NCERT Solutions Class 7 Maths
Chapter – 2 (Fractions and Decimals)
The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 2 Fractions and Decimals Exercise 2.1 has been provided here to help the students in solving the questions from this exercise.
Chapter : 2 Fractions and Decimals
- NCERT Solution Class 7 Maths Exercise – 2.2
- NCERT Solution Class 7 Maths Exercise – 2.3
- NCERT Solution Class 7 Maths Exercise – 2.4
- NCERT Solution Class 7 Maths Exercise – 2.5
- NCERT Solution Class 7 Maths Exercise – 2.6
- NCERT Solution Class 7 Maths Exercise – 2.7
Exercise – 2.1
1. Solve:
(i) 2 –
(ii) 4 +
(iii)
(iv)
(v)
(vi)
(vii)
Solution –
(i) 2 –
=
LCM of 1, 5 = 5
=
=
=
(ii) 4 +
=
LCM of 1, 8 = 8
=
=
=
(iii)
LCM of 5, 7 = 35
=
=
=
(iv)
LCM of 11, 15 = 165
=
=
=
(v)
LCM of 10, 5 and 2 = 10
=
=
=
(vi)
=
LCM of 3, 2 = 6
=
=
=
(vii)
=
LCM of 2, 8 = 8
=
=
=
2. Arrange the following in descending order:
(i)
(ii)
Solution –
(i)
LCM of 9, 3, 21 = 63
Clearly,
Hence,
(ii)
LCM of 5, 7, 10 = 70
Clearly,
Hence,
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 the first row =
Sum along the second row =
Sum along the third row =
Sum along the first column =
Sum along the second column =
Sum along the third column =
Sum along the first diagonal =
Sum along the second diagonal =
Since the sum of the numbers in each row, in each column, and along the diagonals is the same, it is a magic square.
4. A rectangular sheet of paper is
Solution –
Length of sheet =
Breadth of sheet =
We know that,
Perimeter of the rectangle = 2 × (length + breadth)
= 2 ×
LCM of 2, 3 = 6
= 2 ×
= 2 ×
= 2 ×
=
Hence, the perimeter of the sheet of paper is
5. Find the perimeters of (i) ΔABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
Solution –
Given AB =
(i) Perimeter of ∆ABE
Perimeter of the triangle = Sum of all sides
Perimeter of ∆ABE = AB + BE + EA
=
The LCM of 2, 4 and 5 = 20
=
=
=
(ii) Perimeter of the rectangle BCDE
Perimeter of the rectangle = 2 × (length + breadth)
Perimeter of rectangle BCDE = 2 × (BE + ED)
= 2 × [
The LCM of 4, 6 = 12
= 2 ×
= 2 ×
= 2 ×
Perimeter of triangle ABE =
Perimeter of rectangle BCDE =
LCM of 20 and 6 = 60
Clearly,
Hence,
∴ Perimeter of ∆ABE > Perimeter of Rectangle BCDE
6. Salil wants to put a picture in a frame. The picture is
Solution –
The width of Picture =
Required width of a Frame =
∴ The picture should be trimmed by =
The LCM of 5, 10 = 10
=
=
=
Thus, the picture should be trimmed by
7. Ritu ate
Solution –
Part of apple eaten by Ritu is =
Part of apple eaten by Somu is = 1 – Part of apple eaten by Ritu
= 1 –
The LCM of 1, 5 = 5
=
=
∴ Part of apple eaten by Somu is
So,
Now, the difference between the two shares =
=
Thus, Ritu’s share is larger than share of Somu by
8. Michael finished colouring a picture in
Solution –
Time taken by Michael to colour the picture is =
Time taken by the Vaibhav to colour the picture is =
The LCM of 12, 4 = 12
Clearly,
Hence,
Thus, Vaibhav worked for longer time.
So, Vaibhav worked longer time by =
=
=
=
=