RS Aggarwal Class 6 Maths Solutions Chapter 10- Ratio, Proportion And Unitary Method Exercise 10B

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Rs Aggarwal Class 6 Math Solution Chapter 10- Ratio, Proportion And Unitary Method

Exercise 10B

1. Determine if the following numbers are in proportion:

    (i) 4, 6, 8, 12

    (ii) 7, 42, 13, 78

    (iii) 33, 121, 9, 96

    (iv) 22, 33, 42, 63

    (v) 32, 48, 70, 210

    (vi) 150, 200, 250, 300

Solution: 

    (i) 4, 6, 8, 12

    `frac\{4}{6}` = `frac\{cancel 4^2}{cancel 6^3}` = `frac\{2}{3}`

    `frac\{8}{12}` = `frac\{cancel 8^2}{cancel 12^3}` = `frac\{2}{3}`

    ∵ `frac\{2}{3}` =  `frac\{2}{3}`

    Hence, 4 : 6 : : 8 : 12 are in proportion.

    (ii) 7, 42, 13, 78

    `frac\{7}{42}` = `frac\{cancel 7^1}{cancel 42^6}` = `frac\{1}{6}`

    `frac\{13}{78}` = `frac\{cancel 13^1}{cancel 78^6}` = `frac\{1}{6}`

    ∵ `frac\{1}{6}` =  `frac\{1}{6}`

    Hence, 7 : 42 : : 13 : 78 are in proportion.

    (iii) 33, 121, 9, 96

    `frac\{33}{121}` = `frac\{cancel 33^3}{cancel 121^11}` = `frac\{3}{11}`

    `frac\{9}{96}` = `frac\{cancel 9^3}{cancel 96^32}` = `frac\{3}{32}`

    ∵ `frac\{3}{11}` ≠  `frac\{3}{32}`

    Hence, 33 : 121 : : 9 : 96 are not in proportion.

    (iv) 22, 33, 42, 63

    `frac\{22}{33}` = `frac\{cancel 22^2}{cancel 33^3}` = `frac\{2}{3}`

    `frac\{42}{63}` = `frac\{cancel 42^2}{cancel 63^3}` = `frac\{2}{3}`

    ∵ `frac\{2}{3}` =  `frac\{2}{3}`

    Hence, 22 : 33 : : 42 : 63 are in proportion.

    (v) 32, 48, 70, 210

    `frac\{32}{48}` = `frac\{cancel 32^2}{cancel 48^3}` = `frac\{2}{3}`

    `frac\{70}{210}` = `frac\{cancel 70^1}{cancel 210^3}` = `frac\{1}{3}`

    ∵ `frac\{2}{3}` ≠  `frac\{1}{3}`

    Hence, 32 : 48 : : 70 : 210 are not in proportion.

    (vi) 150, 200, 250, 300

    `frac\{150}{200}` = `frac\{cancel 150^3}{cancel 200^4}` = `frac\{3}{4}`

    `frac\{250}{300}` = `frac\{cancel 250^5}{cancel 300^6}` = `frac\{5}{6}`

    ∵ `frac\{3}{4}` ≠  `frac\{5}{6}`

    Hence, 150 : 200 : : 250 : 300 are not in proportion.

2. Verify the following:

    (i) 60 : 105 : : 84 : 147

    (ii) 91 : 104 : : 119 : 136

    (iii) 108 : 72 : : 129 : 86

    (iv) 39 : 65 : : 141 : 235

Solution: 

    (i) 60 : 105 : : 84 : 147

    `frac\{60}{105}` = `frac\{cancel 60^4}{cancel 105^4}` = `frac\{4}{7}`

    `frac\{84}{147}` = `frac\{cancel 84^4}{cancel 147^7}` = `frac\{4}{7}`

    ∵ `frac\{4}{7}` =  `frac\{4}{7}`

    Hence, 150 : 200 : : 250 : 300 are in proportion.

    (ii) 91 : 104 : : 119 : 136

    `frac\{91}{104}` = `frac\{cancel 91^7}{cancel 104^8}` = `frac\{7}{8}`

    `frac\{119}{136}` = `frac\{cancel 119^7}{cancel 136^8}` = `frac\{7}{8}`

    ∵ `frac\{7}{8}` =  `frac\{7}{8}`

    Hence, 91 : 104 : : 119 : 136 are in proportion.

    (iii) 108 : 72 : : 129 : 86

    `frac\{108}{72}` = `frac\{cancel 108^3}{cancel 72^2}` = `frac\{3}{2}`

    `frac\{129}{8}` = `frac\{cancel 129^3}{cancel 86^2}` = `frac\{3}{2}`

    ∵ `frac\{3}{2}` =  `frac\{3}{2}`

    Hence, 108 : 72 : : 129 : 86 are in proportion.

    (iv) 39 : 65 : : 141 : 235

    `frac\{39}{65}` = `frac\{cancel 39^3}{cancel 65^5}` = `frac\{3}{5}`

    `frac\{141}{235}` = `frac\{cancel 141^3}{cancel 86^5}` = `frac\{3}{5}`

    ∵ `frac\{3}{5}` =  `frac\{3}{5}`

    Hence, 39 : 65 : : 141 : 235 are in proportion.

3. Find the value of x in each of the following proportions: 

    (i) 55 : 11 : : x : 6

    (ii) 27 : x : : 63  :84

    (iii) 51 : 85 : : 57 : x 

    (iv) x : 92 : : 87 :116

Solution: 

    (i) 55 : 11 : : x : 6

    Product of extremes = Product of means

      55 × 6 = 11 × x

⇒       11x = 330

⇒           x =  `frac\{cancel 330^3}{cancel 11}`  = 30

    (ii) 27 : x : : 63 : 84

     Product of extremes = Product of means

        27 ​× 84 = x ​× 63

 ⇒         63x = 2268

 ⇒             x =  `frac\{cancel 2268^36}{cancel 63}`  = 36

    (iii) 51: 85 : : 57 : x 

    Product of extremes = Product of means

         51 × x = 85 × 57

 ⇒         51x = 4845

 ⇒            x =   `frac\{cancel 4845^95}{cancel 51}`   = 95

    (iv) x : 92 : : 87 : 116

    Product of extremes = Product of means

          x ×  116 = 92 ​× 87

 ⇒          116x = 8004

 ⇒               x  = `frac\{cancel 8004^69}{cancel 116}` = 69

4. Write (T) for true and (F) for false in case of each of the following:

    (i) 51 : 68 : : 85 : 102

    (ii) 36 : 45 : : 80 : 100

    (iii) 30 bags :18 bags : : Rs 450 : Rs 270

    (iv) 81 kg : 45 kg : : 18 men : 10 men

    (v) 45 km : 60 km : : 12 h : 15 h

    (vi) 32 kg : Rs 36 : : 8 kg : Rs 9

Solution: 

    (i) 51 : 68 : : 85 : 102

   Product of means = 68 × 85 = 5780

   Product of extremes = 51 × 102 = 5202

   Product of means ≠ Product of extremes

    Hence, (F).

    (ii) 36 : 45 : : 80 : 100

  Product of means = 45 ​× 80 = 3600

  Product of extremes = 36 × 100 = 3600

  Product of means = Product of extremes 

   Hence, (T).

    (iii) 30 bags : 18 bags : : Rs 450 : Rs 270

       or 30 :18 : : 450 : 270

     Product of means = 18 × 450 = 8100

     Product of extremes = 30 ​× 270 = 8100

     Product of means = Product of extremes 

     Hence, (T).

    (iv) 81 kg : 45 kg : : 18 men : 10 men

      or 81 : 45 : : 18 : 10

     Product of means = 45 × 18 = 810

     Product of extremes = 81 × 10 = 810

     Product of means = Product of extremes

      Hence, (T).

    (v) 45 km : 60 km : : 12 h : 15 h

     or,45 : 60 : : 12 : 15

     Product of means = 60 × 12 = 720

     Product of extremes = 45 × 15 = 675

     Product of means ≠ Product of extremes 

      Hence, (F).

    (vi) 32 kg : Rs 36 : : 8 kg : Rs 9

     Product of means = 36 × 8 = 288

     Product of extremes = 32 × 9 = 288

     Product of means = Product of extremes

     Hence, (T).

5. Determine if the following ratios form a proportion:

    (i) 25 cm :1 m and Rs 40 : Rs 160

    (ii) 39 litres : 65 litres and 6 bottles : 10 bottles

    (iii) 200 mL : 2.5 L and Rs 4 : Rs 50

    (iv) 2 kg : 80 kg and 25 g : 625 kg 

Solution: 

    (i) 25 cm : 1 m and Rs 40 : Rs 160 

    (or) 25 cm:100 cm and Rs 40 : Rs 160

      `frac\{25}{100}`  =  `frac\{25 ​÷ 25}{100​ ÷ 25}`  = `frac\{1}{4}`  

    and  `frac\{40}{160}`  =  `frac\{40 ÷ 40}{160 ÷ 40}`  = `frac\{1}{4}`  

           Hence, they are in proportion.

    (ii) 39 litres : 65 litres and 6 bottles : 10 bottles

       `frac\{39}{65}`  =  `frac\{39 ​÷ 13}{65​ ÷ 13}`  = `frac\{3}{5}`  

    and  `frac\{6}{10}`  =  `frac\{6 ÷ 2}{10 ÷ 2}`  = `frac\{3}{5}`  

      Hence they are  in proportion.

    (iii) 200 mL : 2.5 L and Rs 4 : Rs 50 

    (or) 200 mL : 2500 mL and Rs 4 : Rs 50

        `frac\{200}{2500}`  =  `frac\{200 ​÷ 100}{2500​ ÷ 100}`  = `frac\{2}{25}`  

    and  `frac\{4}{50}`  =  `frac\{4 ÷ 2}{50 ÷ 2}`  = `frac\{2}{25}`  

     Hence, they are in proportion.

    (iv) 2 kg : 80 kg and 25 g : 625 kg  

    (or)  2 kg : 80 kg and 25 g : 625000 g

        `frac\{2}{80}`  =  `frac\{2 ​÷ 2}{80​ ÷ 2}`  = `frac\{1}{40}`  

    and  `frac\{25}{625000}`  =  `frac\{25 ÷ 25}{625000 ÷ 25}`  = `frac\{1}{25000}`  

        Hence, they are not in proportion.

6. In a proportion, the 1st, 2nd and 4th terms are 51, 68 and 108 respectively. Find the 3rd term.

Solution: 

    Let the 3rd term be x.

    Thus, 51: 68 : : x : 108

     We know:

     Product of extremes = Product of means

            51 × 108 = 68 × x

    ⇒          5508 = 68x

    ⇒          x =  `frac\{cancel 5508^81}{cancel 68}`  = 81

    Hence, the third term is 81.

7. The 1st, 3rd and 4th terms of a proportion are 12, 8 and 14 respectively. Find the 2nd term.

Solution: 

    Let the second term be x.

    Then. 12 : x : : 8 : 14

    We know:

      Product of extremes = Product of means

                     12 × 14 = 8x

      ⇒                 168 = 8x

​      ⇒                    x =   `frac\{cancel 168^21}{cancel 8}`  = 21

     Hence, the second term is 21.

8. Show that the following numbers are in continued proportion:

     (i) 48 , 60 , 75

    (ii) 36 , 90 , 225

    (iii) 16 , 84, 441

Solution: 

    (i) 48 : 60, 60 : 75

      Product of means = 60 × 60 = 3600

      Product of extremes = 48 × 75 = 3600

    Product of means = Product of extremes

       Hence, 48 : 60 : : 60 : 75 are in continued proportion.

    (ii) 36 : 90, 90 : 225

     Product of means = 90 × 90 = 8100

     Product of extremes = 36 × 225 = 8100

    Product of means = Product of extremes

      Hence,  36 : 90, 90 : 225 are in continued proportion.

    (iii) 16 : 84, 84 : 441

    Product of means = 84 × 84 = 7056

    Product of extremes = 16 × 441 = 7056

    Product of means = Product of extremes

    Hence, 16 : 84, 84 : 441 are in continued proportion.

                             

9. If 9, x, x 49 are in proportion, find the value of x.

Solution: 

    Given: 9 : x : : x : 49

    We know:

      Product of means = Product of extremes

              `x × x = 9 × 49`

    ⇒    `x^2 = 441`

    ⇒     `x^2 = (21)^2`

    ⇒         `x = 21`

10. An electric pole casts a shadow of length 20 m at a time when a tree 6 m high casts a shadow of length 8 m. Find the height of the pole.

Solution: 

    Let the height of the pole = x m

    Then, we have:

      x : 20 : : 6 : 8

    Now, we know:

   Product of extremes = Product of means

     8x = 20​ × 6

       x =  `frac\{cancel 120^15}{cancel 8}`  = 15

    ​Hence, the height of the pole is 15 m.

11. Find the value of x if 5 : 3 : : x : 6.

Solution: 

    5 :3 : : x : 6 

    We know:

   Product of means = Product of extremes

     3x = 5 ​× 6

  ⇒ x =  `frac\{cancel 30^10}{cancel 3}`  = 10

    ∴ x = 10