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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
Section C me bhi upload kar de solve kar ke
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