are formed.

⇒ at n = k+1 number of subsets = 2^k + 2^k = 2.2^k = 2^k+1

⇒ P(k+1) is true.

∴ By Mathematical Induction number of subsets of a set containing n distinct elements is 2ⁿ, for all n ϵ N.

A. 5

Given; Given; P(n): 10ⁿ + 3.4ⁿ⁺² + k is divisible by 9.

P(1) = 10 +3.4³ + k = 202 + k

For this number to be divisible by 9; sum of its digits should be 9 or multiples of 9.

⇒ 2 + 2 + k = 9

∴ k = 5

True; This is the principle of mathematical induction.

A. 1

Given; P(n): xⁿ – 1 is divisible by x – k

⇒ P(1) : x – 1

⇒ P(2) : x² – 1 = (x−1)(x+1)

⇒ P(3) : x³ – 1 = (x−1)(x²+xy+1)

⇒ P(4) : x⁴ − 1 = (x²−1)(x²+1) = (x−1)(x+1)(x²+1)

∴ The least positive integral value of k is 1.

Given; P(n): 2n < n! n ϵ N

⇒ P(1) : 2×1<1! ⇒ 2<1; it’s not true.

⇒ P(2) : 2×2<2! ⇒ 4<2; it’s not true.

⇒ P(3) : 2×3<3! ⇒ 6<6; it’s not true.

⇒ P(4) : 2×4<4! ⇒ 8<24; it’s true.

⇒ P(3) : 2×5<5! ⇒ 10<120; it’s true.

∴ n≥4

According to the question,

P(n) is n² < 2ⁿ  for n≥5

Let P(k) = k² < 2^k be true;

⇒ P(k+1) = (k+1)²

= k² + 2k + 1

2^k+1 = 2(2^k) > 2k²

Since, n² > 2n + 1 for n ≥3

We get that,

k² + 2k + 1 < 2k²

⇒ (k+1)² < 2^(k+1)

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = n² < 2ⁿ is true for all natural numbers n ≥ 5.

Let P(n) be the given statement, i.e., P(n) : (2n + 1) < 2ⁿ for all natural numbers, n ≥ 3. We observe that P(3) is true, since 

2.3 + 1 = 7 < 8 = 2³ 

Assume that P(n) is true for some natural number k, i.e., 2k + 1 < 2^k 

To prove P(k + 1) is true, we have to show that 2(k + 1) + 1 < 2^k+1. Now, we have 2(k + 1) + 1 = 2 k + 3 

= 2k + 1 + 2 < 2^k + 2 < 2^k . 2 = 2k + 1 . 

Thus P(k + 1) is true, whenever P(k) is true. 

Hence, by the Principle of Mathematical Induction P(n) is true for all natural numbers, n ≥ 3.

tan^-1x + tan^-1y = π/4

⇒ tan^-1((x + y)/(1 - x.y)) = π/4

⇒ (x + y)/(1 - x.y) = tan(π/4) = 1

⇒ x + y = 1 - xy ⇒  x + y + xy = 1

Answer is (a) a³.(a^3x/3 log a) + k

Correct Answer is (B) \(\frac{\pi}{4}\) 

Now \(cot^{-1}9+cosec^{-1}\frac{\sqrt{41}}{4}\) can be written in terms of tan inverse as \(cot^{-1}9+cosec^{-1}\frac{\sqrt{41}}{4}\)= \(cot^{-1}\frac{1}{9}+cosec^{-1}\frac{4}{5}\) 

Since we know that tan^-1 x + tan^-1 y = \(tan^{-1}(\frac{x+y}{1-xy})\) 

⇒ \(cos^{-1}\frac{1}{9}+cosec^{-1}\frac{4}{5}\) = tan^-1\((\frac{\frac{1}{9}+\frac{4}{5}}{1-(\frac{1}{9}\times\frac{4}{5})})\) 

= tan-1\((\frac{41}{41})\) 

= tan-1 (1) = \(\frac{\pi}{4}\)

Correct answer is 

(1) → (d) (2) → (f) (3) → (c) (4) →(b) (5) → (a) (6) → (e)

correct answer is 2/5

Correct answer is 

(a) 90 

(b) 74 

(c) 50 

(c) 90 m

Because,

Let us assume the Breadth of the rectangular field be x m.

Let us assume the length of the rectangular field is thrice its breadth be 3x m

Then,

Perimeter of the rectangular field = 2(l + b)

= 240 = 2(x + 3x)

= 240 = 2(4x)

= 240 = 8x

= x = 240/8

= x = 30

∴The length of the rectangular field is 3x = 3 × 30 = 90 m

Correct answer is Rs 23.15

A carpet is laid on the floor of a room 8 m by 5 m.

Area of the border = 12 m²

Let the width of the carpet be x meter

Area of floor = Length × Breadth

= 8 × 5

= 40 m²

Length without border = 8 m – (x + x) = (8 – 2x) m

Breadth without border = 5 m – (x + x) m = (5 – 2x) m

Area without border = Length without border × Breadth without border

= (8 – 2x) × (5 – 2x)

= 40 – 16x – 10x + 4x²

Area of border = Area of floor – Area without border

12 = 40 – (40 – 16x – 10x + 4x²)

or 4x² – 26x + 12 = 0

Solving above equation, we have

(x– 6) (4x -2) = 0

x = 6 or x = 1/2

Since Border cannot be greater than carpet.

Therefore, width of border is 1/2 m.

Width of the room left uncovered = 0.25 m

Now,

Length of the room to be carpeted = 4.9 - (0.25 + 0.25) = 4.9 - 0.5 = 4.4 m

Breadth of the room be carpeted = 3.5 - (0.25 + 0.25) = 3.5 - 0.5 = 3m

Area to be carpeted = 4.4 x 3 = 13.2m²

Breadth of the carpet 80 cm = 0.8 m.

We know:

Area of the room = Area of the carpet

Length of the carpet = \(\frac{Area\,of\,the\,room}{Breadth\,of\,the\,carpet}\)

= \(\frac{13.5}{0.8}\)

= 16.5m

Cost of 1 m carpet = Rs 80

Cost of 16.5 m carpet = 80 x 16.5 = Rs 1,320

We are given with a right-angled triangle whose measures are

Base = 48 cm

Hypotenuse = 50 cm

Using Pythagoras Theorem:

Hypotenuse² = Base² + Perpendicular²

50² = 48² + Perpendicular²

or Perpendicular² = 2500 – 2304

or Perpendicular = 14 cm

Now,

Area of a triangle = 1/2 × Base × Height

= 1/2 × 48 cm × 14 cm

= 336 cm²

Given : Circumference = 264 cm

Circumference of a circle = 2πr = 2 × 227 x r = 264

2 × 22 × r = 264 × 7

r = \(\frac {264 \times7}{2\times22}\) = 42

∴ radius of the circle = 42 cm.

Let two sides of an isosceles right-angled triangle are of measure “a”.

Area of triangle = 200 cm². (given)

200 = 1/2(a²)

=> a = 20 cm

Now,

Hypotenuse = √(a²+a²) = √(2a²) = √2 a

= 20 √2

= 20 x 1.414

= 28.28

=> Hypotenuse is 28.28 cm

Perimeter of triangle = Sum of all the sides = 2a + Hypotenuse

= 40 + 28.28

= 68.28

=> Perimeter of triangle is 68.28 cm.

(a) 4.8 cm

From the question it is given that,

∆ MNO is a right-angled triangle, its legs are 6 cm and 8 cm long.

By the rule of Pythagoras theorem,

Hypotenuse² = perpendicular² + base²

In the given figure,

MO² = MN² + NO²

MO² = 6² + 8²

MO² = 36 + 64

MO² = 100

MO = √100

MO = 10 cm

Then, consider the triangle MNO,

Area of triangle MNO = ½ × MN × NO 

= ½ × MO × NP

= ½ × 6 × 8 

= ½ × 10 × NP

Therefore, NP = 24/5

NP = 4.8 cm

(b) 13 cm

We know that, area of triangle = ½ × base (small side) × height

30 = ½ × 5 cm × height

Height = (30 × 2)/5

Height = 60/5

Height = 12 cm

By the rule of Pythagoras theorem,

Hypotenuse² = height² + base²

Hypotenuse² = 12² + 5²

Hypotenuse² = 144 + 25

Hypotenuse² = 169

Hypotenuse = √169

Hypotenuse = 13 cm

Area of a square room

= 1 × 1

= 8 × 8

= 64 Sq.mt

The number of granites

= 2m × 1m

= 2m

= 64/2 = 32

(c) 15.7 cm

From the question it is given that, diameter = 5 cm

Radius = diameter/2

Radius = 5/2 = 2.5 cm

Circumference of a circle = 2πr

= 2 × (22/7) × 2.5

= 15.7 cm