GSEB Solutions for Class 7 Mathematics – Revision 2

GSEB Solutions for Class 7 Mathematics – Revision 2 (English Medium)

GSEB SolutionsMathsScience
Revision 2

Solution 1:

1. The percentage of Oxygen is 21%.
2. Percentage of Nitrogen = 78%
   Percentage of Oxygen = 21%
   Hence, there is 57% more Nitrogen than Oxygen.
   (78 – 21)% = 57%
3. The percentage of other gases is 1%.

Solution 2:

When a number is denoted using the modulus symbol, its absolute value is always positive.

1. |0| = 0
2. |10| = 10
3. |-18| = 18
4. |-3| = 3
5. |23| = 23

Solution 3:

1. (-33) + (-17)
   = -33 – 17
   = -(33 + 17)
   = -50
2. (-33) – (-17)
   = -33 + 17
   = -16
3. 0 – 17
   = -17
4.  (-24) + 24
   = -24 + 24
   = 0
5.  (-5) + (+5)
   = -5 + 5
   = 5 – 5
   = 0

Solution 4:

1.  Since 7 is a prime number, it cannot be factorized.
   Hence, multiply 7 by the smallest number 7 to get a perfect square.
   i.e. 7 × 7 = 7² = 49
   √7² = 7
   Thus, 7 should be multiplied by the smallest number 7 to get perfect square 49.
2.  75
   
   75 = 3 × 5 × 5
   The prime factor 3 does not make a pair.
   Hence, to make the product a perfect square, the number 75 should be multiplied by 3.
   
   Hence, 75 × 3 = 3 × 3 × 5 × 5

Solution 5:

Solution 6:

Solution 7:

1.  (a – b + ab) + (b – c + bc) + (c – a + ac)
   = a – b + ab + b – c + bc + c – a + ac
   = a – a – b + b – c + c + ab + bc + ac
   = 0 + 0 + 0 + ab + bc + ac
   = ab + bc + ac
2. (2lm + 2mn) + (-7lm + 2nl)
   = 2lm + 2mn – 7lm + 2nl
   = 2lm – 7lm + 2mn + 2nl
   = (2 – 7)lm + 2mn + 2nl
   = -5lm + 2mn + 2nl

Solution 8:

1. (15x³ + 3x² – 25) – (16x³)
   = 15x³ + 3x² – 25 – 16x³
   = 15x³ – 16x³ + 3x² – 25
   = (15 – 16)x³ + 3x² – 25
   = -x³ + 3x² – 25
2.  (m² + 6mn – 7n²) – (5m² + 3mn + n²)
   = m² + 6mn – 7n² – 5m² – 3mn – n²
   = m² – 5m² + 6mn – 3mn – 7n² – n²
   = (1 – 5)m² + (6 – 3)mn + (-7 – 1)n²
   = -4m² + 3mn – 8n²

Solution 9:

∠XPC and ∠PQA are corresponding angles.
∴ m∠XPC = m∠PQA
But m∠PQA = 140° (given)
∴ m∠XPC = 140°

1.  ∠PQA and ∠DPQ are alternate angles.
   ∴ m∠PQA = m∠DPQ
   But m∠PQA = 14°
   ∴ m∠DPQ = 140°
2. ∠CPQ and ∠PQA are interior angles on the same side of the transversal.
   (Interior angles on the same side of the transversal are supplementary)
   ∴ m∠CPQ + m∠PQA = 180°
   m∠PQA = 140°
   ∴ 140° + m∠CPQ = 180°
   ∴ m∠CPQ = 180° – 140° = 40°
3.  ∠BQY and ∠DPQ are corresponding angles.
   ∴ m∠BQY = m∠DPQ
   But m∠DPQ = 140°
   ∴ m∠BQY = 140°
4.  ∠PQB and ∠DPQ are interior angles on the same side of the transversal.
   ∴ m∠PQB + m∠DPQ = 180°
   But m∠DPQ = 140°
   ∴ m∠PQB + 140° = 180°
   ∴ m∠PQB = 180° – 140° = 40°
5. ∠XPD and ∠PQB are corresponding angles.
   ∴ m∠XPD = m∠PQB
   But m∠PQB = 40°
   ∴ m∠XPD = 40°
6.  ∠CPQ and ∠AQY are corresponding angles.
   ∴ m∠CPQ = m∠AQY
   But m∠CPQ = 40°
   ∴ m∠AQY = 40°