RS Aggarwal 2021-2022 for Class 6 Maths Solutions Chapter 10- Ratio, Proportion And Unitary Method
RS Aggarwal Class 6 Math Solution Chapter 10- Ratio, Proportion And Unitary Method, Exercise 10B is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.
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
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