## NCERT SOLUTIONS FOR CLASS 7TH MATHS CHAPTER 2 FRACTIONS AND DECIMALS

Fractions and Decimals Chapter 2 class 7 Maths NCERT solutions are given. You should study the textbook lesson Fractions and Decimals very well. You should observe and practice the example problems and solutions given in the textbook. Observe the given below solutions and try them in your own method.  You can see. Ncert solutions for class 7 maths chapter 1 Integers Chapter 4 simple equations solutions 3. Rational numbers solutions Ncert solutions for class 6 th maths  Real numbers solutions class 10 FRACTIONS AND DECIMALS 7th class maths CBSE class 7 maths CHAPTER 2 EXERCISE 2.1 Problem 1 1. Solve i. 2 – 3/5 ii. 4 + 7/8 iii. 3/5 + 2/7 iv. 9/11 – 4/15 v. 7/10 + 2/5 + 3/2 vi. 2 2/3 + 3 1/2 vii. 8 1/2 – 3 5/8 Solutions: i. 2 – 3/5 = ( 2 × 5 )/5 – 3/5 = 10/5 – 3/5 = ( 10 – 3 ) ÷ 5 = 7/5 ii. 4 + 7/8 = ( 4 × 8 )/8 + 7/8 = 32/ 8 + 7/8 = ( 32 + 7 )/8 = 39/8 iii. 3/5 + 2/7 = ( 3 × 7 ) ÷ ( 5 × 7 ) + ( 2 × 5 ) ÷ ( 7 × 5 ) = 21/35 + 10/35 = ( 21 + 10 ) ÷ 35 = 31/35 iv. 9/11 – 4/15 = ( 9 × 15 ) ÷ ( 11 × 15 ) – ( 4 × 11 ) ÷ ( 15 × 11 ) = 135/165 – 44/165 = ( 135 – 44 ) ÷ 165 = 91/ 165 v. 7/10 + 2/5 + 3/2 = 7/10 + ( 2 × 2 ) ÷ ( 5 × 2 ) + ( 3 × 5 ) ÷ (2 × 5 ) = 7/ 10 + 4/ 10 + 15/10 = ( 7 + 4 + 15 ) ÷10 = 26/ 10 = 13/5 = 2 3/5 vi. 2 2/3 + 3 1/2 = 8/3 + 7/2 = ( 8 × 2 ) ÷ ( 3 × 2 ) + ( 7 × 3 ) ÷ ( 2 × 3 ) = 16/6 + 21/ 6 = ( 16 + 21 ) ÷ 6 = 37/6 = 6 1/6 vii. 8 1/2 – 3 5/8 = 17/2 – 29/8 = ( 17 × 4 ) ÷ ( 2 × 4 ) – 29/ 8 = 68/8 – 29/8 = ( 68 – 29 ) ÷ 8 = 39/8 = 4 7/8 Problem 2 Arrange the following in descending order : i. 2/9, 2/3, 8/21 ii. 1/5, 3/7, 6/10 Solutions : i. 2/9, 2/3, 8/21 Changing the fractions into like fractions, we get ( 2 × 4 ) ÷ ( 9 × 4 ), ( 2 × 4 ) ÷ ( 3 × 4 ), 8/ 21 8/ 36, 8/12, 8/21 8/12 > 8/21 > 8/36 Therefore, 2/3 > 8/21 > 2/9 ii. 1/5, 3/7, 6/10 Changing the fractions into like fractions we get ( 1 × 7 × 10 ) ÷ ( 5 × 7 × 10 ), ( 3 × 5 × 10 ) ÷ ( 7 × 5 × 10 ), ( 7 × 7 × 10 ) ÷ ( 10 × 7 × 5 ) 70/350, 150/350, 245/350 245/350 > 150/350 > 70/350 Therefore, 7/10 > 3/7 > 1/5 Problem 3 3. In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square ?
Solution: Sum, along first row = 4/11 + 9/11 + 2/11 = ( 4 + 9 + 2 ) ÷ 11 = 15/11 Sum, along second row = 3/11 + 5/11 + 7/11 = ( 3 + 5 + 7 ) ÷ 11 = 15/11 Sum, along third row = 8/11 + 1/11 + 6/11 = ( 8 + 1 + 6 ) ÷ 11 = 15/11 Sum, along first column = 4/11 + 3/11 + 8/11 = ( 4 + 3 + 8 ) ÷ 11 = 15/11 Sum, along second column = 9/11 + 5/11 + 1/11 = ( 9 + 5 + 1 ) ÷ 11 = 15/11 Sum, along third column = 2/11 + 7/11 + 6/11 = ( 2 + 7 + 6 ) ÷ 11 = 15/11 Sum, along first diagonal = 6/11 + 5/11 + 4/11 = ( 6 + 5 + 4 ) ÷ 11 = 15/11 Sum, along second diagonal = 8/11 + 5/11 + 2/11 = ( 8 + 5 + 2 ) ÷ 11 = 15/11 In this square, the sum of each row, column and diagonal are same. So this is a magic square Problem 4 4. A rectangular sheet paper is 12 1/2 cm long and 10 2/3 cm wide. Find its perimeter. Solution : Length of the rectangular sheet of paper = 12 1/2 cm Breadth of the rectangular sheet of paper = 10 2/3 cm Perimeter of the rectangular sheet of paper = 2 × (length + breadth) = 2 × (12 1/2 + 10 2/3 ) = 2 × ( 25/2 + 32/3 ) = 2 × [ ( 25 × 3 ) ÷ ( 2 × 3 ) ] + [ ( 32 × 2 ) ÷ ( 3 × 2 ) ] = 2 × ( 75/6 + 64/6 ) = 2 × [ ( 75 + 64 ) ÷ 6 ] = 2 × ( 139/6 ) = ( 2 × 139 ) ÷ 6 = 139/3 = 46 1/3 Therefore, the perimeter of the sheet of paper is 46 1/3 cm. Problem 5 5. Find the perimeter of i. ∆ ABE, ii. the rectangle BCDE in this figure. Whose perimeter is greater ? Solutions : i. The perimeter of ∆ ABE = AB + BE + EA = 5/2 + 2 3/4 + 3 3/5 = 5/2 + 11/4 + 18/5 = [ ( 5 × 4 × 5 ) ÷ ( 2 × 4 × 5 ) ] + [ ( 11 × 2 × 5 ) ÷ ( 4 × 2 × 5 ) ] + [ ( 18 × 2 × 4 ) ÷ ( 5 × 2 × 4 ) ] = 100/40 + 144/40 + 110/40 = ( 100 + 144 + 110 ) ÷ 40 = 354/40 = 177/20 = 8 17/20 cm Therefore, the perimeter of ∆ ABC is 177/20 cm = 8 17/20. ii. The length of rectangle = 2 3/4 cm The breadth of rectangle = 7/6 cm Perimeter of the rectangle = 2 × ( length + breadth ) = 2 × ( 2 3/4 + 7/6 ) = 2 × (11/4 + 7/6 ) 2 × [ ( 11 × 6 ) ÷ ( 4 × 6 ) ] + [ ( 7 × 4 ) ÷ ( 6 × 4 ) ] = 2 × ( 66/24 + 28/24 ) = 2 × [ ( 66 + 28 ) ÷ 24 ] = 2 × 94/24 = 47/6 = 7 5/6 cm Therefore, the perimeter of the rectangle BCDE is 47/6 cm = 7 5/6 cm Perimeter of the triangle ∆ ABE = 177/20 cm Perimeter of the rectangle BCDE = 47/6 cm Changing the fractions into like fractions, we get 177/20, 47/6 ( 177 × 6 ) ÷ ( 20 × 6 ), ( 47 × 20 ) ÷ ( 6 × 20 ) 1062/120, 940/120 1062/ 120 > 940/120 Therefore, 177/20 or 8 17/20 is greater So, the perimeter of ∆ ABE is greater than the perimeter of rectangle BCDE. Problem 6 6. Salil wants to put a picture in a frame. The picture is 7 3/5 cm wide. To fit in the frame the picture cannot be more than 7 3/10 cm wide. How much should the picture be trimmed ? Solution : Width of picture = 7 3/5 cm Required width of picture = 7 3/10 cm The picture should be trimmed by = 7 3/5 – 7 3/10 = 38/5 – 73/10 = ( 38 × 2 ) ÷ ( 5 × 2 ) – 73/10 = 76/10 – 73/10 = ( 76 – 73 ) ÷ 10 = 3/10 cm Problem 7 7. Ritu ate 3/5 part of an apple, and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat ?Who had the larger share ?By how much? Solution : The part of apple eaten by Ritu = 3/5 The part of apple eaten by Somu = 1 – part of apple eaten by Ritu = 1 – 3/5 = ( 1 × 5 ) ÷ 5 – 3/5 = 5/5 – 3/5 = ( 5 – 3 ) ÷ 5 = 2/5 Therefore, Somu ate 2/5 part of the apple. Since, 3 > 2, Ritu had the larger share. Difference between the two shares 3/2 – 2/5 = (3 – 2) ÷ 5 = 1/5 Therefore, Ritu’s share is larger than Somu’s share. Problem 8 8. Michael finished coloring a picture in 7/12 hour. Vaibhav finished coloring the same picture in 3/4 hour. Who worked longer ? By what fraction was it longer ? Solution: Time taken by Michael = 7/12 hour Time taken by Vaibhav = 3/4 hour Changing the fractions into like fractions, we get 7/12, 3/4 7/12, ( 3 × 3 ) ÷ ( 4 × 3 ) 7/12, 9/12 Since, 9/12 > 7/12, Vaibhav worked longer. Difference = 9/12 – 7/12 = (9 – 7) ÷ 12 = 2/12 = 1/6 hour

#### SOLUTIONS FOR MATHS CLASS 7 CHAPTER 2, SOLUTIONS FOR FRACTIONS AND DECIMALS, NCERT SOLUTIONS FOR MATHS CLASS 7 CHAPTER 2

EXERCISE 2. 2 Problem 1 1. Which of the drawings a) to d) show: i. 2 × (1/5) ii. 2 × (1/2) iii. 3 × (2/3) iv. 3 × ( 1/5 ) Solutions : i – d ii – b iii – a iv – c Problem 2 2. Some pictures a) to c) are given below. Tell which of them show : i. 3 × ( 1/5 ) = 3/5 ii. 2 × ( 1/3 ) = 2/3 iii. 3 × ( 3/4 ) = 2 1/4 Solutions : i – c ii – a iii – b Problem 3 3. Multiply and reduce lowest form and convert into a mixed fraction. i. 7 × ( 3/5 ) ii. 4 × ( 1/3 ) iii. 2 × ( 6/7 ) iv. 5 × ( 2/9 ) v. ( 2/3 ) × 4 vi. ( 5/2 ) × 6 vii. 11 × ( 4/7 ) viii. 20 × ( 4/5 ) ix. 13 × ( 1/3 ) x. 15 × ( 3/5 ) Solutions : i. 7 × ( 3/5 ) = ( 3 × 7 ) ÷ 5 = 21/5 4 1/5 ii. 4 × ( 1/3 ) = ( 4 × 1 ) ÷ 3 = 4/3 = 1 1/3 iii. 2 × ( 6/7 ) = ( 2 × 6 ) ÷ 7 = 12/7 = 1 5/7 iv. 5 × ( 2/9 ) = ( 5 × 2 ) ÷ 9 = 10/9 = 1 1/9 v. ( 2/3 ) × 4 = ( 2 × 4 ) ÷ 3 = 8/3 = 2 2/3 vi. ( 5/2 ) × 6 = ( 5 × 6 ) ÷ 2 = 30/2 = 15 vii. 11 × ( 4/7 ) = ( 11 × 4 ) ÷ 7 = 44/7 = 6 2/7 viii. 20 × ( 4/5 ) = ( 20 × 4 ) ÷ 5 = 80/5 = = 16 ix. 13 × ( 1/3 ) = ( 13 × 1 ) ÷ 3 = 13/3 = 4 1/3 x. 15 × ( 3/5 ) = ( 15 × 3 ) ÷ 5 = 45/5 = 9 Problem 4 4. Shade: i. 1/2 of the circles in box a) ii. 2/3 of the triangles in box b) iii. 3/5 of the squares in box c) Solutions : Problem 5 5. Find a. 1/2 of i. 24, ii. 46 b. 2/3 of i. 18, ii. 27 c. 3/4 of i. 16, ii. 36 d. 4/5 of i. 20, ii. 35 Solutions : a. 1/2 of i. 24, ii. 46 i. 1/2 of 24 = (1/2 ) × 24 = 12 ii. 1/2 of 46 = ( 1/2 ) × 46 = 23 b. 2/3 of i. 18, ii. 27 i. 2/3 of 18 = ( 2/3 ) × 18 = 12 ii. 2/3 of 27 = ( 2/3 ) × 27 = 18 c. 3/4 of i. 16, ii. 36 i. 3/4 of 16 = ( 3/4 ) × 16 = 12 ii. 3/4 of 36 = ( 3/4 ) × 36 = 27 d. 4/5 of i. 20, ii. 35 i. 4/5 of 20 = ( 4/5 ) × 20 = 16 ii. 4/5 of 35 = ( 4/5 ) × 35 = 28 Problem 6 6. Multiply and express as a mixed fraction : a. 3 × 5 1/5 b. 5 × 6 3/4 C. 7 × 2 1/4 d. 4 × 6 1/3 e. 3 1/4 × 6 f. 3 2/5 × 8 Solutions : a. 3 × 5 1/5 = 3 × ( 26/5 ) = 78/5 = 15 3/5 b. 5 × 6 3/4 = 5 × 27/4 = 135/4 = 33 3/4 C. 7 × 2 1/4 = 7 × 9/4 = 63/4 = 15 3/4 d. 4 × 6 1/3 = 4 × 19/3 = 76/3 = 25 1/3 e. 3 1/4 × 6 = 13/4 × 6 = 78/4 = 19 2/4 f. 3 2/5 × 8 = 17/5 × 8 = 136/5 = 27 1/5 Problem 7 7. Find a. 1/2 of i. 2 3/4, ii. 4 2/9 b. 5/8 of i. 3 5/6, ii. 9 2/3 Solutions : a. 1/2 of i. 2 3/4, ii. 4 2/9 i. 1/2 of 2 3/4 = 1/2 × 11/4 = 11/8 = 1 3/8 ii. 1/2 of 4 2/9 = 1/2 × 38/9 = 38/18 = 19/9 = 2 1/9 b. 5/8 of i. 3 5/6, ii. 9 2/3 i. 5/8 of 3 5/6 = 5/8 × 23/6 = 115/48 = 2 19/48 ii. 5/8 of 9 2/3 = 5/8 × 29/3 = 145/24 = 6 1/24 Problem 8 8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed 2/3 of the water. Pratap consumed the remaining water. i. How much water did Vidya drink ? ii. What fraction of the total quantity of water did Pratap drink ? Solution : i. Water consumed by Vidya = 2/5 of 5 litres = 2/5 × 5 = 2 litres ii. Water consumed by Pratap = 1 – 2/5 = ( 5 – 2 ) ÷ 5 = 3/5 of the total water Therefore, the fraction of total quantity of water = 3/5

### NCERT SOLUTIONS FOR MATHS CLASS 7TH CHAPTER 2,FRACTIONS AND DECIMALS

EXERCISE 2.3 Problem 1 1. i. 1/4 of a. 1/4, b. 3/5, c. 4/3 ii. 1/7 of a. 2/9, b. 6/5, c. 3/10 Solutions : 1. i. a. 1/4 of 1/4 1/4 × 1/4 = ( 1 × 1 ) ÷ ( 4 × 4 ) = 1/16 b. 1/4 of 3/5 1/4 × 3/5 = ( 1 × 3 ) ÷ ( 4 × 5 ) = 3/20 c. 1/4 of 4/3 1/4 × 4/3 = ( 1 × 4 ) ÷ ( 4 × 3 ) = 1/3 ii. a. 1/7 0f 2/9 1/7 × 2/9 = ( 1 × 2 ) ÷ ( 7 × 9 ) = 2/63 b. 1/7 of 6/5 = ( 1 × 6 ) ÷ ( 7 × 5 ) = 6/35 c. 1/7 of 3/10 1/7 × 3/10 = ( 1 × 3 ) ÷ ( 7 × 10 ) = 3/70 Problem 2 2. Multiply and reduce to lowest form ( if possible ) ; i. 2/3 × 2 2/3, ii. 2/7 × 7/9, iii. 3/8 × 6/4 iv. 9/5 × 3/5, v. 1/3 × 15/8, vi. 11/2 × 3/10, vii. 4/5 × 12/7 Solutions : i. 2/3 × 2 2/3 2/3 × 8/3 = ( 2 × 8 ) ÷ ( 3 × 3 ) = 16/9 = 1 7/9 ii. 2/7 × 7/9 2/7 × 7/9 = ( 2 × 7 ) ÷ ( 7 × 9 ) = 2/9 iii. 3/8 × 6/4 3/8 × 6/4 = ( 3 × 6 ) ÷ ( 8 × 4 ) = ( 3 × 3 ) ÷ ( 4 × 4 ) = 9/16 iv. 9/5 × 3/5 9/5 × 3/5 = ( 9 × 3 ) ÷ ( 5 × 5 ) = 27/25 = 1 2/25 v. 1/3 × 15/8 1/3 × 15/8 = ( 1 × 15 ) ÷ ( 3 × 8 ) = ( 1 × 5 ) ÷ ( 1 × 8 ) = 5/8 vi. 11/2 × 3/10 11/2 × 3/10 = ( 11 × 3 ) ÷ ( 2 × 10 ) = 33/20 = 1 13/20 vii. 4/5 × 12/7 4/5 × 12/7 = ( 4 × 12 ) ÷ ( 5 × 7 ) = 48/35 = 1 13/35 Problem 3 3. Multiply the following fractions : i. 2/5 × 5 1/4, ii. 6 2/5 × 7/9, iii. 3/2 × 5 1/3 vi. 5/6 × 2 3/7, v. 3 2/5 × 4/7, vi. 2 3/5 × 3, vii. 3 4/7 × 3/5 Solutions : i. 2/5 × 5 1/4 2/5 × 21/4 = ( 2 × 21 ) ÷ ( 5 × 4 ) = ( 1 × 21 ) ÷ ( 5 × 2 ) = 21/10 = 2 1/10 ii. 6 2/5 × 7/9 32/5 × 7/9 = ( 32 × 7 ) ÷ ( 5 × 9 ) = 224/45 = 4 44/45 iii. 3/2 × 5 1/3 3/2 × 5 1/3 = ( 3 × 16 ) ÷ ( 2 × 3 ) = ( 1 × 8) ÷ ( 1 × 1 ) = 8 iv. 5/6 × 2 3/7 5/6 × 17/7 = ( 5 × 17 ) ÷ ( 6 × 7 ) = 85/42 = 2 1/42 v. 3 2/5 × 4/7 17/5 × 4/7 = ( 17 × 4 ) ÷ ( 5 × 7 ) = 68/35 = 1 33/35 vi. 2 3/5 × 3 13/5 × 3 = ( 13 × 3 ) ÷ 5 = 39/5 = 7 4/5 vii. 3 4/7 × 3/5 25/7 × 3/5 = ( 25 × 3 ) ÷ ( 7 × 8 ) = ( 5 × 3 ) ÷ ( 7 × 1 ) = 15/7 = 2 1/7 Problem 4 4. Which is greater : i. 2/7 of 3/4 or 3/5 of 5/8 ii. 1/2 of 6/7 or 2/3 of 3/7 Solutions : i. 2/7 of 3/4 or 3/5 of 5/8 2/7 of 3/4 2/7 × 3/4 = ( 2 × 3 ) ÷ ( 7 × 4 ) = ( 1 × 3 ) ÷ ( 7 × 2 ) = 3/14 3/5 of 5/8 3/5 × 5/8 = ( 3 × 5 ) ÷ ( 5 × 8 ) = ( 3 × 1 ) ÷ ( 1 × 8 ) = 3/8 3/14, 3/8 3/8 > 3/14 Therefore, 3/5 of 5/8 > 2/7 of 3/4 ii. 1/2 of 6/7 or 2/3 of 3/7 1/3 of 6/7 1/2 × 6/7 = ( 1 × 6 ) ÷ ( 2 × 7 ) = 3/7 2/3 of 3/7 2/3 × 3/7 = ( 2 × 3 ) ÷ ( 3 × 7 ) = 2/7 3/7. 2/7 3/7 > 2/7 Therefore, 1/2 of 6/7 > 2/3 of 3/7 Problem 5 5. Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is 3/4 m. Find the distance between the first and the last sapling. Solution : It can be observed that gaps between first and last saplings = 3 Length of one gap = 3/4 m Therefore, distance between first and last sapling = 3 × 3/4 = 9/4 = 2 1/4 m Problem 6 6. Lipika reads a book for 1 3/4 hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book ? Solution : Number of hours Lipika reads the book per day = 1 3/4 Number of days = 6 Total number of hours required by Lipika to read the book = 1 3/4 × 6 = 7/4 × 6 = 42/4 = 21/2 = 10 1/2 hours. Problem 7 7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 3/4 litres of petrol ? Solution : Number of Km a car can run per one litre petrol = 16 Km Number of Km a car can run for 11/4 litres petrol = 11/4 × 16 = 11 × 4 = 44 km Problem 8 8. a. i. Provide the number in the ……, such that 2/3 × …….. = 10/30. ii. The simplest form of the number obtained in …….. is …….. . b. i. Provide the number in the ……, such that 3/5 × …….. = 24/75 ii. The simplest form of the number obtained in …….. is …….. . Solutions : 8. a. i. Provide the number in … is 5/10 such that 2/3 × 5/10 = 10/30. ii. The simplest form of the number obtained in 5/10 is 1/2 b. i. Provide the number in the .. is 8/15 such that 3/5 × 8/15 = 24/75 ii. The simplest form of the number obtained in 8/15 is 8/15.
##### 7 TH CLASS MATHS SOLUTIONS CHAPTER 2 NCERT
EXERCISE 2. 4 Problem 1 1. Find : i. 12 ÷ ( 3/4 ), ii. 14 ÷ ( 5/6 ), iii. 8 ÷ ( 7/3 ) iv. 4 ÷ 8/3, v. 3 ÷ 2 1/3, vi. 5 ÷ 3 4/7 Solutions : i. 12 ÷ ( 3/4 ) 12 × 4/3 = ( 12 × 4 ) ÷ 3 = 4 × 4 = 16 ii. 14 ÷ ( 5/6 ) 14 × 6/5 = ( 14 × 6 ) ÷ 5 = 84 ÷ 5 = 84/5 =16 4/5 iii. 8 ÷ ( 7/3 ) 8 × 3/7 = ( 8 × 3 ) ÷ 7 = 24 ÷ 7 = 24/7 =3 3/7 iv. 4 ÷ 8/3 4 × 3/8 = ( 4 × 3 ) ÷ 8 = 3 ÷ 2 = 3/2 =1 1/2 v. 3 ÷ 2 1/3 = 3 ÷ 7/3 3 × 3/7 = ( 3 × 3 ) ÷ 7 = 9 ÷ 7 = 9/7 =1 2/7 vi. 5 ÷ 3 4/7 = 5 ÷ 25/7 5 × 7/25 = ( 5 × 7 ) ÷ 25 = ( 1 × 7 ) ÷ 5 = 7/5 =1 2/5 Problem 2 2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers. i. 3 ÷ 7, ii. 5 ÷ 8, iii. 9 ÷ 7, iv. 6 ÷ 5 v. 12 ÷ 7, vi. 1 ÷ 8, vii. 1 ÷ 11 Solutions : i. 3 ÷ 7 = 3/7 Reciprocal of 3/7 is 7/3. 7/3 is an improper fraction. ii. 5 ÷ 8 = 5/8 Reciprocal of 5/8 is 8/5. 8/5 is an improper fraction. iii. 9 ÷ 7 = 9/7 Reciprocal of 9/7 is 7/9. 7/9 is a proper fraction. iv. 6 ÷ 5 = 6/5 Reciprocal of 6/5 is 5/6. 5/6 is a proper fraction. v. 12 ÷ 7 = 12/7 Reciprocal of 12/7 is 7/12. 7/12 is a proper fraction. vi. 1 ÷ 8 = 1/8 Reciprocal of 1/8 is 8. 8 is a whole number. vii. 1 ÷ 11 = 1/11 Reciprocal of 1/11 is 11. 11 is a whole number. Problem 3 3. Find : i. 7/3 ÷ 2, ii. 4/9 ÷ 5, iii. 6/13 ÷ 7 iv. 4 1/3 ÷ 3, v. 3 1/2 ÷ 4, vi. 4 3/7 ÷ 7 Solutions : i. 7/3 ÷ 2 7/3 ÷ 2/1 = 7/3 × 1/2 = ( 7 × 1 ) ÷ ( 3 × 2 ) = 7/6 ii. 4/9 ÷ 5 4/9 ÷ 5/1 = 4/9 × 1/5 = ( 4 × 1 ) ÷ ( 9 × 5 ) = 4/45 iii. 6/13 ÷ 7 6/13 ÷ 7/1 = 6/13 × 1/7 = ( 6 × 1 ) ÷ ( 13 × 7 ) = 6/91 iv. 4 1/3 ÷ 3 = 13/3 ÷ 3 13/3 ÷ 3/1 = 13/3 × 1/3 = ( 13 × 1 ) ÷ ( 3 × 3 ) = 13/9 v. 3 1/2 ÷ 4 = 7/2 ÷ 4 7/2 ÷ 4/1 = 7/3 × 1/4 = ( 7 × 1 ) ÷ ( 2 × 4 ) = 7/8 vi. 4 3/7 ÷ 7 = 31/7 ÷ 7 31/7 ÷ 7/1 = 31/7 × 1/7 = ( 31 × 1 ) ÷ ( 7 × 7 ) = 31/49 Problem 4 4. Find ; i. 2/5 ÷ 1/2, ii. 4/9 ÷ 2/3, iii. 3/7 ÷ 8/7 iv. 2 1/3 ÷ 3/5, v. 3 1/2 ÷ 8/3, vi. 2/5 ÷ 1 1/2 vii. 3 1/5 ÷ 1 2/3, viii. 2 1/5 ÷ 1 1/5 Solution : i. 2/5 ÷ 1/2 2/5 × 2/1 = ( 2 × 2 ) ÷ ( 5 × 1 ) = 4//5 ii. 4/9 ÷ 2/3 4/9 × 3/2 = ( 4 × 3 ) ÷ ( 9 × 2 ) = 2/3 iii. 3/7 ÷ 8/7 3/7 × 7/8 = ( 3 × 7 ) ÷ ( 7 × 8 ) = 3/8 iv. 2 1/3 ÷ 3/5 = 7/3 ÷ 3/5 7/3 × 5/3 = ( 7 × 5 ) ÷ ( 3 × 3 ) = 35/9 = 3 8/9 v. 3 1/2 ÷ 8/3 = 7/2 ÷ 8/3 7/2 × 3/8 = ( 7 × 3 ) ÷ ( 2 × 8 ) = 21/16 = 1 5/6 vi. 2/5 ÷ 1 1/2 = 2/5 ÷ 3/2 2/5 × 2/3 = ( 2 × 2 ) ÷ ( 5 × 3 ) = 4//15 vii. 3 1/5 ÷ 1 2/3 = 16/5 ÷ 5/3 16/5 × 3/5= ( 16 × 3 ) ÷ ( 5 × 5) = 48/25 = 1 23/25 viii. 2 1/5 ÷ 1 1/5 = 11/5 ÷ 6/5 11/5 × 5/6 = ( 11 × 5 ) ÷ ( 5 × 6 ) = 11/6 = 1 5/6
###### FRACTIONS AND DECIMALS SOLUTIONS
EXERCISE 2.5 Problem 1 1. Which is greater ? i. 0.5 or 0.05, ii. 0.7 or 0.5, iii. 7 or 0.7 iv. 1.37 or 1.49, v. 2.03 or 2.30, vi. 0.8 or 0.88 Solutions : i. 0.5 or 0.05 0.5 = 5/10 0.05 = 5/100 Therefore, the denominator has less value is the greater one. Hence 0.5 is greater. ii. 0.7 or 0.5 0.7 = 7/10 0.5 = 5/10 Therefore, the numerator has more value is the greater one. Hence 0.7 is greater. iii. 7 or 0.7 7 = 700/100 0.05 = 5/100 Therefore, the numerator has more value is the greater one. Hence 7 is greater. iv. 1.37 or 1.49 1.37= 137/100 1.49= 149/100 Therefore, the numerator has more value is the greater one. Hence 1.49 is greater. v. 2.03 or 2.30 2.03 = 203/100 2.30= 230/100 Therefore, the numerator has more value is the greater one. Hence 2.30 is greater. vi. 0.8 or 0.88 0.8 = 80/100 0.88 = 88/100 Therefore, the numerator has more value is the greater one. Hence 0.88 is greater. Problem 2 2. Express as rupees using decimals : i. 7 paise, ii. 7 rupees 7 paise iii. 77. Rupees 77 paise, iv. 50 paise, v. 235 paise Solutions : i. 7 paise 7 paise = Rs. 7/100 = Rs. 0.07 ii. 7 rupees 7 paise 7 rupees + 7 paise Rs. 7 + Rs. 7/100 = Rs. 7 + Rs. 0.07 = Rs. 7.07 iii. 77. Rupees 77 paise = 77 rupees + 77 paise = Rs. 77 + Rs. 77/100 = Rs. 7 + Rs. 0.77 = Rs. 77.77 iv. 50 paise = Rs.50/100 = Rs. 0.50 v. 235 paise = Rs. 235/100 = Rs. 2.35 Problem 3 3. i. Express 5 cm in metre and kilometre ii. Express 35 mm in cm, m and km. Solutions : i. Express 5 cm in metre and kilometre 5 cm in metre 5 cm = 5/100 m = 0.05 m 5 cm in kilometre 5 cm = 5/1000 = 0.005 km ii. Express 35 mm in cm, m and km. 35 mm in cm 35 mm = 35/10 cm = 3.5 cm 35 mm in m 35 mm = 35/1000 m = 0.035 cm 35 mm in km 35 mm = 35/1000000 cm = 0.000035 km Problem 4 4. Express in Kg : i. 200 g, ii. 3470 g, iii. 4 kg 8 g Solutions: i. 200 g = 200/1000 kg = 0.2 kg ii. 3470 g 3470 g = 3470/1000 kg = 3.470 kg iii. 4 kg 8 g 4 kg + 8 g = 4 kg + 8/1000 g = 4 kg + 0.008 = 4.008 kg Problem 5 5. Write the following decimal numbers in the expanded form : i. 20.03, ii. 2.03, iii. 200.03 Solutions : i. 20.03 20.03 = ( 2 × 10 ) + ( 0 × 1 ) + [ 0 × (1/10) ] + [ 3 × (1/100 ) ] ii. 2.03 2.03 = ( 2 × 1 ) + [ 0 × (1/10) ] + [ 3 × (1/100 ) ] iii. 200.03 200.03 = ( 2 × 100 ) + ( 0 × 10 ) + ( 0 × 1 ) + [ 0 × (1/10) ] + [ 3 × (1/100 ) ] Problem 5 5. Write the place value of 2 in the following decimal numbers: i. 2.56, ii. 21.37, iii. 10.25, iv. 9.42, v. 63.352 Solution: i. 2.56: ones ii. 21.37: tens iii. 10.25: tenths iv. 9.42: hundredths v. 63.352 : thousandths Problem 7 7. Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and how much? Solution : Distance travelled by Dinesh = A to B + B to C = 7.5 km + 12.7 km = 20.2 km Distance travelled by Ayub = A to D + D to C = 9.3 km + 11.8 km = 21.1 km Difference between their traveled distances = 21.1 km – 20.2 km = 0.9 km Therefore, Ayub travelled 0.9 km distance more than Dinesh. Problem 8 8. Shyama bought 5 kg 300 g apples and 3 kg 350 g mangoes. Sarala bought 5 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits? Solution : Total fruits bought by Shyama = Apple + mangoes 5 kg 300 grams + 3 kg 250 g = 5 kg + 300 g + 3 kg + 250 g = 8 kg + 550 g = 8 kg + 550/1000 kg = 8 kg + 0.550 kg = 8.550 kg Total fruits bought by Sarala = oranges + bananas 4 kg 800 grams + 4 kg 150 g = 4 kg + 800 g + 8 kg + 950 g = 8 kg + 950 g = 8 kg + 950/1000 kg = 8 kg + 0.950 kg = 8.950 kg Therefore, Sarala bought more fruits. Problem 9 9. How much less is 28 km than 42.6 km ? Solution: 42.6 km – 28 km = 14.6 km Therefore, 28 km is 14.6 km less than 42.6 km.
###### NCERT SOLUTIONS FOR 7TH CLASS MATHS CHAPTER 2
EXERCISE 2.6 Problem 1 i. 0.2 × 6, ii. 8 × 4.6, iii. 2.71 5, iv. 20.1 × 4 v. 0.05 × 7, vi. 211.02 × 4, vii. 2 × 0.86 Solutions: i. 0.2 × 6 0.2 × 6 = ( 2/10 ) × 6 = ( 2 × 6 ) ÷ 10 = 12/10 = 1.2 ii. 8 × 4.6 8 × 4.6 = 8 × ( 46/10 ) = ( 8 × 46 ) ÷ 10 = 368/10 = 36.8 iii. 2.71 × 5 2.71 × 5 = ( 271/100 ) × 5 = ( 271 × 5 ) ÷ 100 = 1355/100 = 13.55 iv. 20.1 × 4 20.1 × 4 = ( 201/10 ) × 4 = ( 201 × 4 ) ÷ 10 = 804/10 = 80.4 v. 0.05 × 7 0.05 × 7 = ( 5/100 ) × 7 = ( 5 × 7 ) ÷ 100 = 35/100 = 0.35 vi. 211.02 × 4 211.02 × 4 = ( 211.03/100 ) × 4 = ( 211.02 × 4 ) ÷ 100 = 84408/100 = 844.08 vii. 2 × 0.86 2 × 0.86 = 2 × ( 86/100 ) = ( 2 × 86 ) ÷ 100 = 172/100 = 1.72 Problem 3 3. Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm Solution : Length of a rectangle = 5.7 cm Breadth of the rectangle = 3 cm Area of rectangle = length × breadth = 5.7 cm × 3 cm = (57/10) × 3 = (57 × 3) ÷ 10= 171/10 = 17.1 cm.cm Problem 3 i. 1.3 × 10, ii. 36.8 × 10, iii. 153.7 × 10 iv. 168.07 × 10, v. 31.1 × 100, vi. 156.1 × 100 vii. 3.62 × 100, viii. 43.07 × 100, ix. 0.5 × 10 x. 0.08 × 10, xi. 0.9 × 100, xii. 0.03 × 1000 Solutions: i. 1.3 × 10 1.3 × 10 = ( 13/10 ) × 10 = 13 ii. 36.8 × 10 36.8 × 10 = ( 368/10 ) × 10 = 368 iii. 153.7 × 10 153.7 × 10 = ( 1537/10 ) × 10 = 1537 iv. 168.07 × 10 168.07 × 10 = ( 16807/100 ) × 10 = 16807/10 = 1680.7 v. 31.1 × 100 31.1 × 100 = ( 311/10 ) × 100 = 311 × 10 = 3110 vi. 156.1 × 100 156.1 × 100 = ( 1561/10 ) × 100 =1561 × 10 = 15610 vii. 3.62 × 100 3.62 × 100 = ( 362/100 ) × 100 = 362 viii. 43.07 × 100 43.07 × 100 = ( 4307/100 ) × 100 = 4307 ix. 0.5 × 10 0.5 × 10 = ( 5/10 ) × 10 = 5 x. 0.08 × 10 0.08 × 10 = ( 8/100 ) × 10 = 8/10 = 0.8 xi. 0.9 × 100 0.9 × 100 = ( 9/10 ) × 100 = 9 × 10 = 90 xii. 0.03 × 1000 0.03 × 1000 = ( 3/100 ) × 1000 = 3 × 10 =30 Problem 4 4. A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol? The distance covered by a two-wheeler for one litre of petrol = 55.3 km The distance covered by the two-wheeler for ten litres of petrol = 55.3 km × 10 = 553 km Problem 5 5. Find : i. 2.5 × 0.3, ii. 0.1 × 51.7, iii. 0.2 × 316.8 iv. 1.3 × 3.1, v. 0.5 × 0.05, vi. 11.2 × 0.15 vii. 1.07 × 0.02, viii. 10.05 × 1.05 ix. 101.01 × 0.01, x. 100.01 × 1.1 Solutions: i. 2.5 × 0.3 2.5 × 0.3 = ( 25/10 ) × ( 3/10 ) = ( 25 × 3 ) ÷ 100 = 75/100 = 0.75 ii. 0.1 × 51.7 0.1 × 51.7 = ( 1/10 ) × ( 517/10 ) = ( 1 × 517) ÷ 100 = 517/100 = 5.17 iii. 0.2 × 316.8 0.2 × 316.8 = ( 2/10 ) × ( 3168/10 ) = ( 2 × 3168 ) ÷ 100 = 6336/100 = 63.36 iv. 1.3 × 3.1 1.3 × 3.1 = ( 13/10 ) × ( 31/10 ) = ( 13 × 31 ) ÷ 100 = 403/100 = 4.03 v. 0.5 × 0.05 0.5 × 0.05 = ( 5/10 ) × ( 5/100 ) = ( 5 × 5 ) ÷ 1000 = 25/1000 = 0.025 vi. 11.2 × 0.15 11.2 × 0.15 = ( 112/10 ) × ( 15/100 ) = ( 112 × 15 ) ÷ 1000 = 1680/1000 = 1.68 vii. 1.07 × 0.02 1.07 × 0.02 = ( 107/100 ) × ( 2/100 ) = ( 107 × 2 ) ÷ 10000 = 214/10000 = 0.0214 viii. 10.05 × 1.05 10.05 × 1.05 = ( 10.05/100 ) × ( 1.05/100 ) = ( 10.05 × 1.05 ) ÷ 10000 = 105525/100 = 10.5525 ix. 101.01 × 0.01 101.01 × 0.01 = ( 10101/100 ) × ( 1/100 ) = ( 10101 × 1 ) ÷ 10000 = 10101/10000 = 1.0101 x. 100.01 × 1.1 100.01 × 1.1 = ( 10001/100 ) × ( 11/10 ) = ( 10001 × 11 ) ÷ 1000 = 110011/1000 = 110.011
##### NCERT SOLUTIONS FOR CLASS VII MATHS CHAPTER 2
EXERCISE 2.7 Problem 1 1. Find i. 0.4 ÷ 2, ii. 0.35 ÷ 5 , iii. 2.48 ÷ 4, iv. 65.4 ÷ 6, v. 651.2 ÷ 4, vi. 14.49 ÷ 7, vii. 3.96 ÷ 4, viii. 0.80 ÷ 5 Solutions: i. 0.4 ÷ 2 0.4 ÷ 2 = 4/10 ÷ 2 = 4/10 × 1/2 = 2/10 = 0.2 ii. 0.35 ÷ 5 0.35 ÷ 5 = 35/100 ÷ 5 = 35/100 × 1/5 = 7/100 = 0.07 iii. 2.48 ÷ 4 2.48 ÷ 4 = 247/100 ÷ 4 = 248/100 × 1/4 = 62/100 = 0.62 iv. 65.4 ÷ 6 65.4 ÷ 6 = 654/10 ÷ 6 = 654/10 × 1/6 = 109/10 = 10.9 v. 651.2 ÷ 4 651.2 ÷ 4 = 6512/10 ÷ 4 = 6512/10 × 1/4 = 1628/10 = 162.8 vi. 14.49 ÷ 7 1449 ÷ 7 = 1449/100 ÷ 7 = 1449/100 × 1/7 = 207/100 = 2.07 vii. 3.96 ÷ 4 3.96 ÷ 4 = 396/100 ÷ 4 = 396/100 × 1/4 = 99/100 = 0.99 viii. 0.80 ÷ 5 0.80 ÷ 5 = 80/100 ÷ 5 = 80/100 × 1/5 = 16/100 = 0.16 Problem 2 2. Find i. 4.8 ÷ 10, ii. 52.5 ÷ 10, iii. 0.7 ÷ 10, vi. 33.1 ÷ 10, v. 272.23 ÷ 10, vi. 0.56 ÷ 10, vii. 3.97 ÷ 10 Solutions: i. 4.8 ÷ 10 4.8 ÷ 10 = 48/10 ÷ 10 = 48/10 × 1/10 = 48/100 = 0.48 ii. 52.5 ÷ 10 52.5 ÷ 10 = 525/10 ÷ 10 = 525/10 × 1/10 = 525/100 = 5.25 iii. 0.7 ÷ 10 0.7 ÷ 10 = 7/10 ÷ 10 = 7/10 × 1/10 = 7/100 = 0.07 vi. 33.1 ÷ 10 33.1 ÷ 10 = 331/10 ÷ 10 = 331/10 × 1/10 = 331/100 = 3.31 v. 272.23 ÷ 10 272.23 ÷ 10 = 27223/100 ÷ 10 = 27223/100 × 1/10 = 27223/1000 = 27.223 vi. 0.56 ÷ 10 0.56 ÷ 10 = 56/100 ÷ 10 = 56/100 × 1/10 = 56/1000 = 0.056 vii. 3.97 ÷ 10 3.97 ÷ 10 = 397/100 ÷ 10 = 397/100 × 1/10 = 397/1000 = 0.397 Problem 3 3. Find i. 2.7 ÷ 100, ii. 0.3 ÷ 100, iii. 0.78 ÷ 100 iv. 432.6 ÷ 100, v. 23.6 ÷ 100, vi. 98.53 ÷ 100 Solutions: i. 2.7 ÷ 100 2.7 ÷ 100 = 27/10 ÷ 100 = 27/10 × 1/100 = 27/1000 = 0.027 ii. 0.3 ÷ 100 0.3 ÷ 100 = 3/10 ÷ 100 = 3/10 × 1/100 = 3/1000 = 0.003 iii. 0.78 ÷ 100 0.78 ÷ 100 = 78/100 ÷ 100 = 78/100 × 1/100 = 78/10000 = 0.0078 iv. 432.6 ÷ 100 432.6 ÷ 100 = 4326/10 ÷ 100 = 4326/10 × 1/100 = 4326/1000 = 4.326 v. 23.6 ÷ 100 23.6 ÷ 100 = 236/10 ÷ 100 = 236/10 × 1/100 = 236/1000 = 0.236 vi. 98.53 ÷ 100 98.53 ÷ 100 = 9853/100 ÷ 100 = 9853/100 × 1/100 = 9853/10000 = 0.9853 Problem 4 4. Find 1. 7.9 ÷ 1000, ii. 26.3 ÷ 1000, iii. 38.53 ÷ 1000 iv. 128.9 ÷ 1000, v. 0.5 ÷ 1000 Solutions: 1. 7.9 ÷ 1000 7.9 ÷ 1000 = 79/10 ÷ 1000 = 79/10 × 1/1000 = 79/10000 = 0. 0079 ii. 26.3 ÷ 1000 26.3 ÷ 1000 = 263/10 ÷ 1000 = 263/10 × 1/1000 = 263/10000 = 0. 0263 iii. 38.53 ÷ 1000 38.53 ÷ 1000 = 3853/100 ÷ 1000 = 3853/100 × 1/1000 = 3853/100000 = 0. 03853 iv. 128.9 ÷ 1000 128.9 ÷ 1000 = 1289/10 ÷ 1000 = 1289/10 × 1/1000 = 1289/10000 = 0. 1289 v. 0.5 ÷ 1000 0.5 ÷ 1000 = 5/10 ÷ 1000 = 5/10 × 1/1000 = 5/10000 = 0. 0005 Problem 5 5. Find i 7 ÷ 3.5, ii. 36 ÷ 0.2, iii. 3.25 ÷ 0.5 iv. 30.94 ÷ 0.7, v. 0.5 ÷ 0.25, vi. 7.75 ÷ 0.25 vii. 76.5 ÷ 0.15, viii. 37.8 ÷ 1.4, ix. 2.73 ÷ 1.3 Solutions: i 7 ÷ 3.5 7 ÷ 3.5 = 7 ÷ 35/10 = 7 × 10/35 = 10/5 = 2 ii. 36 ÷ 0.2 36 ÷ 0.2 = 36 ÷ 2/10 = 36 × 10/2 = 18 × 10 = 180 iii. 3.25 ÷ 0.5 3.25 ÷ 0.5 = 325/100 ÷ 5/10 = 325/100 × 10/5 = 65/10 = 6.5 iv. 30.94 ÷ 0.7 30.94 ÷ 0.7 = 3094/100 ÷ 7/10 = 3094/100 × 10/7 = 442/10 = 44.2 v. 0.5 ÷ 0.25 0.5 ÷ 0.25 = 5/10 ÷ 25/100 = 5/10 × 100/25 = 10/5 = 2 vi. 7.75 ÷ 0.25 7.75 ÷ 0.25 = 775/100 ÷ 25/100 = 775/100 × 100/25 = 31 vii. 76.5 ÷ 0.15 76.5 ÷ 0.15 = 765/10 ÷ 15/100 = 775/10 × 100/15 = 51 × 10 = 510 viii. 37.8 ÷ 1.4 37.8 ÷ 1.4 = 378/10 ÷ 14/10 = 378/10 × 10/14 = 27 ix. 2.73 ÷ 1.3 2.73 ÷ 1.3 = 273/100 ÷ 13/10 = 273/100 × 10/13 = 21/10 = 2.1 Problem 6 6. A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol? Solution: Distance covered by a vehicle = 43.2 km Petrol required to cover this distance = 2.4 litres So, distance covered by the vehicle in one litre of petrol = 43.2/2.4 = 18 km Note: Observe the solutions of fractions and decimals and try them in your own methods. 1. Integers 2. Fractions and decimals 3. Simple equations 4. Rational numbers You can also see Ncert solutions for maths class 6 some chapters Ncert solutions for maths class 7 some chapters Ncert solutions for maths class 8 some chapters Intermediate maths solurions for some chapters Solutions for properties of triangles intermediate. Ncert solutions for class 8 chapter 12 Nios solutions for maths 311 book 1 Sets 1.1