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GSEB Solutions for Class 7 Mathematics – Revision 1

GSEB Solutions for Class 7 Mathematics – Revision 1 (English Medium)

GSEB SolutionsMathsScience
Revision 1

Solution 1:

1. Total number of saplings planted
= Saplings of Amla + Saplings of Gulmohar + Saplings of Asopalav + Saplings of Neem
= 6 + 18 + 21 + 15
= 60
2. 6 Saplings of Amla were planted.
3. Number of Asopalav Saplings = 21
gseb-solutions-class-7-mathematics-revision-1-1.1
4. Number of Gulmohar saplings = 18
Number of Neem saplings = 15
Thus, the number of Gulmohar saplings exceeding the number of Neem saplings
= Number of Gulmohar saplings – Number of Neem saplings
= 18 – 15
= 3
5. Number of Gulmohar saplings = 18
gseb-solutions-class-7-mathematics-revision-1-1.2

Solution 2:

1. Total number of runs scored
= Runs scored by Sachin + Runs scored by Ravindra + Runs scored by Sehwag + Runs scored by Parthiv
= 75 + 70 + 110 + 45
= 300
2. Sehwag scored 110 runs.
3. Number of runs scored by Sachin = 75
gseb-solutions-class-7-mathematics-revision-1-2

Solution 3:

1. ‘| |’ is the symbol to show the absolute value of a number.
2. (-3) + 6 = 6 + (-3)
3. 0 × (-100) = 0
Negative integer × Zero = Zero
4. (-4) × (-4) = 16
Negative integer × Negative integer = Positive integer
5. Absolute value of (-70) is 70.
6. 7 + (-7) = 0
Addition of an integer with its additive inverse is zero.
7. 1 × 15 = 15
8. 3 × (-4) × 5 = 3 × 5 × (-4)

Solution 4:

1.
gseb-solutions-class-7-mathematics-revision-1-4.1
24 = 2 × 2 × 2 × 3
Here, the prime factors 2 and 3 do not form pairs.
Hence, 24 is not a perfect square.
2.
gseb-solutions-class-7-mathematics-revision-1-4.2
81 = 3 × 3 × 3 × 3
Here, the prime factor 3 forms two pairs.
∴ 81 = 32 × 32
gseb-solutions-class-7-mathematics-revision-1-4.3
Hence, 81 is a perfect square.
3.
gseb-solutions-class-7-mathematics-revision-1-4.4
111 = 3 × 37
Here, the prime factors 3 and 11 do not form pairs.
Hence, 111 is not a perfect square.
4.
gseb-solutions-class-7-mathematics-revision-1-4.5
576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
Here, the prime factors 2 and 3 form pairs.
∴ 576 = 22 × 22 × 22 × 32
gseb-solutions-class-7-mathematics-revision-1-4.6
Hence, 576 is a perfect square.

Solution 5:

  1. Square of 20 = 202 = 20 × 20 = 400
  2. Square of 25 = 252 = 25 × 25 = 625
  3. Square of 12 = 122 = 12 × 12 = 144
  4. Square of 10 = 102 = 10 × 10 = 100
  5. Square of 16 = 162 = 16 × 16 = 256

Solution 6:

gseb-solutions-class-7-mathematics-revision-1-6
112 = 2 × 2 × 2 × 2 × 7
Here, the prime factor 7 does not form a pair.
Hence, to make a pair of 7, 112 should be multiplied by 7.
Thus, 112 × 7 = 2 × 2 × 2 × 2 × 7 × 7
By multiplying 112 by 7, we get pairs of all the prime factors.
Hence, 112 should be multiplied by 7 so that the product becomes a perfect square.

Solution 7(1):

gseb-solutions-class-7-mathematics-revision-1-7(1)

Solution 7(2):

gseb-solutions-class-7-mathematics-revision-1-7(2)

Solution 7(3):

gseb-solutions-class-7-mathematics-revision-1-7(3)

Solution 7(4):

gseb-solutions-class-7-mathematics-revision-1-7(4)

Solution 7(5):

gseb-solutions-class-7-mathematics-revision-1-7(5)

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