Class 9 MAHARASHTRA STATE TEXTBOOK BUREAU Solutions Maths Chapter 4 - Ratio and Proportion

(i)

(ii)

(iii)

(i)

(ii)

Rs. 14 = 14 × 100 paise = 1400 paise

Rs. 12 and 40 paise = 12×100 paise + 40 paise = 1240 paise

(iii)

5 litres = 5 × 1000 ml = 5000 ml

(iv)

3 years 4 months = 3×12 months + 4 months = (36 + 4) months = 40 months

5 years 8 months = 5×12 months + 8 months = (60 + 8) months = 68 months

(v)

3.8 kg = 3.8 × 1000 gm = 3800 gm

(vi)

7 minutes 20 seconds = 7 × 60 seconds + 20 seconds = (420 + 20) seconds = 440 seconds

5 minutes 6 seconds = 5 x 60 seconds + 6 seconds = (300 + 6) seconds = 306 seconds

(i)

(ii)

(iii)

(iv)

(v)

Let x persons are required to build a house in 6 days.

Days required to build a house and number of persons are in inverse proportion.

∴ 6 × x = 8 × 3

∴ 6x = 24

∴ x = 4

∴ 4 persons are required to build the house in 6 days.

(i)

Let 15 : 25 = x%

∴ 15 : 25 = 60%

(ii)

Let 47 : 50 = x%

∴ 47 : 50 = 94%

(iii)  

(iv)

Let   

(v)

Let  

As the ages of Abha and her mother are in the ratio. 2 : 5.

So, there will be a common multiple and let that common multiple be x.

∴ Present age of Abha = 2x years and

Present age of Abha's mother = 5x years

By the condition given in the question, we have

∴ 5x - 2x = 27

∴ 3x = 27

∴ x = 9

∴ Present age of Abha = 2x = 2 × 9 = 18 years

∴ Present age of Abha's mother = 5x = 5 × 9 = 45 years

∴ The present ages of Abha and her mother are 18 years and 45 years respectively.

Present age of Vatsala = 14 years

Present age of Sara = 10 years

After x years,

Vatsala's age = (14 + x) years

Sara's age = (10 + x) years

By the condition given in the question, we have

∴ 4(14 + x) = 5(10 + x)

∴ 56 + 4x = 50 + 5x

∴ 56 - 50 = 5x - 4x

∴ 6 = x

∴ x = 6

∴ After 6 years, the ratio of their ages will become 5 : 4.

As the ages of Rehana and her mother are in the ratio 2 : 7.

So, there will be a common multiple and let that common multiple be x.

∴ Present age of Rehana = 2x years and

Present age of Rehana's mother = 7x years

After 2 years,

Rehana's age = (2x + 2) years

Age of Rehana's mother = (7x + 2) years

By the condition given in the question, we have

∴ 3(2x + 2) = 1(7x + 2)

∴ 6x + 6 = 7x + 2

∴ 6 - 2 = 7x - 6x

∴ 4 = x

∴ x = 4

∴ Rehana's present age = 2x = 2 x 4 = 8 years

∴ Rehana's present age is 8 years.

(i)

(ii)

(i)

Radius of a circle = r

Circumference of a circle = 2πr

∴ The ratio of radius to circumference of the circle is 1 : 2π.

(ii)

Radius of a circle is r.

Circumference of a circle = 2πr

Area of a circle = πr2

∴ The ratio of circumference of circle with radius r to its area is 2 : r.

(iii)

Length of the sides of a square = 7 cm

Diagonal of a square

∴ The ratio of diagonal of a square to its side is

(iv)

Length of rectangle = (l) = 5 cm

Breadth of rectangle = (b) = 3.5 cm

Now,

Perimeter of the rectangle = 2(l + b)

= 2(5 + 3.5)

= 2 × 8.5

= 17 cm

Area of the rectangle = l × b

= 5 × 3.5

= 17.5 cm2

∴ Ratio of perimeter of rectangle to the area of rectangle is 34 : 35.

(i)

Here,

Since, 35 < 81

(ii)

Here,

Since, 75 < 105

(iii)

Here,

5 × 121 = 605

17 × 18 = 306

Since, 605 > 306

(iv)

Here,

Since, 2160 = 2160

(v)

Here,

9.2 × 7.1 = 65.32

3.4 × 5.1 = 17.34

Since, 65.32 > 17.34

(i)

Since ∠A and ∠B for given parallelogram are in the ratio 5 : 4.

So, there will be a common multiple and let that be x.

∴ m ∠A = 5x°and m ∠B = 4x°

Now, m ∠A + m ∠B = 180° … [Adjacent angles of a parallelogram are supplementary]

∴ 5x° + 4x°= 180°

∴ 9x° = 180°

∴ x° = 20°

∴ m ∠B = 4x° = 4 × 20° = 80°

∴ The measure of ∠B is 80°.

(ii)

As the present ages of Albert and Salim are in the ratio 5 : 9.

So, there will be a common multiple and let that be x.

∴ Present age of Albert = 5x years and

Present age of Salim = 9x years

After 5 years,

Albert's age = (5x + 5) years

Salim's age = (9x + 5) years

By the condition given in the question, we have

∴ 5(5x + 5) = 3(9x + 5)

∴ 25x + 25 = 27x + 15

∴ 25 - 15 = 27 x - 25 x

∴ 10 = 2x

∴ x = 5

∴ Present age of Albert = 5x = 5 × 5 = 25 years

Present age of Salim = 9x = 9 × 5 = 45 years

∴ The present ages of Albert and Salim are 25 years and 45 years respectively.

(iii)

As length and breadth of a rectangle are in the ratio 3 : 1.

So, there will be a common multiple and let that be x.

Length of the rectangle (l) = 3x cm

and Breadth of the rectangle (b) = x cm

Perimeter of the rectangle = 36 cm … (Given)

Since, Perimeter of the rectangle = 2(l + b)

∴ 36 = 2(3x + x)

∴ 36 = 2(4x)

∴ 36 = 8x

∴ x = 4.5

∴ Breadth of the rectangle = 4.5 cm

Length of the rectangle = 3x = 3 × 4.5 = 13.5 cm

∴ The length and breadth of the rectangle are 13.5 cm and 4.5 cm respectively.

(iv)

The ratio of two numbers is 31 : 23.

So, there will be a common multiple and let that be x.

∴ First number = 31x and Second number = 23x

By the condition given in the question, we have

31x + 23x = 216

∴ 54x = 216

∴ x = 4

∴ First number = 31x = 31 × 4 = 124

Second number = 23x = 23 × 4 = 92

∴ The two numbers are 124 and 92.

(v)

The ratio of two numbers is 10 : 9.

So, there will be a common multiple and let that be x.

∴ First number = 10x and Second number = 9x

By the condition given in the question, we have

∴ (10x) (9x) = 360

∴ 90x2 = 360

∴ x2 = 4

∴ x = 2 …. [Taking positive square root on both sides]

∴ First number = 10x = 10 × 2 = 20

Second number = 9x = 9 × 2 = 18

∴ The two numbers are 20 and 18.

Given: a : b = 3 : 1

∴ a = 3b … (I)

Also, b : c = 5 : 1

∴ b = 5c … (II)

Substituting the value of b from (II) in (I)

∴ a = 3(5c)

∴ a = 15c … (III)

(i)

(ii)

Given:

Squaring both the sides, we get

0.04 × 0.4 × a = (0.4)2 × (0.04)2 × b

Given: (x + 3) : (x + 11) = (x - 2) : (x+ 1)

∴ (x + 3)(x +1) = (x - 2)(x + 11)

∴ x(x +1) + 3(x + 1) = x(x + 11) - 2(x + 11)

∴ x2 + x + 3x + 3 = x2 + 11x - 2x - 22

∴ x2 + 4x + 3 = x2 + 9x - 22

∴ 4x + 3 = 9x - 22

∴ 3 + 22 = 9x - 4x

∴ 25 = 5x

∴ x = 5

Given:

Let

∴ a = 7k, b = 3k

Given:

Squaring both the sides

Given:

Taking cube on both the sides

Given:

Let

∴ a = 7k, b = 3k

… (Given)

… (Given)

… (Given)

… (Given)

Given:

By componendo-dividendo, we have

Given:

This equation is true for x = 0

∴ x = 0 is one of the solutions.

If x ≠ 0, then x2 ≠ 0

Dividing both sides by x2

∴ 8x + 12 = 12x - 20

∴ 12 + 20 = 12x - 8x

∴ 32 = 4x

∴ x = 8

∴ x = 0 or x = 8 are the solutions of the given equation.

Given:

If x = 0, then

which is a contradiction

∴ x ≠ 0

Now,

∴ 21(x - 5) = 4(2x + 3)

∴ 21x - 105 = 8x + 12

∴ 21x - 8x = 12 + 105

∴ 13x = 117

∴ x = 9

∴ x = 9 is the solution of the given equation.

Given:

are the solutions of the given equation.

Given:

∴ 9(4x + 1) = 25(x + 3)

∴ 36x + 9 = 25x + 75

∴ 36x - 25x = 75 - 9

∴ 11x = 66

∴ x = 6

∴ x = 6 is the solution of the given equation

Given:

∴ 6(4x + 1) = 5(2x + 3) or 6(4x + 1) = -5(2x + 3)

∴ 24x + 6 = 10x + 15 or 24x + 6 = -10x - 15

∴ 24x - 10x = 15 - 6 or 24x + 10x = - 15 - 6

∴ 14x = 9 or 34x = -21

are the solutions of the given equation.

Given:

∴ 4(3x - 4) = 5(x + 1)

∴ 12x - 16 = 5x + 5

∴ 12x - 5x = 16 + 5

∴ 7x = 21

∴ x = 3 is the solution of the given equation.

Also,

Also,

Given: 5m - n = 3m + 4n

∴ 5m - 3m = 4n + n

∴ 2m = 5n

Squaring both the sides, we get

Given: 5m - n = 3m + 4n

∴ 5m - 3m = 4n + n

∴ 2m = 5n

Since, out of a, b, c no two of them are equal.

∴ The values of (b - c), (c - a) and (a - b) are not zero.

As a(y + z) = b(z + x) = c(x + y)

Let

Let

Thus, each ratio is equal to 1.

Let

Thus, each ratio is equal to

Let … (I)

From (II), (III) and (IV), we get

Let … (I)

Multiplying numerator and denominator of third ration by 5

∴ Every ratio

Given:

If x = 0, then

∴ x ≠ 0

Let … (I)

Multiplying numerator and denominator of second ratio 4x

∴ 7(4x - 5)3(2x + 3)

∴ 28x - 35 = 6x + 9

∴ 28x - 6x = 9 + 35

∴ 22x = 44

∴ x = 2

∴ x = 2 is the solution of the given equation.

Given:

If y = 0, then

∴ y ≠ 0

Let … (I)

Multiplying numerator and denominator of second ratio 5y

∴ y + 8 = 3(1 + 2y)

∴ y + 8 = 3 + 6y

∴ 8 - 3 = 6y - y

∴ 5 = 5y

∴ y = 1

∴ y = 1 is the solution of the given equation.

Let x be subtracted from 12, 16 and 21 such that resultant numbers are in continued proportion.

(12 - x), (16 - x), (21 - x) are in continued proportion.

∴ 4(21 - x) = 5(16 - x)

∴ 84 - 4x = 80 - 5x

∴ 5x - 4x = 80 - 84

∴ x = -4

∴ -4 should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion.

As (28 - x) is the mean proportional of (23 - x) and (19 - x).

∴ -5(19 - x) = 9(28 - x)

∴ -95 + 5x = 252 - 9x

∴ 5x + 9x = 252 + 95

∴ 14x = 347

Let the first number be x.

∴ Third number = 26 - x

Since, 12 is the mean proportional of x and (26 - x).

∴ x(26 - x) = 12 x 12

∴ 26x - x2 = 144

∴ x2 - 26x + 144 = 0

∴ x2 - 18x - 8x + 144 = 0

∴ x(x - 18) - 8(x - 18) = 0

∴ (x - 18) (x - 8) = 0

∴ x = 18 or x = 8

∴ Third number = 26 - x = 26 - 18 = 8 or 26 - x = 26 - 8 = 18

∴ The numbers are 18, 12, 8 or 8, 12, 18.

Given: (a + b + c)(a - b + c) = a² + b² + c²

∴ a(a - b + c) + b(a - b + c) + c(a - b + c) = a² + b² + c²

∴ a² - ab + ac + ab - b² + be + ac - be + c² = a² + b² + c²

∴ a² + 2ac - b² + c² = a² + b² + c²

∴ 2ac - b² = b²

∴ 2ac = 2b²

∴ ac = b²

∴ b² = ac

∴ a, b, c are in continued proportion.

Let

∴ b = ck … (I)

∴ a = bk = (ck)k

∴ a = ck² … (II)

L.H.S = (a + b + c) (b - c)

= [ck² + ck + c] [ck - c] … [From (I) and (II)]

= c(k² + k + 1) c (k - 1)

= c² (k² + k + 1) (k - 1)

R.H.S = ab - c²

= (ck²) (ck) - c² … [From (I) and (II)]

= c²k³ - c²

= c²(k³ - 1)

= c² (k - 1) (k² + k + 1) … [a³ - b³ = (a - b) (a² + ab + b²]

∴ L.H.S = R.H.S

∴ (a + b + c) (b - c) = ab - c²

Let

∴ b = ck … (I)

∴ a = bk = (ck)k

∴ a = ck² … (II)

L.H.S = (a² + b²) (b² + c²)

= [(ck²)² + (ck)²] [(ck)² + c²] … [From (I) and (II)]

= [c²k⁴ + c²k²] [c²k² + c²]

= c²k² (k² + 1) c² (k² + 1)

= c⁴k² (k² + 1)²

R.H.S = (ab + bc)²

= [(ck²) (ck) + (ck)c]² …[From (I) and (II)]

= [c²k³ + c²k]²

= [c²k (k² + 1)]²

= c⁴(k² + 1)²

∴ L.H.S = R.H.S

∴ (a² + b²) (b² + c²) = (ab + bc)²

Let

∴ b = ck … (I)

∴ a = bk = (ck)k

∴ a = ck² … (II)

Let a be the mean proportional of

∴ Mean proportional of is

(B)

Given: 6 : 5 = y : 20

(C)

1 cm = 10 mm

∴ Ratio

(B)

(D)

The ratio of bananas distributed between Shubham and Anil is 3 : 5.

Let the number bananas Shubham got = 3x and number of bananas Anil got = 5x

By the given condition, we have

3x + 5x = 24

∴ 8x = 24

∴ x = 3

∴ Number of bananas Shubham get = 3×3 = 9

(C)

Let x be the mean proportional of 4 and 25

Let r be radius of the circle.

Diameter = 2r

∴ Radius to the diameter of a circle.

Length of the rectangle = 4 cm

Breadth of the rectangle = 3 cm

∴ The ratio of diagonal to the length of a rectangle is 5 : 4.

Perimeter of a square = 4 × (length of side) = 4 × 4 cm = 16 cm

Area of a square = (length of side)2 = (4 cm)2 = 16 cm2

∴ The ratio of numbers denoting perimeter to area of a square is 1 : 1.

When a, b, c are in continued proportion, then b² = ac.

Let, a = 2, b = 4 and c = 8

Here, b² = 42 = 16

Now, ac = 2 × 8 = 16

∴ b² = ac

∴ 2, 4, 8 are in continued proportion.

When a, b, c are in continued proportion, then b² = ac.

Let, a = 1, b = 2 and c = 3

Here, b² = 22 = 4

Now, ac = 1 × 3 = 3

∴ b² ≠ ac

∴ 1, 2, 3 are not in continued proportion.

When a, b, c are in continued proportion, then b² = ac.

Let, a = 9, b = 12 and c = 16

Here, b² = (12)2 = 144

Now, ac = 9 × 16 = 144

∴ b² = ac

∴ 9, 12, 16 are in continued proportion.

When a, b, c are in continued proportion, then b² = ac.

Let, a = 3, b = 5 and c = 8

Here, b² = 52 = 25

Now, ac = 3 × 8 = 24

∴ b² ≠ ac

∴ 3, 5, 8 are not in continued proportion.

Given: a = 3 and c = 27

Also, a, b, c are in continued proportion.

∴ b² = ac

∴ b² = 3 × 27

∴ b² = 81

Taking square root of both sides

∴ b = 9

Let 37 : 500 = x%

∴ 37 : 500 = 7.4%

Let = x%

∴ x = 62.5%

∴ = 62.5%

Let = x%

∴ = 73.33%

Let = x%

∴ = 31.25%

Let = x%

∴ = 12%

1024 MB = 1 GB

∴ Ratio = 5 : 6

17 Rupees = 17 × 100 = 1700 paise

25 Rupees 60 paise = (25 × 100) paise + 60 paise

= 2500 paise + 60 paise

= 2560 paise

∴ Ratio = 85 : 128

5 dozen = 5 × 12 units = 60 units

∴ Ratio = 1 : 2

4 sq. m = 4 × 10000 sq. cm = 40000 sq. cm

∴ Ratio = 50 : 1

1.5 kg = 1.5 × 1000 gm = 1500 gm

∴ Ratio = 3 : 5

Given:

Let

∴ a = 2k, b = 3k

Given:

Squaring both the sides

Given:

Taking cube on both the sides

Given:

Taking cube on both the sides

Given: a, b, c, d are in proportion

Let

∴ a = bk and c = dk … (I)

Given: a, b, c, d are in proportion

Let

∴ a = bk and c = dk … (I)

Given: a, b, c, d are in proportion

Let

∴ a = bk and c = dk … (I)

Given: a, b, c are in continued proportion

Let

∴ b = ck … (I)

And, a = bk

∴ a = (ck)k = ck² … (II)

Given: a, b, c are in continued proportion

Let

∴ b = ck … (I)

And, a = bk

∴ a = (ck)k = ck² … (II)

Given:

If x = 0 then

∴ Which is a contradiction.

∴ x ≠ 0

Let … (I)

Multiplying numerator and denominator of the second ratio by 6x, as x ≠ 0

∴ 29(2x + 3) = 21 (3x + 2)

∴ 5 + 87= 63x + 42

∴ 87 - 42 = 63x - 58x

∴ 45 = 5x

∴ x = 9

∴ x = 9 is the solution of the given equation.

Let

Multiplying numerator and denominator of second ratio by -3

∴ Each ratio =

Let … (I)

From (I), we have

Again from (I), we have

From (II), (III) and (IV), we get