Mathematics (English) - For Class SSC - Part 1 (Science Group) - Solved Model papers 2020 -2021- AS PER CONDENSED SYLLABUS

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Mathematics (English)
For Class SSC - Part 1 (Science Groups)
Solved Model papers 2020 -2021
As Per condensed Syllabus

SECTION "B"
(Short Answer Question)

Q.2: If S = {10, 20, 30, 40}, then find P (S)
Solution:-
|P (S)| = 2ⁿ = 2⁴ = 2 x 2 x 2 x 2 = 16 (where 'n' is number of elements in set S.)
(Hence, Set S has 16 subsets or power sets.)
|P (S)| = { { }, {10}, {20}, {30}, {40}, {10, 20}, {10, 30}, (10, 40), {20, 30}, {20, 40}, {30, 40}, {10, 20, 30), {10, 20, 40}, {10, 30, 40}, {20, 30, 40}, {10, 20, 30, 40} } Ans.

Q.3: If A = {a, b, c, d} and B = {b, d, e, f}, then
Prove that (AUB) - (A∩B) = (A ∆ B)
Solution:
Taking R.H.S:
AΔB = {a, b, c, d} Δ {b, d, e, f}
(By selecting uncommon members among both sets)
We get AΔB = {a, c, e, f} ....... (1)

Taking L.H.S:
AUB = {a, b, c, d} U {b, d, e, f}
AUB = (a, b, c, d, e, f)

A∩B = {a, b, c, d} ∩ {b, d, e, f}
A∩B = (b, d)
AUB - A∩B = (a, b, c, d, e, f) - (b, d)
We get AUB - A∩B = {a, c, e, f} ....... (2)
By (1) and (2)

AΔB = AUB - A∩B Hence Proved

Q.4: Find the value of a²+b²+c² when a+b+c = √ 17 and ab+bc+ca = 2.
Solution:
Given that:
(a+b+c) = √ 17
By squaring on both side
(a+b+c)² = (√ 17)²
From Formula:

 (a+b+c)² = a²+b²+c² + 2ab + 2bc + 2ca 

(a+b+c)² = a²+b²+c² + 2ab + 2bc + 2ca = 17
(a+b+c)² = a²+b²+c² + 2 (ab + bc + ca) = 17
(a+b+c)² = a²+b²+c² + 2 (2) = 17
(a+b+c)² = a²+b²+c² + 4 = 17
(a+b+c)² = a²+b²+c² = 17 - 4

(a+b+c)² = a²+b²+c² = 13 ANSWER

Q.5: Find the value of 85.7 x 2.47 / 8.89 with the help of logarithmic table.

Q.6: What should be added to x⁴ + 4x³ + 10x² + 5, so that it may be a perfect square?

Q.7: If x + y = 7 , xy = 11, then find the value of x - y
Solution:
Data:
x + y = 7
xy = 11
x - y = ?

Using Formula
4ab = (a + b)² - (a - b)²

⇒ 4xy = (x + y)² - (x - y)²
⇒ 4(11) = (7)² - (x - y)²
⇒ 44 = 49 - (x - y)²
⇒ (x - y) ² + 44 = 49
⇒ (x - y) ² = 49 - 44
⇒ (x - y) ² = 5

By taking square root on both side
⇒ √ (x - y) ² = √ 5
⇒ x - y = √ 5 Ans



Q.8: Simplify:

Q.9: 

Q.10: Find A^-1

SECTION "C"
(Descriptive Answer Question)

Q.11: 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) 4a⁴ + b⁴
By applying perfect square formula
(a + b)² = a² + 2ab + b²
⇒ [(2a²)² + 2(2a²)(b²) + (b²)² - 4a²b²]
⇒(2a² + b²)² - (2ab)²
By applying Formula
a² - b² = (a-b)(a+b)
⇒ (2a² + b² - 2ab) (2a² + b² + 2ab)
⇒ Or (2a² -2ab + b²) (2a² + 2ab + b²) Ans.

II) x³ + 64y³ + z³ - 12xyz
By applying formula
a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc -ac)
⇒ (x)³ + (4y)³ + (z)³ - 3(x)(4y)(z)
⇒ {(x) + (4y) + (z)}{(x)² + (4y)² + (z)² -(x)(4y) - (4y)(z) - (z)(x)}
⇒ (x + 4y + z)(x² + 16y² + z² -4xy - 4yz - xz) Ans.

III) 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.

IV) 5x² - 13x - 6
By applying formula
Middle term breaking
⇒ 5x² - 15x + 2x - 6
⇒ 5x(x - 3) + 2(x - 3)
⇒ (5x + 2)(x - 3) Ans.

Q.12: 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.13: If a transversal intersect two parallel lines, the alternate angles so formed are congruent.

Q 14: The sum of the measures of the angles of a triangle is 180°.

Special Thanks To Sir Sajjad Akber Chandio