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Thursday, 27 May 2021

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


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