ML Aggarwal Solution Class 10 Chapter 7 Ratio and Proportion Exercise 7.1

 Exercise 7.1

Question 1

An alloy consists of $27 \frac{1}{2}$ kg of copper and $2 \frac{3}{4}$ kg of tin. Find the ratio by weight of tin to the alloy

Sol :

Copper $=27 \frac{1}{2} \mathrm{~kg}=\frac{55}{2} \mathrm{~kg}$

Tin $=2 \frac{3}{4} \mathrm{~kg}=\frac{11}{4} \mathrm{~kg}$

Total alloy$=\frac{55}{2}+\frac{11}{4}=\frac{110+11}{4}=\frac{121}{4}$ kg

Now Ratio between tin and alloy$=\frac{11}{4} \mathrm{~kg}: \frac{121}{4} \mathrm{~kg}$

=11:121

=1:11

Question 2

Find the compounded ratio of:

(i) 2 : 3 and 4 : 9

(ii) 4 : 5, 5 : 7 and 9 : 11

(iii) (a – b) : (a + b), $(a+b)^{2}:\left(a^{2}+b^{2}\right)$ and $\left(a^{4}-b^{4}\right):\left(a^{2}-b^{2}\right)^{2}$

Sol :

(i) 2 : 3 and 4 : 9

Compound ratio $=\frac{2}{3} \times \frac{4}{9}$

$=\frac{8}{27}$ or 8: 27

(ii) 4 : 5, 5 : 7 and 9 : 11

Compound ratio $=\frac{4}{5} \times \frac{5}{7} \times \frac{9}{11}=\frac{36}{77}$

or 36: 77

(iii) $(a-b):(a+b),(a+b)^{2}:\left(a^{2}+b^{2}\right)$

and $\left(a^{4}-b^{4}\right):\left(a^{2}-b^{4}\right)^{2}$

Compound ratio

$=\frac{a-b}{a+b} \times \frac{(a+b)^{2}}{a^{2}+b^{2}} \times \frac{a^{4}-b^{4}}{\left(a^{2}-b^{2}\right)^{2}}$

$=\frac{a-b}{a+b} \times \frac{(a+b)(a+b)}{a^{2}+b^{2}}\times \frac{\left(a^{2}+b^{2}\right)(a+b)(a-b)}{(a+b)^{2}(a-b)^{2}}$

$=\frac{1}{1}$ or 1: 1

Question 3

Find the duplicate ratio of

(i) 2 : 3

(ii) √5 : 7

(iii) 5a : 6b

Sol :

(i) Duplicate ratio of 2 : 3 $=(2)^{2}:(3)^{2}=4 \cdot 9$

(ii) Duplicate ratio of √5 : 7 $=(\sqrt{5})^{2}:(7)^{2}=5: 49$

(iii) Duplicate ratio of 5a : 6b $=(5 a)^{2}:(6 b)^{2}=25 a^{2}: 36 b^{2}$

Question 4

Find the triplicate ratio of

(i) 3 : 4

(ii) $\frac{1}{2}: \frac{1}{3}$

(iii) $1^{3}: 2^{3}$

Sol :

(i) Triplicate ratio of 3 : 4

$=(3)^{3}:(4)^{3}$

=27 : 64

(ii) Triplicate ratio of $\frac{1}{2}: \frac{1}{3}=\left(\frac{1}{2}\right)^{3}:\left(\frac{1}{3}\right)^{3}$

$=\frac{1}{8}: \frac{1}{27}$

=27: 8

(iii) Triplicate ratio of $1^{3}: 2^{3}=\left(1^{3}\right)^{3}:\left(2^{3}\right)^{3}$

$=(1)^{3}:(8)^{3}$

=1 : 512

Question 5

Find the sub-duplicate ratio of

(i) 9 : 16

(ii) $\frac{1}{4}: \frac{1}{9}$

(iii) $9 a^{2}: 49 b^{2}$

Sol :

(i) Sub-duplicate ratio of 9 : 16

= √9 : √16

= 3 : 4

$=\sqrt{9}: \sqrt{16}=3: 4$

(ii) Sub-duplicate ratio of $\frac{1}{4}: \frac{1}{9}=\sqrt{\frac{1}{4}}: \sqrt{\frac{1}{9}}$

$=\frac{1}{2}: \frac{1}{3}=3: 2$

(iii) Sub-duplicate ratio of $9 a^{2}: 49 b^{2}$

$=\sqrt{9 a^{2}}: \sqrt{49 b^{2}}$

=3 a: 7 b

Question 6

Find the sub-triplicate ratio of

(i) 1 : 216

(ii) $\frac{1}{8}: \frac{1}{125}$

(iii) $27 a^{3}: 64 b^{3}$

Sol :

(i) Sub-triplicate ratio of 1 : 216

$=\sqrt[3]{1}: \sqrt[3]{216}$

$=\left(1^{3}\right)^{\frac{1}{3}}:\left(6^{3}\right)^{\frac{1}{3}}=1: 6$

(ii) Sub-triplicate ratio of $\frac{1}{8}: \frac{1}{125}$

$=\left(\frac{1}{8}\right)^{\frac{1}{3}}:\left(\frac{1}{125}\right)^{\frac{1}{3}}$

$=\left[\left(\frac{1}{2}\right)^{3}\right]^{\frac{1}{3}}:\left[\left(\frac{1}{5}\right)^{3}\right]^{\frac{1}{3}}$

$=\frac{1}{2}: \frac{1}{5}$

=5: 2

(iii) Sub-triplicate ratio of $27 a^{3}: 64 b^{3}$

$=\left[(3 a)^{3}\right]^{\frac{1}{3}}:\left[(4 b)^{3}\right]^{\frac{1}{3}}$

=3a: 4b

Question 7

Find the reciprocal ratio of

(i) 4 : 7

(ii) $3^{2}: 4^{2}$

(iii) $\frac{1}{9}: 2$

Sol :

(i) Reciprocal ratio of 4 : 7 = 7 : 4

(ii) Reciprocal ratio of $3^{2}: 4^{2}=4^{2}: 3^{2}$

=16: 9

(iii) Reciprocal ratio of $\frac{1}{9}: 2=2: \frac{1}{9}=18: 1$

Question 8

Arrange the following ratios in ascending order of magnitude:

2 : 3, 17 : 21, 11 : 14 and 5 : 7

Sol :

Writing the given ratios in fraction

$\frac{2}{3}, \frac{17}{21}, \frac{11}{14}, \frac{5}{7}$

L.C.M of 3,21,14,7=42

Converting the given ratio as equivalent

$\frac{2}{3}=\frac{2 \times 14}{3 \times 14}=\frac{28}{42$

$\frac{17}{21}=\frac{17 \times 2}{21 \times 2}=\frac{34}{42}$

$\frac{11}{14}=\frac{11 \times 3}{14 \times 3}=\frac{33}{42}$

$\frac{5}{7}=\frac{5 \times 6}{7 \times 6}=\frac{30}{42}$

From above, writing in ascending order;

$\frac{28}{42}, \frac{30}{42}, \frac{33}{42}, \frac{34}{42}$ or 

$\frac{2}{3}, \frac{5}{7}, \frac{11}{14}, \frac{17}{21}$

or 2:3 ; 5:7; 11:14 and 17:21

Question 9

(i) If A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7, find A : D

(ii) If x : y = 2 : 3, and y : z = 4 : 7, find x : y : z

Sol :

(i)

Let A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7

$\frac{A}{B}=\frac{2}{3}, \frac{B}{C}=\frac{4}{5}, \frac{C}{D}=\frac{6}{7}$

Multiplying $\frac{A}{B} \times \frac{B}{C} \times \frac{C}{D}=\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7}$

$\therefore \frac{\mathrm{A}}{\mathrm{D}}=\frac{16}{35}$

$\Rightarrow \mathrm{A}: \mathrm{D}=16: 35 \mathrm{Ans}$

(ii) L.C.M of y's terms 3 and 4=12

Making equals of y as 12

$\frac{x}{y}=\frac{2}{3}=\frac{2 \times 4}{3 \times 4}=\frac{8}{12}$ or 8: 12

$\frac{y}{z}=\frac{4}{7} \times \frac{3}{3}=\frac{12}{21}$ or 12: 21

Then x : y : z=8:12:21

Question 10

(i) If $A: B=$\frac{1}{4}: \frac{1}{5}$and B : C$\frac{1}{7}: \frac{1}{6}$, find A : B : C

(ii) If 3A = 4B = 6C, find A : B : C

Sol :

$A: B=\frac{1}{4} \times \frac{5}{1}=\frac{5}{4}$

$B: C=\frac{1}{7} \times \frac{6}{1}=\frac{6}{7}$

LCM of B's terms 4 and 6=12

 Making terms of B's; as 12

$\frac{A}{B}=\frac{5 \times 3}{4 \times 3}=\frac{15}{12}=15: 12$

$\frac{B}{C}=\frac{6 \times 2}{7 \times 2}=\frac{12}{14}=12: 14$

∴ A: B: C=15: 12: 14

(ii) 3A=4B

$\Rightarrow \frac{A}{B}=\frac{4}{3}$

or A : B= 4 : 3

and 4B=6C

$\Rightarrow \quad \frac{B}{C}=\frac{6}{4}=\frac{3}{2}$ 

or B: C=3: 2

∴A : B : C=4 : 3 : 2

Question 11

(i) If $\frac{3 x+5 y}{3 x-5 y}=\frac{7}{3}$, Find x : y

(ii) If a : b=3 : 11, find (15a-3b) : (9a+5b).a

Sol :

(i) $\frac{3 x+5 y}{3 x-5 y}=\frac{7}{3}$

⇒ 9x + 15y = 21x – 35y [By cross multiplication]

⇒ 21x – 9x = 15y + 35y

⇒12 x=50y 

$\Rightarrow \frac{x}{y}=\frac{50}{12}=\frac{25}{6}$

Hence, x : y=25 : 6

(ii) a: b=3: 11 or $\frac{a}{b}=\frac{3}{11}$

Now $\frac{15 a-3 b}{9 a+5 b}=\frac{\frac{15 a}{b}-\frac{3 b}{b}}{\frac{9 a}{b}+\frac{5 b}{b}}$

[Dividing by b]

$=\frac{\frac{15 a}{b}-3}{\frac{9 a}{b}+5}=\frac{15 \times \frac{3}{11}-3}{9 \times \frac{3}{11}+5}$

$\left(\right. \text{ Substituting the value of  }\left.\frac{a}{b}\right)$

$=\frac{\frac{45}{11}-3}{\frac{27}{11}+5}$

$=\frac{\frac{45-33}{11}}{\frac{27+55}{11}}$

$=\frac{\frac{12}{11}}{\frac{82}{11}}$

$=\frac{12}{11} \times \frac{11}{82}$

$=\frac{12}{82}=\frac{6}{41}$

∴(15a-3b) : (9a+5b)

=6:41

Question 12

(i) If (4x² + xy) : (3xy – y²) = 12 : 5, find (x + 2y) : (2x + y).

(ii) If y (3x – y) : x (4x + y) = 5 : 12. Find (x² + y²) : (x + y)².

Sol :

(4x² + xy) : (3xy – y²) = 12 : 5

$\Rightarrow \frac{4 x^{2}+\infty y}{3 x y-y^{2}}=\frac{12}{5}$

⇒ 20x² + 5xy = 36xy – 12y²

$\Rightarrow 20 x^{2}+5 x y-36 x y+12 y^{2}=0$

$\Rightarrow \quad 20 x^{2}-31 x y+12 y^{2}=0$

$\Rightarrow \quad \frac{20 x^{2}}{y^{2}}-\frac{31 x y}{y^{2}}+\frac{12 y^{2}}{y^{2}}=0$ (Dividing by $y^{2}$ )

$\Rightarrow 20\left(\frac{x}{y}\right)^{2}-31\left(\frac{x}{y}\right)+12=0$

$\Rightarrow 20\left(\frac{x}{y}\right)^{2}-15\left(\frac{x}{y}\right)-16\left(\frac{x}{y}\right)+12=0$

$\Rightarrow 5\left(\frac{x}{y}\right)\left[4\left(\frac{x}{y}\right)-3\right]-4\left[4\left(\frac{x}{y}\right)-3\right]=0$

$\Rightarrow\left[4\left(\frac{x}{y}\right)-3\right]\left[5\left(\frac{x}{y}\right)-4\right]=0$

Either $4\left(\frac{x}{y}\right)-3=0,$ then $4\left(\frac{x}{y}\right)=3$

$\Rightarrow \frac{x}{y}=\frac{3}{4}$

or $5\left(\frac{x}{y}\right)-4=0$

then $5\left(\frac{x}{y}\right)=4$

$\Rightarrow \quad \frac{x}{y}=\frac{4}{5}$

Now $\frac{x+2 y}{2 x+y}=\frac{\frac{x}{y}+2}{2 \frac{x}{y}+1}$ (Dividing by y)

(a) When $\frac{x}{y}=\frac{3}{4},$ then

$=\frac{\frac{x}{y}+2}{2 \frac{x}{y}+1}$

$=\frac{\frac{3}{4}+2}{2 \times \frac{3}{4}+1}$

$=\frac{\frac{11}{4}}{\frac{3}{2}+1}$

$=\frac{\frac{11}{4}}{\frac{5}{2}}$

$=\frac{11}{4} \times \frac{2}{5}=\frac{11}{10}$

∴(x+2y) : (2x+y)=11 : 10

(b) When $\frac{x}{y}=\frac{4}{5},$ then

$\frac{x+2 y}{2 x+y}$

$=\frac{\frac{x}{y}+2}{2 \frac{x}{y}+1}$

$=\frac{\frac{4}{5}+2}{2 \times \frac{4}{5}+1}$

$=\frac{\frac{14}{5}}{\frac{8}{5}+1}$

[Dividing by y]

$=\frac{\frac{14}{5}}{\frac{13}{5}}=\frac{14}{5} \times \frac{5}{13}=\frac{14}{13}$

Hence $\frac{x+2 y}{2 x+y}=\frac{11}{10}$ or $\frac{14}{13}$

$\therefore \quad \frac{x+2 y}{2 x+y}$

=11: 10 or 14: 13

(ii) If y(3 x-y): x(4 x+y)=5: 12

Find $\left(x^{2}+y^{2}\right):(x+y)^{2}$

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

$\Rightarrow 36 x y-12 y^{2}=20 x^{2}+5 x y$

$\Rightarrow 20 x^{2}+5 x y-36 x y+12 y^{2}=0$

$\Rightarrow 20 x^{2}-31 x y+12 y^{2}=0$

$\Rightarrow 20 \frac{x^{2}}{y^{2}}-31 \frac{x y}{y^{2}}+\frac{12 y^{2}}{y^{2}}=0$

[Dividing by $y^{2}$]

$\Rightarrow 20\left(\frac{x^{2}}{y^{2}}\right)-31\left(\frac{x y}{y^{2}}\right)+12=0$

$\Rightarrow 20\left(\frac{x}{y}\right)^{2}-15\left(\frac{x}{y}\right)-16\left(\frac{x}{y}\right)+12=0$

$\Rightarrow 5\left(\frac{x}{y}\right)\left[4\left(\frac{x}{y}\right)-3\right]-4\left[4\left(\frac{x}{y}\right)-3\right]=0$

$\Rightarrow\left[4\left(\frac{x}{4}\right)-3\right]\left[5\left(\frac{x}{y}\right)-4\right]=0$

Either $\left[4\left(\frac{x}{y}\right)-3\right]=0$

then $4\left(\frac{x}{y}\right)=3 \Rightarrow \frac{x}{y}=\frac{3}{4}$

or $\left[5\left(\frac{x}{y}\right)-4\right]=0$

then $5\left(\frac{x}{y}\right)=4 \Rightarrow \frac{x}{y}=\frac{4}{5}$

(a) When $\frac{x}{y}=\frac{3}{4}$

then $\left(x^{2}+y^{2}\right):(x+y)^{2}$

$=\frac{x^{2}+y^{2}}{(x+y)^{2}}=\frac{\frac{x^{2}}{y^{2}}+\frac{y^{2}}{y^{2}}}{\frac{1}{y^{2}}(x+y)^{2}}$

(Dividing by $y^{2}$)

$=\frac{\frac{x^{2}}{y^{2}}+1}{\left(\frac{x}{y}+1\right)^{2}}$

$=\frac{\left(\frac{3}{4}\right)^{2}+1}{\left(\frac{3}{4}+1\right)^{2}}=\frac{\frac{9}{16}+1}{\left(\frac{7}{4}\right)^{2}}$

$=\frac{\frac{25}{16}}{\frac{49}{16}}=\frac{25}{16} \times \frac{16}{49}=\frac{25}{49}$

$\because\left(x^{2}+y^{2}\right):(x+y)^{2}=25: 49$

(b) $\because\left(x^{2}+y^{2}\right):(x+y)^{2}=25: 49$

$\frac{x^{2}+y^{2}}{(x+y)^{2}}$

$=\frac{\frac{x^{2}}{y^{2}}+1}{\left(\frac{x}{y}+1\right)^{2}}$

$=\frac{\left(\frac{x}{y}\right)^{2}+1}{\left(\frac{x}{y}+1\right)^{2}}$

$=\frac{\left(\frac{4}{5}\right)^{2}+1}{\left(\frac{4}{5}+1\right)^{2}}$

$=\frac{\frac{16}{25}+1}{\left(\frac{9}{5}\right)^{2}}$

$=\frac{\frac{41}{25}}{\frac{81}{25}}$

$=\frac{41}{25} \times \frac{25}{81}=\frac{41}{81}$

$\because\left(x^{2}+y^{2}\right):(x+y)^{2}=41: 81$

Question 13

(i) If (x – 9) : (3x + 6) is the duplicate ratio of 4 : 9, find the value of x.

(ii) If (3x + 1) : (5x + 3) is the triplicate ratio of 3 : 4, find the value of x.

(iii) If (x + 2y) : (2x – y) is equal to the duplicate ratio of 3 : 2, find x : y.

Sol :

(i) $\frac{x-9}{3 x+6}=\left(\frac{4}{9}\right)^{2}$

$\Rightarrow \frac{x-9}{3 x+6}=\frac{16}{81}$

$\Rightarrow 81 x-729=48 x+96$

$\Rightarrow 81 x-48 x=96+729$

$\Rightarrow 33 x=825$

$\Rightarrow x=\frac{825}{33}=25$

(ii) If (3 x+1):(5 x+3) is the triplicate ratio of 3: 4

then $\frac{3 x+1}{5 x+3}=\frac{(3)^{3}}{(4)^{3}}=\frac{27}{64}$

$\Rightarrow 64(3 x+1)=27(5 x+3)$

$\Rightarrow 192 x+64=135 x+81$

$\Rightarrow 192 x-135 x=81-64$

$\Rightarrow 57 x=17$

$\Rightarrow x=\frac{17}{57}$

Hence $x=\frac{17}{57}$

(iii) If (x+2 y):(2 x-y) is the duplicate ratio of 3: 2

then $\frac{x+2 y}{2x-y}=\frac{(3)^{2}}{(2)^{2}}=\frac{9}{4}$

$\Rightarrow 9(2 x-y)=4(x+2 y)$

$\Rightarrow 18 x-9 y=4 x+8 y$

$\Rightarrow 18 x-4 x=8 y+9 y$

$\Rightarrow 14 x-17 y$

$\Rightarrow \quad \frac{x}{y}=\frac{17}{14}$

$\therefore x: y=17: 14$

Question 14

(i) Find two numbers in the ratio of 8 : 7 such that when each is decreased by $12 \frac{1}{2}$ , they are in the ratio 11 : 9.

(ii) The income of a man is increased in the ratio of 10 : 11. If the increase in his income is Rs 600 per month, find his new income.

Sol :

(i) The ratio is 8 : 7

Let the numbers be 8x and 7x,

According to condition,

$\frac{8 x-\frac{25}{2}}{7 x-\frac{25}{2}}=\frac{11}{9}$

$\Rightarrow \frac{\frac{16 x-25}{2}}{\frac{14 x-25}{2}}=\frac{11}{9}$

$\Rightarrow \frac{(16 x-25) \times 2}{2(14 x-25)}=\frac{11}{9}$

$\Rightarrow \frac{16 x-25}{14 x-25}=\frac{11}{9}$

$\Rightarrow 154 x-275=144 x-225$

$\Rightarrow 154 x-144 x=275-225$

$\Rightarrow 10 x=50$

$\therefore x=\frac{50}{10}=5$

∴ Numbers are 8x=8×5=40

and 7x=7×5=35

(ii) Let the present income=10x

then increased income=11x

∴increased per month=11x-10x=x

∴x=600

Now his new income=11x=11×600

=6600

Question 15

(i) A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 91 kg.

(ii) A school collected Rs 2100 for charity. It was decided to divide the money between an orphanage and a blind school in the ratio of 3 : 4. How much money did each receive?

Sol :

(i) Ratio between the original weight and reduced weight = 7 : 5

Let original weight = 7x

then reduced weight = 5x

If original weight = 91 kg.

then weight $=\frac{91 \times 5 x}{7 x}=65$ kg

(ii) Total amount to be distributed=2100

Ratio between orphanage and a blind school

=3 : 4

Sum of ratio=3+4=7

∴Orphanage school's share

$=₹ 2100 \times \frac{3}{7}$

=900

Blind school's share

$=₹ 2100 \times \frac{4}{7}=₹ 1200$

Question 16

(i) The sides of a triangle are in the ratio 7 : 5 : 3 and its perimeter is 30 cm. Find the lengths of sides.

(ii) If the angles of a triangle are in the ratio 2 : 3 : 4, find the angles.

Sol :

(i) Perimeter of a triangle = 30 cm.

Ratio among sides = 7 : 5 : 3

Sum of ratios 7 + 5 + 3 = 15

Length of first side $=30 \times \frac{7}{15}=14 \mathrm{~cm}$

Length of second side $=30 \times \frac{5}{15}=10 \mathrm{~cm}$

Length of third side $=30 \times \frac{3}{15}=6 \mathrm{~cm}$

∴ Sides are 14 cm, 10 cm, 6 cm

(ii) Sum of angles of a triangle=180°

Ratio among angles=2 : 3 : 4

Sum of ratios=2+3+4=9

∴First angle $=180^{\circ} \times \frac{2}{9}=40^{\circ}$

Second angle $=180^{\circ} \times \frac{3}{9}=60^{\circ}$

Third angle $=180^{\circ} \times \frac{4}{9}=80^{\circ}$

∴Angles are 40°, 60° and 80°

Question 17

Three numbers are in the ratio $\frac{1}{2}: \frac{1}{3}: \frac{1}{4}$ If the sum of their squares is 244, find the numbers.

Sol :

The ratio of three numbers $\frac{1}{2}: \frac{1}{3}: \frac{1}{4}$

$=\frac{6: 4: 3}{12}$

=.6 : 4 : 3

Let first number 6x, second 4x and third 3x

∴ According to the condition

$(6 x)^{2}+(4 x)^{2}+(3 x)^{2}=244$

$\Rightarrow 36 x^{2}+16 x^{2}+9 x^{2}=244$

⇒$61x^{2}=244$

$\Rightarrow x^{2}=\frac{244}{61}=4=(2)^{2}$

∴ x=2

∴ first number =6x=6×2=12

second number =4 x=4×2=8

and third number =3x=3×2=6

Question 18

(i) A certain sum was divided among A, B and C in the ratio 7 : 5 : 4. If B got Rs 500 more than C, find the total sum divided.

(ii) In a business, A invests Rs 50000 for 6 months, B Rs 60000 for 4 months and C, Rs 80000 for 5 months. If they together earn Rs 18800 find the share of each.

Sol :

(i) Ratio between A, B and C = 7 : 5 : 4

Let A’s share = 7x

B’s share = 5x

and C’s share = 4x

Total sum = 7x + 5x + 4x = 16x

Now, according to the condition

5x-4x=500

x=500

∴Total sum=16x=16×500=8000

(ii) A's 6 months investment =₹ 50000

∴ A's 1 month investment

=₹ 50000 × 6=₹ 300000

B's 4 month's investment =₹ 60000

∴ B's 1 month investment

=60000×4=₹ 240000

C's 5 months investment =₹ 80000

∴ C's 1 month investment

=₹ 80000×5=₹ 400000

∴ Ratio between their investments

=300000: 240000: 400000

=30: 24: 40

Sum of ratios =30+24+40=94 

Total earnings=₹ 18800

∴ A's share $=\frac{30}{94} \times 18800=₹ 6000$

B's share $=\frac{24}{94} \times 18800=₹ 4800$

C's share $=\frac{40}{94} \times 18800=₹ 8000$

Question 19

(i) In a mixture of 45 litres, the ratio of milk to water is 13 : 2. How much water must be added to this mixture to make the ratio of milk to water as 3 : 1 ?

(ii) The ratio of the number of boys to the number of girls in a school of 560 pupils is 5 : 3. If 10 new boys are admitted, find how many new girls may be admitted so that the ratio of the number of boys to the number of girls may change to 3 : 2.

Sol :

(i) Mixture of milk and water = 45 litres

Ratio of milk and water =13 : 2

Sum of ratio = 13 + 2 = 15

∴ Quantity of milk $=\frac{45 \times 13}{15}=39$ litres

and quantity of water $=45 \times \frac{2}{15}=6$ litres

Let x litre of water be added, then water =(6+x) litres

Now new ratio=3 : 1

∴39 : (6+x)=3 : 1

$\frac{39}{6+x}=\frac{3}{1}$

39=18+3x

3x=39-18=21

$\therefore x=\frac{21}{3}=7$ litres.

∴7 litres of water is to be added

(ii) ratio between boys and girls= 5 : 3

No. of pupils=560

Sum of ratios=5+3=8

∴No. of boys $=\frac{5}{8} \times 560=350$

and no. of girls $=\frac{3}{8} \times 560=210$

No. of new boys admitted =10

∴ Total number of boys =350+10=360

Let the no. of girls admitted =x

∴ Total number of girls =210+x

Now according to the condition,

360: 210+x=3: 2 

$\Rightarrow \frac{360}{210+x}=\frac{3}{2}$

⇒ 630+3 x=720

⇒ 3 x=720-630=90

$\therefore x=\frac{90}{3}=30$

∴No of girls to be admitted =30

Question 20

(i) The monthly pocket money of Ravi and Sanjeev are in the ratio 5 : 7. Their expenditures are in the ratio 3 : 5. If each saves Rs 80 every month, find their monthly pocket money.

(ii) In class X of a school, the ratio of the number of boys to that of the girls is 4 : 3. If there were 20 more boys and 12 less girls, then the ratio would have been 2 : 1, How many students were there in the class?

Sol :

(i) Let the monthly pocket money of Ravi and Sanjeev be 5x and 7x respectively.

Also, let their expenditure be 3y and 5y respectively.

So, 5 x-3 y=80..(i)

and 7 x-5 y=800..(ii)

Multiplying (i) by 7 and (ii) by 5 and subtracting, we get

$\begin{array}{r} 35 x-21 y=560 \\ 35 x-25 y=400 \\-\phantom{35 x}+\phantom{25 y}-\phantom{400 } \\ \hline 4 y=160 \end{array}$

From (i), 5 x=80+3×40=200

⇒ x=40

So, monthly pocket money of Ravi

=5×40=200

(ii) Let the number of students in the class =x

Ratio of boys and girls =4 : 3

∴No. of boys $=\frac{4 x}{7}$

and no. of girls $=\frac{3 x}{7}$

According to the problem,

$\left(\frac{4 x}{7}+20\right):\left(\frac{3 x}{7}-12\right)=2: 1$

$\frac{4 x+140}{7}: \frac{3 x-84}{7}: 2: 1$

$\Rightarrow \frac{4 x+140}{7} \times \frac{7}{3 x-84}=\frac{2}{1}$

$\Rightarrow \frac{4 x+140}{3 x-84}=\frac{2}{1}$

$\Rightarrow 6 x-168=4 x+140$

$\Rightarrow 6 x-4 x=140+168$

$\Rightarrow 2 x=308$

$\Rightarrow   x=\frac{308}{2}=154$

Hence, number of students =154

Question 21

In an examination, the ratio of passes to failures was 4 : 1. If 30 less had appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. How many students appeared for the examination

Sol :

Let the number of passes = 4x

and number of failures = x

The total number of students appeared = 4x + x = 5x

In the second case, the number of students appeared = 5x – 30

and number of passes = 4x – 20

∴No. of failures=(5x-30)-(4x-20)

=5x-30-4x+20

=x-10

According to the condition

$\frac{4 x-20}{x-10}=\frac{5}{1}$

$\Rightarrow 5 x-50=4 x-20$

$\Rightarrow 5 x-4 x=-20+50$

$\Rightarrow x=30$

∴Number of students appeared=5x

=5×30=150