WBBSE Solutions For Class 10 Maths Algebra Chapter 2 Ratio And Proportion

Algebra Chapter 2 Ratio And Proportion Different Types Of Ratio

Ratio of equality 

If the terms of a ratio are equal, then it is called a ratio of equality.

For example, 2: 2; a: a; b: b are all ratios of equality.

Ratio of inequality 

If the terms of a ratio are unequal, then it is called a ratio of inequality.

For example, 3:5; 6:7; a: b; c : d are all ratios of inequality.

Ratio of greater inequality 

If the antecedent of a ratio is greater than its consequent, then the ratio is called a ratio of greater inequality.

For example, 6: 5 (6 >5), 11: 10 (11 > 10), etc.

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WBBSE Solutions for Class 10 Maths

Ratio of less inequality 

If the antecedent of a ratio is less than its consequent, then the ratio is called a ratio of less inequality.

For example, 3: 4 (3 < 4), 6: 7 (6 < 7), etc.

Inverse ratio 

If in two ratios, the antecedent of one is consequent of the other and the consequent of one is the antecedent of the other, then the ratios are called inverse ratios of each other.

For example, 4 : 3 is the inverse ratio of 3: 4. b: a is the inverse ratio of a: b and vice versa. Any one of two inverse ratios is called the reciprocal of the other ratio.

[The product of a ratio with its inverse ratio is always 1.]

Simple and Compound or mixed ratios 

Simple ratios 

The ratios of the type ₹ 4 :  ₹5; c : d; x: y are called simple ratios.

Mixed or Compound ratios 

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The ratio that is generated by taking the products of the antecedents of all the ratios as the antecedent and the product of all the consequents as the consequent, is called the mixed or compound ratio of that ratio.

For example, the mixed or compound ratio of the ratios a: b and c : d is ac: bd.

In general, the mixed or compound ratio of the ratios a1:b1;a2:b2;a3:b3;…..;an: bn is (a1a2a3…….an)(b1b2b3……bn) is

Duplicate ratio 

The ratio which is made by taking the square of the antecedent of a given ratio as the antecedent and the square of the consequent of the given ratio as the consequent is called the duplicate ratio of the given ratio.

For example, the duplicate ratio of a given ratio a: b is a2:b2

In other words, if two ratios be the same, then the mixed or compound ratio of them is called the duplicate ratio of that ratios.

Sub-duplicate ratio 

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The ratio which is made by taking the square root of the antecedent of a given ratio as the antecedent and the square root of the consequent of the given ratio as the consequent is called the subduplicate ratio of the given ratio.

For example, the sub-duplicate ratio of the ratio (a: b) is the ratio (√a: √b)

Triplicate ratio 

The ratio which is made by taking the cube of the antecedent of a given ratio as the antecedent and the cube of the consequent of the given ratio as the consequent is called the triplicate ratio of the given ratio.

For example, the triplicate ratio of the given ratio a: b is a: b

In other words, the mixed or compound ratio of three same given ratios is called the triplicate ratio of the ratios.

For example, a: b; a: b; a: b are three same ratios. The mixed or compound ratio is

a x a x a :b x b x b = a3: b3

∴ triplicate ratio of a: b is a3: b3

Sub-triplicate ratio 

The ratio which is made by taking the cube-root of the antecedent of a given .ratio as the antecedent and the cube-root of the consequent of the given ratio as the consequent is called the sub-triplicate ratio of the ratios.

For example, the sub-triplicate ratio of 1: 8 is \(\sqrt[3]{1}: \sqrt[3]{8}\) = 1:2

Similarly, the sub-triplicate ratio of a: b is \(\sqrt[3]{a}: \sqrt[3]{b}\)

Class 10 Maths Solutions Wbbse

Algebra Chapter 2 Ratio And Proportion Examples

Example 1. Calculate what ratio and x2: yz will form the mixed ratio xy: z2.

Solution: Let a : b and x2 : yz will form the mixed ratio xy : z2.

∴ a x x2 : b x yz = xy : z2

or, \(\frac{a x^2}{b y z}=\frac{x y}{z^2}\)

or, \(\frac{a}{b}=\frac{x y \times y z}{x^2 \times z^2}=\frac{y^2}{x z}\)

∴ a:b =y2 :xz

Hence the required ratio is y2 :xz

Example 2. Find the compound ratio of the inverse ratios of the ratios \(x^2: \frac{y z}{x}, y^2: \frac{z x}{y} \text { and } z^2: \frac{x y}{z}\)

Solution: The inverse ratios of the ratios \(x^2: \frac{y z}{x}, y^2: \frac{z x}{y} \text { and } z^2: \frac{x y}{z}\) respectively.

∴ their compound ration

= \(\left(\frac{y z}{x} \times \frac{z x}{y} \times \frac{x y}{z}\right):\left(x^2 \times y^2 \times z^2\right)\)

=x y z:\(x^2 y^2 z^2\)

= \(\frac{x y z}{x^2 y^2 z^2}=1: x y z\)

Hence the required compound ration = 1:xyz

Example 3. Find the compound ratio of the ratios (x + y) : (x – y), (x2 + y2) : (x + y)2 and (x2 – y2)2 : (x4 – y4).

Solution: The compound ratio of the ratios (x + y):(x-y),(x2+y2):(x + y)2 and (x2-y2) :(x4-y4).

= \((x+y)\left(x^2+y^2\right)\left(x^2-y^2\right)^2:(x-y)(x+y)^2\left(x^4-y^4\right)\)

= \(\frac{(x+y)\left(x^2+y^2\right)\left(x^2-y^2\right)^2}{(x-y)(x+y)^2\left(x^4-y^4\right)}\)

= \(\frac{(x+y)\left(x^2+y^2\right)\{(x+y)(x-y)\}^2}{(x-y)(x+y)^2\left(x^2+y^2\right)\left(x^2-y^2\right)}\)

= \(\frac{(x+y)^2(x-y)^2}{(x-y)(x+y)\left(x^2-y^2\right)}=\frac{(x+y)(x-y)}{x^2-y^2}\)

= \(\frac{x^2-y^2}{x^2-y^2}=\frac{1}{1}=1: 1\)

Hence the required compound ratio = 1:1

Example 4. If A: B = 2 : 3, B: C = 4: 5, and C : D = 6: 7, then find A : D.

Solution: We have, A:B = 2:3

⇒ \(\frac{A}{B}\) = \(\frac{2}{3}\) ….(1)

B: C = 4: 5

⇒ \(\frac{B}{C}\) = \(\frac{4}{5}\) ….(2)

C: D = 6: 7

⇒ \(\frac{C}{D}\) = \(\frac{6}{7}\) ….(3)

Now applying (1) x (2) x (3) we get

\(\frac{A}{B} \times \frac{B}{C} \times \frac{C}{D}\)

 

= \(\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7}\)

⇒ \(\frac{A}{D}=\frac{16}{35}\)

⇒ A: D = 16: 35

∴ A: D = 16: 35

Example 5. If a: b = 2 : 3, b: c = 4:7, then find a: b: c.

Solution: We have, a : b =(2 : 3) x 4 = 8 : 12 ….(1)

b : c = (4 : 7) x 3 = 12 : 21 …(2)

[Here b has been made equal.]

∴ a:b:c = 8 : 12: 21.

Example 6. If p: q = 5: 7 and p – q = – 4, then find the value of (3p+ 4q).

Solution: We have, p: q = 5: 7

or, \(\frac{p}{q}\) = \(\frac{5}{7}\) = \(\frac{5k}{7k}\)[k≠0]

∴ Let p = 5k and q = 7k.

Also, p- q =-4 or, 5k – 7k = – 4

or, -2k =-4 or, k = \(\frac{-4}{-2}\)= 2.

∴ p = 5k = 5 x 2 = 10 and q = 7k = 7 x 2 = 14.

∴ 3p+ 4q= 3 x 10 + 4 x 14 = 30 + 56 = 86.

Hence the required value = 86.

Example 7.  If x: y = 5: 7, then find (2y – x) : (3x + y).

Solution: We have, x : y = 5 : 7

or, \(\frac{x}{y}\) = \(\frac{5}{7}\)

⇒ \(\frac{x}{y}\) = \(\frac{5k}{7k}\)

∴ Let x = 5k and y =7k

Now, (2y-x):(3x+ y) =

= \(\frac{2 y-x}{3 x+y}=\frac{2 \times 7 k-5 k}{3 \times 5 k+7 k}\)

= \(\frac{14 k-5 k}{15 k+7 k}=\frac{9 k}{22 k}=\frac{9}{22}=9: 22\)

Hence (2y-x):(3x+y) = 9:22

Maths Solutions Class 10 Wbbse

Algebra Chapter 2 Ratio And Proportion Short Answer Type Questions

Example 1. If (5x- 3y) : (2x+ 4y) = 11 : 12, then find x: y

Solution: Given that, (5x – 3y) : (2x+ 4y)= 11 : 12

or, \(\frac{5 x-3 y}{2 x+4 y}=\frac{11}{12}\)

or, 60x – 36y = 22x + 44y

or, 60x – 22x = 44y + 36y

or, 38x = 80y

or, \(\frac{x}{y}=\frac{80}{38} \quad \text { or, } \quad \frac{x}{y}=\frac{40}{19}\)

∴x: y = 40: 19

Example 2. What term should be subtracted from each term of the ratio a : b to make the ratio mini

Solution: Let x should be subtracted from each term.

As per question, (a – x) : (b – x) = m : n

or, \(\frac{a-x}{b-x}=\frac{m}{n}\)

or, mb – mx = na – nx .

or, nx – mx = na – mb

or, x (n – m) = na – mb

or, \(x=\frac{n a-m b}{n-m}=\frac{b m-a n}{m-n}\)

Here \(\frac{b m-a n}{m-n}\) should be subtracted

Example 3. What term should be added to the antecedent and subtracted from the consequent of ratio 4:7. to make a compound ratio of 2 : 3 and 5:4?

Solution: Let the number be x

As per question, (4 + x) : (7 – x) = 2 x5 : 3 x 4.

[the compound ratio of the ratios 2 : 3 and 5:4 = 2×5:3×4]

or, \(\frac{4+x}{7-x}=\frac{2 \times 2}{3 \times 4}\)

or, \(\frac{4+x}{7-x}=\frac{5}{6}\)

or, 24+6x = 35-5x

or, 6x+5x = 35-24 or, 11x = 11

or, \(\frac{11}{11}\) = 1

Hence the required number = 1

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Example 4. If (10x + 3y) : (5x + 2y)= 9:5, then show that (2x+y) : (x + 2y) = 11 : 13.

Solution: (10x + 3y6) : (5x + 2y)= 9:5

or, \(\frac{10 x+3 y}{5 x+2 y}=\frac{9}{5}\)

or, 50x+ 15y = 45x + 18y

or, 50x-45x = 18y – 15y

or, 5x = 3y or, \(\frac{x}{y}\)= \(\frac{3}{5}\)

or, \(\frac{x}{y}\)= \(\frac{3k}{5k}\) [k≠0] .

∴ let x = 3k and y = 5k.

Now, (2x+y):(x+2y) = \(\frac{2 x+y}{x+2 y}\)

= \(\frac{2 \times 3 k+5 k}{3 k+2 \times 5 k}\)

= \(\frac{6 k+5 k}{3 k+10 k}=\frac{11 k}{13 k}=\frac{11}{13}\)

Hence (2x+y):(x+2y)  = 11: 13

 

Algebra Chapter 2 Ratio And Proportion Long Answer Type Questions

Example 1. If a: b = 3: 4 and x: y – 4: 5, then find the value of \(\frac{3 a x-b y}{5 b y-7 a x}\)

Solution: Given that, a: b = 3: 4,

∴ \(\frac{a}{b}\) = \(\frac{3}{4}\)

Again, x: y = 4: 5 or, \(\frac{x}{y}\) = \(\frac{4}{5}\)

Now, Given quantity

= \(\frac{3 a x-b y}{5 b y-7 a x}=\frac{\frac{3 a x}{b y}-1}{5-\frac{7 a x}{b y}}\) [Dividing by]

= \(\frac{3 \cdot \frac{a}{\dot{b}} \cdot \frac{x}{y}-1}{5-7 \cdot \frac{a}{b} \cdot \frac{x}{y}}\)

= \(\frac{3 \times \frac{3}{4} \times \frac{4}{5}-1}{5-7 \times \frac{3}{4} \times \frac{4}{5}}\)

= \(\frac{\frac{9}{5}-1}{5-\frac{21}{5}}=\frac{\frac{9-5}{5}}{\frac{25-21}{5}}=\frac{\frac{4}{5}}{\frac{4}{5}}=1\)

Hence the requires value = 1

Example 2. What number should be added to each term of the ratio 2a: 3b so that the ratio of the additions will be c : d?

Solution: Let the number x should be added to each term.

As per question, (2a + x) : (3b + x) = c : d

or, \(\frac{2 a+x}{3 b+x}=\frac{c}{d}\)

or, 2ad + dx = 3 bc + cx

or, dx – cx = 3 bc – 2ad

or, x (d – c) = 3 bc – 2ad

or, \(x=\frac{-(2 a d-3 b c)}{d-c}\)

or, x= \(\frac{-(2 a d-3 b c)}{-(c-d)}\)

or, \(x=\frac{2 a d-3 b c}{c-d}\)

Hence the required number = \(x=\frac{2 a d-3 b c}{c-d}\)

Example 3. For what value of x, the duplicate ratio of \(\frac{x+a}{x+b} \text { is } \frac{a}{b}\)

Solution: Duplicate ratio of

\(\frac{x+a}{x+b} \text { is } \frac{(x+a)^2}{(x+b)^2}=\frac{x^2+2 x a+a^2}{x^2+2 x b+b^2}\)

Maths Solutions Class 10 Wbbse

As per question, \(\frac{x^2+2 x a+a^2}{x^2+2 x b+b^2}\) = \(\frac{a}{b}\)

or, ax2 + 2 abx + ab2 = bx2 + 2abx +a2b

or, ax2 – bx2 = a2b – ab2 or, x2 (a – b) = ab (a – b)

or, \(x^2=\frac{a b(a-b)}{(a-b)}=a b\)

∴ x2= ab

∴ x = ±√ab

Example 4. If the triplicate of the ratio (x + a): (x + b) is a : 6, then prove that x3– 3abx – ab (a + b) = 0.

Solution: The triplicate of (x + a): (x+b) is \(\frac{(x+a)^3}{(x+b)^3}\)

As per question, \(\frac{(x+a)^3}{(x+b)^3}\) = \(\frac{a}{b}\)

or, a(x + b)3 =b(x + a)3

or, a(x3 +3x2b + 3xb2 + b3)= b(x3 +3x2a + 3xa2 +a3)

or, ax3 + 3 abx2 + 3 ab2x + ab3 = bx3 + 3 abx2 + 3 a2bx + a3b

or, ax3 -bx3 + 3ab2x – 3a2bx + ab3 -a3b = 0.

or, x3(a-b)-3abx(a-b)-ab(a2-b2) = 0

or, x3 (a – b) – 3abx (a-b)- ab (a +b)(a -b) = 0

or, (a-b){x3 – 3abx – ab (a + b)} = 0

or, x3 – 3abx-ab(a + b) = [a≠b, a-b≠0]

∴ x3 – 3abx – ab(a + b) = 0. (Proved)

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Algebra Chapter 2 Ratio And Proportion Proportion

If four quantities be such that the ratio between first and second is equal to the ratio between the third and fourth, then the four quantities form a proportion and the quantities are called proportionate,

i.e., if four quantities a, b, c, and d be such that \(\frac{a}{b}\) = \(\frac{c}{d}\) or, a: b = c : d, then a, b, c, d are called proportional.

The first and fourth terms of this proportion are called extreme terms and the second and the third terms are called middle terms or means. The fourth term is also known as fourth proportional.

∴ \(\frac{a}{b}\)= \(\frac{c}{d}\) or, ad = bc, the product of the extreme terms of four proportional quantities is equal to the product of its middle terms.

This mathematical operation is known as cross-multiplication.

It is easy to calculate the rest term among the four quantities if any three terms are given.

Since any ratio when multiplied or divided by any non-zero real number does not change its value, so the value of any proportion does not change if it is multiplied or divided by any non-zero real number.

Algebra Chapter 2 Ratio And Proportion Continued Proportion

If amongst the three quantities, the ratio between the first and the second quantity is equal to the ratio between the second and the third quantity, then the quantities are called continued proportional.

Mathematically, three quantities a, b, and c will be continued proportional if a: b = b: c or, \(\frac{a}{b}\)= \(\frac{c}{d}\) be

or, b2 = ac.

Here the middle term b is said to be the mean proportional of the other two terms a and c.

Also, the third term c is said to be third proportional of the others two.

Algebra Chapter 2 Ratio And Proportion Formulae Related To Proportion

Alternendo 

If a, b, c and d are in continued proportional,

i.e., if  \(\frac{a}{b}\)= \(\frac{c}{d}\) =  then \(\frac{a}{c}\)= \(\frac{b}{d}\)

This process is known as alternendo.

Invertendo 

If a, b, c, and d are in continued proportional,

i.e., if \(\frac{a}{b}\)= \(\frac{c}{d}\) then \(\frac{b}{a}\)= \(\frac{d}{c}\)

This process is known as invertendo.

Proof: a, b, c, d are in continued proportion,

∴ \(\frac{a}{b}\)= \(\frac{c}{d}\)

or, \(1 \div \frac{a}{b_i}=1 \div \frac{c}{d} \text { or, } 1 \times \frac{b}{a}=1 \times \frac{d}{c} \text { or, } \frac{b}{a}=\frac{d}{c}\)

Componendo 

If a, b, c, d are in continued proportion or if \(\frac{a}{b}\) = \(\frac{c}{d}\)

then \(\frac{a+b}{b}=\frac{c+d}{d}\)

This process is known as componendo.

Proof:

\(\frac{a}{b}=\frac{c}{d} \text { or, } \frac{a}{b}+1=\frac{c}{d}+1 \text { or, } \frac{a+b}{b}=\frac{c+d}{d}\)(Proved)

 

Dividendo 

If a, b, c, d are in continued proportion or if \(\frac{a}{b}\) = \(\frac{c}{d}\)

then \(\frac{a-b}{b}=\frac{c-d}{d}\)

This process is known as dividendo.

Proof:

\(\frac{a}{b}=\frac{c}{d} \text { or, } \cdot \frac{a}{b}-1=\frac{c}{d}-1 \text { or, } \frac{a-b}{b}=\frac{c-d}{d}\)(Proved)

 

Componendo and Dividendo 

If a, b, c, d are in continued proportion or if \(\frac{a}{b}\) = \(\frac{c}{d}\),

then \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)

This process is known as componendo and dividendo.

Proof: \(\frac{a}{b}=\frac{c}{d}\), ∴ \(\frac{a+b}{b}=\frac{c+d}{d}\) (by componendo)

Again, \(\frac{a-b}{b}=\frac{c-d}{d}\)[by dividendo)

Now, \(\frac{a+b}{b} \div \frac{a-b}{b}=\frac{c+d}{d} \div \frac{c-d}{d}\)

or, \(\frac{a+b}{b} \times \frac{b}{a-b}=\frac{c+d}{d} \times \frac{d}{c-d}\)

or, \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)

∴ \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\) (Proved)

Algebra Chapter 2 Ratio And Proportion Formulae Related To Proportion Multiple Choice Questions

Example 1. The fourth proportion of 2, 3, and 4 is

  1. 6
  2. 8
  3. 10
  4. 12

Solution: Let the fourth proportional = x.

∴ \(\frac{2}{3}=\frac{4}{x}\)

or, 2 x=12

or, \(x=\frac{12}{2}\)

∴ 1. 6

Example 2. Third proportional of 4 and 6 is

  1. 6
  2. 9
  3. 12
  4. 18

Solution: Let the third proportional = x.

∴ 4, 6, and x are in continued proportion.

∴ \(\frac{4}{6}\) = \(\frac{6}{x}\)

or, 4x = 36

or, x = \(\frac{36}{4}\)

or, x = 9

∴ 2. 9

Example 3. The mean-proportional of 32 and 50 is

  1. 35
  2. 40
  3. 45
  4. 48

Solution: Let the mean-proportional = x

∴ 32, x, and 50 are in continued proportion.

∴ \(\frac{32}{x}\) = \(\frac{x}{50}\)

or, x2= 1600

or, x = √1600 = 40

∴ 2. 40

Example 4. a is a positive number and if a: \(\frac{27}{64}\) = \(\frac{3}{4}\): a, then the value of a will be

  1. 9
  2. \(\frac{9}{16}\)
  3. \(\frac{16}{9}\)
  4. None of these

Solution: Given that a: \(\frac{27}{64}\) = \(\frac{3}{4}\): a

or, \(\frac{a}{\frac{27}{64}}=\frac{\frac{3}{4}}{a}\)

or, \(a^2=\frac{27}{64} \times \frac{3}{4}\)

or, \(a^2=\frac{81}{256}\)

or, \(a=\sqrt{\frac{81}{256}}=\sqrt{\left(\frac{9}{16}\right)^2}=\frac{9}{16}\)

∴ 2.

  1. \(\frac{9}{16}\)

Wbbse Class 10 Maths Solutions

Example 5. If 2a = 3b = 4c, then a: b: c will be

  1. 3 : 4: 6
  2. 4: 3: 6
  3. 3 : 6: 4
  4. 6: 4: 3

Solution: Given that 2a = 3b = 4c

or, \(\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{3}}=\frac{c}{\frac{1}{4}}\)

∴ a: b: c = \(\frac{1}{2}\): \(\frac{1}{3}\): \(\frac{1}{4}\)

[multiplying by 12]

∴ 4. 6: 4: 3

Example 6. If A: B = 5: 7 and B: C = 8: 9, then a: b: c will be

  1. 40: 56: 63
  2. 56: 40: 63
  3. 63: 56: 40

Solution: Two values of B are 7 and 8, their L.C.M. is 56

A:B = 5:7 = 5×8:7×8 = 40:56

B : C = 8 : 9 = 8 x 7 : 9×7 = 56 : 63

∴ A : B : C = 40 : 56 : 63

 

Algebra Chapter 2 Ratio And Proportion Formulae Related To Proportion State Whether True Of False

Example 1.The compound or mixed ratio of the ratios ab: c2, bc: a2, and ca: b2 = 1: 2.

Solution: The compound ratio of the ratios ab: c2, bc: a2, and ca: b2

= (ab x bc x ca) : (c2 x a2 x b2)

= a2b2c2 : a2b2 c2 = 1:1

Hence the statement is false.

Example 2. x3y, x2y2and xy2 are in continued proportional.

Solution: The given three quantities are x3y, x2y2and xy3.

Now, \(\frac{x^3 y}{x^2 y^2}=\frac{x}{y}\)

Also, \(\frac{x^2 y^2}{x y^3}=\frac{x}{y}\)

i.e., \(\frac{x^3 y}{x^2 y^2}=\frac{x^2 y^2}{x y^3}\)

∴ The given quantities are in continued proportional.

Hence the statement is true

Example 3.  If (2x + 3y) : (6x -3y)= \(\frac{5}{2}\) then the value of x: y = 21: 26

Solution : (2x + 3y) : (6x -3y)= \(\frac{5}{2}\)

or, \(\frac{2 x+3 y}{6 x-3 y}=\frac{5}{2}\)

or, 2 (2x + 3y) = 5 (6x – 3y)

or, 4x+6y=30x-15y

or, 4x – 30x = – 15y – 6y

or, -26x = -21y

or, \(\frac{x}{y}\) = \(\frac{21}{26}\)

∴ x: y = 21: 26

Hence the statement is true

 

Algebra Chapter 2 Ratio And Proportion Formulae Related To Proportion Fill In The Blanks

Example 1. If x,y, and z are in continued proportion, then y is called the ____ proportion of x and z.

Solution: If x, y, and z are in continued proportion,

\(\frac{x}{y}\) = \(\frac{y}{z}\) ∴ y2 = xz

Here y is called the mean proportional of x and z.

Example 2. If x : 2= y : 5 = z : 8, then 50% of x = 20% of y = — % of z.

Solution: Given that x : 2 = y : 5 = z : 8.

or, \(\frac{x}{2}=\frac{y}{5}=\frac{z}{8}\)

or, \(\frac{x}{2} \times 100 \%=\frac{y}{5} \times 100 \%=\frac{z}{8} \times 100 \%\)

or, 50% of x = 20% of y = \(\frac{25}{2}\)% of z.

or, 50% of x = 20% of y = \(12 \frac{1}{2} \%\) of z.

Example 3. If the mean-proportional of (x-4) and (x-5) be then the value of x is _____

Solution: The- mean-proportional of (x – 4) and (x – 5) is x

∴ by definition, \(\frac{x-4}{x}=\frac{x}{x-5}\)

or, x2-4x -5x + 20 = x2

or, – 9x + 20 = 0

or, -9x = -20

or, x = \(\frac{-20}{-9}=2 \frac{2}{9}\)

Hence the value of x is \(2 \frac{2}{9}\)

Example 4. 3rd proportional of 3 and 6 is ____

Solution: The 3rd proportional be x

∴ 3, 6, and x are in continued proportion.

So, \(\frac{3}{6}\) = \(\frac{6}{x}\)

or, 3x = 6 x 6 or, x = \(\frac{6 \times 6}{3}\)

∴ x = 12

Hence the 3rd proportion of 3 and 6 is 12.

Algebra Chapter 2 Ratio And Proportion Formulae Related To Proportion Short Answer Type Questions

Example 1. If \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{2 a-3 b+4 c}{p}\) then find the value of P.

Solution: \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)

= \(\frac{a \times 2-b \times 3+c \times 4}{2 \times 2-3 \times 3+4 \times 4}\)

= \(\frac{2 a-3 b+4 c}{4-9+16}=\frac{2 a-3 b+4 c}{11}\)

∴ \(\frac{a}{2}\) = \(\frac{b}{3}\) = \(\frac{c}{4}\)

= \(\frac{2 a-3 b+4 c}{11}=\frac{2 a-3 b+4 c}{p}\)

∴ p=11 .

Example 2. If \(\frac{3 x-5 y}{3 x+5 y}=\frac{1}{2}\) then find the value of \(\frac{3 x^2-5 y^2}{3 x^2+5 y^2}\)

Solution: Given that \(\frac{3 x-5 y}{3 x+5 y}=\frac{1}{2}\)

or, 6x = 10y = 3x + 5y or, 6x – 3x = 5y + 10y

or, 3x = 15y   or, x = \(\frac{15y}{3}\) or, x = 5y

∴ Given qunatity

= \(\frac{3 x^2-5 y^2}{3 x^2+5 y^2}\)

= \(\frac{3 \times(5 y)^2-5 y^2}{3 \times(5 y)^2+5 y^2}\)

= \(\frac{75 y^2-5 y^2}{75 y^2+5 y^2}=\frac{70 y^2}{80 y^2}=\frac{7}{8}\)

Hence the required value = \(\frac{7}{8}\)

Wbbse Class 10 Maths Solutions

Example 3. If x,12,y,27 are in continued proportion then find the positive value of x and y.

Solution: x,12,y,27 are in continued proportion

∴ \(\frac{x}{12}=\frac{12}{y}\) or, xy = 144 …(1)

Again, 12, y, 27 are in continued proportion,

∴ \(\frac{12}{y}=\frac{y}{27}\)

or, y2 = 12 x 27

or, y2= 324 or, y = √324 = ±18

But y>0, y = 18

∴ from(1) we get, x X 18 = 144

or, \(\frac{144}{18}\) = 8

Hence x = 8 and y = 18

Example 4. If a: b = 3: 2 and b: c=3:2 then find the value of the ration(a+b): (b+c)

Solution: a: b = 3: 2

⇒ \(\frac{a}{b} ={3}{2}\)

∴ Let a = 3 k1 and b= 2k1 [k1≠ 0]

Also, b: c =3: 2

⇒ \(\frac{b}{c} ={3}{2}\)

∴ Let b = 3 k2 and c= 2k2 [k2≠ 0]

Now, (a+b):(b+c)

= \(\frac{a+b}{b+c}=\frac{3 k_1+2 k_1}{3 k_2+2 k_2}\)

= \(\frac{5 k_1}{5 k_2}=\frac{k_1}{k_2}\)….(1)

Again, \(\frac{a}{b}=\frac{3}{2}\)

⇒ \(\frac{3 k_1}{3 k_2}=\frac{3}{2}\)

⇒ \(\frac{k_1}{k_2}=\frac{3}{2}\)

∴ from (1) we get, \(\frac{a+b}{b+c}=\frac{3}{2}\)

Hence (a+b) : (b+c) = 3: 2

Example 5. If a : b = 3 : 4 and x: y = 5 : 7, then find the value of (3ax-by) : (4by -7ax)

Solution: Given that a : b = 3 : 4

Wbbse Class 10 Maths Solutions

or, \(\frac{a}{b}=\frac{3}{4}\) or, \(a=\frac{3b}{4}\) …(1)

Again, x:y = 5: 7

or, \(\frac{x}{y}=\frac{5}{7}\)

or, \(x=\frac{5 y}{7}\)…(2)

Now, (3ax-by):(4 b y-7 a x)

= \(\frac{3 a x-b y}{4 b y-7 a x}\)

= \(\frac{3 \times \frac{3 b}{4} \times \frac{5 y}{7}-b y}{4 b y-7 \times \frac{3 b}{4} \times \frac{5 y}{7}}\)

= \(\frac{\frac{45}{28} b y-b y}{4 b y-\frac{15 b y}{4}}\)

=\(\frac{\frac{45 b y-28 b y}{28}}{\frac{16 b y-15 b y}{4}}\)

= \(\frac{\frac{17 b y}{28}}{\frac{b y}{4}}\)

= \(\frac{17 b y}{28} \times \frac{4}{b y}=\frac{17}{7}\)

Hence the required value = \(\frac{17}{7}\)

Aliter: Given that a : b = 3 : 4

∴ Let a = 3k1, b = 4k1

Again, x : y = 5 : 1

∴ Let x = 5k2, y= 7k2. [k1, k2 ≠0]

∴ Given quantity = (3 ax – by) : (4by – 7ax)

= \(\frac{3 a x-b y}{4 b y-7 a x}\)

= \(\frac{3 \times 3 k_1 \times 5 k_2-4 k_1 \times 7 k_2}{4 \times 4 k_1 \times 7 k_2-7 \times 3 k_1 \times 5 k_2}\)

= \(\frac{45 k_1 k_2-28 k_1 k_2}{112 k_1 k_2-105 k_1 k_2}=\frac{17 k_1 k_2}{7 k_1 k_2}=\frac{17}{7}\)

Hence the required value = \(\frac{17}{7}\)

Example 6. Find the mean-proportional of (a + b)2 and (a – b)2.

Solution: Let the mean-proportional of (a+b)2 and (a -b)2 be x. 

∴ \(\frac{(a+b)^2}{x}=\frac{x}{(a-b)^2}\)

or, x2 = (a + b)2 x (a – b)2

or, x2 = {{a + b)(a – b)}2 or, x2 = (a2 – b2)2

or, x = ± (a2 – b2)

Hence the required mean proportional = ± (a2 – b2)

 

Algebra Chapter 2 Ratio And Proportion Formulae Related To Proportion Long Answer Type Questions

Example 1. If x: a = y: b = z: c, then prove that 

1. \(\frac{x^3}{a^2}+\frac{y^3}{b^2}+\frac{z^3}{c^2}=\frac{(x+y+z)^3}{(a+b+c)^3}\)

Solution: Given that x: a = y: b = z: c

or, \(\frac{x}{a}\)= \(\frac{y}{b}\)= \(\frac{z}{c}\) =k (k≠0) (let)

∴ x = ak, y= bk and z = ck

L.H.S \(=\frac{x^3}{a^2}+\frac{y^3}{b^2}+\frac{z^3}{c^2}\)

= \(\frac{(a k)^3}{a^2}+\frac{(b k)^3}{b^2}+\frac{(c k)^3}{c^2}\)

= \(\frac{a^3 k^3}{a^2}+\frac{b^3 k^3}{b^2}+\frac{c^3 k^3}{c^2}\)

= \(a k^3+b k^3+c k^3\)

= \(k^3(a+b+c)\)

RHS = \(\frac{(x+y+z)^3}{(a+b+c)^2}\)

= \(\frac{(a k+b k+c k)^3}{(a+b+c)^2}\)

= \(\frac{\{k(a+b+c)\}^3}{(a+b+c)^2}\)

= \(\frac{k^3(a+b+c)^3}{(a+b+c)^2}=k^3(a+b+c)\)

∴ L.H.S = R.H.S (proved)

2. (a2 + b2 + c2) (x2 + y2 + z2) = (ax + by + cz)2

Solution: Given that x: a = y: b = z: c

or, \(\frac{x}{a}\) = \(\frac{y}{b}\) = \(\frac{z}{c}\) =k (k≠0) (let)

∴ x = ak, y= bk and z = ck

LHS = (a2 + b2 + c2) (x2 + y2 + z2)

= (a2 +b2 +c2){(ak)2 + (bk)2 + (ck)2}

= (a2 +b2 +c2)(a2k2+b2k2 +c2k2)

= (a2 + b2 +c2).k2(a2 +b2 + c2) = k2(a2 +b2 + c2)

RHS = (ax + by + cz)2 = (a.ak + b.bk + c.ck)2

= (a2k + b2k + c2k)2  ={k (a2 +b2 + c2)}

= k2(a2+b2+c2)

∴ LHS = RHS

Example 2. If a : b = c : d, then show that \(\sqrt{a^2+c^2}: \sqrt{b^2+d^2}\)= (pa+ qc): (pb + qd)

Solution: Given that a: b = c: d

⇒ \(\frac{a}{b}\) = \(\frac{c}{d}\) = k(Let) [k≠0]

∴ a = bk and c=dk

Now, LHS = \(\sqrt{a^2+c^2}: \sqrt{b^2+d^2}\)

= \(\frac{\sqrt{a^2+c^2}}{\sqrt{b^2+d^2}}=\frac{\sqrt{(b k)^2+(d k)^2}}{\sqrt{b^2+d^2}}\)

=\(\frac{\sqrt{b^2 k^2+d^2 k^2}}{\sqrt{b^2+d^2}}\)

= \(\frac{\sqrt{k^2\left(b^2+d^2\right)}}{\sqrt{b^2+d^2}}\)

= \(\frac{k \sqrt{b^2+d^2}}{\sqrt{b^2+d^2}}\)=k

RHS = \((p a+q c):(p b+q d)=\frac{p a+q c}{p b+q d}\)

= \(\frac{p \times b k+q \times d k}{p b+q d}=\frac{k(p b+q d)}{(p b+q d)}\)=k .

∴ LHS = RHS

Example 3. If a: b = c: d = e: f then prove that (a2 + c2 + e2)(b2 + d 2 + f2)= (ab + cd + ef)2

Wbbse Class 10 Maths Solutions

Solution: Given that a: b = c: d = e: f

or, \(\frac{a}{b}\) = \(\frac{c}{d}\) = \(\frac{e}{f}\) = k(let) (k≠0)

∴ a =bk, c= dk and e = fk

Now, LHS = (a2+c2+e2)(b2+d2+f2)

= {(bk)2+(dk)2+(fk)2}(b2+d2+f2)

= (b2k2 +d2k2 + f2k2)(b2 +d2 +f2)

= k2(b2 +d2 + f2)(b2 +d2 +f2)

= k2(b2+d2+f2)2.

RHS= (ab + cd + ef)2 = (bk.b + dk.d+ fk.f)2

= (b2k + d2k + f2k)2 ={k(b2+d2+f2)}2

= k2(b2+d2+f2)2

LHS = RHS (proved)

Example 4. If a: b= b: c, then prove that 

1. \(a^2 b^2 c^2=\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=a^3+b^3+c^3\)

Solution: Given that a: b = b :c

⇒ \(\frac{a}{b}\) = \(\frac{a}{b}\) =k(let), [k≠0]

∴ a=bk = ck.k = ck  and b= ck2 and b = ck

LHS = \(a^2 b^2 c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)

= \(\frac{b^2 c^2}{a}+\frac{c^2 a^2}{b}+\frac{a^2 b^2}{c}\)

= \(\frac{(c k)^2 \cdot c^2}{c k^2}+\frac{c^2\left(c k^2\right)^2}{c k}+\frac{\left(c k^2\right)^2 \cdot(c k)^2}{c}\)

= c3+ c3k3+c3k6 = c3(1-k+k6)

RHS = a3 +b3 + c3 = (ck2)3 + (ck)3 +c3

= c3 +c3k3 +c3k6 =c3(1+ k3+k6) 

∴ LHS = RHS (proved)

2. \(\frac{a b c(a+b+c)^3}{(a b+b c+c a)^3}\) = 1

Solution: Given that a: b = b :c

⇒ \(\frac{a}{b}\) = \(\frac{a}{b}\) =k(let), [k≠0]

∴ a=bk = ck.k = ck  and b= ck2 and b = ck

LHS = \( \frac{a b c(a+b+c)^3}{(a b+b c+c a)^3}=\frac{c k^2 \cdot c k \cdot c\left(c k^2+c k+c\right)^3}{\left(c k^2 \cdot c k+c k \cdot c+c \cdot c k^2\right)^3}\)

= \(\frac{c^3 k^3\left\{c\left(k^2+k+1\right)\right\}^3}{\left(c^2 k^3+c^2 k+c^2 k^2\right)^3}=\frac{c^6 k^3\left(k^2+k+1\right)^3}{c^6 k^3\left(k^2+k+1\right)^3}\)=1

RHS = 1

∴ LHS = RHS(proved)

Example 5. If a,b,c,d are in continued proportion, then prove that (b – c)2 + (c – a)2 + (b-d)2 = (a-d)2

Solution: Since a,b,c,d are in continued proportion,

\(\frac{a}{b}\) = \(\frac{b}{c}\) = \(\frac{c}{d}\) = k(let) [k≠0]

∴ a = bk,

= ck.k

= ck2

= dk.k2

= dk3

∴ b = ck,

= dk.k

= dk2

∴ c = dk

Now, LHS = (b- c)2+ (c- a)2 + (b – d)2

= (dk2 – dk)2+ (dk – dk3)2 + (dk2 – d)2

= d2k4 – 2d2k3 + d2k2 + d2k2– 2d2k4 + d2k6 + d2k4 – 2d2k2+ d2 = d2k6 – 2d2k3 + d2

RHS =(a-d)2

= (dk3 -d)2

= d2k6 -2d2k3 + d2

∴ LHS = RHS. (Proved)

Example 6. 1. If \(\frac{a}{b}\) = \(\frac{x}{y}\), then show that (a +b)(a2 + b2)x3 = (x + y)(x2 + y2) a3

Solution: \(\frac{a}{b}\) = k(let)[k≠0]

∴ a = bk and x = yk

Now, LHS = (a + b)(a2 + b2)x3

= (bk + b){(bk)2 + b2}.(yk)3 = b(k+ 1). b2 (k2 + 1).y3k3

= b3 (k + 1)(k2 + 1). y3k3

RHS = (x + y)(x2 + y2)a3

= (yk + y){(yk)2+y2}.(bk)3

= y(k + 1).y2(k2 + 1) .b3k3

= y3(k + 1)(k2 + 1) .b3k3

= b3(k + 1)(k2 + 1). y3k3

∴ LHS = RHS. (Proved)

2. If \(\frac{x}{l m-\dot{n}^2}=\frac{y}{m n-l^2}=\frac{z}{n l-m^2}\) then prove that lx + my + nz = 0

Solution: \(\frac{x}{l m-\dot{n}^2}=\frac{y}{m n-l^2}=\frac{z}{n l-m^2}\) = k(let)[k≠0]

∴ x = k (lm – n2), y=k(mn – l2), z = k(nl – m2)

or, lx – k(l2m – n2l), my = k (m2n – l2m), nz = k (n2l – m2n)

or, Ix + my + nz = k(l2m-n2l + m2n – l2m + n2l-m2n)

= k x 0 = 0

∴ lx + my + nz = 0(proved)

3. If \(\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}\) then show that (b-c)x + (c- a) y + (a – b) z = 0.

Solution: \(\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}\) = k(let)[k≠0]

∴ x = k(b + c- a), or, (b -c)x= k(b -c)(b + c-a)

or, (b – c)x= k (b2 – c2) – ak(b – c) ….(1)

y = k(c + a – b), or, (c- a)y = k(c – a)(c + a – b)

or, (c-a)y = k(c2– a2) -bk(c- a) …(2)

z = k(a + b-c), or, (a-b)z= k(a – b)(a + b-c)

or, (a- b)z = k (a2– b2) -ck(a- b) …(3)

∴ adding (1), (2) and (3) we get,

(b – c)x -K (c-a)y + (a – b)z = k(b2 -c2 + c2 -a2 + a2 – b2)- k (ab -ac + bc-ab+ ca- bc)

= k x 0 – k x 0 =0-0=0

∴ (b – c)x + (c- a)y + (a – b)z = 0 (Proved)

Class 10 Maths Wbbse Solutions

4. If \(\frac{x}{y}\) = \(\frac{a + 2}{a – 2}\) then show that \(\frac{x^2-y^2}{x^2+y^2}=\frac{4 a}{a^2+4}\)

Solution: Given that, \(\frac{x}{y}\) = \(\frac{a + 2}{a – 2}\)

or, \(\frac{x^2}{y^2}=\frac{(a+2)^2}{(a-2)^2}\) (Squaring)

or, \(\frac{x^2-y^2}{x^2+y^2}=\frac{(a+2)^2-(a-2)^2}{(a+2)^2+(a-2)^2}\) [Dividendo-componendo process]

or, \(\frac{\dot{x}^2-y^2}{x^2+y^2}=\frac{4 \cdot a \cdot 2}{2\left(a^2+2^2\right)}\)

or, \(\frac{x^2-y^2}{x^2+y^2}=\frac{4 a}{a^2+4}\)

∴ \(\frac{x^2-y^2}{x^2+y^2}=\frac{4 a}{a^2+4}\)(Proved)

Wbbse Class 10 Maths Solutions

5. If x = \(\frac{8ab}{a+b}\) then find the value of \(\left(\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}\right)\)

Solution: Given that x = \(\frac{8ab}{a+b}\)

or, \(\frac{x}{4a}\) = \(\frac{2b}{a+b}\)

or, \(\frac{x+4 a}{x-4 a}\)=\(\frac{2 b+a+b}{2 b-a-b}\)=\(\frac{a+3 b}{b-a}\) …(1)

Again, \(x=\frac{8 a b}{a+b}\)

or, \(\frac{x}{4 b}=\frac{2 a}{a+b}\)

or, \(\frac{x+4 b}{x-4 b}=\frac{2 a+a+b}{2 a-a-b}=\frac{3 a+b}{a-b}\)…(2)

∴ adding (1) and (2) we get,

\(\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}=\frac{a+3 b}{b-a}+\frac{3 a+b}{a-b}\)

Class 10 Maths Wbbse Solutions

= \(\frac{-a-3 b+3 a+b}{a-b}=\frac{2 a-2 b}{a-b}=\frac{2(a-b)}{(a-b)}=2\)

∴ \(\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}=2\).

6. 1. If \(\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}\) then prove that a+b+c = 0 = pq+qb+rc

Solution: \(\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}\) = k(let) [k≠0]

∴ a – k(q – r); b = k (r -p); c – k(p-q)

∴ a + b + c=k(q-r) + k(r-p) + k(p-q)

= k(q-r+r-p+p-q) = k x 0 = 0

Again, pa + qb + rc – p. k(q – r) + q. k (r – p) + r.k (p – q)

= k (pq – pr + qr – pq + pr – qr)

= k x 0 = 0.

∴ a + b + c = 0 = pa + qb + rc (Proved)

2. If \(\frac{ax + by}{a}\) = \(\frac{bx-ay}{b}\) then show that each ratio is equal to x.

Solution: Given that, \(\frac{ax + by}{a}\) = \(\frac{bx-ay}{b}\)

= \(\frac{a^2 x+a b y+b^2 x-a b y}{a^2+b^2}\) [by addendo process]

= \(\frac{x\left(a^2+b^2\right)}{\left(a^2+b^2\right)}=x\)

∴ each ratio = x

Example 8. 1. If \(\frac{a+b}{b+c}\) = \(\frac{c+d}{d+a}\) then prove that c =a or, a + b + c + d = 0

Solution: Given that \(\frac{a+b}{b+c}\)

or, \(\frac{a+b-b-c}{b+c}=\frac{c+d-d-a}{d+a}\) [by dividendo]

or, \(\frac{a-c}{b+c}=\frac{c-a}{d+a}\)

or, \(\frac{-(c-a)}{b+c}=\frac{c-a}{d+a}\)

or, \(\frac{-(c-a)}{b+c}-\frac{c-a}{d+a}=0\)

or, \(-(c-a)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)=0\)

or, \(-(c-a)\left\{\frac{d+a+b+c}{(b+c)(d+a)}\right\}=0\)

or, \(\quad-(c-a)\left\{\frac{a+b+c+d}{(b+c)(d+a)}\right\}=0\)

∴ either c-a=0 or, \(\frac{a+b+c+d}{(b+c)(d+a)}=0\)

⇒ c = a or, a + b + c + d= 0 .

∴ either c = a or, a + b + c + d=0

Class 10 Maths Wbbse Solutions

2. If \(\frac{x}{b+c}\) = \(\frac{y}{c+a}\) = \(\frac{z}{a+b}\) then show that \(\frac{a}{y+z-x}\) = \(\frac{b}{z+x-y}\) = \(\frac{c}{x+y-z}\)

Solution: \(\frac{x}{b+c}\) = \(\frac{y}{c+a}\) = \(\frac{z}{a+b}\)

= \(\frac{y+z-x}{c+a+a+b-b-c}\) [applying addebdo process]

= \(\frac{y+z-x}{2a}\)

∴ each ratio = \(\frac{y+z-x}{2a}\) …(1)

Again, \(\frac{x}{b+c}\) = \(\frac{y}{c+a}\) = \(\frac{z}{a+b}\)

= \(\frac{z+x-y}{a+b+b+c-c-a}\) [applying addebdo process]

= \(\frac{z+x-y}{2b}\)

∴ each ratio = \(\frac{z+x-y}{2b}\)…(2)

Also, \(\frac{x}{b+c}\) = \(\frac{y}{c+a}\) = \(\frac{z}{a+b}\)

= \(\frac{x+y-z}{b+c+c+a-a-b}\) [applying addebdo process]

= \(\frac{x+y-z}{2c}\) …(3)

Now, from(1),(2) and(3) we get, \(\frac{y+z-x}{2a}\)=\(\frac{z+x-y}{2b}\)=\(\frac{x+y-z}{2c}\)

or, \(\frac{y+z-x}{a}\)=\(\frac{z+x-y}{b}\)=\(\frac{x+y-z}{c}\)

∴ \(\frac{a}{y+z-x}\) = \(\frac{b}{z+x-y}\) = \(\frac{c}{x+y-z}\) (proved)

3. If \(\frac{x+y}{3 a-b}=\frac{y+z}{3 b-c}=\frac{z+x}{3 c-a}\) then show that \(\frac{x+y+z}{a+b+c}=\frac{a x+b y+c z}{a^2+b^2+c^2}\)

Solution: \(\frac{x+y}{3 a-b}=\frac{y+z}{3 b-c}=\frac{z+x}{3 c-a}\)

= \(\frac{x+y+y+z+z+x}{3 a-b+3 b-c+3 c-a}\) [by addendo process]

= \(\frac{2(x+y+z)}{2(a+b+c)}=\frac{x+y+z}{a+b+c}\)

∴ each ratio \(\frac{x+y+z}{a+b+c}\) …(1)

Again, \(\frac{x+y}{3 a-b}=\frac{y+z}{3 b-c}=\frac{z+x}{3 c-a}\)

= \(\frac{x+y-y-z+z+x}{3 a-b-3 b+c+3 c-a}\) [by addendo process]

= \(\frac{2 x}{2 a-4 b+4 c}=\frac{x}{a-2 b+2 c}\)

= \(\frac{a x}{a^2-2 a b+2 c a}\) …(2)

Similarly, \(\frac{x+y}{3 a-b}=\frac{y+z}{3 b-c}=\frac{z+x}{3 c-a}\)

= \(\frac{b y}{2 a b + b^2-2 b c }\) …(3)

and \(\frac{x+y}{3 a-b}=\frac{y+z}{3 b-c}=\frac{z+x}{3 c-a}\)

= \(\frac{c z}{c^2-2 c a+2 b a}\) …(4)

Now, applying on (2) (3) and (4) we get

\(\frac{x+y}{3 a-b}=\frac{y+z}{3 b-c}=\frac{z+x}{3 c-a}\)

= \(\frac{a x+b y+c z}{a^2-2 a b+2 c a+2 a b+b^2-2 b c+c^2-2 c a+2 b c}\)..(5)

So, from (1) and (5) we get

\(\frac{x+y+z}{a+b+c}=\frac{a x+b y+c z}{a^2+b^2+c^2}\) (proved)

4. If \(\frac{x}{b}=\frac{y}{b}={z}{c}\) then show that \(\frac{x^2-y z}{a^2-b c}=\frac{y^2-z x}{b^2-c a}=\frac{z^2-x y}{c^2-a b}\)

Solution: \(\frac{x}{b}=\frac{y}{b}={z}{c}\) = k(let) [k≠0]

∴ x = ak, y = bk and z =ck

Now, \(\frac{x^2-y z}{a^2-b c}=\frac{(\dot{a} k)^2-(b k)(c k)}{a^2-b c}\)

= \(\frac{a^2 k^2-b c k^2}{a^2-b c}=\frac{k^2\left(a^2-b c\right)}{\left(a^2-b c\right)}=k^2\)….(1)

\( \frac{y^2-z x}{b^2-c a}=\frac{(b k)^2-c k \cdot a k}{b^2-c a}\)

= \(\frac{b^2 k^2-k^2 c a}{b^2-c a}=\frac{k^2\left(b^2-c a\right)}{b^2-c a}=k^2\) ….(2)

= \(\frac{z^2-x y}{c^2-a b}=\frac{(c k)^2-(a k)(b k)}{c^2-a b}\)

= \(\frac{c^2 k^2-a b k^2}{c^2-a b}=\frac{k^2\left(c^2-a b\right)}{\left(c^2-a b\right)}=k^2\)…(3)

∴ We get from (1), (2), and (3)

\(\frac{x^2-y z}{a^2-b c}=\frac{y^2-z x}{b^2-c a}=\frac{z^2-x y}{c^2-a b}\) (proved)

Class 10 Maths Wbbse Solutions

Example 9. If (a + b + c + d): (a + b – c – d) = (a- b + c – d) : (a – b – c + d), then prove that a: b = c:d.

Solution: (a + b + c + d) : (a + b – c-d)=(a – b + c – d) : (a – b – c + d)

or, \(\frac{a+b+c+d}{a+b-c-d}=\frac{a-b+c-d}{a-b-c+d}\)

or, \(\frac{a+b+c+d+a+b-c-d}{a+b+c+d-a-b+c+d}=\frac{a-b+c-d+a-b-c+d}{a-b+c-d-a+b+c-d}\)

[by componendo and dividendo process]

or, \(\frac{2(a+b)}{2(c+d)}=\frac{2(a-b)}{2(c-d)}\)

or, \(\frac{a+b}{c+d}=\frac{a-b}{c-d}\)

or, \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\) [by alternative process]

or, \(\frac{a+b+a-b}{a+b-a+b}=\frac{c+d+c-d}{c+d-c+d}\) [by componendo and dividendo process]

or, \(\frac{2 a}{2 b}=\frac{2 c}{2 d}\)

or, \(\frac{a}{b}=\frac{c}{d}\)

∴ a: b = c: d(proved)

Example 10. 1. If \(\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1\) then show that \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1\)

Solution: \(\frac{a^2}{b+c}\) = 1

or, \(\frac{a}{b+c}\) = \(\frac{1}{a}\) [Dividing by a]

or, \(\frac{a}{a+b+c}\) = \(\frac{1}{a}\) [by componendo process]

∴ \(\frac{1}{1+a}  =\frac{a}{a+b+c}\) …(1)

Similarly, \(\frac{1}{1+b} =\frac{b}{a+b+c}\) …(2)

and, \( \frac{1}{1+c}=\frac{c}{a+b+c}\)….(3)

Now, adding (1), (2) and (3) we get

\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\)

Class 10 Maths Wbbse Solutions

= \(\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\)

= \(\frac{a+b+c}{a+b+c}=1\)

∴ \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\) = (proved)

2. If x2: (by + cz) = y2: (cz + ax) = z2: (ax + by) = 1 then show that \(\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}=1\)

Solution: x2: (by + cz) = y2: (cz + ax) = z2: (ax + by) = 1

or, \(\frac{x^2}{b y+c z}=\frac{y^2}{c z+a x}=\frac{z^2}{a x+b y}=1\)

∴ \(\frac{x^2}{b y+c z}=1\) ⇒\(\frac{x}{b y+c z}=\frac{1}{x}\) [Dividing by x]

⇒ \(\frac{a x}{b y+c z}=\frac{a}{x}\)[Multiplying by a]

⇒ \(\frac{a x}{a x+b y+c z}=\frac{a}{a+x}\) [by componendo process]

∴ \(\frac{a x}{a x+b y+c z}=\frac{a}{a+x}\) ….(1)

Similarly, it can be proved that

\(\frac{b y}{a x+b y+c z}=\frac{b}{b+y}\) ….(2)

and \(\frac{c z}{a x+b y+c z}=\frac{c}{c+z}\) ….(3)

Now, adding (1), (2) and(3) we get

\(\frac{a x}{a x+b y+c z}+\frac{b y}{a x+b y+c z}+\frac{c z}{a x+b y+c z}\)

Class 10 Maths Wbbse Solutions

= \(\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}\)

\(\begin{aligned}
& \frac{a x+b y+c z}{a x+b y+c z}=\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z} \\
& 1=\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z} \\
& \frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}=1 . \quad \text { (Proved) }
\end{aligned}\)

 

Example 11. 1. If \(\frac{x}{x a+y b+z c}=\frac{y}{y a+z b+x c}=\frac{z}{z a+x b+y c}\) and x+y+z ≠0, then show that each ratio = \(\frac{1}{a + b + c }\)

Solution: \(\frac{x}{x a+y b+z c}=\frac{y}{y a+z b+x c}=\frac{z}{z a+x b+y c}\)

= \(\frac{x+y+z}{x a+y b+z c+y a+z b+x c+z a+x b+y c}\)

= \(\frac{x+y+z}{x(a+b+c)+y(a+b+c)+z(a+b+c)}\)

= \(\frac{x+y+z}{(a+b+c)(x+y+z)}\)

= \(\frac{1}{a+b+c}\) [because x+y+z ≠0]

Hence each ratio = \(\frac{1}{a + b + c }\) (proved)

Class 10 Maths Wbbse Solutions

2. If \(\frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}\) then prove that \(\frac{a(b-c)}{\dot{y}^2-z^2}=\frac{b(c-a)}{z^2-\dot{x}^2}=\frac{c(a-b)}{x^2-y^2}\)

Solution: \(\frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}\)

or, \(\frac{a}{y+z}=\frac{b-c}{z+x-x-y}\)

=\(\frac{b-c}{z-y}\)

∴ \(\frac{a}{y+z}=\frac{b-c}{z-y}\) ….(1)

Similarly, it can be proved that \(\frac{b}{z+x}=\frac{c-a}{x-z}\) ….(2)

and, \(\frac{c}{x+y}=\frac{a-b}{y-x}\) ….(3)

Now, expressioris on the LHS (1), (2) and (3) are equal (Given).

∴ All the 6 expressions of (1), (2) and (3) of both the sides are equal.

∴ \(\frac{a}{y+z} \times \frac{b-c}{z-y}=\frac{b}{z+x} \times \frac{c-a}{x-z}=\frac{c}{x+y} \times \frac{a-b}{y-x}\)

or, \(\frac{a(b-c)}{-(y+z)(y-z)}=\frac{b(c-a)}{-(z+x)(z-x)}=\frac{c(a-b)}{-(x+y)(x-y)}\)

or, \(\frac{a(b-c)}{y^2-z^2}=\frac{b(c-a)}{z^2-x^2}=\frac{c(a-b)}{x^2-y^2}\)

∴ \(\frac{a(b-c)}{y^2-z^2}=\frac{b(c-a)}{z^2-x^2}=\frac{c(a-b)}{x^2-y^2}\) (proved)

3. If \(\frac{b}{a+b}=\frac{a+c-b}{b+c-a}=\frac{a+b+c}{2 a+b+2 c}\)(where a+b+c≠0) then prove that \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)

Solution: \(\frac{b}{a+b}=\frac{a+c-b}{b+c-a}=\frac{a+b+c}{2 a+b+2 c}\)

or, \(\frac{2 b}{2(a+b)}=\frac{a+c-b}{b+c-a}=\frac{2(a+b+c)}{2(2 a+b+2 c)}\)

= \(\frac{2 b+a+c-b+2 a+2 b+2 c}{2 a+2 b+b+c-a+4 a+2 b+4 c}\)

= \(\frac{3 a+3 b+3 c}{5 a+5 b+5 c}\)

= \(\frac{3(a+b+c)}{5(a+b+c)}=\frac{3}{5}\) [because a+b+c ≠0]

∴ \(\frac{b}{a + b}\)=\(\frac{3}{5}\)

or, 5b = 3a+3b or, 5b-3b = 3a

or, 2b = 3a

or, \(\frac{a}{b}\) = \(\frac{2}{3}\)

or, \(\frac{a}{2}\) = \(\frac{b}{3}\) ….(1)

Again, \(\frac{a + c – b}{b + c – a}\) = \(\frac{3}{5}\)

or, 5a + 3a -5b -3b = 3c -5c

or, 8a – 8b = – 2c or, 8a – 12a = – 2c [2b=3a]

or, -4ac = -2c

or, \(\frac{a}{-2}\) = \(\frac{c}{-4}\) or, \(\frac{a}{2}\) = \(\frac{c}{4}\) ….(2)

From (1), (2) we get \(\frac{a}{2}\) = \(\frac{b}{3}\) = \(\frac{c}{4}\)

Hence \(\frac{a}{2}\) = \(\frac{b}{3}\) = \(\frac{c}{4}\) (proved)

4. If \(\frac{x}{y+z}=\frac{y}{z+x}=\frac{z}{x+y}\) then prove that the value of each ration is equal to \(\frac{1}{2}\) or, (-1)

Solution: \(\frac{x}{y+z}=\frac{y}{z+x}=\frac{z}{x+y}\) = k(let) [k≠0]

∴ x = k (y + z); y = k (z + x); z = k (x + y).

Now, x + y + z = k (y + z) + k (z + x) + k (x + y) = k(y + z+ z+ x +x + y)

= k (2x + 2y + 2z) = 2 k (x + y + z)

or, x + y + z – 2k (x + y + z) = 0

or, (x + y + z)( 1 – 2k) = 0.

∴ either x + y + z = 0 or, 1 – 2k = 0

⇒ 2k = 1 ⇒ k = \(\frac{1}{2}\)

∴ each ratio = \(\frac{1}{2}\)

Again, x + y + z ⇒ y + z = -x

⇒ \(\frac{x}{y+z}=\frac{x}{-x}=-1\)

Similarly, z + x = -y

⇒ \(\frac{y}{z+x}=\frac{y}{-y}=-1\)

and x + y = -z

⇒ \(\frac{z}{x+y}=\frac{z}{-z}=-1\)

∴ \(\frac{x}{y+z}=\frac{y}{z+x}=\frac{z}{x+y}\) implies that the value of each ratio is equal to \(\frac{1}{2}\) or (-1) (proved)

 

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