Max. Marks: 80
Class- X
Mathematics-Basic (241)
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Marking Scheme SQP-2020-21
Duration: 3hrs
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156 = 22 x 3 x 13
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Quadratic polynomial is given by x2 - (a +b) x +ab
x2-2x -8
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HCF X LCM =product of two numbers
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LCM (96,404) = 96 X 404
4
LCM = 9696
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OR
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Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the factors occur.
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x – 2y =0
3x + 4y -20 =0
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1/3 ? -2/4
As, a1/a2 ? b1/b2 is one condition for consistency.
Therefore, the pair of equations is consistent.
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? = 60°
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Area of sector = ?/360° ?r2
A = 60°/360° X 22/7 X (6)2 cm2
A= 1/6 X 22/7 X36 cm2
= 18.86cm2
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OR
Another method-
Horse can graze in the field which is a circle of radius 28 cm.
So, required perimeter = 2?r= 2.?(28) cm
=2 x 22/7 X (28)cm
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= 176 cm
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By converse of Thale's theorem DE II BC
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?ADE = ? ABC = 70°
Given ? BAC = 50°
?ABC + ? BAC +?BCA =180° (Angle sum prop of triangles)
70° + 50° + ? BCA = 180°
?BCA = 180° - 120° = 60°
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OR
EC = AC – AE = (7- 3.5) cm 3.5 cm
AD/BD = 2/3 and AE/EC= 3.5/3.5 = 1
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So, AD/BD ? AE/EC
Hence, By converse of Thale's Theorem, DE is not Parallel to BC.
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Length of the fence = Total cost/Rate
= Rs.5280/Rs.24/metre = 220 m
So, length of fence = Circumference of the field
: 220m= 2 ?r=2X22/7X r
So, r = 220 x 7/2 x 22 m =35 m
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- //Question 15
This download link is referred from the post: CBSE Class 10 Marking Scheme 2021 Model Question Paper || CBSE Board Exams 2021 Marking Scheme
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