Open App

Class 6 Exam > Class 6 Notes > Mathematics (Maths) Class 6 > RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3)

Ratio, Proportion & Unitary Method (Exercise 9.3) RD Sharma Solutions | Mathematics (Maths) Class 6 PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


 
 
 
 
 
 
Exercise 9.3                                                                               page: 9.14 
1. Which of the following statements are true? 
(i) 16: 24 = 20: 30 
(ii) 21: 6 = 35: 10 
(iii) 12: 18 = 28: 12 
(iv) 51: 58 = 85: 102 
(v) 40 men: 200 men = Rs 5: Rs 25 
(vi) 99 kg: 45 kg = Rs 44: Rs 20 
Solution: 
 
(i) 16: 24 = 20: 30 
It can be written as 
16/24 = 20/30 
Dividing 16/24 by 4/4 and 20/30 by 5/5 
(16/24) ÷ (4/4) = (20/30) ÷ (5/5) 
On further calculation 
4/6 = 4/6 
We get 
2/3 = 2/3  
 
Hence, 16: 24 = 20: 30 is true. 
 
(ii) 21: 6 = 35: 10 
It can be written as 
21/6 = 35/10 
Dividing 21/6 by 3/3 and 35/10 by 5/5 
(21/6) ÷ (3/3) = (35/10) ÷ (5/5) 
On further calculation 
7/2 = 7/2 
 
Hence, 21: 6 = 35: 10 is true. 
 
(iii) 12: 18 = 28: 12 
It can be written as 
12/18 = 28/12 
On further calculation 
6/9 ? 14/6 
 
Hence, 12: 18 = 28: 12 is false. 
 
(iv) 51: 58 = 85: 102 
It can be written as 
51/58 = 85/102 
On further calculation 
51/58 ? 5/6 
 
Hence, 51: 58 = 85: 102 is false. 
 
(v) 40 men: 200 men = Rs 5: Rs 25 
Page 2


 
 
 
 
 
 
Exercise 9.3                                                                               page: 9.14 
1. Which of the following statements are true? 
(i) 16: 24 = 20: 30 
(ii) 21: 6 = 35: 10 
(iii) 12: 18 = 28: 12 
(iv) 51: 58 = 85: 102 
(v) 40 men: 200 men = Rs 5: Rs 25 
(vi) 99 kg: 45 kg = Rs 44: Rs 20 
Solution: 
 
(i) 16: 24 = 20: 30 
It can be written as 
16/24 = 20/30 
Dividing 16/24 by 4/4 and 20/30 by 5/5 
(16/24) ÷ (4/4) = (20/30) ÷ (5/5) 
On further calculation 
4/6 = 4/6 
We get 
2/3 = 2/3  
 
Hence, 16: 24 = 20: 30 is true. 
 
(ii) 21: 6 = 35: 10 
It can be written as 
21/6 = 35/10 
Dividing 21/6 by 3/3 and 35/10 by 5/5 
(21/6) ÷ (3/3) = (35/10) ÷ (5/5) 
On further calculation 
7/2 = 7/2 
 
Hence, 21: 6 = 35: 10 is true. 
 
(iii) 12: 18 = 28: 12 
It can be written as 
12/18 = 28/12 
On further calculation 
6/9 ? 14/6 
 
Hence, 12: 18 = 28: 12 is false. 
 
(iv) 51: 58 = 85: 102 
It can be written as 
51/58 = 85/102 
On further calculation 
51/58 ? 5/6 
 
Hence, 51: 58 = 85: 102 is false. 
 
(v) 40 men: 200 men = Rs 5: Rs 25 
 
 
 
 
 
 
It can be written as 
40/200 = 5/25 
We get 40/200 = 1/5 and 5/25 = 1/5 
 
Hence, 40 men: 200 men = Rs 5: Rs 25 is true. 
 
(vi) 99 kg: 45 kg = Rs 44: Rs 20 
It can be written as 
99/45 = 44/20 
Dividing the fraction by 9 
(99/45) ÷ (9/9) = (44/20) ÷ (9/9) 
On further calculation 
11/5 = 11/5 
 
Hence, 99 kg: 45 kg = Rs 44: Rs 20 is true. 
 
2. Find which of the following are in proportion: 
(i) 8, 16, 6, 12 
(ii) 6, 2, 4, 3 
(iii) 150, 250, 200, 300 
Solution: 
 
(i) 8, 16, 6, 12 
We know that 
8: 16 = 8/16 = 1/2 
6: 12 = 6/12 = 1/2 
So we get 8/16 = 6/12 
 
Therefore, 8, 16, 6, 12 are in proportion. 
 
(ii) 6, 2, 4, 3 
We know that 
6: 2 = 6/2 = 3/1 
4: 3 = 4/3  
So we get 3/1 ? 4/3 
 
Therefore, 6, 2, 4, 3 are not in proportion. 
 
(iii) 150, 250, 200, 300 
We know that 
150: 250 = 150/250 = 3/5 
200: 300 = 200/300 = 4/6 = 2/3 
So we get 3/5 ? 2/3  
 
Therefore, 150, 250, 200, 300 are not in proportion. 
 
3. Find x in the following proportions: 
(i) x: 6 = 55: 11 
(ii) 18: x = 27: 3 
(iii) 7: 14 = 15: x 
Page 3


 
 
 
 
 
 
Exercise 9.3                                                                               page: 9.14 
1. Which of the following statements are true? 
(i) 16: 24 = 20: 30 
(ii) 21: 6 = 35: 10 
(iii) 12: 18 = 28: 12 
(iv) 51: 58 = 85: 102 
(v) 40 men: 200 men = Rs 5: Rs 25 
(vi) 99 kg: 45 kg = Rs 44: Rs 20 
Solution: 
 
(i) 16: 24 = 20: 30 
It can be written as 
16/24 = 20/30 
Dividing 16/24 by 4/4 and 20/30 by 5/5 
(16/24) ÷ (4/4) = (20/30) ÷ (5/5) 
On further calculation 
4/6 = 4/6 
We get 
2/3 = 2/3  
 
Hence, 16: 24 = 20: 30 is true. 
 
(ii) 21: 6 = 35: 10 
It can be written as 
21/6 = 35/10 
Dividing 21/6 by 3/3 and 35/10 by 5/5 
(21/6) ÷ (3/3) = (35/10) ÷ (5/5) 
On further calculation 
7/2 = 7/2 
 
Hence, 21: 6 = 35: 10 is true. 
 
(iii) 12: 18 = 28: 12 
It can be written as 
12/18 = 28/12 
On further calculation 
6/9 ? 14/6 
 
Hence, 12: 18 = 28: 12 is false. 
 
(iv) 51: 58 = 85: 102 
It can be written as 
51/58 = 85/102 
On further calculation 
51/58 ? 5/6 
 
Hence, 51: 58 = 85: 102 is false. 
 
(v) 40 men: 200 men = Rs 5: Rs 25 
 
 
 
 
 
 
It can be written as 
40/200 = 5/25 
We get 40/200 = 1/5 and 5/25 = 1/5 
 
Hence, 40 men: 200 men = Rs 5: Rs 25 is true. 
 
(vi) 99 kg: 45 kg = Rs 44: Rs 20 
It can be written as 
99/45 = 44/20 
Dividing the fraction by 9 
(99/45) ÷ (9/9) = (44/20) ÷ (9/9) 
On further calculation 
11/5 = 11/5 
 
Hence, 99 kg: 45 kg = Rs 44: Rs 20 is true. 
 
2. Find which of the following are in proportion: 
(i) 8, 16, 6, 12 
(ii) 6, 2, 4, 3 
(iii) 150, 250, 200, 300 
Solution: 
 
(i) 8, 16, 6, 12 
We know that 
8: 16 = 8/16 = 1/2 
6: 12 = 6/12 = 1/2 
So we get 8/16 = 6/12 
 
Therefore, 8, 16, 6, 12 are in proportion. 
 
(ii) 6, 2, 4, 3 
We know that 
6: 2 = 6/2 = 3/1 
4: 3 = 4/3  
So we get 3/1 ? 4/3 
 
Therefore, 6, 2, 4, 3 are not in proportion. 
 
(iii) 150, 250, 200, 300 
We know that 
150: 250 = 150/250 = 3/5 
200: 300 = 200/300 = 4/6 = 2/3 
So we get 3/5 ? 2/3  
 
Therefore, 150, 250, 200, 300 are not in proportion. 
 
3. Find x in the following proportions: 
(i) x: 6 = 55: 11 
(ii) 18: x = 27: 3 
(iii) 7: 14 = 15: x 
 
 
 
 
 
 
(iv) 16: 18 = x: 96 
Solution: 
 
(i) x: 6 = 55: 11 
It can be written as 
x/6 = 55/11 
We get 
x/6 = 5/1 
On further calculation  
x = 5 (6) = 30 
 
(ii) 18: x = 27: 3 
It can be written as 
18/x = 27/3 
We get  
18/x = 9/1 
On further calculation 
x = 18/9 = 2 
 
(iii) 7: 14 = 15: x 
It can be written as 
7/14 = 15/x 
We get 
1/2 = 15/x 
On further calculation 
x = 15 (2) = 30 
 
(iv) 16: 18 = x: 96 
It can be written as 
16/18 = x/96 
We get 
8/9 = x/96 
On further calculation 
x = 8/9 (96) = 256/3 
 
4. Set up all proportions from the numbers 9, 150, 105, 1750. 
Solution: 
 
The proportions from the numbers are 
9: 150 = 3: 50 
9: 105 = 3: 35 
9: 1750  
150: 9 = 50: 3 
150: 105 = 10: 7 
150: 1750 = 3: 35 
105: 9 = 35: 3 
105: 150 = 7: 10 
105: 1750 = 3: 50 
1750: 9  
1750: 150 = 35: 3 
Page 4


 
 
 
 
 
 
Exercise 9.3                                                                               page: 9.14 
1. Which of the following statements are true? 
(i) 16: 24 = 20: 30 
(ii) 21: 6 = 35: 10 
(iii) 12: 18 = 28: 12 
(iv) 51: 58 = 85: 102 
(v) 40 men: 200 men = Rs 5: Rs 25 
(vi) 99 kg: 45 kg = Rs 44: Rs 20 
Solution: 
 
(i) 16: 24 = 20: 30 
It can be written as 
16/24 = 20/30 
Dividing 16/24 by 4/4 and 20/30 by 5/5 
(16/24) ÷ (4/4) = (20/30) ÷ (5/5) 
On further calculation 
4/6 = 4/6 
We get 
2/3 = 2/3  
 
Hence, 16: 24 = 20: 30 is true. 
 
(ii) 21: 6 = 35: 10 
It can be written as 
21/6 = 35/10 
Dividing 21/6 by 3/3 and 35/10 by 5/5 
(21/6) ÷ (3/3) = (35/10) ÷ (5/5) 
On further calculation 
7/2 = 7/2 
 
Hence, 21: 6 = 35: 10 is true. 
 
(iii) 12: 18 = 28: 12 
It can be written as 
12/18 = 28/12 
On further calculation 
6/9 ? 14/6 
 
Hence, 12: 18 = 28: 12 is false. 
 
(iv) 51: 58 = 85: 102 
It can be written as 
51/58 = 85/102 
On further calculation 
51/58 ? 5/6 
 
Hence, 51: 58 = 85: 102 is false. 
 
(v) 40 men: 200 men = Rs 5: Rs 25 
 
 
 
 
 
 
It can be written as 
40/200 = 5/25 
We get 40/200 = 1/5 and 5/25 = 1/5 
 
Hence, 40 men: 200 men = Rs 5: Rs 25 is true. 
 
(vi) 99 kg: 45 kg = Rs 44: Rs 20 
It can be written as 
99/45 = 44/20 
Dividing the fraction by 9 
(99/45) ÷ (9/9) = (44/20) ÷ (9/9) 
On further calculation 
11/5 = 11/5 
 
Hence, 99 kg: 45 kg = Rs 44: Rs 20 is true. 
 
2. Find which of the following are in proportion: 
(i) 8, 16, 6, 12 
(ii) 6, 2, 4, 3 
(iii) 150, 250, 200, 300 
Solution: 
 
(i) 8, 16, 6, 12 
We know that 
8: 16 = 8/16 = 1/2 
6: 12 = 6/12 = 1/2 
So we get 8/16 = 6/12 
 
Therefore, 8, 16, 6, 12 are in proportion. 
 
(ii) 6, 2, 4, 3 
We know that 
6: 2 = 6/2 = 3/1 
4: 3 = 4/3  
So we get 3/1 ? 4/3 
 
Therefore, 6, 2, 4, 3 are not in proportion. 
 
(iii) 150, 250, 200, 300 
We know that 
150: 250 = 150/250 = 3/5 
200: 300 = 200/300 = 4/6 = 2/3 
So we get 3/5 ? 2/3  
 
Therefore, 150, 250, 200, 300 are not in proportion. 
 
3. Find x in the following proportions: 
(i) x: 6 = 55: 11 
(ii) 18: x = 27: 3 
(iii) 7: 14 = 15: x 
 
 
 
 
 
 
(iv) 16: 18 = x: 96 
Solution: 
 
(i) x: 6 = 55: 11 
It can be written as 
x/6 = 55/11 
We get 
x/6 = 5/1 
On further calculation  
x = 5 (6) = 30 
 
(ii) 18: x = 27: 3 
It can be written as 
18/x = 27/3 
We get  
18/x = 9/1 
On further calculation 
x = 18/9 = 2 
 
(iii) 7: 14 = 15: x 
It can be written as 
7/14 = 15/x 
We get 
1/2 = 15/x 
On further calculation 
x = 15 (2) = 30 
 
(iv) 16: 18 = x: 96 
It can be written as 
16/18 = x/96 
We get 
8/9 = x/96 
On further calculation 
x = 8/9 (96) = 256/3 
 
4. Set up all proportions from the numbers 9, 150, 105, 1750. 
Solution: 
 
The proportions from the numbers are 
9: 150 = 3: 50 
9: 105 = 3: 35 
9: 1750  
150: 9 = 50: 3 
150: 105 = 10: 7 
150: 1750 = 3: 35 
105: 9 = 35: 3 
105: 150 = 7: 10 
105: 1750 = 3: 50 
1750: 9  
1750: 150 = 35: 3 
 
 
 
 
 
 
1750: 105 = 50: 3 
 
Hence, the proportions that are formed are  
9: 150 :: 105: 1750 
150: 9 :: 1750: 105 
1750: 150 :: 105: 9 
9: 105 :: 150: 1750 
 
5. Find the other three proportions involving terms of each of the following: 
(i) 45: 30 = 24: 16 
(ii) 12: 18 = 14: 21 
Solution: 
 
(i) 45: 30 = 24: 16 can be written as 3: 2 in simplest form 
So the other three proportions involving terms are 
45: 24 = 30: 16 can be written as 15: 8 in simplest form 
30: 45 = 16: 24 can be written as 2: 3 in simplest form 
16: 30 = 24: 45 can be written as 8: 15 in simplest form 
 
(ii) 12: 18 = 14: 21 can be written as 2: 3 in simplest form 
So the other three proportions involving terms are 
12: 14 = 18: 21 can be written as 6: 7 in simplest form 
21: 18 = 14: 12 can be written as 7: 6 in simplest form 
18: 12 = 21: 14 can be written as 3: 2 in simplest form 
 
6. If 4, x, 9 are in continued proportion, find the value of x. 
Solution: 
 
We know that 4, x, 9 are in continued proportion 
It can be written as 
4: x :: x: 9 
We get 
4/x = x/9 
On further calculation 
x
2
 = 9 (4) = 36 
So we get 
x = 6 
 
7. If in a proportion, the first, second and fourth terms are 32, 112 and 217 respectively, find the third 
term. 
Solution: 
 
It is given that in a proportion the first, second and fourth terms are 32, 112 and 217 
Consider x as the third term 
We can write it as 
32: 112 :: x: 217 
On further calculation 
32/112 = x/217 
So we get 
x = 32/112 (217) = 62 
Page 5


 
 
 
 
 
 
Exercise 9.3                                                                               page: 9.14 
1. Which of the following statements are true? 
(i) 16: 24 = 20: 30 
(ii) 21: 6 = 35: 10 
(iii) 12: 18 = 28: 12 
(iv) 51: 58 = 85: 102 
(v) 40 men: 200 men = Rs 5: Rs 25 
(vi) 99 kg: 45 kg = Rs 44: Rs 20 
Solution: 
 
(i) 16: 24 = 20: 30 
It can be written as 
16/24 = 20/30 
Dividing 16/24 by 4/4 and 20/30 by 5/5 
(16/24) ÷ (4/4) = (20/30) ÷ (5/5) 
On further calculation 
4/6 = 4/6 
We get 
2/3 = 2/3  
 
Hence, 16: 24 = 20: 30 is true. 
 
(ii) 21: 6 = 35: 10 
It can be written as 
21/6 = 35/10 
Dividing 21/6 by 3/3 and 35/10 by 5/5 
(21/6) ÷ (3/3) = (35/10) ÷ (5/5) 
On further calculation 
7/2 = 7/2 
 
Hence, 21: 6 = 35: 10 is true. 
 
(iii) 12: 18 = 28: 12 
It can be written as 
12/18 = 28/12 
On further calculation 
6/9 ? 14/6 
 
Hence, 12: 18 = 28: 12 is false. 
 
(iv) 51: 58 = 85: 102 
It can be written as 
51/58 = 85/102 
On further calculation 
51/58 ? 5/6 
 
Hence, 51: 58 = 85: 102 is false. 
 
(v) 40 men: 200 men = Rs 5: Rs 25 
 
 
 
 
 
 
It can be written as 
40/200 = 5/25 
We get 40/200 = 1/5 and 5/25 = 1/5 
 
Hence, 40 men: 200 men = Rs 5: Rs 25 is true. 
 
(vi) 99 kg: 45 kg = Rs 44: Rs 20 
It can be written as 
99/45 = 44/20 
Dividing the fraction by 9 
(99/45) ÷ (9/9) = (44/20) ÷ (9/9) 
On further calculation 
11/5 = 11/5 
 
Hence, 99 kg: 45 kg = Rs 44: Rs 20 is true. 
 
2. Find which of the following are in proportion: 
(i) 8, 16, 6, 12 
(ii) 6, 2, 4, 3 
(iii) 150, 250, 200, 300 
Solution: 
 
(i) 8, 16, 6, 12 
We know that 
8: 16 = 8/16 = 1/2 
6: 12 = 6/12 = 1/2 
So we get 8/16 = 6/12 
 
Therefore, 8, 16, 6, 12 are in proportion. 
 
(ii) 6, 2, 4, 3 
We know that 
6: 2 = 6/2 = 3/1 
4: 3 = 4/3  
So we get 3/1 ? 4/3 
 
Therefore, 6, 2, 4, 3 are not in proportion. 
 
(iii) 150, 250, 200, 300 
We know that 
150: 250 = 150/250 = 3/5 
200: 300 = 200/300 = 4/6 = 2/3 
So we get 3/5 ? 2/3  
 
Therefore, 150, 250, 200, 300 are not in proportion. 
 
3. Find x in the following proportions: 
(i) x: 6 = 55: 11 
(ii) 18: x = 27: 3 
(iii) 7: 14 = 15: x 
 
 
 
 
 
 
(iv) 16: 18 = x: 96 
Solution: 
 
(i) x: 6 = 55: 11 
It can be written as 
x/6 = 55/11 
We get 
x/6 = 5/1 
On further calculation  
x = 5 (6) = 30 
 
(ii) 18: x = 27: 3 
It can be written as 
18/x = 27/3 
We get  
18/x = 9/1 
On further calculation 
x = 18/9 = 2 
 
(iii) 7: 14 = 15: x 
It can be written as 
7/14 = 15/x 
We get 
1/2 = 15/x 
On further calculation 
x = 15 (2) = 30 
 
(iv) 16: 18 = x: 96 
It can be written as 
16/18 = x/96 
We get 
8/9 = x/96 
On further calculation 
x = 8/9 (96) = 256/3 
 
4. Set up all proportions from the numbers 9, 150, 105, 1750. 
Solution: 
 
The proportions from the numbers are 
9: 150 = 3: 50 
9: 105 = 3: 35 
9: 1750  
150: 9 = 50: 3 
150: 105 = 10: 7 
150: 1750 = 3: 35 
105: 9 = 35: 3 
105: 150 = 7: 10 
105: 1750 = 3: 50 
1750: 9  
1750: 150 = 35: 3 
 
 
 
 
 
 
1750: 105 = 50: 3 
 
Hence, the proportions that are formed are  
9: 150 :: 105: 1750 
150: 9 :: 1750: 105 
1750: 150 :: 105: 9 
9: 105 :: 150: 1750 
 
5. Find the other three proportions involving terms of each of the following: 
(i) 45: 30 = 24: 16 
(ii) 12: 18 = 14: 21 
Solution: 
 
(i) 45: 30 = 24: 16 can be written as 3: 2 in simplest form 
So the other three proportions involving terms are 
45: 24 = 30: 16 can be written as 15: 8 in simplest form 
30: 45 = 16: 24 can be written as 2: 3 in simplest form 
16: 30 = 24: 45 can be written as 8: 15 in simplest form 
 
(ii) 12: 18 = 14: 21 can be written as 2: 3 in simplest form 
So the other three proportions involving terms are 
12: 14 = 18: 21 can be written as 6: 7 in simplest form 
21: 18 = 14: 12 can be written as 7: 6 in simplest form 
18: 12 = 21: 14 can be written as 3: 2 in simplest form 
 
6. If 4, x, 9 are in continued proportion, find the value of x. 
Solution: 
 
We know that 4, x, 9 are in continued proportion 
It can be written as 
4: x :: x: 9 
We get 
4/x = x/9 
On further calculation 
x
2
 = 9 (4) = 36 
So we get 
x = 6 
 
7. If in a proportion, the first, second and fourth terms are 32, 112 and 217 respectively, find the third 
term. 
Solution: 
 
It is given that in a proportion the first, second and fourth terms are 32, 112 and 217 
Consider x as the third term 
We can write it as 
32: 112 :: x: 217 
On further calculation 
32/112 = x/217 
So we get 
x = 32/112 (217) = 62 
 
 
 
 
 
 
8. Show that the following numbers are in continued proportion: 
(i) 36, 90, 225 
(ii) 48, 60, 75 
(iii) 16, 84, 441 
Solution: 
 
(i) 36, 90, 225 
Consider the fraction 36/90 
By dividing the fraction by 18 
We get 
36/90 = 2/5 
 
Consider the fraction 90/225 
By dividing the fraction by 45 
We get  
90/225 = 2/5 
 
Hence, 36: 90 :: 90: 225. 
 
(ii) 48, 60, 75 
Consider the fraction 48/60 
By dividing the fraction by 12 
We get 
48/60 = 4/5 
 
Consider the fraction 60/75 
By dividing the fraction by 15 
We get  
60/75 = 4/5 
 
Hence, 48: 60 :: 60: 75. 
 
(iii) 16, 84, 441 
Consider the fraction 16/84 
By dividing the fraction by 4 
We get 
16/84 = 4/21 
 
Consider the fraction 84/441 
By dividing the fraction by 21 
We get  
84/441 = 4/21 
 
Hence, 16: 84 :: 84: 441. 
 
9. The ratio of the length of a school ground to its width is 5: 2. Find its length if the width is 40 metres. 
Solution: 
 
It is given that 
Ratio of length of a school ground to its width = 5: 2

	Mathematics (Maths) Class 6 92 videos\|348 docs\|54 tests

Mathematics (Maths) Class 6

92 videos|348 docs|54 tests

Join Course for Free

Top Courses for Class 6

View all

Related Exams

Class 6

About this Document

	4.86/5 Rating
	Dec 23, 2024 Last updated

Document Description: RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) for Class 6 2024 is part of Mathematics (Maths) Class 6 preparation. The notes and questions for RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) have been prepared according to the Class 6 exam syllabus. Information about RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) covers topics like and RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) Example, for Class 6 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3).

Introduction of RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) in English is available as part of our Mathematics (Maths) Class 6 for Class 6 & RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) in Hindi for Mathematics (Maths) Class 6 course. Download more important topics related with notes, lectures and mock test series for Class 6 Exam by signing up for free. Class 6: Ratio, Proportion & Unitary Method (Exercise 9.3) RD Sharma Solutions | Mathematics (Maths) Class 6

Description

Full syllabus notes, lecture & questions for Ratio, Proportion & Unitary Method (Exercise 9.3) RD Sharma Solutions | Mathematics (Maths) Class 6 - Class 6 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 6 | Best notes, free PDF download

Information about RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3)

In this doc you can find the meaning of RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) defined & explained in the simplest way possible. Besides explaining types of RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) tests, examples and also practice Class 6 tests

	Mathematics (Maths) Class 6 92 videos\|348 docs\|54 tests

Mathematics (Maths) Class 6

92 videos|348 docs|54 tests

Join Course for Free

Download as PDF

Explore Courses for Class 6 exam

Top Courses for Class 6

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

Free

Summary

pdf

Important questions

Proportion & Unitary Method (Exercise 9.3) RD Sharma Solutions | Mathematics (Maths) Class 6

practice quizzes

Ratio

Viva Questions

Proportion & Unitary Method (Exercise 9.3) RD Sharma Solutions | Mathematics (Maths) Class 6

past year papers

Semester Notes

study material

Previous Year Questions with Solutions

mock tests for examination

Objective type Questions

Sample Paper

video lectures

Exam

Extra Questions

Proportion & Unitary Method (Exercise 9.3) RD Sharma Solutions | Mathematics (Maths) Class 6

Ratio

ppt

shortcuts and tricks

MCQs

;

Additional Information about RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) for Class 6 Preparation

RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) Free PDF Download

The RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) is an invaluable resource that delves deep into the core of the Class 6 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) now and kickstart your journey towards success in the Class 6 exam.

Importance of RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3)

The importance of RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) cannot be overstated, especially for Class 6 aspirants. This document holds the key to success in the Class 6 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) Notes

RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3). It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) Notes on EduRev are your ultimate resource for success.

RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) Class 6 Questions

The "RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) Class 6 Questions" guide is a valuable resource for all aspiring students preparing for the Class 6 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) on the App

Students of Class 6 can study RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3), students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of RD Sharma Solutions: Ratio, Proportion & Unitary Method (Exercise 9.3) is prepared as per the latest Class 6 syllabus.

Education Revolution