Mathematics - For Class SSC - Part 2 (Science Group) - Solved Model papers 2020 - 2021- AS PER CONDENSED SYLLABUS

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Mathematics
For Class X Or SSC - Part 2 (Science Groups)
Solved Model papers 2020 - 2021
As Per condensed Syllabus

SECTION "B"

Q.2 (i) If A = {1, 2, 3, 4} and B = {2, 4, 6, 8} then
Prove that A∆B = (AUB) - (A∩B)
Solution:
Taking R.H.S:
AΔB = {1, 2, 3, 4} Δ {2, 4, 6, 8}
(By selecting uncommon members among both sets)
We get AΔB = {1, 3, 6, 8} ....... (1)

Taking L.H.S:
AUB = {1, 2, 3, 4} U {2, 4, 6, 8}
AUB = (1, 2, 3, 4, 6, 8)

A∩B = {1, 2, 3, 4} ∩ {2, 4, 6, 8}
A∩B = (2, 4)
AUB - A∩B = (1, 2, 3, 4, 6, 8) - (2, 4)
We get AUB - A∩B = {1, 3, 6, 8} ....... (2)
By (1) and (2)

AΔB = AUB - A∩B Hence Proved

Q.2 (ii) Simplify

Q.3 (i): Find the value of 85.7 x 2.47 / 8.89 with the help of logarithmic table.

Q.3 (ii) If y - z = , yz = -5, then find the value of y³ - z³
Solution:
Data:
y³ - z³ = ?
yz = -5
y - z = 4

Calculation:
⇒ y - z = 4
Taking cube on both side
⇒ (y -z)³ = (4)³

Using Formula
(a - b)³ = a³ - 3a²b + 3ab² - b³

⇒ y³ - 3y²z + 3yz² - z³ = 64
⇒ y³ - 3yz(y - z) - z³ = 64
⇒ y³ - 3(-5)(4) - z³ = 64
⇒ y³ - (-60) - z³ = 64
⇒ y³ + 60 - z³ = 64
⇒ y³ - z³ = 64 -60
⇒ y³ - z³ = 4 Ans

Q.4 (i): For what value of "p", 4a⁴ + 4a³ - 3a² - pa + 1 will be a perfect square?

Q.4(ii): Find The solution set of x+5 / 10 < 25-4x / 5 , ∀ x ∈ N

Q.5(i) Solve the equation with the help of Cramer's rule:
5x - 2y = 1,
2x - y = 0

Q.5(ii) Find relation independent of "x" from the following equation:
x - ¹/_x = 2a
x² + ¹/_x² = b²

Q.6(i): Prove that:

Q.6(ii): Solve the following equation by using quadratic formula:
3a² - 12a - 15 = 0

Q.7(i) Find the solution set of:

Q.7(ii): Find all the Values of the trigonometric ratios of 45°.

SECTION "C"

Q.8: If a side of a triangle is extended, the exterior angle so formed is, in measure, greater than either of the two interior opposite angles.

Q.9: Factorization
Sequence to apply formula for factorization
1) Try Common
2) Perfect square (a + b)² = a² + 2ab + b²
3) a² - b² = (a-b)(a+b)
4) Middle term breaking
5) (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
6) (a³ + b³) = (a + b)(a² + ab + b²)
7) a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc -ac)

I) a⁴ +2 b⁴
Solution:
By applying perfect square formula
(a + b)² = a² + 2ab + b²
⇒ [(a²)² + 2(a²)(2b²) + (2b²)² - 4a²b²]
⇒(a² + 2b²)² - (2ab)²
By applying Formula
a² - b² = (a-b)(a+b)
⇒ (a² + 2b² - 2ab) (a² + 2b² + 2ab)
⇒ Or (a² -2ab + 2b²) (a² + 2ab + 2b²) Ans.

II) x² + 11x + 10
By applying formula
Middle term breaking
⇒ x² + x + 10x + 10
⇒ x(x + 1) + 10(x + 1)
⇒ (x + 10)(x + 1) Ans.

III) a³ - a² + 2
⇒ a³ - a² + 1 + 1
⇒ a³ + 1 - a² + 1
⇒ a³ + 1 - (a² - 1)
⇒ Or (a³ + 1³) - (a² - 1²)
By applying formula
(a³ + b³) = (a + b)(a² + ab + b²)
and a² - b² = (a-b)(a+b)
⇒ {(a)³ + (1)³} - {(a)² - (1)²}
⇒ {(a + 1)[(a)² - (a)(1) + (1)²]} - {(a + 1)(a - 1)}
⇒ {(a + 1) (a² - a + 1} - {(a + 1)(a - 1)}
By applying formula
Try Common
⇒ (a + 1) (a² - a + 1) - (a - 1)
⇒ (a + 1) (a² - a + 1 - a + 1)
⇒ (a + 1) (a² - 2a + 2) Ans.

IV) 27x³ + 8y⁶ - 1 + 18xy²
By applying formula
a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc -ac)
⇒ (3x)³ + (2y²)³ + (-1) - 3(3x)(2y²)(-1)
⇒ {(3x) + (2y²) + (-1)}{(3x)² + (2y²)² + (-1)² -(3x)(2y²) - (2y²)(-1) - (-1)(3x)}
⇒ (3x + 2y² -1)(9x² + 4y⁴ + 1 -6xy² + 2y² + 3x) Ans.

(V) a² + 6ac + 9c² - b²
Solution:
⇒ a² + 6ac + 9c² - b²
⇒ (a)² + 2(a)(3c) + (3c)² - (b)²
By applying perfect square formula
(a + b)² = a² + 2ab + b²
⇒ (a + 3c)² - (b)²
By applying formula
a² - b² = (a - b)(a + b)
⇒ (a + 3c + b) (a + 3c -b)
⇒ (a + b + 3c) (a -b + 3c)Ans.

VI) x³ + 64
By applying formula
(a³ + b³) = (a + b)(a² + ab + b²)
⇒ (x)³ + (4)³
⇒ (x + 4) {(x)² + (x)(4) + (4)²}
⇒ (x + 4) (x² + 4x + 16)Ans.

Q.10: Construct a ∆PQR in which mPR = 5cm, mQR = 4.5cm, and m ∠Q = 65° and write down the steps of construction.
Solution:

Steps of Construction:

1. Draw a line segment mQR = 4.5cm.
2. At point Q draw an angle ∠RQX of measure 65°.
3. With centre R draw an arc of length 5cm which cuts Qx at point P.
4. Join R to P. such that mPR = 5cm.
5. So, we get ∆PQR.

Q 11: The sum of the measures of the angles of a triangle is 180°.

Special Thanks To Sir Sajjad Akber Chandio