# Chapter – 5 : Ratio And Proportion | Chapter Solution Class 10

 Book Name : Ganit Prakash Subject : Mathematics Class : 10 (Madhyamik) Publisher : Prof Nabanita Chatterjee Chapter Name : Ratio And Proportion (5th Chapter)

Application 1.

Let us see the ratios are given below and let us write in the blank places.

Solution:

 Ratio Ratio of equality / Ratio of inequality Ratio of greater in equality / ratio of less inequality Inverse Ratio 7 : 5 Ratio of equality Ratio of greater in equality 5 : 7 6 : 6 Ratio of equality 6 : 6 1 : 4 Ratio of equality Ratio of greater in equality 4 : 1 Compound ratio of 9 : 2 Ratio of equality Ratio of greater in equality 2 : 9 Compound ratio of 7 : 5 1 : 4 & 7 : 5 = \frac{7 × 1 × 9}{5 × 4 × 2}=\frac{63}{40} Ratio of equality Ratio of greater in equality 40 : 63

Application 4

Let us write the inverse ratio of lowest form of x2y p : xy2 p [Let me do it myself]

Solution:

x2y p : xy2p = \frac{x}{y} = x : y

∴ Inverse ratio of the lowest form of x2y p : x2p is y : x .

Application 6

If the ratio of two numbers is 2 : 3 and their H.C.F is 7, let us write two numbers. [Let me do it myself]

Solution:

Numbers are (2 × 7) = 14 and (3 × 7) = 21.

Application 9

Let us write the inverse ratio of the mixed ratio of three ratios p2 q : r, q2 r : p and r2 p : q [Let me do it myself].

Solution:

Compound ratio of p2q : r, q2r : p and r2p : q is p2q2 r2 : pqr : 1

∴ The inverse ratio = 1 : pqr.

Application 11

If A : B =3 : 7 and B : C = 8 : 5, let us find A : C [Let me do it myself].

Solution:

A : B = 4 : 5 and B : C = 6 : 7

\frac{A}{B} × \frac{B}{C}=\frac{4}{5} × \frac{6}{7}=\frac{24}{35} \\

∴ A : C = 24 : 35

Application 13

If A : B = 5 : 9 and B : C = 4 : 5, let us write the value of A : B : C [Let me do it myself].

Solution:

If A : B = 5 : 9 and B : C = 4 : 5 , the value of B in two cases 9 & 4 , their LCM = 9 × 4 = 36

A : B = 5 : 9 = 5 × 4 : 9 × 4 = 20 : 36 B : C = 4 : 5 = 4 × 9 : 5 × 9 = 36 : 45

∴ A : B : C = 20 : 36 : 45

2nd way :

A : B = 5 : 9 & B : C = 4 : 5

B : C = 4 : 5 = 1 : \frac{5}{4} = 9 : \frac{45}{4} \\

A : B : C = 5 : 9 : \frac{45}{4} \\

= 20 : 36 : 45

Application 15

If x : y = 7 : 4, let us show that (5x – 6y) : (3x + 11y) = 11 : 65 [Let me do it myself].

Solution:

x : y = 7 : 4

Let, x = 7p & y =4p

\frac{5 x-6 y}{3 x+11 y}=\frac{5 × 7 p-6 × 4 p}{3 × 7 p+11 × 4 p}=\frac{35 p-24 p}{21 p+44 p}=\frac{11 p}{65 p}=\frac{11}{65} \\

∴ (5x – 6y) : (3x + 11y) = 11 : 65 ∴

Alternative method:

\frac{5 x-6 y}{3 x+11 y}=\frac{\frac{5 x-6 y}{y}}{\frac{3 x+11 y}{y}}

=\frac{5 \frac{x}{y}-6}{3 \frac{x}{y}+11}=\frac{5 × \frac{7}{4}-6}{3 × \frac{7}{4}+11} \\

=\frac{\frac{35}{4}-6}{\frac{21}{4}+11}=\frac{\frac{35-24}{4}}{\frac{41+44}{4}}

=\frac{11}{65}

= 11 : 65

Application 17

If (2x + 5y) : (5x – 7y) = 5 : 3, let us find x : y [Let me do it myself]

Solution:

(2x + 5y) : (5x – 7y) = 5 : 3

or, \frac{2 x+5 y}{5 x-7 y}=\frac{5}{3}

or, 5(5x – 7y) = 3(2x + 5y)

or, 25x – 35y = 15y + 35y

or, 19x = 50y

or, \frac{x}{y}=\frac{50}{19}

∴x : y = 50 : 19

Application 19

If (7x – 5y) : (3x + 4y) = 7 : 11, let us find the value of (5x – 3y) : (6x + 5y)[ Let me do it myself].

Solution:

(7x – 5y) : (3x + 4y) = 7 : 11

or, \frac{7 x-5 y}{3 x+4 y}=\frac{7}{11}

or, 11(7x – 5y) = 7(3x + 4y)

or, 77x – 55y = 55y + 28y

or, 77x – 21x = 55y + 28y

or, 56x = 83y

or, \frac{x}{y}=\frac{83}{56}

or, x : y = 83 : 56

∴ Let x = 83p & y = 56p

(56x – 3y) : (6x + 5y)

(5x – 3y) : (6x + 5y)

= (5 × 83p – 3 × 56p):(6 × 83p +5 × 56p)

= (415p – 168p) : (498p + 280p)

= 247p : 778p

= 247 : 778

Application 21

Let us write what should be added to each term of the ratio 5 : 3 to make the ratio 7 : 6 [Let me do it myself]

Solution:

Let the required number be x, which should be added to the both terms.

\frac{5+x}{3+x}=\frac{7}{6}

or, 7(3 + x) = 6(5 + x)

or, 21 + 7x = 30 + 6x

or, 7x – 6x = 30 – 21

∴ x = 9

∴ The required number = 9

## LET US WORK OUT – 5.1

Question 1. (i)

Let us express the followings as ratio and let us write by understanding ratio of equality, ratio of leser inequality, ratio of greater inequality in each case :

4 months and 1 year 6 months

Solution:

4 months : 1 year 6 months

= 4 months : (12 + 6) months = 4 months : 18 months = 2 : 9

The ratio is of leser inequality.

Question (ii)

75 paise and Re. 1 and 25 paise

Solution:

75 p & Re. 1 & 25 p

= 75p : 125p = 3 : 5

This ratio is of leser inequality.

Question (iii)

60 cm and 0.6 metre

Solution:

60 cm. & 0.6 m.= 60 cm : (0.6 × 100) cm = 60 cm : 60 cm = 1 : 1

This is an equality.

Question (vi)

1.2 kg and gram.

Solution:

1.2 kg & 60gm

= 1.2 kg : 60 gm = (1.2 × 1000) gm: 60 gm

= 1200 gm : 60 gm = 20 : 1

This ratio is of greater inequality.

Question 2. (i)

Let us write the ratio of p kg and q gram.

Solution:

p kg & q gram

= (p × 1000) gm : q gram = 1000p : q

Question (ii)

Let us write when it is possible to find ratio of x days z months.

Solution:

It is possible if both the quantities are expressed in same unit.

Question (iii)

Let us write what type of mixed ratio of a ratio and its inverse ratio.

Solution:

It will be an equality.

Question (iv)

Let us find mixed ratio of \frac{a}{b}: c, \frac{b}{c}: a, \frac{c}{a}: b.

Solution:

The compound ratio of (\frac{a}{b}: c) ; (\frac{b}{c}: a) \ \& \ (\frac{c}{a}: b)

is (\frac{a}{b} × \frac{b}{c} × \frac{c}{a}):(c × a × b) \\

= 1 : abc

Question (v)

Let us write by calculating what ratio and x2: y z will form with the mixed ratio xy : z2.

Solution:

Let the required ratio = p : q

∴ (x2.p) : (yz.q) = xy : z2

or, \frac{p x^2}{q y z}=\frac{x y}{z^2}

\frac{p}{q}=\frac{x y \cdot y z}{x2 \cdot x2} \\

=\frac{y2}{x z} = y2 : xz

Question (iv)

Let us calculate the compound ratio of inverse ratios of x^2 : \frac{y z}{x}, y^2: \frac{z x}{y}, x^2: \frac{y x}{z}

Solution:

The inverse ratios of (x^2: \frac{y z}{x}),(y^2: \frac{z x}{y}) \ \& \ (z^2: \frac{x y}{z})

are (\frac{y z}{x}: x^2),(\frac{z x}{y}: y^2) \ \& \ (\frac{y x}{z}: z^2)

The required compoud ratio is

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

= xyz : x2y2z2= 1 : xyz.

Question 3. (i)

Let us find mixed ratio or compound ratio of the following ratios :

4 : 5, 5 : 7 and 9 : 11

Solution:

The compound ratio of (4 : 5), (5 : 7) & (9 : 11) is

(4 × 5 × 9) : (5 × 7 × 11) = 36 : 77

Question (ii)

(x + y) : (x – y), (x2 + y2) (x + y)2 and (x2 – y2) 2 : (x4 – y4)

Solution:

The compound ratio of (x + y) : (x – y), (x2 + y2) (x + y)2 and (x2 – y2) 2 : (x4 – y4)

=\frac{(x+y) (x2+y2) (x2-y2)2}{(x-y) (x+y)2 \cdot(x^{4}-y^{4})} \\

=\frac{(x+y) (x2+y^{2}) (x+y)2(x2-y2)2}{(x-y) (x+y) (x+y) (x2+y2) (x-y) (x+y)} \\

=\frac{(x+y)^{3}(x-y)2}{(x+y)^{3}(x-y)2}=\frac{1}{1} = 1 : 1

Question 4. (i)

If A : B = 6 : 7 and B : C = 8 : 7, let us find A : C

Solution:

Find A : C, of A : B = 6 : 7 & B : C = 8 : 7

\frac{A}{C}=\frac{A}{B} × \frac{B}{C}=\frac{6}{7} × \frac{8}{7}=\frac{48}{49}

∴ A : C = 48 : 49 Ans.

Question  (ii)

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

Solution:

Find A : C, if A : B = 6 : 7 & B : C = 8 : 7

\frac{A}{D}=\frac{A}{B} × \frac{B}{C} × \frac{C}{D}=\frac{2}{3} × \frac{4}{5} × \frac{6}{7}=\frac{16}{35}

∴ A : D = 16 : 35 Ans.

Question  (iii)

If A : B = 3 : 4 and B : C = 2 : 3, let us find A : B : C.

Solution:

Find A : B : C, if A : B = 3 : 4 & B : C = 2 : 3

A : B = 3 : 4 B : C = 2 : 3 = 4 : 6

∴ A : B : C = 3 : 4 : 6 Ans.

Question  (iv)

If x : y =2 : 3 and y : z = 4 : 7 let us find x : y : z.

Solution:

x : y = 2: 3 = 8 : 12

y : z = 4 : 7 = 12 : 21

x : y : z = 8 : 12 : 21

Question 5. (i)

If x : y = 3 : 4, let us find (3y – x) : (2x + y)

Solution:

If x : y = 3 : 4, find (3y – x) : (2x + y)

Let x = 3k & y = 4k (where k is a real no & k \pm 0 )

\frac{3 y-x}{2 x+y}=\frac{3 × 4 k-3 k}{2 × 3 k+4 k}=\frac{12 k-3 k}{6 k+4 k}=\frac{9 k}{10 k}=\frac{9}{10}=9: 10

∴(3y – x) : (2x + y) = 9 : 10 Ans.

Question (ii)

If a : b = 8 : 7, let us show that (7a – 3b) : (11a – 9b) = 7 : 5

Solution:

If a : b = 8 : 7; show that, (7a – 3b) : (11a – 9b) = 7 : 5

Let a = 8k & b = 7k (where k is a real number & k \neq 0 )

\frac{7 a-3 b}{11 a-9 b}=\frac{7 × 8 k-3 × 7 k}{11 × 8 k-9 × 7 k}=\frac{56 k-21 k}{88 k-63 k}=\frac{35 k}{25k}=\frac{7}{5} = 7 : 5

∴ (7a – 3b) : (11a – 9b) = 7 : 5. Ans.

Question (iii)

If p : q = 5 : 7 and p – q = -4, let us find the value of 3p + 4q.

Solution:

If p : q = 5 : 7 & p – q = -4; find 3p + 4q.

Let p = 5kq & q = 7k, (where k is a real number & k \neq 0 )

∴ 5k – 7k = -4

or, -2k = -4

∴ k = 2.

∴ 3p + 4q = 3 × 5k + 4 × 7k

= 15k + 28k

= 43k

= 43 × 2 = 86

3p + 4q = 86 Ans.

Question 6. (i)

If (5x – 3y) : (2x + 4y) = 11 : 12, let us find x : y

Solution:

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

\frac{5 x-3 y}{2x + 4y} = \frac{11}{12}

or, 12x (5x – 3y) = 11 × (2x + 4y)

or, 60x – 36y = 22x + 44y

or, 60x – 22x = 44y + 36y

or, 38x = 80y

or, \frac{x}{y} = \frac{80}{38} = \frac{40}{19}

∴ x : y = 40 : 19 ans.

Question (ii)

If (3a + 7b) : (5a – 3b) = 5 : 3, let us find a : b.

Solution:

If (3a + 7b) : (5a – 3b) = 5 : 3, find a : b

\frac{3a + 7b}{5a - 3b} = \frac{5}{3}

or, 5(5a – 3b) = 3(3a + 7b)

or, 25a – 15b = 9a + 21b

or, 25a – 9a = 21b + 15b

or, 16a = 36b

\frac{a}{b} = \frac{36}{16} = \frac{9}{4}

∴a : b = 9 : 4 ans.

Question 7. (i)

If (7x – 5y) : (3x + 4y) = 7 : 11, let us show that (3x – 2y) : (3x + 4y) = 127 : 473.

Solution:

If (7x – 5y) : (3x + 4y) = 7 : 11, show that (3x – 2y) : (3x + 4y) = 127 : 473

\frac{7x - 5y}{3x + 4y} = \frac{7}{11}

or, 11 ×(7x – 5y) = 7 ×(3x + 4y)

or, 77x – 55y = 21x + 28y

or, 77x – 21x = 28y + 55y

or, 56x = 83y

\frac{x}{y} = \frac{83}{56}

Let x = 83k & y = 56k, (where k is a real number & k \neq 0 )

∴(3x – 2y) : (3x + 4y)

= (3 × 83k – 2 × 56k) : (3 × 83k + 4 × 56k)

= (249 – 112k) : (249k + 224k)

= 137k : 473k = 137 : 473

∴(3x – 2y) : (3x + 4y) = 137 : 473 Proved. ans.

Question (ii)

If (10x + 3y) : (5x + 2y) = 9 : 5, let us show that (2x +y) : (x + 2y) = 11 : 13

Solution:

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

\frac{10x + 3y}{5x + 2y} = \frac{9}{5}

or, 5(10x + 3y) = 9(5x + 2y)

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

or, 50x – 45x = 18y – 15y

\frac{x}{y} = \frac{3}{5}

Let x = 3k & y = 5k, (where k is a real number & k \neq 0 )

∴(2x +y) : (x + 2y)

= (2 × 3k + 5k) : (3k + 2 × 5k) = (6k + 5k) : (3k + 10k)

= 11k : 13k = 11 : 13

∴(2x +y) : (x + 2y) = 11 : 13 Proved.

Question 8. (i)

Let us calculate what term should be added to both terms of the ratio 2 : 5 to make the ratio 6 : 11.

Solution:

Let x be the number which should be added with both the terms of 2 : 5 to get the ratio of 6 : 11.

\frac{2 + x}{5 + x} = \frac{6}{11}

or, 6(5 + x) = 11 \quad(2 + x)

or, 30 – 22 = 11x – 6x

or, + 8 = + 5x

\text{ x } = \frac{8}{5}

∴ The required number = \frac{8}{5}.ans.

Question (ii)

Let us calculate what term should be subtracted from each term of the ratio a : b to make the ratio m : n.

Solution:

Let the number = x.

\frac{a - x}{b - x} = \frac{m}{n} or, an – nx = bm – mx

or, mx – nx = bm – an or, x(m – n) = bm – an

\text{ x }\frac{bm -a n}{m - n} \quad ∴ The required number = \frac{bm -a n}{m - n}ans.

Question (iii)

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 required number = x

According to the problem,

\frac{4 + x}{7 - x} = \frac{2}{3} × \frac{5}{4} \quad \text{ or, } \frac{4 + x}{7 - x} = \frac{5}{6} \text { or } 6 ×(4 + x) = 5 ×(7 - x)

or, 24 + 6x = 35 – 5x

or, 6x + 5x = 35 – 24

11x = 11 \quad \thereforex = \frac{11}{11} = 1

∴ The required number = 1.

Application 23.

Let us see whether four numbers 2, 3, 4 and 6 are in proportion or not :

Solution:

2 × 6 = 3 × 4 and 3 × 4 = 2 × 6

∴ 2 : 3 : : 4 : 6

Application 24. Let us see whether four numbers 2.5, – 2, – 5 and 4 are in proportion or not.

Solution : 2.5 × 4 = ( – 2) ×( – 5) and ( – 2) ×( – 5) = 2.5 × 4

Application 25.

Let us see whether four the numbers 2, 7, 12 and 42 are in proportion or not.

Solution:

2 × 42 = 84 & 7 × 12 = 84

∴ 2 : 7 : : 12 : 42 (Are in proportion)

Application 26.

Let us see whether four terms - \sqrt{2}, 6, 1 and - \sqrt{18}are in proportion or not.

Solution:

( - \sqrt{2}) ×( - \sqrt{18}) = \sqrt{36} = 6. \ 6 \quad 1 = 6

( - \sqrt{2}) : 6 : : 1 : ( - \sqrt{18}) (Are in proportion).

Application 28

Let us see whether 2a, 3b, 6a cand 9b care proportional or not.

Solution:

2a × 9bc = 18abc & 3b × 6ac = 18abc

Application 29

Let us see whether 8x, 5yz, 40 q x and 25qyx are proportional or not.

Solution:

8x, 5yz, 40 qx & 25qyz

8x × 25qyz = 200 qxyz & 5 yz × 40qx = 200qxyz

∴ 8x : 5 y z : : 40 qx : 25qyz (Are in proportion).

Application 31

If 8 : y : : 2 : 21, let us write by calculating the value of y. [Let do itmyself]

Solution:

If 8 : y : : 2 : 21, find the value of y.

Ans. 8 × 21 = y × 2 or, 2 y = 168 ∴ y = \frac{168}{2} = 84.

Application 32

Let us find the fourth proportional of 6, 9, 12.

Solution:

6x = 9 × 12 \quad \thereforex = \frac{9 × 12}{6} = 18.ans.

Application 33

Let us write by calculating the fourth proportional of 5, 4, 25 [Let me do it myself].

Solution:

Let the 4th proportional = x

∴ 5 : 4 : : 25 : x

or, 5x = 4 × 25 \quad \thereforex = \frac{4 × 25}{5} = 20.

∴ Fourth proportional = 20 ans.

Application 34

Let us write by calculating how many independent proportions can be obtained from the proportional numbers 5, 6, 10and 12 and mention those proportions. [Let me do itmyself]

Solution:

5, 6, 10, 12are proportional.

∴ 5 : 6 : : 10 : 12

(i) 5, 10, 6, 12; (ii) 6, 5, 12, 10; (iii) 10, 5, 12, 6

These 3 are the different forms of proportional.

Application 36

Let us write by calculating which number is to be added to each of 12, 22, 42and 72 to make the sums proportional. [Let me do itmyself]

Solution:

12, 22, 42, 72

Let x be the number which should be added with each term so that the sums will be proportional.

∴(12 + x) : (22 + x) : : (42 + x) : (72 + x)

or, \frac{12 + x}{22 + x} = \frac{42 + x}{72 + x}

or, 864 + 72x + 12x + x 2 = 924 + 42x + 22x + x 2

or, 84x – 64x = 924 – 864

or, 20x = 60 \quad \therefore \text{ x } = \frac{60}{20} = 3.

∴ The required number = 3

Application 38.

Let us write by calculating which number is to be added to each of 3.6, 7 and 10 to make the sums proportional.

Solution:

Let the required number = x

∴ (3 + x) : (6 + x) : : (7 + x) : (10 + x)

\frac{3 + x}{6 + x} = \frac{7 + x}{10 + x}

or, 30 + 10x + 3x + x² = 42 + 7x + 6x + x²

30 + 13x = 42 + 13x

∴ If any real number (\neq 0) is added, the sums will be in proportional.

Application 39.

Let us find the 3rd proportional of 9 and 15.

Solution:

x = \frac{15 × 15}{9} \thereforex = 25

Application 40.

Let us find 3rd proportional of Rs. 3 and Rs. 12. [Let me do it myself]

Solution:

Let the 3rd proportional = Rs x.

\frac{3}{12} = \frac{12}{x} \text { or, } 3x = 12 × 12 \quad \therefore \text{ x } = \frac{12 × 12}{3} = 48

∴ 3rd proportional = Rs. 48 .

Application 44.

Let us write by calculating the mean proportional of 0.5 and 4.5. [Let me do it myself]

Solution:

Let the 3rd proportional = x

\frac{9 pq}{12pq^2} = \frac{12 q^2}{x} \quad \therefore \text{ x }= \frac{144 p^2 q^4}{9 p q} = 16pq Ans.

Application 45.

If the extreme terms of three positive continued proportional numbers are pqr, \frac{p r}{q}, let us find the mean proportional.

Solution:

Mean proportional = \sqrt{0.5 × 4.5} = \sqrt{2.25} = 1.5 Ans.

Application 46.

Let us find the mean proportional of positive numbers xy² and xz² [Let me do it myself].

Solution:

Mean proportional = \sqrt{x y^2 × x z^2} = \sqrt{x² y² z²} = xyz Ans.

## LET US SEE BY CALCULATING – 5.2

Question 1

Let us find the value of x for the following proportions :

(i) 10 : 35 : : x : 42

Solution:

10 : 35 : : x : 42

or, \frac{10}{35} = \frac{ x}{42}

or, 35 x = 10 × 42

x = \frac{10 × 42}{35} = 12

(ii) x : 50 : : 3 : 2

Solution:

x : 50 : : 3 : 2

or, 2 x = 50 × 3

x = \frac{50 × 3}{2} = 75 ∴

Question 2

Let us find the fourth proportional of the followings :

(i) \frac{1}{3}, \frac{1}{4}, \frac{1}{5}

Solution:

\frac{1}{3}, \frac{1}{4}, \frac{1}{5}

∴ 4th proportional = \frac{\frac{1}{4} × \frac{1}{5}}{1 / 3} = \frac{3}{20}

(ii) 9.6kg, 7.6kg, 28.8kg.

Solution:

9.6kg, 7.6kg, 28.8kg.

∴ 4th proportional = \frac{7.6kg × 28.8kg}{9.6kg} = 22.8kg.

(iii) x²y², y² z, z² x

Solution:

x²y², y² z, z² x

∴ 4th proportional = \frac{y^2z × z^2x}{ x^2y} = \frac{y z³}{ x^2y}

(iv) (p – q), (p² – q²), p² – pq + q²

Solution:

(p – q) ;(p² – q²), (p^2 – p q + q²)

∴ 4th proportional = \frac{(p^2 - q^2) (p^2 - p q + q^2)}{(p - q)} = (p + q) (p^2 - p q + q^2) = p3 + q3 Ans.

Question 3

Let us find the 3rd proportional of the following positive numbers :

(i) 5, 10

Solution:

5, 10

3rd proportional = \frac{10 × 10}{5} = 20

(ii) 0.24, 0.6

Solution:

0.24, 0.6

3rd proportional = \frac{0.6 × 0.6}{0.24} = \frac{36}{24} = \frac{3}{2} = 1.5

(iii) p3 q², q² r

Solution:

p3 q², q² r

3rd proportional = \frac{q² r × q² r}{p³ q²} = \frac{q^4 r}{p³ q²} = \frac{q² r²}{p³}

(iv) ( x -y)², ( x² -y²)2

Solution:

( x -y)², ( x² -y²)2

∴ 3rd proportional = \frac{( x^2 -y^2)^2( x^2 -y^2)^2}{( x -y)^2}

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

= ( x +y)² ( x -y)² Ans.

Question 4

Let us find the mean proportional of the following positive numbers :

(i) 5 and 80

Solution:

5 and 80

Required mean proporitnal = \sqrt{5 × 80} = \sqrt{400} = 20. Ans.

(ii) 8.1 & 2.5

Solution:

8.1 & 2.5

Mean proportional = \sqrt{8.1 × 2.5} = \sqrt{20.25} = 4.5. Ans.

(iii) x3y and xy3

Solution:

x3y and xy3

Mean proportional \sqrt{ x³y \cdot xy³} = \sqrt{ x^4y^4} = x²y². Ans.

(iv) ( x -y)², ( x +y)²

Solution:

( x -y)², ( x +y)²

∴ Mean proportional = \sqrt{( x -y)² \cdot( x +y)²} = ( x -y) ( x +y). Ans.

Question 5

If the two ratios a : b and c : d e xpress mutually opposite relations, let us write what relation will be e xpressed by their inverse relation.

Solution:

Each other are inverse proportion.

Question 6

Let us write how many continued proportions can be constructed by three numbers in continued proportions.

Solution:

Two.

Question 7

If the first and second of the five numbers in continued proportion are 2 and 6 respectively, let us find the fifth number.

Solution:

Let 5 consecutive proportional numbers are a, b, c, d, e.

\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{d}{e}. Here, a = 2 & b = 6

c = \frac{6 × 6}{2} = 18

\frac{6}{18} = \frac{18}{d} \therefore d = \frac{18 × 18}{6} = 54

Again, \frac{c}{d} = \frac{d}{e} \quad \text { or, } \frac{18}{54} = \frac{54}{e}

e = \frac{54 × 54}{18} = 54 × 3 = 162

Question 8

Let us write by calculating what should be added to each of 6, 15, 20 and 43 to make the sums proportional.

Solution:

6, 15, 20 & 43

Let x be the required number, which should be added to each term.

Then (6 + x) : (15 + x) : : (20 + x) : (43 + x)

or, \frac{6 + x}{15 + x} = \frac{20 + x}{43 + x}

or, (6 + x) (43 + x) = (15 + x) (20 + x)

or, 258 + 43x + 6x + x2 = 300 + 20x + 15x + x2

or, 49x – 35x = 300 – 258

or, 14x = 42 \quad \therefore x = \frac{42}{14} = 3

∴ The required number = 3. Ans.

Question 9

Let us find what should be subtracted from each of 23, 30, 57 and 78 to make the results proportional.

Solution:

Let x be the number which should be subtracted from each of the terms & then they will be proportional.

∴ (23 -x) : (30 -x) : : (57 – x) : (78 -x)

or, \frac{23 -x}{30 -x} = \frac{57 -x}{78 -x}

or, (30 – x) (57 – x) = (23 – x) (78 – x)

or, 1710 – 57x – 30x + x2 = 1794 – 78x – 23x + x 2

or, 87x + 101x = 1794 – 1710

or, 14x = 84 \quad \therefore x = \frac{84}{14} = 6

∴ The required number = 6.

Question 10

Let us find what should be subtracted from each of p, q, r, s to have four numbers in proportion.

Solution:

p, q, r, s.

Letx be the number which should be subtracted from each of the terms so that they will be proportional.

∴(p -x) : (q -x) : : (r -x) : (s -x)

or, \frac{p -x}{q -x} = \frac{r -x}{s - r}

or, (q – x) (r – x) = (p – x) (s – x)

or, q r – rx – qx + x2 = ps – ps – sx + x 2

or, px + sx – rx – qx = ps – qr

or, x(p + s – r – q) = ps – qr

x = \frac{p s - q r}{p + s - r - q}

∴ The required number = \frac{p s - q r}{p + s - r - q}. Ans.

Question 13

We understand, if 2 : 3 : : 10 : 15 then 2 : 10 : : 3 : _______ [Let me do it myself]

Solution:

If 2 : 3 : : 10 : 15

then 2 : 10 : : 3 : 15

Question 15

We understand, if 6 : 10 : : 9 : 15 then 10 : 6 : : ______ : _______ [Let me do it myself]

Solution:

If 6 : 10 : : 9 : 15

then 10 : 6 : : 15 : 9

Question 17

We understand, if 4 : 5 : : 8 : 10, then with the help of componendo of propor tion, we get (4 + 5) : 5 = ____ : 10 [Let me do it myself].

Solution:

If 5 : 4 = 8 : 10, with the help of componendo, we get (4 + 5) : 5 = (8 + 10) : 10

Question 19

We understand, if 5 : 4 = 10 : 8, then with the help of dividendo of proportion we get (5 – 4) : 4 = _____ : 8 [Let us write it by verifying].

Solution:

If 5 : 4 = 10 : 8, with the help of componendo we get (5 – 4) : 4 = (10 – 8) : 8

Application 49.

If 5 : 4 : : 10 : 8, then with the help of componendo and dividendo of proportion, we get, (5 + 4) : (5 – 4) : : (10 + 8) : ______ [Let me do it myself].

Solution:

(5 + 4) : (5 – 4) = (10 + 8) : (10 – 8)

Application 51.

If a : b : : c : d, let us prove that (4a + 7b) : (4a – 7b) : : (4c + 7d) : (4c – 7d) [Let me do it myself].

Solution:

If a : b : : c : d, prove that (4a + 7b) : (4a – 7b) : : (4c + 7d) : (4c – 7d)

Ans. Let \frac{a}{b} = \frac{c}{d} =k[k \neq 0]

∴ a = bk and c = dk

\text { L.H.S } = \frac{4 a + 7 b}{4 a - 7 b} = \frac{4 × bk + 7 b}{4 × dk - 7 b} = \frac{4 bk + 7 b}{4 dk - 7 b} = \frac{b(4k + 7)}{b(4k - 7)} = \frac{4k + 7}{4k - 7}

\text { R.H.S } = \frac{4c + 7 d}{4c - 7 d} = \frac{4 × dk + 7 d}{4 × dk - 7 d} = \frac{d(4k + 7)}{d(4k - 7)} = \frac{4k + 7}{4k - 7}

\text { L.H.S = R.H.S }

Application 52.

We can write by applying the addendo property of proportion, \frac{2}{3} = \frac{6}{9} = \frac{8}{12} = \frac{2 + 6 + 8}{ + + }[Let me do it myself]

Solution:

\frac{2}{3} = \frac{6}{9} = \frac{8}{12} = \frac{2 + 6 + 8}{3 + 9 + 12} = \frac{16}{24}

Application 58.

If \frac{x}{a + b + c} \frac{y}{b + c - a} \frac{z}{c + a - b}, let us prove that each ratio = \frac{x + y + z}{a + b + c}. [Let me do it myself]

Solution:

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

\text { Each ratio } = \frac{x + y + z}{a + b - c + b + c - a + c + a - b} = \frac{x + y + z}{a + b + c}

Application 66.

If (4a + 5b) (4c – 5d) = (4 a – 5b) (4c + 5 d), let us prove that a, b, c and d are in proportion. [Let me do it myself]

Solution:

(4a + 5b) (4c – 5d) = (4a – 5b) (4c + 15d)

or, 16ac + 20bc – 20ad – 25bd = 16ac – 20bc + 20ad – 25bd

or, bc = a d

\frac{a}{b} = \frac{c}{d}

∴ a, b, c, & d are in proportion.

## LET US WORK OUT – 5.3

Question 1

If a : b = c : d, let us show that :

(i) (a² + b²) : (a2 – b²) = (ac + bd) : (ac – bd)

Solution:

(a² + b²) : (a² – b²) = (ac + bd) : (ac – bd)

Let \frac{a}{b} = \frac{c}{d} = k[k \neq 0]

∴ a = b k, c = d k

L.H.S = \frac{a² + b²}{a² - b²} = \frac{(b k)² + b²}{(b k)² - b²} = \frac{b² k² + b²}{b² k² - b²} = \frac{b²(k² + 1)}{b²(k² - 1)} = \frac{k² + 1}{k² - 1}

\text { R.H.S } = \frac{ac + bd}{ac - bd} = \frac{b k \cdot d k + b \cdot d}{b k \cdot d k - b \cdot d} = \frac{bd k² + bd}{bd k² - bd} = \frac{bd(k² + 1)}{bd(k² - 1)} = \frac{k^2 + 1}{k² - 1}

∴ L.H.S = R.H.S

i.e., (a² + b²) : (a^2 – b²) = (ac + bd) : (ac – bd) Proved.

(ii) (a² + ab + b²) : (a² – ab + b²) = (c² + cd + d²) : (c² – cd + d²)

Solution:

(a² + ab + b²) : (a² – ab + b²) = (c² + c d + d²) : (c² – c d + d²).

\text { L.H.S } = \frac{a² + ab + b²}{a² - ab + b²}

= \frac{(b k)² - bd \cdot b + b²}{(b k)² - b k \cdot b + b²} = \frac{b² k² + b² k + b²}{b² k² - b² k + b²} = \frac{b²(k² + k + 1)}{b²(k² - k + 1)} = \frac{k² + k + 1}{k² - k + 1}

\text { R.H.S } = \frac{c² + c d + d²}{c² - c d + d²}

= \frac{(d k)² + d k \cdot d + d²}{(d k)² - d k \cdot d + d²} = \frac{d² k² + d² k + d²}{d² k² - d² k + d²} = \frac{d²(k² + k + 1)}{d²(k² - k + 1)} = \frac{k² + k + 1}{k² - k + 1}

∴ L.H.S = R.H.S., i.e, (a² + ab + b²) : (a² – ab + b²) = (c² + c d + d²) : (c² – c d + d²) Proved

(iii) \sqrt{a² + b²} : \sqrt{b² + d²} = (pa + qc) : (pb + qd)

Solution:

\sqrt{a² + b²} : \sqrt{b² + d²} = (p a + q c) : (p b + q d)

\text { L.H.S } = \frac{\sqrt{a² + c²}}{\sqrt{b² + d²}} = \frac{\sqrt{b² k² + d² k²}}{\sqrt{b² + d²}} = \frac{k(\sqrt{b² + d²})}{(\sqrt{b² + d²})} = k

\text { R.H.S } = \frac{p a + q c}{p b - q d} = \frac{p \cdot d k + q \cdot d k}{p b - q d} = \frac{k(p d + q d)}{p b - q d} = k

\text { L.H.S } = \text { R.H.S }

\sqrt{a² + c²} : \sqrt{b² + d²} = (p a + q c) : (p d – q d) Proved.

Question 2

If x : a = y : b = z : c, let us prove that :

Let \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k( where k \neq 0)

∴ x = ak, y = bk & z = ck.

(i) \frac{x<sup>3</sup>}{a²} + \frac{y<sup>3</sup>}{b<sup>2</sup>} + \frac{z<sup>3</sup>}{c²} = \frac{(x + y + z)<sup>3</sup>}{(a + b + c)²}

Solution:

\frac{x³}{a²} + \frac{y³}{b^2} + \frac{z³}{c²} = \frac{(x + y + z)³}{(a + b + c)²}

To prove \frac{x³}{a²} + \frac{y³}{b^2} + \frac{z³}{c²} = \frac{(x + y + z)³}{(a + b + c)²}

L.H.S. = \frac{x³}{a²} + \frac{y³}{b²} + \frac{z³}{c²}

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

=\frac{a³ k³}{a²} + \frac{b³ k³}{b²} + \frac{c³ k³}{b²} = ak3 + bk3 + ck3 = k3(a + b + c)

R.H.S. =\frac{(x + y + z)³}{(a + b + c)²} = \frac{(a k + b k + c k)³}{(a + b + c)²} = \frac{k³(a + b + c)³}{(a + b + c)²} = k³(a + b + c)

∴ L.H.S. = R.H.S.

\frac{x³}{a²} + \frac{y³}{b^2} + \frac{z³}{c²} = \frac{(x + y + z)³}{(a + b + c)²} Proved.

(ii) \frac{x³ + y³ + z³}{a³ + b³ + c³} = \frac{x y z}{ab c}

Solution:

To prove, \frac{x³ + y³ + z³}{a³ + b³ + c³} = \frac{x y z}{ab c}

L.H.S = \frac{x³ + y³ + z³}{a³ + b³ + c³} = \frac{(a k)³ + (b k)³ + (c k)³}{a³ + b³ + c³} = \frac{a³ k³ + b³ k³ + c³ k³}{a³ + b³ + c³}

= \frac{k 3(a³ + b³ + c³)}{(a³ + b³ + c³)} = k3

R.H.S. = \frac{x y z}{ab c} = \frac{a k \cdot b k \cdot c k}{a \cdot b \cdot c} = k3

∴ L.H.S. = R.H.S. i.e., \frac{x³ + y³ + z³}{a³ + b³ + c³} = \frac{x y z}{ab c} Proved.

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

Solution:

To prove, (a² + b² + c²) (x² + y² + z²) = (ax + by + cz)²

L.H.S = (a² + b² + c²) (x² + y² + z²)

= (a² + b² + c²) (a² k² + b² k² + c² k²) = k²(a² + b² + c^2) (a² + b² + c²)

= k²(a² + b² + c²)²

R.H.S. = (ax + by + cz)²

= (a \cdot a k + b \cdot b k + c . c k)² = \{k(a² + b² + c^2)\}²

= k²(a² + b² + c²)²

∴ L.H.S = R.H.S

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

Question 3

If a : b = c : d = e : f, let us prove that,

(i) Each ratio = \frac{5 a - 7 c - 13 e}{5 b - 7 d - 13 f}

Solution:

Each ratio = \frac{5 a - 7 c - 13 e}{5 b - 7 d - 13 f}

Let, \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k( \text{ where } k \neq 0)

∴ a = bk ; c = dk ; e = fk

\frac{5 a - 7 c - 13 e}{5 b - 7 d - 13 f} = \frac{5 b k - 7 d k - 13 f k}{5 b - 7 d - 13 f} = \frac{k(5 b - 7 d - 13 f)}{(5 b - 7 d - 13 f)} = k

\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k = \frac{5 a - 7 c - 13 e}{5 b - 7 d - 13 f}

(ii) (a² + c² + e²) (b² + c² + e²) (b² + d² + f²) = (ab + c d + e f)²

Solution:

To prove (a² + c² + e²) (b² + c² + e²) (b² + d² + f²) = (ab + c d + e f)²

L.H.S. = (a² + c² + e²) (b² + d² + f²)

= (b² k² + d² k² + d² k²) (b² + d² + f²)

= k²(b² + d² + f²) (b² + d² + f²) = k²(b² + d² + f²)²

R.H.S. = (ab + cd + ef)2

= (bk.b + dk.d + fk.f)2

= \{k(b² + d² + f²)\}²= k²(b² + d² + f²)²

L.H.S. = R.H.S.

(a² + c² + e²) (b² + d² + f²) = (ab + cd + ef)² Proved.

Question 4

If a : b = b : c, let us prove that :

(i) (\frac{a + b}{b + c})^2 = \frac{a² + b²}{b² + c²}

Solution:

Prove that (\frac{a + b}{b + c})² = \frac{a² + b²}{b² + c²}

Let \frac{a}{b} = \frac{b}{c} = k( where k \neq 0)

\mathrm{b} = \mathrm{ck}, \& \mathrm{a} = \mathrm{bk} = c k \cdot k = c k²

\text { L.H.S } = (\frac{a + b}{b + c})² = (\frac{c k² + c k}{c k + c})² = \{\frac{c k(k + 1)}{c(k + 1)}\}² = k²

∴ L.H.S. = R.H.S. \quad \therefore(\frac{a + b}{b + c})² = \frac{a² + b²}{b^2 + c²} Proved.

(ii) a² b² c²(\frac{1}{a³} + \frac{1}{b³} + \frac{1}{c³}) = a3 + b3 + c3

Solution:

To prove, a² b² c²(\frac{1}{a³} + \frac{1}{b³} + \frac{1}{c³}) = a3 + b3 + c3

\text { L.H.S. } = {a}² b² c²(\frac{1}{a³} + \frac{1}{b³} + \frac{1}{c³})

= (c k²)² \cdot(c k)² \cdot c²[\frac{1}{(c k²)³} + \frac{1}{(c k)³} + \frac{1}{c³}]

= c^6 k^6[\frac{1}{c³ k^6} + \frac{1}{c³ k³} + \frac{1}{c³}]

= c3 + c3 k3 + c3 k6

R.H.S. = a3 + b³ + c3

= (ck²)3 + (c k)3 + c3

= c3k6 + c³ k3 + c3

= c3 + c3 k3 + c3k6

∴ L.H.S. = R.H.S.

a² b² c²(\frac{1}{a³} + \frac{1}{b³} + \frac{1}{c³}) = a³ + b³ + c³ \text { Proved. }

\text { (iii) } \frac{a b c(a + b + c)³}{(a b + b c + c a)³} = 1

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

Solution:

To prove \frac{a b c(a + b + c)³}{(a b + b c + c a)³} = 1

L.H.S. = \frac{a b c(a + b + c)³}{(a b + b c + c a)³}

= \frac{c k² \cdot c k \cdot c(c² + c k + c)³}{(c k² \cdot c k + c k \cdot c + c \cdot c k²)³}

= \frac{c³ k³ \cdot c³(k² + k + 1)³}{(c² k³ + c² k + c² k²)³} = \frac{c^6 k³ \cdot c³(k² + k + 1)³}{\{c² k(k² + k + 1)\}³}

= \frac{c^6 k³(k² + k + 1)³}{c^6 k³(k² + k + 1)³} = 1 = R.H.S.

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

Question 5 (i)

If a, b, c, d are in continued proportion, let us prove that (i) (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²

Solution:

(a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)² as a, b, c, d are continued proportional.

Let \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k( \text{ where } k \neq 0).

= c = d k ; b = c k = d k \cdot k = d k² ; a = b k = d k² \cdot k = d k²

L.H.S. = (a² + b² + c²) (b² + c² + d²)

= \{(d k³)² + (d k²)² + (d k)²\}\{(d k²)² + (d k)² + (d)²\}

= (d² k6 + d² k4 + d² k²) (d² k4 + d² k² + d²)

= d² k²(k4 + k² + 1) d²(k4 + k² + 1)

= d4 k²(k4 + k² + 1)²

R.H.S = (ab + bc + cd)²

= (d k^2 \cdot d k^2 + d k^2 \cdot d k + d k \cdot d)^2

= (d²k5 + d²k³ + d²k)²

= \{d^2 k(k^4 + k^2 + 1)\}^2

= d4 k²(k4 + k² + 1)²

L.H.S = R.H.S Proved.

(ii) (b – c)² + (c – a)² + (b – d)² = (a – d)²

Solution:

To prove, (b – c)² + (c – a)² + (b – d)² = (a – d)²

L.H.S. = (b – c)² + (c – a)² + (b – d)²

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

= \{d k(k - 1)\}² + \{d k(1 - k²)\}² + \{d(k² - 1)\}²

= d²k²(k² – 2 k + 1) + d² k²(1 – 2k² + k4) + d²(k4 – 2 k3 + 1)

= d²(k4 – 2 k3 + k² + k² – 2 k4 + k6 + k4 – 2 k² + 1)

= d²(k6 – 2 k3 + 1) = d²(k3 – 1)²

R.H.S = (a – d)² = (d k² – d)² = d²(k³ – 1)²

L.H.S. = R.H.S Proved.

∴ L.H.S. = R.H.S Proved.

Question 6

(i) If \frac{m}{a} = \frac{n}{b}, let us show that (m² + n²) (a2 + b²) = (am + bn)².

Solution:

If \frac{m}{a} = \frac{n}{b}, prove that (m² + n²) (a2 + b²) = (am + bn)².

Let, \frac{m}{a} = \frac{n}{b} = k(where \ k \ \neq 0)

∴ m = ak, n = bk

L.H.S. = (m² + n²) (a2 + b²)

= (a² k² + b² k²) (a² + b²) = k²(a² + b²) (a² + b²) = \{k(a² + b²)\}²

R.H.S. = (am + bm)²

= (a.ak + b.bk)² = (a²k + b²k)² = \{k(a² + b²)\}²

∴ L.H.S. = R.H.S. Proved.

(ii) If \frac{a}{b} = \frac{x}{y}, let us show that (a + b) (a² + b²) x3 = (x + y) (x² + y²) a3

Solution:

If \frac{a}{b} = \frac{x}{y}, prove that (a + b) (a² + b²) x3 = (x + y) (x² + y²) a3

Let, \frac{\mathrm{a}}{\mathrm{b}} = \frac{\mathrm{x}}{\mathrm{y}} = \mathrm{k}( where \mathrm{k} \neq 0)

∴ a = bk & x = yk

L.H.S. = (a + b) (a² + b²) x³

= (bk + b) (b²k² + b²) (yk)³ = b(k + 1) b²(k² + 1) y³k³

= b³ k³ y³(k + 1) (k² + 1)

R.H.S. = (x + y) (x² + y²) a³

= (yk + y) (y² k² + y²) (b³ k³) = y(k + 1) y²(k² + 1) b³ k³

= b³ k³ y³(k + 1) (k² + 1)

∴ L.H.S. = R.H.S. Proved.

(iii) If, \frac{x}{l m - n²} = \frac{y}{m n - l²} = \frac{z}{n l - m²}, let us show that lx + my + nz = 0.

Solution:

If \frac{x}{lm - n²} = \frac{y}{m n - l²} = \frac{z}{nl - m²}, prove that, mid x + my + nz = 0

Let, \frac{x}{lm - n²} = \frac{y}{m n - l²} = \frac{z}{n l - m²} = k( where \ k \neq 0)

∴ x = k(lm – n²) ; y = k(m n – l²) ; z = k(nl – m²)

Now, lx + my + nz

= lk(lm – n²) + m k(m n – l²) + n k(nl – m²)

= k[ l² m – l² + m² n – lm + ln² – m² n]

= k × 0 = 0 Proved.

(iv) If \frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}, let us show that (b – c)x + (c – a)y + (a – b)x = 0

Solution:

If \frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c} prove that (b – c)x + (c – a) + y + (a – b) z = 0

Let \frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c} = k( where \ k \neq 0).

∴ x = k(b + c – a)

y = k(c + a – b)

z = k(a + b – c)

L.H.S. = (b – c) x + (c – a) y + (a – b) z

= (b – c) (b + c – a) k + (c – a) (c + a – b) k + (a – b) (a + b – c) k

= k[b² – c² – a b + a c + c² – a² – b c + a b + (a² – b²) – a c + b c]

= k × 0 = 0 = R.H.S. Proved.

(v) If \frac{x}{y} = \frac{a + 2}{a - 2}, let us show that \frac{x² - y²}{x² + y²} = \frac{4 a}{a² + 4}.

Solution:

If \frac{x}{y} = \frac{a + 2}{a - 2}, prove that \frac{x² - y²}{x + y²} = \frac{4 a}{a² + 4}.

Squaring both sides,

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

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

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

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

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

(vi) If x=\frac{8 a b}{a+b}, let us write by calculating the value of \left(\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}\right)

Solution:

x =\frac{8 a b}{a+b} , find the value of \frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}

Given, \frac{x}{1}=\frac{8 a b}{a+b}

\frac{x}{4 a} =\frac{2 b}{a+b}

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

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

Again, \frac{x}{1}=\frac{8 a b}{a+b}

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

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

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

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

=\frac{3 b+a-3 a-b}{b-a}

= \frac{2 b - 2 a}{b - a}

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

\frac{x + 4 a}{x - 4 a} + \frac{x + 4 b}{x - 4 b} = 2 Ans.

Question 7

(i) If \frac{a}{3} = \frac{b}{4} = \frac{c}{7}, let us show that \frac{a + b + c}{c} = 2.

Solution:

If \frac{a}{3} = \frac{b}{4} = \frac{c}{7}, show that \frac{a + b + c}{c} = 2

Let \frac{a}{3} = \frac{b}{4} = \frac{c}{7} = k( where \ k \neq 0).

∴ a = 3k ; b = 4k, c = 7k

Now, \frac{a + b + c}{c} = \frac{3 k + 4 k + 7 k}{7 k} = \frac{14 k}{7 k} = 2 Proved.

(ii) If \frac{a}{q - r} = \frac{b}{r - p} = \frac{c}{p - q}, let us show that a + b + c = 0 = pa + qb + rc.

Solution:

If \frac{a}{q - r} = \frac{b}{r - p} = \frac{c}{p - q}, show that a + b + c = 0 = pa + qb + rc

Let \frac{a}{q - r} = \frac{b}{r - p} = \frac{c}{p - q} = k( where k \neq 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 \times 0 = 0 Proved.

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

= k [pq – pr + qr – pq + pr – qr]

= k × 0 = 0 Proved.

(iii) If \frac{a x + b y}{a} = \frac{b x + a y}{b}, let us show that each ratio is equal to x.

Solution:

If \frac{a x + b y}{a} = \frac{b x + a y}{b}, then show that each ratio = x.

\frac{a x + b y}{a} = \frac{b x + a y}{b}

\frac{a b x + a b y}{a^2} = \frac{b^2 x + a b y}{b^2} = \frac{a^2 x + a b y + b^2 x - a b y}{a^2 + b^2} = \frac{x\left(a^2 + b^2\right)}{\left(a^2 + b^2\right)} = x \text { Proved. }

Question 8

(i) If \frac{a + b}{b + c} = \frac{c + d}{d + a} , let us prove that c = a or, a + b + c + d = 0 .

Solution:

\frac{a + b}{b + c} = \frac{c + d}{d + a}

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

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

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

(a + b + c + d)\left(\frac{1}{c + d} - \frac{1}{a + d}\right)

∴ Either a + b + c + d = 0 ……………… (ii)

or, \frac{1}{c + d} = \frac{1}{a + d}

or, c + d = a + d \quad \therefore c = a ……………… (i)

(ii) If \frac{x}{b + c} = \frac{y}{c + a} = \frac{z}{a + b}, let us show that \frac{a}{y + z - x} = \frac{b}{z + x - y} = \frac{c}{x + y - z}

Solution:

If \frac{x}{b + c} = \frac{y}{c + a} = \frac{z}{a + b} = k

∴ x = k(b + c), y = y = k(c + a), z = k(a + b)

Now, \frac{a}{y + z - x} = \frac{a}{k(c + a + b - b - c)} = \frac{a}{k \cdot 2 a} = \frac{1}{2 k}

\frac{b}{z + x - y} = \frac{b}{k(a + b + b + c - c - a)} = \frac{b}{k \cdot 2 b} = \frac{1}{2 k}

\frac{c}{x + y - z} = \frac{c}{k(b + c + c + a - a - b)} = \frac{c}{k \cdot 2 c} = \frac{1}{2 k}

\frac{a}{y + z - x} = \frac{b}{z + x - y} = \frac{c}{x + y - z} Proved.

(iii) If \frac{x + y}{3 a - b} = \frac{y + z}{3 b - c} = \frac{z + x}{3 c - a}, let us show that \frac{x + y + z}{a + b + c} = \frac{a x + b y + c z}{a^{2} + b^{2} + c^{2}}

Solution:

If \frac{x + y}{3 a - b} = \frac{y + z}{3 b - c} = \frac{z + x}{3 c - a}, prove that \frac{x + y + z}{a + b + c} = \frac{a x + b y + c z}{a^{2} + b^{2} + c^{2}}

Let \frac{x + y}{3 a - b} = \frac{y + z}{3 b - c} = \frac{z + x}{3 c - a} = k( where \ k \neq 0).

∴ x + y = k(3a – b)

x + z = k (3b – c)

z + x = k (3c – a)

Adding, 2 (x + y + z) = k (3a – b + 3b – c + 3c – a)

2 (x + y + z) = k.2 (a + b + c)

∴ x + y + z = k (a + b + c)

Subtracting (2) from (4), we get

x = k (a – 2b +  c)

Similarly, y = k (b – 2c + 2a) and z = k (c – 2a + 2b)

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

Again, \frac{a x + b y + c z}{a^2 + b^2 + c^2} = \frac{a k (a - 2 b + 2 c) + b k (b - 2 c + 2 a) + c k (c - 2 a + 2 b)}{\left (a^2 + b^2 + c^2\right)}

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

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

\frac{x + y + z}{a + b + c} = \frac{a x + b y + c z}{a^2 + b^2 + c^2} Proved.

(iv) If \frac{x}{a} = \frac{y}{b} = \frac{z}{c}, let us 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:

If \frac{x}{a} = \frac{y}{b} = \frac{z}{c}, prove 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}

Let \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k ( where \ k \neq 0).

∴ x = ak ; y = bk ; & = ck

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

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

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

\frac{x^2 - y z}{a - b c} = \frac{y^2 - z x}{b^2 - c a} = \frac{z^2 - x y}{c^2 - a b} Proved.

Question 9

(i) If \frac{3 x + 4 y}{3 u + 4 v} = \frac{3 x + 4 y}{3 u - 4 v}, let us show that \frac{x}{y} = \frac{u}{v}.

Solution:

If \frac{3 x + 4 y}{3 u + 4 v} = \frac{3 x + 4 y}{3 u - 4 v} prove that \frac{x}{y} = \frac{u}{v}

Ans. \frac{3 x + 4 y}{3 u + 4 v} = \frac{3 x + 4 y}{3 u - 4 v}

or, \frac{3 x + 4 y}{3 x - 4 y} = \frac{3 u + 4 v}{3 u - 4 v}

or, \frac{3 x + 4 y + 3 x - 4 y}{3 x + 4 y - 3 x + 4 y} = \frac{3 u + 4 v + 3 u - 4 v}{3 u + 4 v - 3 u + 4 v}

or, \frac{6 x}{8 y} = \frac{6 u}{8 v}

\frac{x}{y} = \frac{u}{v} Proved.

(ii) If (a + b + c + d) : (a + b – c – d) = (a – b + c – d) : (a – b – c + d), let us prove that a : b = c : d.

Solution:

If (a + b + c + d) : (a + b – c – d) = (a – b + c – d) : (a – b – c + d), prove that a : b = c : d

Ans. \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}

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

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

or, ac – ad + bc – bd = ac + ad – bc – bd

\frac{a}{b} = \frac{c}{d}, i.e., a : b = c : d Proved.

Question 10

(i) If \frac{a^2}{b + c} = \frac{b^2}{c + a} = \frac{c^2}{a + b} = 1, let us show that \frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c} = 1.

Solution : If \frac{a^2}{b + c} = \frac{b^2}{c + a} = \frac{c^2}{a + b} = 1, prove that \frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c} = 1

\frac{a^2}{b + c} = \frac{b^2}{c + a} = \frac{c^2}{a + b} = 1

∴ a2 = b + c ; b2 = c + a ; c2 = a + b

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

= \frac{a}{a + a^2} + \frac{b}{b + b^2} + \frac{1}{c + c^2}

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

= \frac{a + b + c}{a + b + c} = 1 Proved

(ii) If x2 : (b y + c z) = y2 : (c z + a x) = z2 : (a x + b y) = 1, let us prove that \frac{a}{a + x} + \frac{b}{b + y} + \frac{c}{c + z} = 1

Solution:

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

\frac{x^2}{b y + c z} = \frac{y^2}{c z + a x} = \frac{z^2}{a x + b y} = 1

x^2 = b y + c z ; y^2 = c z + a x ; z^2 = ax + by

L.H.S = \frac{a}{a + x} + \frac{b}{b + y} + \frac{c}{c + z}

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

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

= \frac{a x + b y + c z}{a x + b y + c z} = 1 Proved.

Question 11

(i) 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 \neq 0, let us show that each ratio is equal to \frac{1}{a + b + c}.

Solution:

Given, \frac{x}{x a + y b + z c} = \frac{y}{y a + z b + x c} = \frac{z}{z a + x b + y c} & x + y + z \neq 0. Prove that each ratio = \frac{1}{a + b + c}

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

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

(ii) If \frac{x^{2} - y z}{a} = \frac{y^{2} - z x}{b} = \frac{z^{2} - x y}{c}, let us prove that (a + b + c) (x + y + z) = ax + by + cz.

Solution:

If \frac{x^{2} - y z}{a} = \frac{y^{2} - z x}{b} = \frac{z^{2} - x y}{c} , prove that (a + b + c) (x + y + z) = ax + by + cz .

Let \frac{x^{2} - y z}{a} = \frac{y^{2} - z x}{b} = \frac{z^{2} - x y}{c} = \frac{1}{k}(\text { where } k \neq 0) .

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

L.H.S. = (a + b + c) (x + y + z) = \left\{k\left(x^{2} - y z + y^{2} - z x + z^{2} - x y\right)\right\}(x + y + z)

= k(x + y + z)\left(x^{2} + y^{2} + z^{2} - x y - y z - z x\right) = k\left(x^{3} + y^{3} + z^{3} - 3 x y z\right)

R.H.S. = ax + by + cz = k\left(x^{2} - y z\right) x + k\left(y^{2} - z x\right) y + k\left(z^{2} - x y\right) z

= k\left(x^{3} - x y z + y^{3} - x y z + z^{3} - x y z\right)

= k\left(x^{3} + y^{3} + z^{3} - x y z\right)

∴ L.H.S = R.H.S. Proved.

(iii) If \frac{a}{y + z} = \frac{b}{z + x} = \frac{c}{x + y} , let us prove that \frac{a(b - c)}{y^{2} - z^{2}} = \frac{b(c - a)}{z^{2} - x^{2}} = \frac{c(a - b)}{x^{2} - y^{2}}

Solution:

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

Let, \frac{a}{y + z} = \frac{b}{z + x} = \frac{c}{x + y} = k( where k \neq 0) \therefore a = k(y + z) ; b = k(z + x) ; c = k(x + y)

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

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

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

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

## 12. Very short answer type questions (V.S.A) :

Question 1

The fourth proportional of 3, 4 and 6 is

1. 8
2. 10
3. 12
4. 24

Solution:

The 4th proportional of 3, 4 & 6 = \frac{4 × 6}{3} = 8 – – – – – – – – – – – – (a)

Question 2

The 3rd proportional of 8 and 12 is

1. 12
2. 16
3. 18
4. 20

Solution:

The 3rd proportional of 8 & 12 = \frac{12 × 12}{8} = 18 – – – – – – – – – – – – (c)

Question 3

The mean proportional of 16 and 25 is

1. 400
2. 100
3. 20
4. 40

Solution:

The mean proportional of 16 & 25 = \sqrt{16 × 25} = 4 × 5 = 20 – – – – – – – – – – – – (c)

Question 4

a is a positive number and if a : \frac{27}{64} = \frac{3}{4} : a, then the value of a is

1. \frac{81}{256}
2. 9
3. \frac{9}{16}
4. \frac{16}{9}

Solution:

If a : \frac{27}{64} = \frac{3}{4} : a

a = \sqrt{\frac{27}{64} × \frac{3}{4}} = \sqrt{\frac{81}{256}} = \frac{9}{16} – – – – – – – – – – – – (c)

Question 5

If 2a = 3b = 4c, then a : b : c is

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

Solution:

If 2a = 3b = 4c, find a : b : c

or, \frac{2 a}{12} = \frac{3 b}{12} = \frac{4 c}{12}

or, \frac{a}{6} = \frac{b}{4} = \frac{c}{c} \quad \therefore a : b : c = 6 : 4 : 3 – – – – – – – – – – – – (d)

## B. Let us write whether true of false :

(i) Compound ratio of ab : c2, bc : a2 and ca : b2 is 1 : 1

Solution:

\frac{a b}{c^2} × \frac{b c}{a^2} × \frac{c a}{b^2} = 1 : 1 – – – – – – – – – True.

(ii) x3y, x2y2 and xy3 are continued proportional.

Solution:

If \frac{x^3 y}{x^2 y^2} = \frac{x^2 y^2}{x y^3}

or, \frac{x}{y} = \frac{x}{y} – – – – – – – – – – – –  True.

## C. Let us fill in the blanks :

(i) If the product of three positive consecutive numbers is 64 , then their mean proportional is _______.

Solution:

4.

(ii) If a : 2 = b : 5 = c : 8, then 50% of a = 20% of b = % of c.

Solution:

12.5%

(iii) If mean proportional of (x – 2) and (x – 3) is x, then the value of x is

Solution:

x = 6/5

## 13. Short Answers (S.A.) :

(i) If \frac{a}{2} = \frac{b}{3} = \frac{c}{4} = \frac{2 a - 3 b + 4 c}{p}, let us find the value of p.

Solution:

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

∴ a = 2k, b = 3k & c = 4k

∴ 2a – 3b + 7c = pk

or, 2.2k – 3.3k + 4.4k =k

11k = p.k \quad \therefore p = 11.

(ii) If \frac{3 x - 5 y}{3 x + 5 y} = \frac{1}{2}, let us find the value of \frac{3 x^2 - 5 y^2}{3 x^2 + 5 y^2}.

Solution:

If \frac{3 x - 5 y}{3 x + 5 y} = \frac{1}{2} or, 6x – 10y = 3x + 5y

or, 6x – 3x = 10y + 5y \quad or, 3x + 5y \quad \therefore x = 5y

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

(iii) If a : b = 3 : 4 and x : y = 5 : 7, let us find the value of (3ax – by) : (4by – 7ax)

Solution:

\frac{a}{b} = \frac{3}{4} \& \frac{x}{y} = \frac{5}{7}

∴ a = 3k, b = 4k & x = 5p, y = 7p

(3ax – by) : (4by – 7ax)

= (3.3k × 5p – 4k × 7p) : (4.4k × 7p – 7.3k × 5p)

= (4kp – 28kp) : (112kp – 105kp)

= 17kp : 7kp = 17 : 7

(iv) If x, 12, y, 27 are in continued proportion, let us find the positive value of x and

Solution:

\frac{x}{12} = \frac{12}{y} = \frac{y}{27}, \quad \therefore y = \sqrt{12 × 27} = \sqrt{324} = 18

x = \frac{144}{18} = 6

∴ x = 6 & y = 18 .

(v) If a : b = 3 : 2, and b : c = 3 : 2  let us find the value of (a + b) : (b + c).

Solution:

a : b = 3 : 2 & b : c = 3 : 2

= 9 : 6 = 6 : 4

∴ a = 9k, b = 6k, c = 4k

a + b : b + c = (9k + 6k) : (6k + 4k)

= 15k : 10k

= 3 : 2

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