Ratio And Proportion

(i) 72,60

Reduced form of ratio of first number to second number:

(Break the expression in order to simplify it further)

(ii) 38,57

Reduced form of ratio of first number to second number:

(Break the expression in order to simplify it further)

(iii) 52, 78

Reduced form of ratio of first number to second number:

(Break the expression in order to simplify it further)

(i) Reduced form of the ratio of 700Rs and 308Rs is:

(Break the expression in order to simplify it further)

(ii) Reduced form of the ratio of 14Rs and 12.40Rs is:

Multiply denominator and numerator by 100:

Divide numerator and denominator by 10:

(Break the expression in order to simplify it further)

(iii) 5 litre = 5000 ml

∴ Reduced form of the ratio of 5000 ml and 2500 ml is:

(Break the expression in order to simplify it further)

(iv) 3 years = 3 × 12 = 36 months

∴ 3 years 4 months = 40 months

5 years = 5 × 12 = 60

∴ 5 years 8 months = 68 months

∴ Reduced form of the ratio of 40 months and 68 months is:

(Break the expression in order to simplify it further)

(v) 3.8 kg = 3.8 × 1000 = 3800 gm

∴ Reduced form of the ratio of 3800 gm and 1900 gm is:

(Break the expression in order to simplify it further)

(vi) 7 minutes = 7 × 60 = 420 seconds

∴ 7 minutes 20 seconds = 440 seconds

5minutes = 5 × 60 = 300 seconds

∴ 5 minutes 6 seconds = 306 seconds

∴ Reduced form of the ratio of 440 seconds and 306 seconds is:

(Break the expression in order to simplify it further)

(i) Reduced form of the ratio of 75:100 is:

(ii) Reduced form of the ratio of 44:100 is:

(iii) Reduced form of 6.25% is:

(iv) Reduced form of the ratio of 52:100 is:

(v) Reduced form of 0.64% is:

No. of persons required to build a house in 8 days =3
No. of persons required to build a house in 1 day =3×8 = 24
No. of persons required to build a house in 6 days = 24/6 = 4

∴ 4 persons are required to build the same house.

(i) 15 : 25

= 15/25

= ((15/25) × 100)%

= (15 × 4) %

= 60 %

(ii) 47 : 50

= 47/50

= ((47/50) × 100)%

= (47 × 2) %

= 94 %

(iii) 7/10

= ((7/10) × 100)%

= (7 × 10) %

= 70 %

(iv) 546/600

= ((546/600) × 100)%

= (546/6) %

= 91 %

(v) 7/16

= ((7/16) × 100)%

= (7 × 6.25) %
= 43.75 %

Let age of Abha = x years

∴ Age of Abha’s mother at her birth = (x + 27) years

Ratio of Abha’s and her mother’s age = 2:5

Cross multiply and get:

5x = 2(x + 27)

⇒ 5x = 2x + 54

⇒ 5x – 2x = 54

⇒ 3x = 54

⇒ x = 54/3

⇒ x = 18

∴ Age of Abha = 18 years

Age of Abha’s mother = 18 + 27 = 45 years

Given: Present age of Vatsala = 14 years

Present age of Sara = 10 years

Let after x years, the ratio of their ages will be 5:4.

∴ Age of Vatsala after x years = (14 + x) years

Age of Sara after x years = (10 + x) years

Ratio of their ages = 5:4

On cross multiplying, we get:

56 + 4x = 50 + 5x

⇒ 5x – 4x = 56 – 50

⇒ x = 6

∴ After 6 years, their ages will be 20 years and 16 years and ratio

of their ages will be 5:4.

Given: Ratio of present age of Rehana and her mother = 2:7

So, let present age of Rehana = 2x

∴ Present age of her mother = 7x

After 2 years, age of Rehana = (2x + 2) years

After 2 years, age of her mother = (7x + 2) years

Now, given that after two years ratio of their ages will be 1:3.

On cross multiplying, we get:

6x + 6 = 7x + 2

⇒ 7x – 6x = 6 – 2

⇒ x = 4

∴ Present age of Rehana = 2x = (2 × 4) years = 8 years

(i) Let

∴ on comparing first two equalities, we get:

5/7 = x/28

Cross multiply and get:

7x = 28 × 5

⇒ x = 4 × 5 = 20

Now, compare the first and third equalities and get:

5/7 = 35/y

Cross multiply and get:

5y = 7 × 35

⇒ y = 7 × 7 = 49

Now, compare the first and fourth equalities and get:

5/7 = z/3.5

Cross multiply and get:

7z = 5 × 3.5

⇒ 7z = 5 × (35/10)

⇒ z = 5 × (5/10)

⇒ z = 25/10 = 2.5

∴

(ii) Let

∴ On comparing first two equalities, we get:

9/14 = 4.5/x

Cross multiply and get:

9x = 14 × 4.5

⇒ x = 14 × 0.5 = 7

Now, compare the first and third equalities and get:

9/14 = y/42

Cross multiply and get:

14y = 9 × 42

⇒ y = 9 × 3 = 27

Now, compare the first and fourth equalities and get:

9/14 = z/3.5

Cross multiply and get:

14z = 9 × 3.5

⇒ z = 9 × (3.5/14)

⇒ z = 9 × (0.25)

⇒ z = 2.25

∴

(i) Let r be the radius of a circle.

Circumference of circle = 2πr

Ratio of radius to circumference of circle = r/2πr

= 1/2π

= 1 : 2π

(ii) Let r be the radius of a circle.

Circumference of circle = 2πr

Area of the circle =2πr²

Ratio of radius to circumference of circle = 2πr/πr²

= 2/r

= 2 : r

(iii) Side of square = 7 cm

Diagonal of square = √2 × side = 7√2 cm

Ratio of diagonal of a square to its side = 7/7√2

= 1/√2

= 1 : √2

(iv) Length of rectangle = 5 cm

Breadth of rectangle = 3.5 cm

Perimeter of rectangle = 2(Length + Breadth)

= 2(5+3.5)

= 2(8.5)

= 17 cm

Area of rectangle = Length × Breadth

= 5 × 3.5

= 16.5 cm²

Ratio of Perimeter to area of rectangle

= 17/16.5

= 170/165

= 34/33

(i)

Given ratios are

Step I: Make the second term of both the ratios equal.

Multiply and divide first ratio by √7:

Multiply and divide second ratio by 3:

Step II: Compare the first terms (numerators) of the new ratios.

Since the denominators of new ratios are equal, compare the numerators of the new ratios:

Since, 9>√35, therefore .

Therefore the second ratio is greater than the first ratio according to the ratio comparison rules.

⇒

(ii)

Given ratios are

Step I: Make the second term of both the ratios equal.

Multiply and divide first ratio by √5:

Multiply and divide second ratio by √7:

Step II: Compare the first terms (numerators) of the new ratios.

Since the denominators of new ratios are equal, compare the numerators of the new ratios:

Since, 21>15, therefore .

Therefore the second ratio is greater than the first ratio according to the ratio comparison rules.

⇒

(iii)

Given ratios are

Step I: Make the second term of both the ratios equal.

Multiply and divide first ratio by 121:

Multiply and divide second ratio by 18:

Step II: Compare the first terms (numerators) of the new ratios.

Since the denominators of new ratios are equal, compare the numerators of the new ratios:

Since, 605 <306, therefore .

Therefore, the first ratio is greater than the second ratio according to the ratio comparison rules.

⇒

(iv)

Given ratios are

Simplifying the ratios, we get:

Since, the denominators of both the terms are same; compare the first terms (numerators) of the new ratios.

Since the denominators of new ratios are equal, compare the numerators of the new ratios:

Since, √5 = √5, therefore .

Therefore, both the ratios are equal, according to the ratio comparison rules.

⇒

(v)

Given ratios are

Simplifying the ratios, we get:

(Multiply the numerator and denominator of both the ratios by 10)

Step I: Make the second term of both the ratios equal.

Multiply and divide first ratio by 71:

Multiply and divide second ratio by 51:

Step II: Compare the first terms (numerators) of the new ratios.

Since the denominators of new ratios are equal, compare the numerators of the new ratios:

Since, 6532 > 1734, therefore .

Therefore the first ratio is greater than the second ratio according to the ratio comparison rules.

⇒

Given: ∠A : ∠B = 5:4

Let the measure of ∠A = 5x

Then ∠B = 4x

Since the adjacent angles of a parallelogram are complementary, therefore ∠A + ∠B = 180°

⇒ 5x + 4x = 180°

⇒ 9x = 180°

⇒ x = 20°

∴ ∠ A = 5x = 5 × 20° = 100°

∴ ∠ B = 4x = 4 × 20° = 80°

Given: Ratio of present age of Albert and Salim=5:9

So, let present age of Albert = 5x years

∴ Present age of Salim = 9x years

Five years later, age of Albert = (5x + 5) years

Five years later, age of Salim = (9x + 5) years

Given that this ratio of their ages is 3:5.

∴ According to the given problem:

⇒ 25x + 25 = 27x + 15

⇒ 27x – 25x = 25 – 15

⇒ 2x = 10

⇒ x = 5

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

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

Let l and b be the length and breadth of rectangle.

Given that Length : Breadth = 3:1

Perimeter of rectangle = 36

Let length of rectangle = l = 3x cm

∴ Breadth of the rectangle = b = x cm

Perimeter of rectangle = 2(l + b) = 36 cm

∴ 2(3x + x) = 36

⇒ 2(4x) = 36

⇒ 8x = 36

⇒ x = 36/8 = 4.5

∴ Breadth of rectangle = x = 4.5 cm

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

Let a and b be two numbers.

Given that a:b = 31:23

Sum of a and b = a + b = 216

Let the first number ‘a’ = 31x

Then the second number ‘b’ = 23x

Consider a + b = 216

⇒ 31x + 23x = 216

⇒ 54x = 216

⇒ x = 216/54 = 4

∴ The first number ‘a’ = 31x = 31 × 4 = 124

The second number ‘b’ = 23x = 23 × 4 = 92

Let a and b be two numbers.

Given that a:b = 10:9

Product of a and b = ab = 360

Let the first number ‘a’ = 10x

Then the second number ‘b’ = 9x

Consider ab = 360

⇒ 10x × 9x = 360

⇒ 90x² = 360

⇒ x² = 4

⇒ x = 2

If x = 2, then first number ‘a’ = 10x = 10 × 2 = 20

The second number ‘b’ = 9x = 9 × 2 = 18

If x = -2, then first number ‘a’ = 10x = 10 × (-2) = -20

The second number ‘b’ = 9x = 9 × (-2) = -18

(i) Given that a:b = 3:1

∴ a/b = 3

⇒ a = 3b ………………… (1)

Also, b:c = 5:1

∴ b/c = 5

⇒ c = b/5 ………………… (2)

Consider

=(9)³

=729

(ii)

Given that a:b = 3:1

∴ a/b = 3

⇒ a = 3b ………………… (1)

Also, b:c = 5:1

∴ b/c = 5

⇒ c = b/5 ………………… (2)

Consider

Given:

Squaring both sides:

0.04 × 0.4 × a = 0.16 × 0.0016 × b

⇒ 0.016 × a = 0.000256 × b

⇒ a/b = 0.000256/0.016

⇒ a/b = 0.016 = 16/1000

⇒ a/b = 2/125

∴ a:b = 2:125

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

⇒ (x + 3)/(x + 11) = (x - 2)/(x + 1)

Cross multiply and obtain:

⇒ (x + 3)(x + 1) = (x - 2)(x + 11)

⇒ x² + 4x +3 = x² + 9x - 22

⇒ 4x + 3 = 9x - 22

⇒ 9x – 4x = 3 + 22

⇒ 5x = 25

⇒ x = 5

∴ x = 5

Given: a/b = 7/3

∴ a = (7/3)b

(i)

(ii)

(iii)

(iv)

(i)

Given:

Apply componendo and dividendo, i.e., :

Take square root on both sides:

(ii)

Given:

Apply componendo and dividendo, i.e., :

Again apply componendo and dividendo, i.e., :

(iii) Given:

Apply componendo and dividendo, i.e., :

Again apply componendo and dividendo, i.e., :

(iv) Given:

Apply componendo and dividendo, i.e., :

Again apply componendo and dividendo, i.e., :

Given:

Apply componendo and dividendo, i.e., :

∴ a/b = 49/3

⇒ a²/b² =2401/9

⇒ 2a²/7b² =686/9

Apply componendo and dividendo, i.e., :

Given:

We first put x = 0 in the expression and obtain:

⇒ -20/-5 = 12/3

⇒ 4=4 which holds true.

∴ x=0 is a solution.

Now, Consider ……………… (1)

Multiply both sides by �:

……………….(2)

Apply dividendo:

Cross multiply and get:

12x – 20 = 8x + 12

⇒ 12x – 8x = 12+ 20

⇒ 4x = 32

⇒ x = 8

∴ x=0, 8 are the solutions.

Given:

We first put x = 0 in the expression and obtain:

⇒ 63/12 = 3/(-5) which does not hold true.

∴ x=0 is not a solution.

Now, let ……………… (1)

Multiply numerator and denominator of second expression by 5x:

……………….(2)

∴ From equation (1) and (2), we get:

Cross multiply and obtain:

8x + 12 = 21x – 105

⇒ 21x – 8x = 12 + 105

⇒13x = 107

⇒ x = 9

∴ x=2 is the solution.

Given:

We first put x = 0 in the expression and obtain:

⇒ 2/0 = 17/8 which does not hold true.

∴ x=0 is not a solution.

Now, Consider

Apply componendo and dividendo:

Take square roots on both sides:

Case 1:

Cross multiply and obtain:

6x + 3 = 10x – 5

⇒ 4x = 8

⇒ x = 2

Case 2:

Cross multiply and obtain:

6x + 3 = -10x + 5

⇒ 16x = 2

⇒ x = 1/8

∴ x = 2, 1/8 are the solutions.

Given:

We first put x = 0 in the expression and obtain:

⇒ (1+√3)/( 1-√3)=4/1 which does not hold true.

∴ x=0 is not a solution.

Now, Consider

Apply componendo and dividendo:

Squaring both sides:

Cross multiply and get:

36x + 9 = 25x + 75

⇒ 36x -25x = 75 - 9

⇒ 11x = 66

⇒ x = 6

∴ x = 6 is the solution.

Given:

We first put x = 0 in the expression and obtain:

⇒ 10/9 = 61/36 which does not hold true.

∴ x=0 is not a solution.

Consider the given equation:

Apply dividendo:

Take square root on both sides:

Case 1:

Cross multiply and obtain:

24x + 6 = 10x + 15

⇒ 14x = 9

⇒ x = 9/14

Case 2:

Cross multiply and obtain:

24x + 6 = -10x - 15

⇒ 34x = -21

⇒ x = -21/34

∴ x = 9/14 and x = -21/34 are the solutions.

Given:

Apply componendo and dividendo:

Take cube roots on both sides:

Cross multiply and get:

12x - 16 = 5x + 5

⇒ 12x – 5x = 5 + 16

⇒ 7x = 21

⇒ x = 3

∴ x = 3 is the solution.

(i)

∴

∴

Thus,

(ii)

∴

∴

Thus,

(i) Given: 5m – n = 3m + 4m

⇒ 5m – 3m = 4n + n

⇒ 2m = 5n

⇒ m/n = 5/2

⇒ m²/n² = 25/4

Apply componendo and dividendo:

= 29:21

(ii)

Given: 5m – n = 3m + 4m

⇒ 5m – 3m = 4n + n

⇒ 2m = 5n

⇒ m/n = 5/2

⇒ 3m/4n = 15/8

Apply componendo and dividendo:

= 23: 7

Given: a(y+z) = b(z+x) = c(x+y)

Divide all be ‘abc’:

Cancel out the common factor:

Rearrange the terms:

Now, subtract third term from first term, subtract first term from second term and subtract second term from third term and obtain the equivalent:

Solve and cancel the opposite terms:

Hence, proved.

Given:

Add all the terms and obtain the equivalent:

∴ Each ratio is equal to 1.

Given:

Add all the terms and obtain the equivalent:

∴ Each ratio is equal to (a+b)/2.

Given: ……………… (1)

Each ratio = ………………… (2)

Each ratio = ………………… (3)

Each ratio = ………………… (4)

∴ from equation (2), (3) and (4), we get:

Hence, showed.

Given:

(Multipliy and divide the last term by 5)

Each ratio =k = (Adding all the terms)

∴ Each ratio equals x/y.

(i)

Given:

We first put x = 0 in the expression and obtain:

⇒ 9/21 = -5/3 which does not hold true.

∴ x=0 is not a solution.

Now, let ……………… (1)

Multiply numerator and denominator of second expression by 4x:

……………….(2)

∴ From equation (1) and (2), we get:

Cross multiply and obtain:

⇒ 28x – 35 = 6x + 9

⇒ 28x – 6x = 9 + 35

⇒ 22x = 44

⇒ x = 44/22

⇒ x = 2

∴ x=2 is the solution.

(ii)

Given:

We first put y = 0 in the expression and obtain:

⇒ (-12)/(-4) = 8/1 which does not hold true.

∴ y=0 is not a solution.

Now, let ……………… (1)

Multiply numerator and denominator of second expression by 5y:

……………….(2)

∴ From equation (1) and (2), we get:

Cross multiply and obtain:

⇒ y + 8 = 3 + 6y

⇒ 6y – y = 8 - 3

⇒ 5y = 5

⇒ y = 1

∴ y=1 is the solution.

Let x be the number that should be subtracted from 12, 16, 21 so that the numbers remain in continued proportion.

Numbers a, b, c are said to be continued proportion if b² = ac.

∴ From the definition of continued proportion, we get:

⇒ (16 - x)² = (12 - x)(21 - x)

⇒ 256 + x² – 32x = 252 - 33x + x²

⇒-32x + 33x = 252 – 256

⇒ x= -4

∴ -4 should be subtracted from 12, 16, 21 so that the numbers remain in continued proportion.

A number b is said to be mean proportional of two numbers a and c if

b² = ac.

∴ From the definition of mean proportion, we get:

(28 - x)² = (23 - x)(19 - x)

⇒ 784 + x² – 56x = 437 - 42x + x²

⇒-56x + 42x = 437 – 784

⇒ -14x= -347

⇒ x = 347/14

Let the numbers be x, y, z.

As the numbers are in continued proportion, therefore

y² = xz ……………… (1)

Also, the mean proportion = 12

∴ y = √xz = 12

⇒ xz = 144 …………… (2)

It is given that the sum of remaining two numbers = 26

∴ x + z = 26

⇒ x = 26 – z

Put the value of x in equation (2):

(26 – z)z = 144

⇒ 26z – z² = 144

⇒ z² – 26z + 144 = 0

⇒ z² – 8z – 18z + 144 = 0

⇒ z(z - 8) – 18(z – 8) = 0

⇒ (z - 8)(z - 18) = 0

⇒ z = 8 or z = 18

∴ x = 26 – 8 or x = 26 – 18

⇒ x = 18 or x = 8

y = 12

∴ The numbers in proportion be 8, 12, 18 or 18, 12, 8.

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

⇒ a² –ab + ac + ab - b² + bc + ca – bc + c² = a² + b² + c²

⇒ a² –ab + ac + ab - b² + bc + ca – bc + c² - a² - c²= b² + b²

⇒ 2ac = 2b²

⇒ b² = ac

∴ a, b, c are in continued proportion.

(i)

Given: a/b = b/c

⇒ b² = ac

Consider (a + b+ c)(b – c) = ab – ac + b² – bc + cb – c²

= ab – ac + ac – c² (∵ b² = ac)

= ab – c²

(ii)

Given:

a/b = b/c

⇒ b² = ac

Consider (a² + b²)(b² + c²) = a²b² + a² c² + b² b² + b²c²

= a²b² + ac(ac)+ b²(ac)+ b²c² (∵ b² = ac)

= a²b² + b²(ac)+ b²(ac)+ b²c² (∵ b² = ac)

= a²b² + 2b²(ac)+ b²c²

= a²b² + 2ab²c+ b²c²

= (ab + bc)²

(iii)

Given: a/b = b/c

⇒ b² = ac

Consider (a² + b²)/ab = (a² + ac)/ab (∵ b² = ac)

= (a + c)/b

Mean proportion of two numbers is the square root of their product.

∴ Mean proportion of is:

Given: 6:5 = y:20

⇒ 6/5 = y/20

Cross multiply and get:

5y = 6 × 20

⇒ y = 6 × 4 = 24

∴ Option B is correct.

1 cm = 100 mm

∴ 1mm : 1cm

⇒ 1mm : 100mm

= 1 : 100

∴ Option A is correct.

Given: Nitin’s age = 24 years

Mohasin’s age = 36 years

∴ Ration of Nitin’s age to Mohasin’s age = 24:36

= 24/36

= 2/3

= 2:3

∴ Option B is correct.

Total bananas = 24

Ratio in which the bananas are divided = 3:5

Let number of bananas Shubham got = 3x

∴ Number of bananas Anil got = 5x

∴ 3x+5x = 24

⇒ 8x = 24

⇒ x = 3

∴ Shubham got (3 × 3) = 9 bananas.

Thus, option D is correct.

Mean proportional of two numbers a and b = √(ab)

∴ Mean proportional of 4 and 25 = √(4 × 25)

= √100 = 10

(i) Ratio of 21 and 48 in the reduced form:

(To simplify, break the numbers in simpler form)

∴ Ratio of 21 and 48 in reduced form is 1:4.

(ii) Ratio of 36 and 90 in the reduced form:

(To simplify, break the numbers in simpler form)

∴ Ratio of 36 and 90 in reduced form is 2:5.

(iii) Ratio of 65 and 117 in the reduced form:

(To simplify, break the numbers in simpler form)

∴ Ratio of 65 and 117 in reduced form is 5:9.

(iv) Ratio of 138 and 161 in the reduced form:

(To simplify, break the numbers in simpler form)

∴ Ratio of 138 and 161 in reduced form is 6:7.

(v) Ratio of 114 and 133 in the reduced form:

(To simplify, break the numbers in simpler form)

∴ Ratio of 114 and 133 in reduced form is 6:7.

(i) Let r be the radius of the circle.

Let d be the diameter of the circle.

Diameter = 2 × Radius

∴ Ratio of radius to diameter in the reduced form = Radius:Diameter

∴ Ratio of radius to diameter in the reduced form = 1:2

(ii) Given: Length of rectangle = l = 4 cm

Breadth of rectangle = b = 3 cm

Diagonal of rectangle = √(l² + b²)

= √(16 + 9)

= √25 = 5

∴ Diagonal of rectangle = 5 cm

Ratio of diagonal to the length of a rectangle = 4:5

(iii) Given: Side of square = 4 cm

Perimeter of square = 4 × Side = 4 × 4 = 16 cm²

Area of the square = (Side)² = (4)² = 14 cm²

The ratio of perimeter to area of a square = 16:14 = 8:7

(i) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion,

i.e. a:b∷b:c

or b² = ac

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

∴ (4)² = 2 × 8

⇒ 16 = 16, which holds true.

∴ 2, 4, 8 are in continued proportion.

(ii) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion,

i.e. a:b∷b:c

or b² = ac

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

∴ (2)² = 1 × 3

⇒ 4 = 3, which does not hold true.

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

(iii) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion,

i.e. a:b∷b:c

or b² = ac

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

∴ (12)² = 9 × 16

⇒ 144 = 144, which holds true.

∴ 9, 12, 16 are in continued proportion.

(iv) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion,

i.e. a:b∷b:c

or b² = ac

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

∴ (5)² = 3 × 8

⇒ 25 = 24, which does not hold true.

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

Given: a, b, c are in continued proportion.

Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion,

i.e. a:b∷b:c

or b² = ac

Here, a = 3, c = 27

∴ (b)² = 3 × 27

⇒ b² = 81

⇒ b = √81 = 9

∴ b = -9 or 9

(i) 37: 500 = 37/500

= ((37/500) × 100)%

= (37/5) %

= 7.4 %

(ii) 5/8 = ((5/8) × 100)%

= (5 × 12.5) %

= 62.5 %

(iii) 22/30 = ((22/30) × 100)%

= (220/3) %

= 73.33 %

(iv) 144/1200 = ((144/1200) × 100)%

= (144/12) %

= 12 %

(i) 1024MB = 1 GB

Reduced form of the ratio of 1 GB and 1.2 GB is:

∴ The ratio in reduced form is 5:6.

(ii) 60 paise = 0.60 Rupees

∴ 25 Rupees and 60 paise = 25.60 Rupees

Reduced form of the ratio of 17 Rupees and 25.60 Rupees is:

(Break the numbers in simpler form)

∴ The ratio in reduced form is 85:128.

(iii) 1 dozen = 12 units

∴ 5 dozens = 5 × 12 = 60 units

Reduced form of the ratio of 5 dozens (= 60 units) and 120 units is:

∴ The ratio in reduced form is 1:2.

(iv) 4 sq m = 4 m²=4(100cm)²

∴ 4 sq m = 40000 sq cm

Reduced form of the ratio of 4 sq m(=40000 sq cm) and 800 sq cm is:

(Break the numbers in simpler form)

∴ The ratio in reduced form is 50:1.

(v) 1 kg = 1000 gm

∴ 1.5 kg = 1500 gm

Reduced form of the ratio of 1500 gm and 2500 gm is:

∴ The ratio in reduced form is 3:5.

Given: a/b = 2/3

∴ a = (2/3)b

(i)

(ii)

(iii)

(iv)

(i)

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

a, b, c, d are in proportion a : b : : c : d ad = bc

i.e. Product of extremes = product of means.

∴ ad = bc

If then:

(11a² + 9ac)(b² + 3bd) = (a² + 3ac)(11b² + 9bd)

⇒ 11a²b² + 33a²bd + 9ab²c + 27abcd = 11a²b² +9a²bd + 33ab²c + 27abcd

⇒ 33a²bd + 9ab²c = 9a²bd + 33ab²c

⇒ 24a²bd = 24ab²c

⇒ a²bd = ab²c

⇒ ad = bc, which holds true as the numbers are in continued proportion.

∴

(ii)

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

a, b, c, d are in proportion a : b : : c : d ad = bc

i.e. Product of extremes = product of means.

∴ ad = bc

If

then:

(a² + 5c²)(b²) = (b² + 5d²)(a²)

⇒ a²b² + 5c²b² = b²a² + 5a²d²

⇒ 5b²c² = 5a²d²

⇒ b²c² = a²d²

⇒ bc = ad

⇒ ad = bc, which holds true as the numbers are in continued proportion.

∴

(iii)

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

a, b, c, d are in proportion a : b : : c : d ad = bc

i.e. Product of extremes = product of means.

∴ ad = bc

If then:

(a² + ab + b²)(c² – cd + d²) = (a² - ab + b²)(c² + cd + d²)

⇒ a²c² - a²cd + a²d² + abc² – abcd + abd² + b²c² – b²cd + b²d² = a²c² + a²cd + a²d² - abc² – abcd - abd² + b²c² + b²cd + b²d²

⇒ -2a²cd + 2abc² + 2abd² -2b²cd = 0

⇒ 2abc² - 2b²cd = 2a²cd – 2abd²

⇒ 2bc[ac – bd] = 2ad[ac - bd]

⇒ 2bc = 2ad

⇒ ad = bc, which holds true as the numbers are in continued proportion.

∴

(i) Given: a, b, c are in continued proportion.

∴ b² = ac

If then:

a(a - 4c) = (a – 2b)(a + 2b)

⇒ a² – 4ac = a² – 4b²

⇒ -4ac = -4b²

⇒ b² = ac, which holds true as the numbers are in continued proportion.

∴

(ii) Given: a, b, c are in continued proportion.

∴ b² = ac

If then:

b(a - c) = (a – b)(b + c)

⇒ ab – bc = ab + ac – b² - bc

⇒ ab – bc - ab - ac + bc = -b²

⇒ -ac = -b²

⇒ b² = ac, which holds true as the numbers are in continued proportion.

∴

Given:

We first put x = 0 in the expression and obtain:

⇒ 42/58 = 3/2 which does not hold true.

∴ x=0 is not a solution.

Now, let ……………… (1)

Multiply numerator and denominator of second expression by 6x:

……………….(2)

∴ From equation (1) and (2), we get:

Cross multiply and get:

58x + 87 = 63x + 42

⇒ 63x - 58x = 87 – 42

⇒ 5x = 45

⇒ x = 9

∴ x=9 is the solution.

Given:

(Multiply and divide the middle term by -3)

Each ratio =k = (Adding all the terms)

∴ Each ratio equals x/y.

Take one term and subtract other two other from it:

Hence, proved.