Model Mathematics 2012-13
February 16, 2013
1. This is an open question .
2. Yes , ABCD is cyclic as the sum of the opposite angles is 180◦
The circle with AC as diameter will not pass through A. Because angle A is not 90
degree .A will be outside the circle drawn with BC as diameter
3. The coordinates of other two vertices are (7,2) and (3,-4)
4. The first term is 15 and common difference is 18. The n th term of this AP is 18n-3.
the sum of the first n terms is (15+18n−3) × n. On adding 1 to this sum we get
2
(6 + 9n)n + 1. It is 9n2 + 6n + 1. Its is equal to (3n + 1)2
5. Just produce PS to meet the circle at A.∠P AR = ∠P QR.(Reason : angle in the
same arc)
P AR is a right triangle because PA is the diameter.∠P RA = 90◦
∠RP S + ∠P AR + 90 = 180 so ,∠RP S + ∠P AR = 90
6. e, a , l form a right triangle .e2 = 72 + l2 ,l = 24
√
2
h, l, a form a right angled triangle. h2 + 72 = 242 , h = 527
2
7. Draw rough figure.Let CD be the altitude to AB and let it be h.sin 80 = h . Therefore
8
h = 8 × .98 = 7.84
Area of the triangle A = 1 × 15 × 7.84 = 58.8
2
8. A)Let x be the width . So length is x+6 . The product x(x+6) = 135, x2 +6x−135 =
0, x = 9 .Length = 15 , width 9
B)Let x be the length and y the width. 2x + 2y = 40, x+y=20 . Therefore y=20-x
. The product x × y = 120 , x(20 − x) = 120, x2 − 20x + 120 = 0, The discrimenent
of the equation (−20)2 − 4 × 1 × 120 = −80. No solution for this eqaution . Such a
rectangle cannot be constructed.
9. There are 10 balls in the first bag.10 balls in the second bag. The total number of
outcomes is 10 × 10 = 100.The number of favourable outcomes is 3 × 4 + 7 × 6. The
probability of getting the balls of same colour is 3×4+7×6 = 100 .The probability of
54
100
getting balls of different colours is 3×6+4×7 = 100 46
100
10. Draw the coordinate axes and mark the points in the plane
11. The remainder obtained by dividing the given polynomial by (x-2) is P(2) = 48+4k.
The ramainder obtained by dividing the given polynomial by (x-3) is 103+9k.
103 + 9k = 48 + 4k implies k=-11
1
The polynomial is 2x2 − 11x2 + 17x − 2. P( 2 ) is 4, not Zero . So (2x-1) is not a
factor
√ √
12. AB= (2 −− 1)2 + (1 − 5)2 = 25 √ 5 BC= (7 − 2)2 + (−11 − 1)2 = 169 =
=
13,AC= (7 −− 1)2 + (−11 − 5)2 = 320
The sum of any two sides is more than the third side . So AB , AC and BC can be
the sides of the triangle
13. The sum of the product of frequency and sectional mean is 3800 , sum of the fre-
quencies is 50 . The mean is 3800 =76
50
14. Draw a circle of radius 3.5 . Divide the angle around the center as 100, 100 160 with
reference to any radius . The arms of the angle meet the circle at the points which
are the verices of the required triangle
15. This is just like a circulat pattern . The angles form an AP :9,15, 21 —.To get the
angle sum 360 , ten terms are needed . This can be find out as follows
360 = 9+(6n+3) × n , n=10 So n radial segments of 10 cm are needed .
2
16. The first term is 4500. common difference is 500. The 15 th term is 4500+14×500 =
11500. The cost for making last metre is 11500, which is 7000 more than the cost of
first metre. Total expense is 120000 .
17. Construct incircle . There is a mistake in the question. How can we construct such
a triangle!
18. 2x2 − 5x − 3 is written in the form of a second degree equation and determine the
solutions. They are 3, −1 . so (x-3)(2x+1) is the factorised form of of the given
2
polynomial
19. PB = PA ( tangents from external point to a circle) , PC = PA , Therefore PB =
PC . Consider the triangles PBA and PAC . They are congruent . so AB = AC
20. Median is 40.4
21. Initial speed of the car is x.Time taken is 300 .When x becomes x+5 , t becomes t-3
x
300
keeping the product (x+5)(t-2) =300 same . so , we camn wriet 300x − 2 = x+5 . It
gives the second degree equation x2 + 5x − 750 = 0 Solving the speed of the car is
25 km/h
22. Let O be the center of the circumcircle of ABC.Drop OP perpendicualr to BC.
5
∠A = 40. ∠BOC = 80, ∠BOP = 40 sin 40 = x . Radius x = 7.8 , Diameter = 15.6.
y AP
Let OP = y . cos 40 = 7.8 y= 5.92 Consider right triangle APB . AB = sin 65,
13.72
AB = .91 . AB =15.07. Similarly we can find AC .
23. common radius is 6 by applying Pythagorus theorem . Adding the volume of come
and hemisphere , then divide it by volume of a small sphere of radius .6 , we get the
number of spheres is 833.
B) r=3 , h= 10-6=4, Volume = 36π CSA of cylinder is 24π, Two hemispheres
become a sphere of surface area 36π. Total area to be painted is 60π=188.4m2 Cost
is 188.4 × 250 = 47100
24. A(4,-2) is on the line 3x+2y-8=0. B(5,3) is on 2x-3y-1=0.
C(2,1) is the point of intersection of the lines .
Slop of line passing through C and A is (−2−1) = −3 2
(4−2)
Slop of CB is (1−3) = −2 = 2 . The product is -1. Since the product of the slopes is
−3 3
(2−5)
-1, the lines are perpendicular.
