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

GSEB SolutionsMathsScience

**Revision 2**

**Solution 1:**

- The percentage of Oxygen is 21%.
- Percentage of Nitrogen = 78%

Percentage of Oxygen = 21%

Hence, there is 57% more Nitrogen than Oxygen.

(78 – 21)% = 57% - The percentage of other gases is 1%.

**Solution 2:**

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

- |0| = 0
- |10| = 10
- |-18| = 18
- |-3| = 3
- |23| = 23

**Solution 3:**

- (-33) + (-17)

= -33 – 17

= -(33 + 17)

= -50 - (-33) – (-17)

= -33 + 17

= -16 - 0 – 17

= -17 - (-24) + 24

= -24 + 24

= 0 - (-5) + (+5)

= -5 + 5

= 5 – 5

= 0

**Solution 4:**

- 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 = 7^{2}= 49

√7^{2}= 7

Thus, 7 should be multiplied by the smallest number 7 to get perfect square 49. - 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:**

- (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 - (2lm + 2mn) + (-7lm + 2nl)

= 2lm + 2mn – 7lm + 2nl

= 2lm – 7lm + 2mn + 2nl

= (2 – 7)lm + 2mn + 2nl

= -5lm + 2mn + 2nl

**Solution 8:**

- (15x
^{3}+ 3x^{2}– 25) – (16x^{3})

= 15x^{3}+ 3x^{2}– 25 – 16x^{3}

= 15x^{3}– 16x^{3 }+ 3x^{2}– 25

= (15 – 16)x^{3 }+ 3x^{2}– 25

= -x^{3 }+ 3x^{2}– 25 - (m
^{2}+ 6mn – 7n^{2}) – (5m^{2}+ 3mn + n^{2})

= m^{2}+ 6mn – 7n^{2}– 5m^{2}– 3mn – n^{2}

= m^{2}– 5m^{2}+ 6mn – 3mn – 7n^{2}– n^{2}

= (1 – 5)m^{2}+ (6 – 3)mn + (-7 – 1)n^{2}

= -4m^{2}+ 3mn – 8n^{2}

**Solution 9:**

∠XPC and ∠PQA are corresponding angles.

∴ m∠XPC = m∠PQA

But m∠PQA = 140° (given)

∴ m∠XPC = 140°

- ∠PQA and ∠DPQ are alternate angles.

∴ m∠PQA = m∠DPQ

But m∠PQA = 14°

∴ m∠DPQ = 140° - ∠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° - ∠BQY and ∠DPQ are corresponding angles.

∴ m∠BQY = m∠DPQ

But m∠DPQ = 140°

∴ m∠BQY = 140° - ∠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° - ∠XPD and ∠PQB are corresponding angles.

∴ m∠XPD = m∠PQB

But m∠PQB = 40°

∴ m∠XPD = 40° - ∠CPQ and ∠AQY are corresponding angles.

∴ m∠CPQ = m∠AQY

But m∠CPQ = 40°

∴ m∠AQY = 40°