RS Aggarwal 2020 2021 Solutions for Class 7 Maths Chapter 8 - Ratio And Proportion

Question 1:

Question 2:

(i) HCF of 24 and 40 is 8.
∴ 24 : 40 = $\frac{24}{40}= \frac{24 ÷ 8}{40 ÷ 8}= \frac{3}{5}$ = 3 : 5

Hence, 24 : 40 in its simplest form is 3 : 5.

(ii) HCF of 13.5 and 15 is 1.5.

 $$\begin{gathered} \frac{13.5}{15}=\frac{135}{150} \\ \mathrm{The} \mathrm{HCF} \mathrm{of} 135 \mathrm{and} 150 \mathrm{is} 15. \\ =\frac{135 ÷ 15}{150 ÷ 15}=\frac{9}{10} \end{gathered}$$

Hence, 13.5 : 15 in its simplest form is 9 : 10.

(iii)  $\frac{20}{3} : \frac{15}{2}=40 : 45$
The HCF of 40 and 45 is 5.

∴ 40 : 45 = $\frac{40}{45}=\frac{40 ÷ 5}{45 ÷ 5}=\frac{8}{9}$ = 8 : 9

Hence, $6\frac{2}{3} : 7\frac{1}{2}$ in its simplest form is 8 : 9

(iv) 9 : 6
The HCF of 9 and 6 is 3.
∴ 9 : 6 = $\frac{9}{6}=\frac{9 ÷ 3}{6 ÷ 3}$= 3 : 2
Hence, $\frac{1}{6}:\frac{1}{9}$ in its simplest form is 3 : 2.

(v) LCM of the denominators is 2.
∴ 4 : 5 : $\frac{9}{2}$ = 8 : 10 : 9
The HCF of these 3 numbers is 1.

∴ 8 : 10 : 9 is the simplest form.

(vi) 2.5 : 6.5 : 8 = 25 : 65 : 80

The HCF of 25, 65 and 80 is 5.
∴ 25 : 65 : 80 = $\frac{25}{{\displaystyle \frac{65}{80}}}= \frac{25 ÷ 5}{{\displaystyle \frac{65 ÷ 5}{80 ÷ 5}}}=\frac{5}{{\displaystyle \frac{13}{16}}}$ = 5 : 13 : 16

Question 3:

(i) Converting both the quantities into the same unit, we have:
   75 paise : (3 $\times$ 100) paise = 75 : 300

= $\frac{75}{300}= \frac{75 ÷ 75}{300 ÷ 75}=\frac{1}{4}$    (∵ HCF of 75 and 300 = 75)
= 1 paise : 4 paise

(ii)  Converting both the quantities into the same unit, we have:
105 cm : 63 cm = $\frac{105}{63}=\frac{105÷21}{63÷21}= \frac{5}{3}$   (∵ HCF of 105 and 63 = 21)
= 5 cm : 3 cm

(iii) Converting both the quantities into the same unit
65 min : 45 min = $\frac{65}{45}= \frac{65÷5}{45÷5}=\frac{13}{9}$  (∵ HCF of 65 and 45 = 5)
= 13 min : 9 min

(iv) Converting both the quantities into the same unit, we get:
8 months : 12 months = $\frac{8}{12}=\frac{8÷4}{12÷4}=\frac{2}{3}$  (∵ HCF of 8 and 12 = 4)
= 2 months : 3 months

(v) Converting both the quantities into the same unit, we get:

2250g : 3000 g = $\frac{2250}{3000}=\frac{2250÷750}{3000÷750}= \frac{3}{4}$    (∵ HCF of 2250 and 3000 = 750)

= 3 g : 4 g

(vi)  Converting both the quantities into the same unit, we get:
1000 m : 750 m =  $\frac{1000}{750}=\frac{1000÷250}{750÷250} = \frac{4}{3}$    (∵ HCF of 1000 and 750 = 250)
= 4 m : 3 m

Question 4:

$\frac{A}{B } = \frac{7}{5} \mathrm{and} \frac{B}{C} = \frac{9}{14}$

Therefore, we have:

$$\begin{gathered} \frac{A}{B}\times \frac{B}{C} = \frac{7}{5}\times \frac{9}{14} \\ \frac{A}{C} = \frac{9}{10} \end{gathered}$$

∴ A : C = 9 : 10

Question 5:

$$\begin{gathered} \frac{A}{B}=\frac{5}{8} \mathrm{and} \frac{B}{C} = \frac{16}{25} \\ \mathrm{Now}, \mathrm{we} \mathrm{have}:\frac{A}{B}\times \frac{B}{C} = \frac{5}{8}\times \frac{16}{25}\Rightarrow \frac{A}{C} = \frac{2}{5} \end{gathered}$$

∴ A : C = 2 : 5

Question 6:

A : B = 3 : 5

B : C = 10 : 13 =  $\frac{10÷2}{13÷2}=5 :\frac{13}{2}$

Now, A : B : C = 3 : 5 : $\frac{13}{2}$

∴ A : B : C = 6 : 10 : 13

Question 7:

We have the following:

A : B = 5 : 6
B : C = 4 : 7  = $\frac{4}{7} = \frac{4\times {\displaystyle \frac{6}{4}}}{7\times {\displaystyle \frac{6}{4}}}= 6 : \frac{21}{2}$

∴ A : B : C =  5 : 6 : $\frac{21}{2}$ =  10 : 12 : 21

Question 8:

Sum of the ratio terms  = 7 + 8 = 15

Now, we have the following:

Kunal's share = Rs 360 $\times \frac{7}{15}= 24\times 7$ = Rs 168

Mohit's share = Rs 360 $\times \frac{8}{15} = 24\times 8$ = Rs 192

Question 9:

Sum of the ratio terms = $\frac{1}{5}+\frac{1}{6}=\frac{11}{30}$

Now, we have the following:
Rajan's share = Rs 880 $\times \frac{{\displaystyle \frac{1}{5}}}{{\displaystyle \frac{11}{30}}} = \mathrm{Rs} 880 \times \frac{6}{11} = \mathrm{Rs} 80\times 6$  =  Rs 480
Kamal's share = Rs 880 $\times \frac{{\displaystyle \frac{1}{6}}}{{\displaystyle \frac{11}{30}}}= \mathrm{Rs} 880 \times \frac{5}{11}= \mathrm{Rs} 80 \times 5$ = Rs 400

Question 10:

Sum of the ratio terms is (1 + 3 + 4) = 8

We have the following:

A's share =  Rs 5600 $\times \frac{1}{8} =\mathrm{Rs} \frac{5600}{8 }= \mathrm{Rs} 700$

B's share =  Rs 5600 $\times \frac{3}{8}= \mathrm{Rs} 700 \times 3$ = Rs 2100

C's share = Rs 5600 $\times \frac{4}{8} =\mathrm{Rs} 700 \times 4$ = Rs 2800

Question 11:

Let x be the required number.
Then, (9 + x) : (16 + x) = 2 : 3

$$\begin{gathered} \Rightarrow \frac{9+x}{16+x} = \frac{2}{3} \\ \Rightarrow 27 + 3x = 32 + 2x\Rightarrow x =5 \end{gathered}$$

Hence, 5 must be added to each term of the ratio 9 : 16 to make it 2 : 3.

Question 12:

Suppose that x is the number that must be subtracted.
Then, (17 − x) : (33 − x) = 7 : 15

$$\begin{gathered} \Rightarrow \frac{17 - x}{33 - x}=\frac{7}{15} \\ \Rightarrow 255 - 15x = 231 - 7x \Rightarrow 8x = 255 - 231 = 24\Rightarrow x = 3 \end{gathered}$$

Hence, 3 must be subtracted from each term of ratio 17 : 33 so that it becomes 7 : 15.

Question 13:

Suppose that the numbers are 7x and 11x.

Then, (7x + 7) : (11x + 7) = 2 : 3
⇒ $\frac{7x + 7}{11x + 7}=\frac{2}{3}$

⇒ 21x + 21 = 22x + 14

⇒ x = 7

Hence, the numbers are (7 $\times$ 7 =) 49 and (11 $\times$ 7 =) 77.

Question 14:

Suppose that the numbers are 5x and 9x.
Then, (5x − 3) : (9x − 3) = 1 : 2

⇒ $\frac{5x - 3}{9x -3}=\frac{1}{2}$

⇒ 10x − 6 = 9x − 3
⇒ x = 3

Hence, the numbers are (5 $\times$ 3 =) 15 and (9 $\times$ 3 =) 27.

Question 15:

Let the numbers be 3x and 4x.
Their LCM is 12x.
Then, 12x = 180
⇒ x = 15

∴ The numbers are (3 $\times$ 15 =) 45 and (4 $\times$ 15 =) 60.

Question 16:

Suppose that the present ages of A and B are 8x yrs and 3x yrs.
Then, (8x  + 6) : (3x + 6) = 9 : 4
⇒ $\frac{8x+6}{3x+6}= \frac{9}{4}$
⇒ 32x + 24 = 27x + 54
⇒ 5x = 30
⇒ x = 6

Now, present age of A = 8 $\times$ 6 yrs = 48 yrs
Present age of  B = 3 $\times$ 6 yrs = 18 yrs

Question 17:

Suppose that the weight of zinc is x g.

Then, 48.6 : x = 9 : 5

⇒ x = $\frac{48.6\times 5}{9}=\frac{243}{9}$ = 27

Hence, the weight of zinc in the alloy is 27 g.

Question 18:

Suppose that the number of boys is x.
Then, x : 375 = 8 : 3

⇒ x = $\frac{8\times 375}{3}=8\times 125$ = 1000

Hence, the number of girls in the school is 1000.

Question 19:

Suppose that the monthly income of the family is Rs x.

Then, x : 2500 = 11 : 2

⇒ x = $\frac{11\times 2500}{2}=11\times 1250$
⇒ x = Rs 13750

Hence, the income is Rs 13,750.
∴ Expenditure = (monthly income − savings)
                         =Rs (13750 − 2500)
                          = Rs 11250

Question 20:

Let the numbers one rupee, fifty paise and twenty-five paise coins be 5x, 8x and 4x, respectively.

Total value of these coins = ($5x \times \frac{100}{100}+ 8x\times \frac{50}{100} + 4x \times \frac{25}{100}$)

 $$\begin{gathered} \Rightarrow 5x + \frac{8x}{2} + \frac{4x}{4} \\ = \frac{20x + 16x + 4x}{4}=\frac{40x}{4}=10x \end{gathered}$$

However, the total value is Rs 750.
∴ 750 = 10x
⇒ x = 75

Hence, number of one rupee coins = 5 $\times$ 75 = 375
Number of fifty paise coins = 8 $\times$ 75 = 600
Number of twenty-five paise coins = 4 $\times$ 75 = 300

Question 21:

(4x + 5) : (3x + 11) = 13 : 17

$$\begin{gathered} \Rightarrow \frac{4x+ 5}{3x + 11}=\frac{13}{17} \\ \Rightarrow 68x + 85 = 39x + 143\Rightarrow 29x = 58\Rightarrow x = 2 \end{gathered}$$

Question 22:

$$\begin{gathered} \frac{x}{y }= \frac{3}{4} \\ \Rightarrow x=\frac{3y}{4} \end{gathered}$$

Now, we have (3x + 4y) : (5x + 6y)
$$\begin{gathered} =\frac{3x +4y}{5x + 6y}=\frac{3\times {\displaystyle \frac{3y}{4}}+4y}{5\times {\displaystyle \frac{3y}{4}}+6y} \\ = \frac{9y+16y}{15y +24y } = \frac{25y}{39y}=\frac{25}{39} \end{gathered}$$

= 25 : 39

Question 23:

$$\begin{gathered} \frac{x}{y} = \frac{6}{11} \\ \Rightarrow x = \frac{6y}{11} \end{gathered}$$

Now, we have:

$$\begin{gathered} \frac{8x -3y}{3x + 2y} \\ = \frac{8\times {\displaystyle \frac{6y}{11}} -3y}{3\times {\displaystyle \frac{6y}{11}}+2y} \\ =\frac{48y-33y}{18y + 22y} \\ =\frac{15y}{40y}=\frac{3}{8} \end{gathered}$$

∴ (8x − 3y) : (3x + 2y) = 3 : 8

Question 24:

Suppose that the numbers are 5x and 7x.
The sum of the numbers is 720.
i.e., 5x + 7x = 720
⇒ 12x = 720
⇒ x = 60

Hence, the numbers are (5 $\times$ 60 =) 300 and (7 $\times$ 60 =) 420.

Question 25:

(i) The LCM of 6 and 9 is 18.

$$\begin{gathered} \frac{5}{6}=\frac{5\times 3}{6\times 3}=\frac{15}{18} \\ \frac{7}{9}=\frac{7\times 2}{9\times 2}=\frac{14}{18}\mathrm{Clearly}, \frac{14}{18}<\frac{15}{18} \end{gathered}$$

∴ (7 : 9) $<$ (5 : 6)

(ii)  The LCM of 3 and 7 is 21.

$$\begin{gathered} \frac{2}{3}=\frac{2\times 7}{3\times 7}=\frac{14}{21} \\ \frac{4}{7}=\frac{4\times 3}{7\times 3}=\frac{12}{21} \\ \mathrm{Clearly}, \frac{12}{21}<\frac{14}{21} \end{gathered}$$

∴ (4 : 7) $<$ (2 : 3)

(iii)  The LCM of 2 and 7 is 14.

$$\begin{gathered} \frac{1\times 7}{2\times 7}=\frac{7}{14} \\ \frac{4\times 2}{7\times 2}= \frac{8}{14} \end{gathered}$$

Clearly, $\frac{7}{14}<\frac{8}{14}$

∴ (1 : 2) $<$ (4 : 7)

(iv) The LCM of 5 and 13 is 65.

$$\begin{gathered} \frac{3}{5}=\frac{3\times 13}{5\times 13}= \frac{39}{65} \\ \frac{8}{13}= \frac{8\times 5}{13\times 5}= \frac{40}{65} \\ \mathrm{Clearly}, \frac{39}{65}<\frac{40}{65} \end{gathered}$$

∴ (3 : 5) $<$ (8 : 13)

Question 1:

(i) We have $\frac{5}{6}, \frac{8}{9} \mathrm{and} \frac{11}{18}.$
$$\begin{gathered} 2 \underline{\overline{) 6,9, 18}} \\ \underline{3\underline{ \overline{)3, 9}}, 9} \\ 3 \overline{)\underline{1,3, 3} } \\ \overline{)\underline{1,1, 1} } \end{gathered}$$

The LCM of 6, 9 and 18 is 18. Therefore, we have:

$$\begin{gathered} \frac{5}{6}= \frac{5\times 3}{6\times 3}=\frac{15}{18} \\ \frac{8}{9}= \frac{8\times 2}{9\times 2}=\frac{16}{18} \frac{11}{18}=\frac{11}{18}\mathrm{Clearly}, \frac{11}{18}<\frac{15}{18}<\frac{16}{18} \end{gathered}$$

Hence, (11 : 18) $<$ (5 : 6) $<$ (8 : 9)

(ii) $\mathrm{We} \mathrm{have} \frac{11}{14}, \frac{17}{21}, \frac{5}{7} \mathrm{and} \frac{2}{3}.$
$$\begin{gathered} 2 \underline{\overline{) 14,21,7, 3}} \\ \underline{7\underline{ \overline{)7,21,7, 3}}, } \\ 3 \overline{)\underline{1,3,1, 3} } \\ \overline{)\underline{1,1,1, 1} } \end{gathered}$$

The LCM of 14, 21, 7 and 3 is 42.

$$\begin{gathered} \frac{11}{14}=\frac{11\times 3}{14\times 3}=\frac{33}{28} \\ \frac{17}{21}=\frac{17\times 2}{21\times 2}=\frac{34}{42} \\ \frac{5}{7}=\frac{5\times 6}{7\times 6}=\frac{30}{42} \\ \frac{2}{3}=\frac{2\times 14}{3\times 14}=\frac{28}{42} \\ \mathrm{Clearly}, \frac{28}{42}<\frac{30}{42}<\frac{33}{28}<\frac{34}{42} \\ \mathrm{Hence}, (2 : 3) < (5 : 7) < (11 : 14) < (17 : 21) \end{gathered}$$

Question 2:

We have:

Product of the extremes = 30 $\times$ 60 = 1800
Product of the means = 40 $\times$ 45 = 1800
Product of extremes = Product of means

Hence, 30 : 40 :: 45 : 60

Question 3:

We have:
Product of the extremes = 36 $\times$ 7 = 252
Product of the means = 49 $\times$ 6 = 294
Product of the extremes $\ne$ Product of the means

Hence, 36, 49, 6 and 7 are not in proportion.

Question 4:

Product of the extremes = 2 $\times$ 27 = 54
Product of the means  = 9 $\times$x = 9x

Since 2 : 9 :: x : 27, we have:
Product of the extremes = Product of the means
⇒ 54 = 9x
⇒ x = 6

Question 5:

Product of the extremes = 8 $\times$ 35 = 280
Product of the means = 16 $\times$x = 16x

Since 8 : x :: 16 : 35, we have:
Product of the extremes = Product of the means
⇒ 280 = 16x
⇒ x = 17.5

Question 6:

Product of the extremes = x $\times$ 60 = 60x
Product of the means = 35 $\times$ 48 = 1680

Since x : 35 :: 48 : 60, we have:
Product of the extremes = Product of the means
⇒ 60x= 1680
⇒ x = 28

Question 7:

(i) Let the fourth proportional be x.
Then, 8 : 36 :: 6 : x

8 $\times x = 36 \times 6$                                   [Product of extremes = Product of means]
⇒ 8x = 216
⇒ x = 27
Hence, the fourth proportional is 27.

(ii) Let the fourth proportional be x.
Then, 5 : 7 :: 30 : x
⇒ $5 \times x = 7 \times 30$                                      [Product of extremes = Product of means]
⇒ 8x = 216
⇒ 5x = 210
⇒ x = 42

Hence, the fourth proportional is 42.

(iii) Let the fourth proportional be x.
Then, 2.8 $\times x =14 \times 3.5$                                [Product of extremes = Product of means]
⇒ 8x = 216
⇒ 2.8x = 49
⇒ x = 17.5
Hence, the fourth proportional is 17.5.

Question 8:

36, 54 and x are in continued proportion.
Then, 36 : 54 :: 54 : x
⇒ $36 \times x =54 \times 54$                                [Product of extremes = Product of means]
⇒ 36x = 2916
⇒ x = 81

Question 9:

27, 36 and x are in continued proportion.
Then, 27 : 36 :: 36 : x
⇒ $27\times x = 36 \times 36$         [Product of extremes = Product of means]
⇒ 27x = 1296
⇒ x = 48

Hence, the value of x is 48.

Question 10:

(i)  Suppose that x is the third proportional to 8 and 12.
Then, 8 :12 :: 12 : x
⇒ 8 $\times x = 12 \times 12$                                            (Product of extremes = Product of means )
⇒ 8x = 144
⇒ x = 18

Hence, the required third proportional is 18.

(ii) Suppose that x is the third proportional to 12 and 18.
Then, 12 : 18 :: 18 : x
⇒ $12 \times x = 18 \times 18$                                       (Product of extremes = Product of means )
⇒ 12x = 324
⇒ x = 27

Hence, the third proportional is 27.

(iii) Suppose that x is the third proportional to 4.5 and 6.
Then, 4.5 : 6:: 6 : x
⇒ $4.5 \times x= 6 \times 6$                                        (Product of extremes = Product of means )
⇒ 4.5x = 36
⇒ x = 8

Hence, the third proportional is 8.

Question 11:

The third proportional to 7 and x is 28.
Then, 7 : x :: x : 28
⇒ 7 $\times$ 28 = $x^2$           (Product of extremes = Product of means)
⇒ x = 14

Question 12:

(i)  Suppose that x is the mean proportional.
 
Then, 6 : x :: x : 24

⇒ $6 \times 24 = x \times x$                                         (Product of extremes = Product of means)
⇒ $x^{2 } = 144$
⇒ x = 12

Hence, the mean proportional to 6 and 24 is 12.

(ii)  Suppose that x is the mean proportional.

Then, 3 : x :: x : 27
$$\begin{gathered} \Rightarrow 3 \times 27 = x \times x \\ \Rightarrow x^2 = 81 \end{gathered}$$                                        (Product of extremes =Product of means)
⇒ x = 9

Hence, the mean proportional to 3 and 27 is 9.

(iii)  Suppose that x is the mean proportional.

Then, 0.4 : x :: x : 0.9

$$\begin{gathered} \Rightarrow 0.4 \times 0.9 = x \times x \\ \Rightarrow x^2 = 0.36 \end{gathered}$$                              (Product of extremes =Product of means)
⇒x = 0.6

Hence, the mean proportional to 0.4 and 0.9 is 0.6.

Question 13:

Suppose that the number is x.
 
Then, (5 + x) : ( 9 + x) :: (7 + x) : (12 + x)

$$\begin{gathered} \Rightarrow (5 + x) \times (12 + x) = (9 + x) \times (7 + x) (\mathrm{Product} \mathrm{of} \mathrm{extremes} = \mathrm{Product} \mathrm{of} \mathrm{means}) \\ \Rightarrow 60 +5 x + 12 x + x^2 = 63 + 9x + 7x + x^2 \\ \Rightarrow 60 + 17x = 63 + 16x \\ \Rightarrow x = 3 \end{gathered}$$

Hence, 3 must be added to each of the numbers: 5, 9, 7 and 12, to get the numbers which are in proportion.

Question 14:

Suppose that x is the number that is to be subtracted.

Then, (10 − x) : (12 − x) :: (19 − x) : (24 − x)

$$\begin{gathered} \Rightarrow (10- x) \times (24 - x) =(12 - x) \times (19 - x) (\mathrm{Product} \mathrm{of} \mathrm{extremes} =\mathrm{Product} \mathrm{of} \mathrm{means}) \\ \Rightarrow 240 - 10x -24x + x^2 = 228 - 12x -19x + x^2 \\ \Rightarrow 240 - 34x = 228 - 31x \\ \Rightarrow 3x = 12 \\ \Rightarrow x = 4 \end{gathered}$$
.
Hence, 4 must be subtracted from each of the numbers: 10, 12, 19 and 24, to get the numbers which are in proportion.

Question 15:

Distance represented by 1 cm on the map = 5000000 cm = 50 km

Distance represented by 3 cm on the map = 50 $\times$ 4 km = 200 km

∴ The actual distance is 200 km.

Question 1:

(Height of tree) : (height of its shadow) = (height of the pole) : (height of its shadow)

Suppose that the height of pole is x cm.

Then, 6 : 8 = x : 20

⇒ x = $\frac{6\times 20}{8} = 15$
∴ Height of the pole = 15 cm

Question 2:

The correct option is (d).


$$\begin{gathered} \frac{a}{c}= \frac{a}{b}\times \frac{b}{c} = \frac{3}{4}\times \frac{8}{9} \\ = \frac{2}{3} \end{gathered}$$

Hence, a : c = 2 : 3

Question 3:

(a) 15 : 8

$$\begin{gathered} \frac{A}{B}= \frac{2}{3} \\ \frac{B}{C}= \frac{4}{5} \\ \mathrm{Then}, \frac{A}{B}\times \frac{B}{C } = \frac{2}{3}\times \frac{4}{5}= \frac{8}{15} \\ \mathrm{Hence}, C : A = 15 : 8 \end{gathered}$$

Question 4:

The correct option is (d).


$$\begin{gathered} A = \frac{3B}{2} \\ C = \frac{4B}{5} \\ \therefore A : C = \frac{A}{C} = \frac{{\displaystyle \frac{3B}{2}}}{{\displaystyle \frac{4B}{5}}}= \frac{15}{8} \end{gathered}$$
Hence, A : C = 15 : 8

Question 5:

The correct option is (b).

$$\begin{gathered} \frac{15}{100}A =\frac{20}{100}B \\ \Rightarrow \frac{A}{B }=\frac{4}{3} \end{gathered}$$

Hence, A : B = 4 : 3

Question 6:

(a)  1 : 3 : 6

$$\begin{gathered} A = \frac{1}{3}B \\ C = 2B \\ \therefore A : B : C = \frac{1}{3}B : B : 2B = 1 : 3 : 6 \end{gathered}$$

Question 7:

(b)  30 : 42 : 77


$$\begin{gathered} \frac{A}{B} = \frac{5}{7} \\ \Rightarrow A = \frac{5B}{7}\frac{B}{C} = \frac{6}{11}\Rightarrow C =\frac{11B}{6} \\ \therefore A : B : C = \frac{5B}{ 7} : B : \frac{11B}{6} = 30 : 42 : 77 \end{gathered}$$

Question 8:

(c)  6 : 4 : 3

$$\begin{gathered} 2A=3B = 4C \\ \mathrm{Then}, A = \frac{3B}{2} \mathrm{and} C = \frac{3B}{4} \\ \therefore A : B : C = \frac{3B}{2} : B : \frac{3B}{4} = 6 : 4 : 3 \end{gathered}$$

Question 9:

(a) 3 : 4 : 5

$$\begin{gathered} A =\frac{3B}{4} \\ C = \frac{5B}{4} \\ \therefore A : B : C = \frac{3B}{4 } : B : \frac{5B}{4} \end{gathered}$$

                 = 3 : 4 : 5

Question 10:

(b)  15 : 10 : 6

$$\begin{gathered} \frac{1}{x }:\frac{1}{y}=2 : 3 \\ \mathrm{Then}, y : x = 2 : 3 \mathrm{and} y = \frac{2}{3}x \\ \frac{1}{y}:\frac{1}{z }= 3 : 5 \\ \mathrm{Then}, z : y =3 : 5 \mathrm{and} z = \frac{3}{5}y \\ \therefore x : y : z = x : \frac{2}{3}x : \frac{3}{5}y = x : \frac{2}{3}x: \frac{3}{5}\times \frac{2}{3}x \\ =x : \frac{2}{3}x : \frac{2}{5}x = 15 : 10 : 6 \end{gathered}$$

Question 11:

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

$$\begin{gathered} x = \frac{3y}{4} \\ \therefore \frac{7x + 3y}{7x - 3y} = \frac{7{\displaystyle \frac{3y}{4}}+3y}{7{\displaystyle \frac{3y}{4}} - 3y} \\ =\frac{21y + 12y}{21y - 12y} = \frac{33y}{9y}= \frac{11}{3} \end{gathered}$$

Hence, (7x + 3y) : (7x − 3y) = 11 : 3

The correct option is (c).