Karnataka Secondary Education Examination Board (KSEEB) for SSLC 2015 Examination
SAMPLE QUESTION PAPERS solutions Mathematics
Class
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10
S O L U T I O N S SAMPLE QUESTION PAPER - 6 Self Assessment ______ _____________ _____________ _____________ _____________ _____________ ___________ ____
Time : 3 Hours 45 Minutes
Maximum Marks : 80
I. Solutions of Multiple Multiple Choice Questions 1. (C) – 1
12. Statemen Statementt : The rectangle contained
by any two sides of a triangle is equal to rectangle contained by altitude drawn to the third side and the circum diameter. 1
1
2. (B) 6
1
3. (D) 0·3
1
4. (A) 30
1
5. (C) a = 1, b = – 3
1
6. (B)
1
⇒ ∠ ABC +
∠ ABC = 180°
7. (D)
1
⇒
∠ ABC = 180°
8. (C) (0, – 7)
1
⇒
∠ ABC =
∠ ABC + ∠BOC = 180°
13.
½
× 180°
II. Solutions of one mark questions
= 72°.
9.
½
14. Volume of double cone (½ + ½ = 1)
=
1 3
2
πr (h + H ) units.
1
III. Solutions of two marks questions 15. 10.
n( A A ∪ B) = n( A A – B) + n(B – A) + n( A A ∩ B)½
= 7 + 11 + 4 11.
½ 2
½
=2×3×5
½
2
f (– (– 1) = (1) + 2(1) – 3(1) + 4
∴ LCM (12, 15, 30) = 2 × 3 × 5
=1+2–3+4 = 4.
= 22 × 3, 15 = 3 × 5 and 30
= 22. 3
12 = 2 × 2 × 3
= 60 1
HCF (12, 15, 30) = 3
½ ½
0 1 – s s a l C r e p a P n o i t s e u Q e l p m a S , ) C L S S ( a k a t a n r a K l a a w s O
S C I T A M E H T A M
S O L U T I O N S SAMPLE QUESTION PAPER - 6 Self Assessment ______ _____________ _____________ _____________ _____________ _____________ ___________ ____
Time : 3 Hours 45 Minutes
Maximum Marks : 80
I. Solutions of Multiple Multiple Choice Questions 1. (C) – 1
12. Statemen Statementt : The rectangle contained
by any two sides of a triangle is equal to rectangle contained by altitude drawn to the third side and the circum diameter. 1
1
2. (B) 6
1
3. (D) 0·3
1
4. (A) 30
1
5. (C) a = 1, b = – 3
1
6. (B)
1
⇒ ∠ ABC +
∠ ABC = 180°
7. (D)
1
⇒
∠ ABC = 180°
8. (C) (0, – 7)
1
⇒
∠ ABC =
∠ ABC + ∠BOC = 180°
13.
½
× 180°
II. Solutions of one mark questions
= 72°.
9.
½
14. Volume of double cone (½ + ½ = 1)
=
1 3
2
πr (h + H ) units.
1
III. Solutions of two marks questions 15. 10.
n( A A ∪ B) = n( A A – B) + n(B – A) + n( A A ∩ B)½
= 7 + 11 + 4 11.
½ 2
½
=2×3×5
½
2
f (– (– 1) = (1) + 2(1) – 3(1) + 4
∴ LCM (12, 15, 30) = 2 × 3 × 5
=1+2–3+4 = 4.
= 22 × 3, 15 = 3 × 5 and 30
= 22. 3
12 = 2 × 2 × 3
= 60 1
HCF (12, 15, 30) = 3
½ ½
0 1 – s s a l C r e p a P n o i t s e u Q e l p m a S , ) C L S S ( a k a t a n r a K l a a w s O
S C I T A M E H T A M
2|
OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics
Class –10
=
½
= 1½
16.
as
5–4 =1
½
21. Let p(x) = 2x3 – 5x2 + x + a and f (x) = ax3 + 2x2 – 3
U = Students appeared from examination M = Maths S = Science Percentage of students failed in both = 1% = ½ 17. (i) Three-digit numbers less than 500 =4×4×3 = 48 ½ H
4
3
P1
=–2+a
½
4
P1
= 8a + 8 – 3
=5
⇒
– 2 + a = 2(8a + 5)
⇒
– 2 + a = 16a + 10
⇒
15a = – 12 a =
½
For degree q(x) = degree r(x), we have
U P1
degree of quotient will be equal to the degree of remainder when quotient is not constant.
∴ Total required numbers = 48 + 20 + 5
= 73.
½
=
Example :
p(x) = 7x3 + 2x + 5, g (x) = 7x2
1
q(x) = x and r(x) = 2x + 5
Clearly, degree q(x) = 1 = degree q(x) =
½
x = 121.
∴
½
19. ( 3 18 + 2 12 ) ( 50 − 27 ) ½
= 9 2 ( 5 2 − 3 3) + 4 3( 5 2 − 3 3 )
½
½
g (x) · q(x) + r(x) = 7x2 (x) + (2x + 5)
= 7x3 + 2x + 5 = p(x) ½
22. Here, a = 1, b = – 3, c = 1
½
45 × 2 − 27 6 + 20 6 − 12 × 3
1
Checking for division algorithm :
Thus, division algorithm is satisfied.
= ( 9 2 + 4 3 ) (5 2 − 3 3 )
=
½
OR
½ 5
½
R1 = 2R2
∴
(iii) One-digit numbers less than 500
½
R2 = f (2) (2) = a(2)3 + 2(2)2 – 3
and
Given,
U
T
=
= 16 – 20 + 2 + a
= 8a + 5
= 5 × 4 = 20
18.
= 2(2)3 – 5(2)2 + 2 + a
P1
(ii) Two-digit numbers less than 500
5P 1
R1 = p(2)
∴
U
T
4P 1
Since R1 and R2 are the remainders r emainders when p(x) and f (x) are divided by ( x – 2).
m + n =
½
mn =
½
= ½
= 5
20.
5
+2
3 −
5
−2
=
5 ( 5 − 2 ) − 3 ( 5 + 2) ( 5 )2
−
( 2 )2
1
∴
= =
½
Sample Mathematics Class –10 OSWAAL Question CBSE (CCE),Papers
| 33 26. Rough sketch :
= =9–2=7 2
½ 2
L.H.S. = sec θ + cosec θ
23.
=
½ sin 2 θ + cos 2 θ
= =
cos2 θ · sin 2 θ
½ ½
1 cos2
θ · sin 2 θ
½
= sec2 θ · cosec2 θ
½
= R.H.S.
½
24.
Point of division P y-coordinate of P =
1 · y + 2( − 1) 1+2
= 2 (given)
y = 2 × 3 + 2 =8
(100 + 150) × 50
½
½ 27.
l = 4 cm 2πR = 18
∴ M bisect the chord AB.
MB =
× 100 × 150 ½
= 7500 + 6250 + 7500 = 21250 sq. m.
Draw OM ⊥ AB. ∴
× 150 × 100 +
+
25. Radius of circle, OB = 13 cm
=
½
=2
⇒ ⇒
Area of field ABCD = Ar (∆ APB) + Ar (trapezium PBCQ) + Ar (∆CQD) ½
½
½
⇒
R =
=
cm
½
AB
2πr = 6 =
× 24
= 12 cm
⇒
½
In rt. ∆OMB, by Pythagoras theorem OM2 = OB2 – MB2
½
= 132 – 122 OM = 5 cm.
=
cm
½
Curved surface area of frustum = πl(R + r) ½ = π × 4 = 4π ×
= 169 – 144 = 25 ∴
r =
½
= 48 cm2.
½
| 4 OSWAAL CBSE (CCE), Mathematics Class –10 4 | OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10 OR Given, r1 = 12 cm, h1 = 20 cm, r2 = 3 cm, h2 = ?
We know,
=
⇒
=
⇒
h2 =
= 5 cm ½ ∴ The cut to be made at 5 cm from the water. ∴ Height of frustum so obtained, h = 20 – 5 = 15 cm ½ Volume of the frustum = = =
2
2
πh(r1 + r2 + r1r2)
½
30.
P(red ball) =
½
P(blue ball) =
½
=2× x = 12 Hence, number of blue balls is 12. IV. Solutions of three marks questions Sn = 3n2 + 4n 31. Given,
½
⇒
½
Sn – 1 = 3(n – 1)2 + 4(n – 1)
∴
= 3(n2 – 2n + 1) + 4n – 4 = 3n2 – 2n – 1
× 189
= 2970 cm . 28. Given, N = 7, A = 10, R = 5 The graph is as follows :
½
According to the question, P(blue ball) = 2 × P(red ball)
× 15(144 + 9 + 12 × 3)
3
Number of red balls = 6 Let number of blue balls = x Total number of balls = 6 + x
½
th
n term, T n = Sn – Sn – 1
½
= (3n2 + 4n) – (3n2 – 2n – 1)
½
= 6n + 1
½
T 25 = 6 × 25 + 1
∴
= 151 2
29. Given, in ∆LMN , ∠LMN = ∠PNK = 46° ∠ M = a, PN = x or N = b, NK = c ½ LM || PN ∴ ½ (If corresponding angles are equal then lines are parallel)
1
32. d2
fd 2
− 10
100
200
185
− 5
25
125
6
252
0
0
0
47
5
235
5
25 5
125
52
2
104
10
100
200
x
f
fx
32
2
64
37
5
42
n = 20
d
=
x−x
Σfx
Σfd 2
= 840
= 650
1
Arithmetic Mean = = 42 S.D. = ∴
= ∴
(corollary of Thales theorem) ½ = ∴
x =
½
1 ½
σ =
= = 5·7
½
33. The given equation is in the form Ax2 + Bx + C = 0, where A = b – c, B = c – a and C = a – b ½
Sample Mathematics Class –10 OSWAAL Question CBSE (CCE),Papers For equal roots, 2
B – 4 AC = 0 ½ ⇒ (c – a) – 4(b – c) (a – b) = 0 ½ 2 2 2 ⇒ c – 2ac + a – 4(ab – b – ca + cb) = 0 ½ 2 2 2 ⇒ c – 2ac + a – 4ab + 4b + 4ca – 4bc = 0 c2 + a2 + 2ca – 4ab – 4bc + 4b2 = 0 ⇒ ⇒ (c + a)2 – 4b(a + c) + (2b)2 = 0 ½ 2 ⇒ [(c + a) – 2b] = 0 ⇒ (c + a) – 2b = 0 c + a = 2b. ½ ∴ OR Let the cost price be ` x. Selling price = ` 18·75 ∴ Loss = cost price – selling price 2
⇒
× x = x – 18·75
1
x2 = 100x – 1875 x2 – 100x + 1875 = 0 ⇒ x2 – 75x – 25x + 1875 = 0 ⇒ ⇒ x(x – 75) – 25(x – 75) = 0 (x – 75)(x – 25) = 0 x – 75 = 0 or x – 25 = 0 ∴ ⇒ x = 75 or x = 25 The cost price is ` 75 or ` 25. 34. Given, ∆ ABC ~ ∆PQR Since triangles are similar, therefore
| 55 OR Given : ∆ ABC ~ ∆PQR ar (∆ ABC) = ar (∆PQR)
½
To prove : ∆ ABC ≅ ∆PQR Proof : We know that areas of similar triangles are proportional to the squares of the corresponding sides. ½ ∴
=
=
=
⇒
1 =
=
=
½
⇒
= ⇒
=
1
[·.· given, areas are equal] ½ 2 2 2 ⇒ AB = PQ , BC = QR and CA = RP AB = PQ, BC = QR and CA = RP ½ ∴ Hence, ∆ ABC ≅ ∆PQR [SSS criteria] ½ 35. Let the height of flagpost, PQ = x Angle of elevation at C = 30° 2
1
= k (say)
…(i) ½
a = kp, b = kq, c = kr
½
2
2
½
When moved 6 m towards the post i.e., at B, angle of elevation = 30° + 15° = 45°
Now,
In ∆ AOB, =
½
=
⇒
In ∆OAC,
=
= k
…(ii) ½
From (i) and (ii), we have =
=
=
·
½
tan 45° = 1 =
= OB = x
1
tan 30° =
⇒
=
⇒
x + 6 =
=6⇒
…(i) 1 =6
| 6 OSWAAL CBSE (CCE), Mathematics Class –10 6 | OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10 x =
∴
Hence, height of the post =
m.
1
OR Given : O is the centre of the circle. P is an external point. PA and PB are the two tangents drawn from an external point P. ½
OR Given, x sin θ = y cos θ ⇒
x =
…(i) 1
and x sin3 θ + y cos3 θ = sin θ cos θ Eliminating x from (i) and (ii), we get .sin3 θ + y cos3 θ = sin θ cos θ 2
…(ii)
½
3
⇒ y cos θ sin θ + y cos θ = sin θ cos θ
y cos θ [sin2 θ + cos2 θ] = sin θ cos θ y cos θ (1) = sin θ cos θ ⇒ y = sin θ ⇒ …(iii) 1 Substituting this value of y in (i), x = cos θ …(iv) ∴ Squaring (iii) and (iv) and then adding, we get x2 + y2 = cos2 θ + sin 2 θ = 1 ½ 36. Given : A and B are the centres of touching circles. P is the point of contact. ½
To prove : ∠ AOB + ∠ APB = 180° Proof : In quadrilateral PAOB,
½
∠PAO + ∠ AOB + ∠OBP + ∠ APB = 360°
½
But
∠PAO = 90° and ∠OBP = 90°
½
∴ 90° + ∠ AOB + 90° + ∠ APB = 360°
½
⇒ ∠ AOB + ∠ APB = 360° – (90° + 90°)
= 360° – 180° = 180°.
½
V. Solutions of four marks questions 37. Three numbers are in G.P.
Let numbers are
a, ar.
Product of numbers = ⇒ ∴
× a × ar
343 = a3 ⇒ a = a =7
1
Sum of the three numbers. 57 =
⇒
+ a + ar
57r = a + ar + ar2 [Multiply by r] 57r = 7 + 7r + 7r2
½
57r – 7r = 7r2 + 7 ∴
To prove : A, B and P are collinear. Construction : Draw the tangent XPY . Join AP and BP. ½ Proof : Since radius drawn at the point of contact is perpendicular to the tangent. ∴
∠ APX = 90°
½
and
∠BPX = 90°
½
⇒
∠ APX = ∠BPX = 90°
7r2 – 50r + 7 = 0 7r2 – 49r – r + 7 = 0
7r(r – 7) – 1(r – 7) = 0 (r – 7) (7r – 1) = 0 r – 7 = 0, 7r = 1 r = 7 or r =
Numbers are
⇒ AP and BP lies on the same line.
½
Hence, A, B and P are collinear.
½
1
½
a, ar
7, 7 × 7 i.e., 1, 7, 49.
1
Sample Question Papers Papers (SA-2)
|| 7
OR
Let r be a common ratio of the G.P. a, b, c, d then b = ar, c = ar2 and d = ar3. ½ 2 2 2 L.H.S. = (b – c) + (c – a) + (d – b) = (ar – ar2)2 + (ar2 – a)2 + (ar3 – ar)3 2 2
2
2
2
2
2 2
2
1 2
= a r (1 – r) + a (r – 1) + a r (r – 1) = a2[r2(r – 2r + r2) + (r4 – 2r2 + 1)
+ r2(r4 – 2r2 + 1)] 1 = a2[r2 – 2r3 + r4 + r4 – 2r2 + 1 + r6 – 2r4 + r2] = a2 (r6 – 2r3 + 1) = a2 (1 – r3)2
1
32
= (a – ar )
= (a – d)2 = R.H.S. 38.
½
2
y = x + 2 The value of y for corresponding values of x are given in the following table : x
− 2
− 1
0
1
2
y
6
3
2
3
6
In ∆ ABD, AD2 = AB2 + BD2 (Pythagoras Theorem) … ½ 2 In ∆ ABC, AC = AB2 + BC2 (Pythagoras theorem) …(ii) ½ From eqn. (i), AD2 = AB2 +
(D is the mid-point of BC) ½ ⇒ 4 AD = 4 AB2 + BC2 BC2 = 4 AD2 – 4 AB2 ⇒ …(iii) ½ Using this in eqn. (ii), AC2 = AB2 + 4 AD2 – 4 AB2 ½ 2 2 2 AC = 4 AD – 3 AB . Proved. ½ C1 = R = 5·5 cm 40. C2 = r = 3·5 cm d = 5·5 + 3·5 = 9 cm C3 = R – r = 5·5 – 3·5 = 2 cm 1 Steps of construction : (a) Draw AB = 9 cm. (b) Construct circles C1 and C2 at A and B, with radius 5·5 cm and 3·5 cm respectively. ½ 2
2
The curve i.e., parabola does not intersect the x-axis. 2 ∴ There is no real value of x for x + 2 = 0. Hence, there are no real roots. 1 39. Given : ABC is right angled at B and D is the mid-point of BC. ½ BD = DC =
BC
½
(c) Draw C3 with centre A and radius R – r = 2 cm. ½ (d) Construct perpendicular bisector to AB bisecting AB at M. (e) With M as centre and radius MA, draw C4. ½ (f) Join AM and AN to intersect C3 at M and N and produce them to meet the circle C1 at P and R. ½ (g) Join BM and BN . (h) With the radius MB with P and R as centres, draw arcs to intersect C2 at Q and S. (j) Join PQ and RS. ½ ∴ PQ and RS are D.C.T. By measurement : PQ = RS = 8·7 cm. ½ ●●
8|
OSWAAL CBSE (CCE), Mathematics
Class –10
SAMPLE QUESTION PAPER - 7
Self Assessment ___________________________________________ 0 1 – s s a l C r e p a P n o i t s e u Q e l p m a S , ) C L S S ( a k a t a n r a K l a a w s O
Time : 2 Hours 45 Minutes
Maximum Marks : 80
I. Solutions of Multiple Choice Questions 1. (C) 24th 1 2. (D) a haptagon 1 3. (C) 0 ≤ r ≤ 1 1 4. (B) 0·7 1 3 5. (B) x 1 6. (D)
1 1 1
7. (B) II quadrant 8. (C) (0, 0) II. Solutions of one mark questions 9. Euclid’s division lemma : Given positive integers a and b there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b. 10. A ∩ (B ∪ C) = ( A ∩ B) ∪ ( A ∩ C). 1 11. Since the curve cut the x -axis at 3 points. ½ ∴ Number of zeroes of f (x) = 3. ½ 12. Two triangles are said to be similar if : (i) their corresponding sides are proportional and ½ (ii) their corresponding angles are equal. ½ 13. There two tangents are parallel. 1 Volume of sphere I 14. = Volume of sphere II
S C I T A M E H T A M
=
½
III. Solutions of two marks questions 15. If the number is divisible by 35, 56 and 91, then it is the LCM of these numbers. By prime factorization, 35 = 5 × 7, 56 = 2 3 × 7 and 91 = 7 × 13 3 ∴ L.C.M. (35, 56, 91) = 2 × 5 × 7 × 13 = 3640 1 The least number divisible by 35, 56 and 91 is 3640. Since it leaves a remainder 7, the required number is 3640 + 7 = 3647. 1 16. Two sets A and B are disjoint sets when A ∩ B = φ. ½ A = {1, 2, 3, 4} and B = {4, 5, 6, 7} ½ A ∩ B = {4} ≠ φ ½ ∴ A and B are not disjoint sets. ½ n C2 = 10 17.
= 10
⇒ ⇒ ⇒
n2 – n – 20 = 0 (n – 5) (n + 4) = 0
⇒ n = 5 or n = – 4, which is not possible.
n = 5.
∴
½
½ ½ ½ 3
18. The 3 particular books can be arranged in P3 ways. ½
3 particular books as one and remaining 4 books can be arranged in 5P5 ways. ½ Total number of ways = 3P3 × 5P5
½
= (3 × 2 × 1) × (5 × 4 × 3 × 2 × 1) ⇒
=
⇒
½
= 6 × 120 = 720
½
Sample Question Papers 9 | 3
=
19.
a 3 · ab +
OSWAAL CBSE (CCE), Mathematics
3
ab · b3
1
=
½
=
½
=
=
= cos2 θ – (1 – cos2 θ)
x =
20.
| 9 Class –10
2
=
⇒
= 2 cos θ – 1 = R.H.S.
½
½
½ ½
24. For line I, slope m1 =
=
⇒
1 2 6
+
5
×
2 6
−
5
2 6
−
5
½
=
=
∴ x +
½
For line II, slope m2 =
=
½
=
½
21. Let f (x) = x3 + x2 – 3x + 5 and g (x) = x – 1 By synthetic division, 1 1 1 –3 5 1 2 –1 1 2 –1 4 2 ∴ Quotient, q(x) = x + 2x – 1 1 and remainder, r(x) = 4 1 OR Here, p(x) = ax2 + bx + c Since zeroes are reciprocal to each other, then
let α and
=1
be the zeroes of p(x).
1
=
½
Since two lines are perpendicualr, then m1m2 = – 1 ∴ 1×
½
=–1
⇒
1 = – m – 2
⇒
m = –1 – 2
=–3
½
25. Steps of construction :
(a) Draw a line segment O1O2 = 5 cm. (b) With O1 and O2 as centres draw two circles C1 and C2 of same radii 3 cm.
∴ Product of zeroes, α ×
= 1 =
⇒
1+1
or a = c,
Which is the required condition. 22. Here, a = 49, b = – k , c = – 81 Let one root is m, then other root is – m. ∴ Sum of roots = ⇒
m + (– m) =
⇒
0 =
⇒
k = 0
23. L.H.S. =
1 26. Scale 25 m = 1 cm ½ ½ ½
1 ½
=
½
10 ||
OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
AP AQ AR AD QE QB RC 27. Given, r =
= 100 m = 4 cm = 150 m = 6 cm = 200 m = 8 cm = 300 m = 12 cm = 50 m = 2 cm = 75 m = 3 cm = 100 m = 4 cm
= πr(2h + l)
1
= 8 cm, h = 15 cm
l =
½
=
[4·8 + 8·5]
=
× 13·3
= 313·5 m2 2 ∴ Cost of canvas used at ` 50/m = 313·5 × 50 = ` 15675. 28. F = 6, E = 12, V = 8
½
½ ½
= = 17 cm
½
T.S.A. = 2πrh + πr2 + πrl = πr(2h + r + l)
½
=
× 8[30 + 8 + 17]
=
× 8 × 55
= 1383 cm2 (Approx.) OR
½
Euler’s formula for polyhedra F + V = E + 2 ⇒ 6 + 8 = 12 + 2 ⇒ 14 = 14, which is true Hence, Euler’s formula is verified. AB2 = AC ·AM 29. (Corollary) 2 AB = AC·16 MC ⇒ ⇒
AB = 4
Again
BC2 = AC ·MC BC = AC·MC
AC·MC
½ ½ ½ ½
…(i) ½
½
2
l =
( 4 )2
15 + 2
⇒
(Corollary) …(ii) ½
From (i) and (ii), we have AB = 4BC ½ 30. If three coins are tossed together, then the sample space is S = {HHH , HHT , HTH , THH , HTT , THT , TTH , TTT } ∴ n(S) = 8 ½ (i) An event of getting atmost two heads is A = {HHT , HTH , THH , HTT , THT , TTH , TTT } ⇒ n( A) = 7 ½ Curved surface area = 2πrh + πrl
∴
P( A) =
=
½
Sample Question Papers 11 |
OSWAAL CBSE (CCE), Mathematics
(ii) An event of getting all heads is B = {HHH } ⇒ n(B) = 1 P(B) =
∴
=
From (ii), 2(14 + 2x2) + 3x2 = 203 ⇒ 28 + 4x2 + 3x2 = 203 ⇒ 7x2 = 203 – 28
½
IV. Solutions of three marks questions 31. Let the three terms in H.P. are a, b, c. a = 2c, ∴ (given) …(i) ½ Now, b = H.M. between a and c = 20, (given) ½ ⇒
= 20
⇒
= 20
⇒
= 20 ⇒
½ 4 3
c =
∴
c = 20
= 15
½
a = 2 × 15 = 30 From (i), Hence, three terms are 30, 20 and 15.
½ ½
32. f
C.I.
20 − 40 2 40 − 600 7 60 − 80 12 80 − 100 19 100 − 120 5 n = 45 Totall
x midpoint 30 50 70 90 110
fx
d
=
x −x
d2
60 30 − 78 = − 48 2304 350 50 − 78 = − 28 784 840 70 − 78 = − 8 64 1710 90 − 78 = − 12 144 550 110 − 78 = − 32 1024 Σfx = 3510
fd 2
4608 5488 768 2736 5120 Σfd 2 = 18720
1
Now, mean
= =
= 78
½
x2 =
⇒
| 11 Class –10
½
= 25
x = 5 cm ½ 2 2 y = 2x + 14 From (i), = 2(5)2 + 14 = 64 y = 8 cm. ∴ ½ Hence, the length of the sides of squares are 5 cm and 8 cm. OR Let one odd positive number be x. then other will be = x + 2 According to the question, x2 + (x + 2)2 = 130 1 2 2 x + x + 4 + 4x = 130 ⇒ ⇒ 2x2 + 4x + 4 = 130 ⇒ 2x2 + 4x – 126 = 0 x2 + 2x – 63 = 0 ⇒ x2 + 9x – 7x – 63 = 0 ⇒ ½ x(x + 9) – 7(x + 9) = 0 ⇒ ⇒ (x – 7)(x + 9) = 0 ½ ⇒ x – 7 = 0 or x + 9 = 0 x = 7 or x = – 9 (not possible) ⇒ x =7 So, Hence, odd positive numbers are 7, 9. 1 34. Given : In ∆ ABC and ∆DEF, ∴
∠BAC = ∠EDF
Variance,
2
σ =
∠ ABC = ∠DEF
and
∠ ACB = ∠DFE
= = 416 S.D., σ =
½
= 20·396 ½ This means, each score on an average deviates from mean value, 78 by 30·396. ½ 33. Let the length of the smaller square = x m and length of the larger square = y cm According to the question, y2 – 2x2 = 14 …(i) ½ and From (i),
2 y2 + 3x2 = 203 y2 = 14 + 2x2
…(ii) ½ ½
To prove :
= Construction : Mark X on AB and Y on AC, such that AX = DE and AY = DF, Join XY . 1 Proof : In ∆ AXY and ∆DEF, ∠ A = ∠D
[Given]
12 ||
OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
∴ ∴
AX = DE,
[By construction]
AY = DF
[By construction]
∆ AXY = ∆DEF
[SAS-criteria]
XY = EF
½
and
∠ AXY = ∠DEF [Congruent triangles]
⇒
∠ AXY = ∠DEF = ∠ ABC [Given] .. ∠ AXY = ∠ ABC [ . ∠ AXY and ∠ ABC
Since ∴ ⇒
are corresponding angles] XY || BC ½ =
½
[Basic proportionality theorem and corollary] ⇒
=
In ∆OAB, tan 60°
Proved. ½
OR In the figure, ABCD is a rhombus. ∴ AB = BC = CD = AD AD || BC and AB || DC
x y
=
⇒
y =
1
∴
x – 24 =
½
½ ⇒
= 24
⇒
= 24 ⇒ x =
Hence, height of the cliff, x =
⇒
=
=4
½
=
m.
OR
In ∆ ADQ and ∆PCQ, AD || PC ∴
½
[·.· AD = BC] ½
⇒
cos θ (1 + sin θ) + cos θ (1 − sin θ) (1 − sin θ)(1 + sin θ)
= 4
1
cos θ + cos θ sin θ + cos θ − cos θ sin θ
⇒
⇒ ∴
=
+1
½
=
+1
[·.· DC = BC] ½
⇒
=4 ½
⇒
=4
⇒
=4
1= =
35. Let the height of the cliff be x from the ground. OD = x – 24 ∴ In ∆OCD,
tan 45° =
1
Proved. ½ ⇒
⇒
1 − sin2 θ
=1
= 1 ⇒ y = x – 24
1
cos θ =
=
= cos 60° ∴ θ = 60°. ½ 36. Let AB be a diameter of a given circle and let CD and EF be the tangent lines drawn to the circle at A and B respectively. Since tangent at a point to a circle is perpendicular to the radius through the point.
Sample Question Papers 13 |
OSWAAL CBSE (CCE), Mathematics
13 Class |–10
The sum of squares of extremes = 58 ∴ (a – r)2 + (a + r)2 = 58 (5 – r)2 + (5 + r)2 = 58 (25 + r2 – 10r) + (25 + r2 + 10r) = 58 2
1
2
25 + r + 25 + r = 58 2r2 = 58 – 50 r2 = AB ⊥ CD and AB ⊥ EF ∴ ∠CAB = 90° and ∠ ABF = 90° ∠CAB = ∠ ABF ∠CAB and ∠ ABF are alternate int. angles. CD || EF ∴ OR AB = BC (a) In ∆ ABC, ∴ ∠C = ∠ A ∴
1 ½ ½ 1
=4
r = ± 2
⇒
1
The terms of AP in order : a – r, a, a + r i.e., 5 – 2, 5, 5 + 2 or 5 + 2, 5, 5 – 2 i.e., 3, 5, 7 or 7, 5, 3.
1
OR
Let, given series be 4 + 12 + 16 + … Here,
a = 4, r =
…(i)
=3=
⇒ Series (i) is geometric series.
Sn =
∴
a =
=5
1
, if r > 1
½
= 52
1
S3 =
∴
Also, ∠ A + ∠B + ∠C = 180° ∠C + 68° + ∠C = 180° 2∠C = 112° ⇒ ∠C = 56° ∴ ∠ ACB = 56° 1 (b) In the figure, ∠ AOB = 2∠ ACB = 2 × 56° = 112° 1 (c) In the quadrilateral OADB, ∠ A = ∠B = 90° (·.· AD and BD are tangents to the circle) ∴ ∠ AOB and ∠ ADB = 180° 112° + ∠ ADB = 180° ∴ ∠ ADB = 180° – 112° = 68° 1 V. Solutions of four marks questions 37. Let the three terms of A.P. are a – r, a and a + r. Sum of these terms = 15, given a – r + a + a + r = 15 ⇒ 3a = 15
½
= and
S6 =
= = 1456 Now,
1
=
S3 : S6 = 1 : 28. 1 2 38. x – x – 2 = 0 ⇒ x – (x + 2) = 0 split the equation y = x2 and y = 2 + x Steps : ∴
2
(a) Prepare the table for corresponding values of x and y satisfying the equation y = x2. x
0
1
y
0
1
−
1
2
1
2
−
4
4
1
( x , y) (0 , 0) ( 1, 1) ( − 1, 1) ( 2, 4) ( − 2, 4)
(b) Prepare the table for corresponding values of x and y satisfying the equation y = 2 + x. x
0
1
2
y
2
3
4
−
1
1
−
2
0
( x , y) ( 0 , 2 ) (1, 3 ) (2 , 4 ) ( − 1 , 1) ( − 2 , 0 )
1
14 ||
OSWAAL OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
In ∆ABC and ∆DAC ,
(c) Choose the scale of x-axis, 1 cm = 1 unit y-axis, 1 cm = 1 unit (d) Plot the points (0, 0); (1, 1); (– 1, 1); (2, 4) and (– 2, 4) on the graph sheet. (e) Join the points by a smooth curve. (f) Plot the points (0, 2); (1, 3), (2, 4); (– 1, 1) and (– 2, 0) on the graph sheet. (g) Join the points to get a straight line. (h) From the intersecting points (– 1, 1) and (2, 4) of curve and the line, draw the perpendiculars to the x-axis.
∠BAC = ∠ ADC = 90° ∠ ACB = ∠ ACD ∴
∆ ABC || ∆DAC
⇒
=
⇒
½
BC·DC = AC2
…(ii) ½
By adding (i) and (ii), we get BC·BD + BC·DC = AB2 + AC2 ⇒ ⇒
BC(BD + DC) = AB2 + AC2 2
½
2
BC·BC = AB + AC ,
(as BD + DC = BC) ∴
2
2
BC = AB + AC2
½
Hence, in a right-angled triangle, square of hypotenuse is equal to the sum of the square of the other two sides. ½
1 40.
C1 = R = 3 cm, C2 = r = 2 cm C3 = R + r = 5 cm
(i) Perpendiculars meet the x-axis at the points A(– 1, 0) and B(2, 0). 2 ∴ Roots of the equation x – x – 2 = 0 are x = – 1 and x = 2. 1 39. Given : In ∆ ABC, ∠BAC = 90° BC2 = AB2 + AC2 To prove : ½ Construction : Draw AD ⊥ BC
Distance of the centres, d = (3 + 2 + 3) cm
= 8 cm
1
Steps of construction :
(a) With ‘ A’ as centre, draw circle C1 with radius 3 cm. With ‘ B’ as centre, draw circle C2 with radius 2 cm. With ‘ A’ as centre draw circle C3 with radius 5 cm. ½
1
Proof : In ∆ ABC and ∆DBA, ∠BAC = ∠BDA = 90° ∠ ABC = ∠ ABD ∴ ∆ ABC || ∆DBA ½ (AA-similarity criteria) ⇒ ⇒
= BC·BD = AB2
…(i) ½
(b) Construct the tangents BM to circle C3 from the external point B. Join AM. ½ (c) Let AM intersect circle C1 at P. (d) From B, draw AP || BR. Join PR. ∴ PR is the transverse common tangent. ½ By measurement, PR = 7 cm. ½ ●●
15 |
OSWAAL CBSE (CCE), Mathematics
Class –10
SAMPLE QUESTION PAPER - 8 Self Assessment ___________________________________________
Time : 2 Hours 45 Minutes I. Solutions of Multiple Choice Questions 1. (B) A, G, H are in G.P. 2. (D) 3
Maximum Marks : 80 13. Since M is mid point of AB, then 1 1
AM =
3. (D)
1
=
4. (D) Coefficient of variation
1
= 8 cm
5. (C) 0
1
AB
× 16
and ∠OMA = 90° ∴
2
½ 2
OM = OA – AM
6. (C)
1
7. (B) 12
1
= 100 – 64
8. (D) x = 2, y = 6
1
= 36
= (10)2 – (8)2
⇒
II. Solutions of one mark questions
L.C.M. (a, b) × H.C.F. ( a, b) = a × b
III. Solutions of two marks questions 15. Let
∴ H.C.F. (36, 32) =
∴
= 4.
½
B – A = {1, 2, 3} ∴ (B – A)′= U – (B – A) = {0, 4, 5, 6, 7, 8, 9} ½
11. Given, product of zeroes = 3
= 3k – 6 = k ⇒ 2k = 6 ⇒ k = 3.
is a rational number say r. =r
S C I 2 = ½ T ⇒ = A ∴ = M which is a contradiction as right hand E side is a rational number while is an H irrational number. ½ T Hence is an irrational number. A ½ M ½
⇒
=
⇒
½
= 2π(R + r) (h + R – r).
½
Here a = 36, b = 32 and L.C.M. (36, 32) = 288
∴
OM = 6 cm.
14. T.S.A. of right circular hollow cylinder
9. We know that
10.
2
½ ½
12. If a straight line is drawn parallel to one side of a triangle, then it divides the other two sides proportionally. 1
0 1 – s s a l C r e p a P n o i t s e u Q e l p m a S , ) C L S S ( a k a t a n r a K l a a w s O
Squaring both sides, we get
5
16 ||
OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
p(x) is zero, when = 0 or
16. (i)
i.e., when x = 1
=0 or x =
∴ Required zeroes are
and
·
1
(ii) OR p(x) = 4x + 2x3 – 3x2 + 8x + 5a 4
Let
By factor theorem, (x + 2) is a factor of p(x) if p(– 2) = 0 1
1
From (i) and (ii),
4
17. Two bowlers out of 5 can be chosen in C2 ways. ½
Remaining 9 players out 17 – 5 = 12 can be chosen in 12C9 ways. ½ ∴ Total number of ways = C2 ×
12
5
C9 = C2 ×
12
C3
½
64 – 16 – 12 – 16 + 5a = 0
⇒
20 + 5a = 0 a =
22. Here,
½
⇒
n(n – 1) = 20 × n
⇒
n – 1 = 20
⇒
n = 21.
=
and
=
∴
=
33
=
12
x2 – (m + n)x + mn = 0
½
½
27
x2 – 4x + 1 = 0.
½
1
4 cos θ = 11 sin θ
23. ⇒
=
1
∴ Required equation is
1
=
1
mn = ( 2 + 3 )( 2 − 3 )
=4–3=1
½ 4 ×3
m =
1
19. L.C.M. of orders 3 and 4 surds = 12 ∴
= – 4. and n =
m + n = =4
and
n
P2 = 20 P1
18.
⇒
∴
n
2
∴
= = 10 × 220 = 2200.
3
⇒ 4(– 2) + 2(– 2) – 3(– 2) + 8(– 2) + 5 a = 0
5
5
p(– 2) = 0
∴
A ∩ (B ∪ C) = ( A ∩ B) ∪ ( A ∪ C)
cos θ =
sin θ
½
∴
20.
=
5 ( 5 + 3 ) − 3( 5 − 3 ) (
= = 21. Let
5 )2 − (
3 )2
1
½
=
½
½
=4
½
p(x) =
=
= 24.
= =
=
1
d = AB =
·
½
Sample Question Papers 17 |
OSWAAL CBSE (CCE), Mathematics
=
17 Class |–10
½
=
BC =
= =
92
( 1)2
+ −
81 + 1
=
½
82
CA =
=
82 + 8 2
=
= 2 cm
AQ = 100 metre =
= 5 cm
AR = 120 metre =
= 6 cm
RC = 60 metre =
= 3 cm
½
Since AB = BC, the triangle is isosceles. 25. Given, OP ⊥ AB and OQ ⊥ CD ⇒ OP bisects AB and OQ bisects CD ⇒ AP =
PB = 40 metre =
AB and CQ =
CD
½
½
1
27. Volume of hemispherical tank
m3
= =
m3 m3
Volume to be emptied = Join OA and OC. In ∆OPA and ∆OQC, OP = OQ (Given) OA = OC (Radii of a same circle) ∠P = ∠Q (each is a rt. angle) ∴ ∆OPA ≅ ∆OQC (RHS property) ½ ⇒ AP = CQ (C.P.C.T.) ½ ⇒
AB =
CD ⇒ AB = CD.
= ∴
= 7 cm
AP = 50 metre =
= 2·5 cm
QE = 80 metre =
= 4 cm
99000 28
litres
½
seconds
= 990 seconds = 16·5 minutes
½
OR Diameter of hemisphere = Side of cubical box ⇒ 2R = 7 cm
½
26. Given, Scale 20 m = 1 cm
AD = 140 metre =
Required time =
1
∴
R =
cm
½
S.A. of solid = S.A. of the cube – area of base of hemisphere + CSA of hemisphere ½ 2 2 = 6l – πR + 2πR2
18 ||
OSWAAL OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
= 6l2 + πR2
½
= 6 × 49 + ½ = 332·5 cm2. 28. A network is traversable if it contains : (i) two odd nodes and any number of even nodes, or 1 (ii) all even nodes. 1 29. (i) In ∆ ABC and ∆ AMP, ∠ A = ∠ A (Common) ∠ ABC = ∠ AMP = 90° (Given)
∴
∠B = 30° + d
= 30° + 30° = 60° and ∠C = 30° + 2d = 30° + 2 × 30° = 90° ∴ ∆ ABC is a right angled triangle. 32. For English : Given, = 56 and σ = 5·75 C.V. =
× 100
=
× 100
= 10·27 C.V. for English = 10·27 Similarly, C.V. for mathematics =
=
∆ ABC ~ ∆ AMP
1 (ii) Since ∆ ABC ~ ∆ AMP (Proved) ∴ Corresponding sides are proportional. ⇒
=
1
30. The total number of mangoes after mixing = 9 + 30 = 39 n(S) = 39 ∴ ½ (a) An event of selecting a good mango, n( A) = 30
P( A) =
½
(b) An event of selecting a rotten mango, n(B) = 9 ∴
P(B) =
·
IV. Solutions of three marks questions 31. In ∆ ABC, let ∠ A = 30° Since ∠ A, ∠B and ∠C are in A.P. ∴ ∠B = 30° + d and ∠C = 30° + 2d But ∠ A + ∠B + ∠C = 180° ⇒ 30° + 30° + d + 30° + 2d = 180° ⇒ 3d = 90° ⇒ d = 30°
½
× 100
x = 3 or x =
(not possible) Hence, the whole number = 3. 1 OR Here, a = 1, b = 1, c = – (a + 2)(a + 1) x =
½ ½ ½
½
= 9·68 ½ On comparison, the co-efficient of variation (C.V.) in mathematics is the least. Hence, the performance in Mathematics is more consistent. 1 33. Let the number be x. According to the question, 3x2 – 4x = 15 ⇒ 3x2 – 4x – 15 = 0 1 2 ⇒ 3x – 9x + 5x – 15 = 0 ⇒ 3x(x – 3) + 5(x – 3) = 0 ⇒ (x – 3)(3x + 5) = 0 1 x – 3 = 0 or 3 x + 5 = 0 ⇒ ⇒
1
1
× 100
= 8·56 and C.V. for science ∴ by AA-criteria,
½ ½
½
= =
½
Sample Question Papers 19 |
=
4 a 2 + 12 a + 9
− 1±
2
=
½
=
½
x =
∴
−
In ∆PQR, ∠PQR PQ2 ∴ PD But PQ2 ∴ ⇒
19 Class |–10
OR = 90° and QP ⊥ PR = PR × PD (corollary) = 4DR (given) = PR × 4DR
PR =
PQ2
…(i) 1
4DR
1 + ( 2 a + 3) 2
= and
OSWAAL CBSE (CCE), Mathematics
= a + 1
½
x =
= – (a + 1)
=
½
34. Given : ABC is a triangle in which DE || BC. To prove :
=
½
Again, ⇒
QR2 = PR × DR (corollary) PR =
…(ii) 1
From equations (i) and (ii), we get
Construction : Draw DN ⊥ AE and EM ⊥ AD. Join BE and CD. ½
= PQ2 = 4QR2 PQ = 2QR. ⇒ Proved. 1 35. Let BD is the height of the tree. It is broken at A such that ⇒
AD = AC = x, say
=
Proof :
=
and
area ( ∆ ADE)
…(i) ½
=
area ( ∆DEB)
In ∆ ABC, ∠ ACB = 60°, ∠B = 90° ∴
tan 60° =
∆DEB and ∆DEC lies on the same base DE and
⇒
AB =
between the same parallel lines DE and BC. ½ ∴ area (∆DEB) = area (∆DEC) ½ Hence from (i) and (ii), we have
By Pythagoras theorem, AC2 = AB2 + BC2 x2 =
=
=
…(ii) ½
½
m
= 1200 + 400
1
20 ||
OSWAAL OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
= 1600
⇒ BPC is a straight line.
x = 40
⇒
1
Height of the tree = BD = AB + x = =
m.
Hence, B, P, C are collinear. ½ OR AB = 10 cm Given, AO = OB = OQ (Radii of a circle) and
1
OR
L.H.S. = =
tan
=
AB
=
× 10
= 5 cm
½
θ + sec θ − (sec 2 θ − tan 2 θ )
tan
θ − sec θ + 1
[·.· 1 = sec 2 θ – tan2 θ] 1 =
(tan
θ + sec θ) − [(sec θ + tan θ)
tan
(sec
θ − tan θ) ]
θ − sec θ + 1
½
= (tan θ + sec θ)
½
= (tan θ + sec θ) ×
½
OC = AC – OA
∴
=6–5 = 1 cm
Let
PC = PQ = x
∴
OP = OQ – PQ
=
(Radii of a circle)
= (5 – x) cm
= R.H.S.
½
36. Given : Circles with centres X and Y touch each other externally at P. Construction : Join AP, BP, PC and PD.
½
Proof : Since AB is the diameter. ∠ APB = 90° (Angle in a semi-circle) i.e., …(i) ½
½
Since OB is a tangent at C to a smaller circle. PC ⊥ CO
∴
In ∆PCO, by Pythagoras theorem, OP2 = OC2 + PC2
(5 – x)2 = (1)2 + x2 2
1
2
25 + x – 10x = 1 + x
x2 – 10x – x2 = 1 – 25 x =
= 2·4 ∴ Radius of the smaller circle = x = 2·4 cm. 1
V. Solutions of four marks questions 37. Given, b = G.M. of a and c =
Again, CD is the diameter. ∠CPD = 90°
∴
(Angle in a semi-circle) …(i) ½ From (i) and (ii), ∠ APB = ∠CPD = 90°
½
…(i) ½
x = A.M. of a and b =
…(ii) ½
y = A.M. of b and c =
…(iii) ½
From (i), b2 = ac ⇒ c =
½
Now,
½
⇒ Vertically opposite angles are equal. ⇒ APD and BPC are two intersecting straight
lines.
½
=
Sample Question Papers 21 |
OSWAAL CBSE (CCE), Mathematics
=
½
=
½
21 Class |–10
1
= ½ OR
Let the three terms of G.P. be
½
The curve is a parabola which intersect the x-axis at the points A(– 3, 0) and B(2, 0).
× a × ar = 216
Given,
3
⇒
a = (6)
⇒
a =6
2
∴ The roots of equation x + x – 6 = 0 are
3
½
and
= 156 2
⇒
a
= 156
⇒
36
= 156 r2 + r + 1 =
⇒
3r2 + 3r + 3 – 13r = 0
⇒
3r2 – 10r + 3 = 0
⇒
3r2 – 9r – r + 3 = 0
⇒
3r(r – 3) – 1(r – 3) = 0
⇒ ⇒
(r – 3) (3r – 1) = 0 r = 3
½
39. Let
1
AD = x
then
CA = 2 AD = 2x
and
BD = 3 AD = 3x
½
r
1
So, by Pythagoras theorem AC2 = AD2 + CD2 ⇒
(2x)2 = x2 + CD2
⇒
CD2 = (2x)2 – x2
= 4x2 – x2 = 3x2
r =
1
6, 6 × 3 i.e., 2, 6, 18. ½
∴ Required terms are
x = – 3 and x = 2.
In ∆ ADC, ∠ ADC = 90°
r =
⇒
or
38.
a, ar.
⇒
CD =
In ∆BDC,
∠D = 90°
BC2 = BD2 + CD2
y = x + x – 6 The values of y for corresponding values of x are given in the following table :
0
1
2
–1
–2
–3
y
– 6
–4
0
–6
–4
0
= = 9x2 + 3x2 = 12x2 BC = Also,
1
1
By Pythagoras theorem,
2
x
3x 2 + x 3
AB = AD + DB
= x + 3x = 4x
1
22 ||
OSWAAL OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
Now, AC2 + BC2 = = 4x2 + 12x2 = 16x2 = AB2
1
∴ By converse of Pythagoras theorem., ∆ ABC is right angle triangle, right angled at C.
Hence, ∠BCA = 90°.
Proved. ½
40. Distance between the centres,
d = 2 cm C1 = R = 5 cm C2 = r = 3 cm C3 = R – r
= 1·5 cm
1
Steps of construction :
(a) With ‘ A’ as centre draw circle C1. With ‘B’ as centre draw circle C2 and with ‘ A’ as centre, draw circle C3. 1
1 (b) Construct the tangents BN and BM to circle C3 from the external point B. Join AN and AM. (c) Let AN produce meet the circle C1 at P and AM produce at R. 1 (d) With centres P and R and radius PB draw two arcs to intersect C2 at Q and S. Join PQ and RS. ∴ PQ and RS are the direct common tangents. 1 ●●
23 |
OSWAAL CBSE (CCE), Mathematics
Class –10
SAMPLE QUESTION PAPER -9 Self Assessment ___________________________________________
Time : 2 Hours 45 Minutes
Maximum Marks : 80
I. Solutions of Multiple Choice Questions 1. (D) 15 2. (A) 24 3. (C) Determining the boiling point of water 4. (A) are equal to one another 5. (D) (B) and (C) are correct 6. (D) 2 7. (A) 14 8. (C)
13. ·.· 1 1 1 1 1 1
OQ ⊥ AB ⇒ ∠OQB = 90° In ∆OPQ, OP = OQ ⇒
= 25° 2
14. Volume = πr h
1 1
=
=
½
A = {1, 2, 3}
10.
and
½ ½
× 7 × 7 × 10
= 1540 cm3.
½
III. Solutions of two marks questions 15.
= which is a irrational number.
∠OPQ = ∠OQP
= 90° – 65°
II. Solution of one mark questions 9.
½
344 = 2 × 2 × 2 × 43 = 23 × 43
½
60 = 2 × 2 × 3 × 5 = 22 × 3 × 5
(·.· x ∈ N )
B = {3, 1}
½
∴ H.C.F. = (344, 60)
= 22 = 4
(·.· 3 y = 9 ⇒ y = 3 and 2 y = 2 ⇒ y = 1) ½
½
But L.C.M. (a, b) × H.C.F. ( a, b) = a × b ½
∴ A ∪ B = {1, 2, 3} ∪ {1, 3} = {1, 2, 3} = A ½
11. Dividend = Divisor × Quotient + Remainder 1 12.
∴ L.C.M. (344, 60) =
= = 5160.
½
16.
BD2 = AD × CD
∴
= 16 × 4 = 64 BD = 4 cm
½
½
1
0 1 – s s a l C r e p a P n o i t s e u Q e l p m a S , ) C L S S ( a k a t a n r a K l a a w s O
S C I T A M E H T A M
24 ||
OSWAAL OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
21. Given, p(x) = x3 – 1, g (x) = x – 1
n( A) = 35, n(B) = 20 + 10
2 + x +1 3 x −1 x
x−
= 30
)
1
x
n( A) only = 35 – 10
3 − x2
( −) ( +)
= 25 ∴
2 −1 2 x −x x
n( A ∪ B) = 25 + 10 + 20
= 55
( −)
½
( +) 1 1 ( −) ( +) 0 x−
Teachers = 6, Doctors = 4 No. of Committees : Possibilities Teachers Doctors No. of ways 6 C = 15 4C = 6 (i) 15 × 6 = 990 4 2 6 C = 20 4C = 4 (ii) 20 × 4 = 80 3 3 6 C = 15 4C = 1 (iii) 15 × 1 = 15 2 4 Total 185
17.
Total = 185 committees
2
(n + 1) ! = 12 × ( n – 1) !
18.
⇒ (n + 1)n × (n – 1) ! = 12 × ( n – 1) ! ⇒
(n + 1)n = 12, as (n – 1) !
⇒
(n + 1)n = 4 × 3
½ ≠
0 ½
= (3 + 1) = 3 n =3
∴ 1/3
19. Let
a =2
½
– 1/3
and b = 2
, then
a3 = 2 and b3 = 2 – 1 = a3 + b3 = 2 +
∴ 2
½
½
=
2
⇒ (a + b) (a + b – ab) =
½
+ 2 – 1/3) [(21/3)2 + (2 – 1/3)2 – 21/3 × 2 – 1/3)] =
⇒
(21/3 + 2 – 1/3) (22/3 + 2 – 2/3 – 1) =
Since
2
∴ Quotient, q(x) = x + x + 1
and remainder, r(x) = 0 1 Verification : We know that p(x) = g (x)·q(x) + r(x) Now, g (x)·q(x) + r(x) = (x – 1)(x2 + x + 1) + 0 = x3 + x2 + x – x2 – x – 1 = x3 – 1 = p(x) Thus, Division Algorithm is verified. 1 OR 3 p(x) = x – 3x2 + ax – 10 Let By factor theorem, if ( x – 5) is a factor of x3 – 3x2 + ax – 10, then p(5) = 0. p(5) = 0 ∴ 3 2 ⇒ (5) – 3(5) + a(5) – 10 = 0 1 ⇒ 125 – 75 + 5a – 10 = 0 ⇒ 40 + 5a = 0
⇒
½
a =
∴
22. Given,
1/3
⇒ (2
x−
∴
½
1
B = a2 = a =±
is a rational number, so R.F. of
(21/3 + 2 – 1/3) is (2 2/3 + 2 – 2/3 – 1).
=–8
=±2 Now, if B =
1
then
20. (3 2 + 2 3 )( 2 3 − 4 2 )
= 3 2 (2 3 − 4 2 ) + 2 3(2 3 − 4 2 ) 1 = 6 6 − 12 × 2 + 4 × 3 − 8 6 = =
1
a = ±2
=±2 =±2×4 = ± 8.
1
Sample Question Papers 25 |
OSWAAL CBSE (CCE), Mathematics
25 Class| –10
26. Scale 20 m = 1 cm 23.
AP = 50 m = ½
=
= 2·5 cm
PB = 100 m =
= 5 cm
AQ = 80 m =
= 4 cm
½
=
=
1
½
= tan A cot B
½
24. Let the required ratio be m : n.
Given,
( x1, y1) = (– 3, 10), (x2, y2) = (6, – 8)
and
(x, y) = (– 1, k )
½
By section formula,
⇒
x =
½
–1 =
½
⇒
– m – n = 6m – 3n
⇒
– m – 6m = – 3n + n
⇒
– 7m = – 2n
⇒
=
∴
m : n = 2 : 7.
QE = 60 m =
= 3 cm
AR = 120 m =
= 6 cm
RC = 70 m =
= 3·5 cm
AD = 200 m =
= 10 cm
1
27. Curved surface area of frustum of a cone ½
25. From construction,
= πl(r1 + r2)
1
=
× 5(10 + 4)
½
=
× 5 × 14
= 220 cm2. ½
½
OR
Given, for melted cone, h = 3·6 cm, r = 1·6 cm For recast cone, ∠ ACB = 90° = ∠ ADB
½
∠ AEB = 90° = ∠ AFB
½
Hence, angles in semi-circle are right angle. ½
R = 1·2 cm, H = ?
According to the question, Vol. of melted cone = Vol. of recasted cone
½
26 ||
OSWAAL OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
2
πr h =
⇒
2
πR H
½
2
½
2
⇒ (1·6) × (3·6) = (1·2) × H ⇒
H =
n(S) = 36 30. Here, (i) Let A be the event of getting an even sum. ∴ A = {(1, 1), (1, 3), (1, 5), (2, 2) (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)} n( A) = 18 ⇒
=
1
(ii) Let B be the event of getting a total of 7. B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} ∴ ⇒ n(B) = 6
½
= 6·4 cm Hence, height of recasted cone = 6·4 cm. 28. N = 8, A = 8, R = 6
,
P( A) =
∴
=
,
∴ P(B) =
1
1 IV. Solutions of three marks questions 31. Let 3 terms of G.P. are
Euler’s formula for a graph N + R = A + 2
∴
+ a + ar = 7
…(i)
and
× a × ar = 8
…(ii) ½
½
a3 = 23 ⇒ a = 2
(ii) ⇒
4+6 =8+2 10 = 10, which is true
From (i),
½
Hence, Euler’s formula is verified by given network.
a, ar.
+ 2 + 2r = 7 2
⇒
½ ½
=7–2
29. In ∆ AEB and ∆CED, ∠ ABE = ∠EDC
(Given)
∠ AEB = ∠CED
(V.O.A.)
⇒ ⇒ ⇒
2
=5
2r2 – 5r + 2 = 0 (r – 2) (2r – 1) = 0 r = 2 or r =
⇒
½
When r = 2, three terms of G.P. are 2, 2 × 2 i.e., 1, 2, 4 , three terms of G.P. are
When r = ∴ by AA-criterion, ∆ AEB ~ ∆CED
½
⇒
=
½
⇒
=
(Given, CD = 4 AB)
⇒
DE =
Now,
BD = BE + ED = BE + 4BE
∴
BD = 5BE
= 4BE
½
½
2 1/ 2
32.
½
, 2, 2 ×
i.e., 4, 2, 1.
Scores x
d = x−x
d2
40
− 8
64
36
− 12
144
64
− 16
256
48
0
0
52
4
16
Σx = 240
Σd 2 = 480
½
1
Sample Question Papers 27 | Here
n =5
Mean
∴
OSWAAL CBSE (CCE), Mathematics
=
½
σ =
1
C.V. =
σ
× 100 =
x
9·79
× 100 =
20·39
48
½
From (i) and (ii), we have 2x2 = 3x + 20 ⇒ 2x2 – 3x – 20 = 0 2 ⇒ 2x – 8x + 5x – 20 = 0 ⇒ 2 x(x – 4) + 5( x – 4) = 0 ⇒ (x – 4) (2 x + 5) = 0
33. Let time taken by tap of larger diameter = x hrs.
Since both the water taps together can fill a tank in 2
hrs. i.e., in
hrs.
∴ In 1 hr. both fill the portion of tank =
=
⇒
½ ½
x = 4 or x =
⇒
and time taken by tap of smaller diameter = (x + 2) hrs.
27 Class| –10
(not possible) ½
x =4 Hence, age of son = 4 years and age of father, y = 2x2 = 2(4)2 = 32 years. ½ 34. Given : In ∆ ABC in which P and Q are points on AB and AC respectively such that PQ || BC and AD is the median, cutting BC at E. ∴
1
= ⇒ ⇒ ⇒
or or ⇒ ⇒ ⇒
35(2x + 2) 12x – 46x – 70 6x2 – 23x – 35 6x2 – 30x + 7x – 35 6x(x – 5) + 7(x – 5) (6x + 7) (x – 5) 2
= 12x(x + 2) =0 =0 =0 =0 =0
x =
1
or x = 5
PE = EQ To prove : 1 Proof : In ∆ ABD, .. PE || BD ( . PQ || BC, given) ∴
=
…(i) 1 (Thales theorem)
Rejecting x = In ∆ ADC,
(because time cannot be negative) x = 5 hrs. ∴ Hence, smaller tap can fill the tank in 5 + 2 = 7 hrs and larger tap in 5 hrs. 1 OR Let the age of son = x years and the age of father = y years y = 2x2 Condition I : …(i) ½ After 8 years, son’s age = x + 8 and father’s age = y + 8 Condition II : y + 8 = 3(x + 8) + 4 ⇒ y = 3x + 24 + 4 – 8 ⇒ y = 3x + 20 ⇒ …(ii) ½
EQ || DC
(·.· PQ || BC, given)
∴ By Thales theorem,
=
…(ii)
From (i) and (ii), we have = But
BD = DC (·.· AD is the median)
∴
PE = EQ.
1
OR Given : ∆ ABC and ∆BDC are on the same base BC. To prove :
=
½
Construction : Draw AE ⊥ BC and DF ⊥ BC ½
28 ||
OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
Proof : In ∆ AOE and ∆DOF ∠ AOE = ∠DOF (Vertically opposite angles) ∠ AEO = ∠DFO = 90° (Construction)
⇒
1 =h
⇒
h =
1
OR
L.H.S. = =
½
=
½
=
tan 2 θ + cot 2 θ + 2 tan θ cot θ
as tan a cot θ = 1 1 ∴
∆ AOE ~ ∆DOF
⇒
=
(AA-Criteria) 1
Now,
=
=
½
·
Proved. ½
=
½
= tan θ + cot θ = R.H.S. ½ 36. Since tangents from an exterior point to a circle are equal in length. ∴ BP = BQ (tangents drawn from B) …(i) CP = CR (tangents drawn from C) …(ii) and AQ = AR (tangents drawn from A)…(iii) 1
35. Let height of aeroplane, PC = h km Distance between two stones A and B = 1 km Let AC = x km ⇒ BC = (1 – x) km
In ∆PCA, tan α = ⇒
x =
1
In ∆PCB, tan β = ⇒
1–x =
⇒
=
⇒
1 =
1
From (iii), we have AQ = AR ⇒ AB + BQ = AC + CR AB + BP = AC + CP ⇒ (iv) 1 Now, Perimeter of ∆ ABC = AB + BC + AC = AB + (BP + PC) + AC = ( AB + BP) + ( AC + PC) = 2( AB + BP), using (iv) = 2( AB + BQ), using (i) = 2 AQ ∴
AQ =
(Perimeter of ∆ ABC)
1
Sample Question Papers 29 |
OSWAAL CBSE (CCE), Mathematics
OR Three circles with centres A, B and C touch each other externally. Let r1, r2 and r3 are the radius of circles C1, C2 and C3 respectively. ½
a =
⇒
29 Class| –10
=
=
=
1
∴ Sum of first mn terms
=
½
= =
Let
AB = 7 cm r1 + r2 = 7
⇒
…(i)
BC = 8 cm r2 + r3 = 8
⇒
and ⇒
…(ii)
r3 + r1 = 9
…(iii) 1
Adding all there, we get 2(r1 + r2 + r3) = 7 + 8 + 9 = 24 ⇒
r1 + r2 + r3 = 12
…(iv) ½
Now apply [(iv) – (i)], we get r3 = 12 – 7
⇒
= 5 cm Apply [(iv) – (ii)], we get r1 = 12 – 8
⇒
= 4 cm Apply [(iv) – (iii)], we get r2 = 12 – 9
⇒
½
OR (i) Here, Arithmetic series is 1800 + 1750 + 1700 + … a = 1800, d = T 2 – T 1 = 1750 – 1800 = – 50, n = 12, S12 = ? ½
Sn =
CA = 9
(mn + 1).
∴ S12 =
[2a + (n – 1)d] [2 × 1800 + (12 – 1)(– 50)]
½ ½
= 6[3600 + 11(– 50)] = 6[3600 – 550] = 6 × 3050 = 18300 1 ∴ Total amount paid in 12 instalments = ` 18,300. (ii) Extra amount paid = ` 18,300 – ` 15,000 = ` 3,300 1 2 y = – x + 8x – 16 38. The values of y for corresponding values of x are given in the following table : 1
= 3 cm So, radii of the circles are 5 cm, 4 cm and 3 cm. 1 V. Solutions of four marks questions 37.
T n = a + (n – 1)d =
…(i) ½
T m = a + (m – 1)d =
…(ii) ½ 2
Apply [(i) – (ii)], we get (n – m)d = ⇒
=
α =
(i) ⇒ a + (n – 1)
1
=
30 ||
OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
The curve i.e., parabola intersects the x-axis at only one point A(4, 0). ∴ The roots of the equation are 4 and 4. 1 39. In right ∆ ABD, by Pythagoras theorem AB2 = AD2 + BD2 …(i) ½ In right ∆ ADC, by Pythagoras theorem AC2 = AD2 + DC2 …(ii) ½
Subtracting (ii) from (i), we get AB2 – AC2 = BD2 – DC2 ∴ =
⇒
CD =
∴
AB2 – AC2 =
BC
½ ½
= Thus 2 AB2 + BC2 = 2 AC2. ½ 40. Steps of construction : (a) C1 and C2 are two concentric circles with common centre O and radii 2 cm and 4 cm respectively. ½ (b) Take point P from O at a distance 8 cm such that OP = 8 cm. ½
½
– DC2, Given, BD =
CD
AB2 – AC2 =
½
Since
=
½
⇒
=
(c) M is mid-point of OP. ½ (d) With M as centre and OM as radius, draw a circle to cut C1 and C2 at A, B, C and D respectively. Join PA, PB, PC and PD. 1½ (e) PA and PB are tangents to C1. ½ (f) PC, PD and tangents to C2. ½ ●●
31 |
OSWAAL CBSE (CCE), Mathematics
Class –10
SAMPLE QUESTION PAPER -10 Self Assessment ___________________________________________
Time : 2 Hours 45 Minutes I. Solutions of Multiple Choice Questions 1. (C) 200
Maximum Marks : 80 ∴ C.S.A. of a cone
1
= πrl
½
2. (C)
1
=
3. (B) 40%
1
= 220 cm2.
4. (A)
1
× 100
1
6. (D) cot 0°
1
7. (C) y = 3x
1
8. (D) None of these
1
½
III. Solutions of two marks questions 15.
5. (B) – 4
× 7 × 10
3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17.
1
II. Solutions of one mark questions 9. We know that
first number × second number = HCF × LCM ½ ∴ HCF × LCM = 50 × 95 = 4750
n( A′) = n(U) – n( A)
10.
½ ½
= 28 – 12 = 16.
½
11. The graph of f (x) intersect the x-axis at two points. ½ ∴ Number of real zeroes of f (x) = 2.
½
12. If a straight line divides two sides of a triangle proportionally, then the straight line is parallel to the third side. 1 13. Since ∴
14. r =
AB || PT and ∠PAB = 60° ∠ APT = ∠PAB = 60°
= 7 cm, l = 10 cm
½ ½
1 16. To prove, A ∪ (B ∩ C) = ( A ∪ B) ∩ ( A ∩ C)
0 1 – s s a l C r e p a P n o i t s e u Q e l p m a S , ) C L S S ( a k a t a n r a K l a a w s O
S C I ={ } T A ∪ (B ∩ C) = {1, 2, 3, 5, 7, 11} ∪ { } = {1, 2, 3, 5, 7, 11} …(i) 1 A Again ( A ∪ B) ∩ ( A ∪ C) M = {1, 2, 3, 4, 5, 6, 7, 8, 10, 11} E ∩ {1, 2, 3, 5, 7, 9, 11} H = {1, 2, 3, 5, 7, 11} …(ii) T From (i) and (ii), we have A A ∪ (B ∩ C) = ( A ∪ B) ∩ ( A ∪ C) 1 M B ∩ C = {2, 4, 6, 8, 10} ∩ {1, 3, 5, 7, 9, 11}
32 ||
OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10 ⇒
n
Cr =
17.
1
⇒
n
Cn – r =
= nCr
=
27a + 27 – 13 = 54 – 12 + a 27a + 14 = 42 + a
⇒
26a = 28
∴
a =
1 ½
=
½
1
OR
Let 18. 9P3 + 3 × 9P2 =
·
p(x) = xn – 1
In order to show that ( x – 1) is a factor of n (x – 1), it is sufficient to show that p(1) = 0 1 ∴
p(1) = 1n – 1 = 1 – 1 = 0 n
∴ (x – 1) is a factor of ( x – 1).
1
22. Here a = 1, b = p, c = q
=
Let one root be m, then other root is 3 m.
=
½
∴
sum of roots =
= 10P3.
½
⇒
m + 3m =
19.
= = =
1
½
p 1
⇒
4m = – p
⇒
m = –
p 4
½
and product of roots =
= ⇒
= =
½
1
=q
⇒
20. R.F. of denominator ( 3 + 6) is ( 6 − 3 ) ∴
m × 3m =
=
½
=
½
=
½
3p2 = 16q.
∴
cos θ =
23.
½ ½
= =
½
21. Let
p(x) = ax3 + 3x2 – 13
g (x) = 2x3 – 4x + a
and
By Pythagoras theorem, AB2 = AC2 – BC2
By remainder theorem, the two remainders are p(3) and g (3).
= (13)2 – (5)2
By the given condition,
= 169 – 25
p(3) = g (3) 3
2
3
⇒ a(3) + 3(3) – 13 = 2(3) – 4(3) + a
= 144 1
∴
AB = 12.
½
Sample Question Papers 33 |
OSWAAL CBSE (CCE), Mathematics
Now,
sin θ =
½
and
tan θ =
½
33 Class| –10
24. AD is median. ∴ D is mid-point of BC.
½
Co-ordinates of D = = (– 2, 3) Now, AD =
= 3 cm
GC = 75 m =
=3m
½ ½
AD = 100 m =
= ∴ Length of median through A = ∠ ADB = ∠ ACB = 70°
25.
AG = 75 m =
units. ½
27. Given,
½
= 4 cm.
1
h = 20 cm
External radius R = 12·5 cm
(Angles in the same segment of a circle)
Internal radius r = 11·5 cm TSA of the pipe = 2πh(R + r) + 2π(R2 – r2)
1
= 2π(R + r) (h + R – r) = 2× =
× 44 × 24 × 21
= 3168 cm2.
In ∆ ABD, ∠DAB + ∠DBA + ∠ ADB = 180° ½ 60° + ∠DBA + 70° = 180° ½ ⇒ ∠DBA = 180° – (60° + 70°) = 180° – 130° = 50° ½ 26. Given, Scale 25 m = 1 cm
(12·5 + 11·5) (20 + 12·5 – 11·5) ½
OR
⇒
AF = 25 m =
= 1 cm
For cone,
r1 = 5 cm and h = 20 cm
Let r2 be the radius of sphere, then Vol. of the recasted sphere = Vol. of the metallic cone 3
FB = 50 m =
πr2 =
= 2 cm ⇒
AH = 50 m =
= 2 cm
HE = 50 m =
= 3 cm
½
⇒ ∴
2
πr1 h
1
4r23 = r12h r23 = r2 = 5 cm.
1
34 ||
OSWAAL OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
28. For hexahedron, F = 6, V = 8, E = 12
1
IV. Solutions of three marks questions n = 25, which is odd 31. Here,
So, middle term =
term
=
Euler’s formula for solid F + V = E + 2
½
6 + 8 = 12 + 2 ⇒
14 = 14, which is true
½
term
= 13th term T 13 = 12 ∴ ⇒ a + (13 – 1)d = 12 a + 12d = 20 ⇒ ∴
Hence, Euler formula is verified.
Sum, S25 =
[2a + (25 – 1)d]
=
29.
=
M ↔ Z, L ↔ Y , K ↔ X
KL ↔ XY , KM ↔ XZ, ML ↔ ZY ½
The yield of ragi =
·
½
n(S) = 64C3
= = 64 × 21 × 31
½
On a chess board, there will be 32 black and 32 white squares. Let A be the event of selecting 2 black and 1 white or 1 black and 2 white, n( A) = 32C2 × 32C1 + 32C1 × 32C2
= 2 × 32C2 × 32C1 × 32
= 32 × 31 × 32 P( A) =
½
× 8100
= 2250 tons The yield sugarcane =
30. There are 64 squares in a chess out of which 3 are to be selected, so
∴
½ ½
× 8100
= 900 tons
Corresponding ratios are
=2×
½
1
Corresponding sides are
=
…(i) 1
[2a + 24d]
= 25[a + 12d] = 25[20] = 500 32. (a) The yield of rice
In ∆KLM and ∆ XYZ, corresponding vertices are
½ (Given)
½
·
1
½
× 8100
= 1800 tons The yield of others=
× 8100
½
= 3150 tons ½ (b) From pie-chart increase in degree of ragi over rice = 100° – 40° = 60° ½ ∴
% increase =
× 100
= 16·66% ½ 33. Let the three consecutive numbers be x, x + 1, x + 2. As per given condition, x2 + (x + 1)2 + (x + 2)2 = 149 ½ 2 2 2 ⇒ x + x + 2x + 1 + x + 4x + 4 = 149 ½ 2 ⇒ 3x + 6x + 5 – 149 = 0 ⇒ 3x2 + 6x – 144 = 0 x2 + 2x – 48 = 0 ⇒ ½ 2 ⇒ x + 8x – 6x – 48 = 0 x(x + 8) – 6(x + 8) = 0 ⇒ ½
Sample Question Papers 35 |
OSWAAL CBSE (CCE), Mathematics
(x + 8) (x – 6) = 0 ⇒ x = – 8 or x = 6 x ≠ – 8 Since x =6 ∴ x + 1 = 7, x + 2 = 8 Hence, the three numbers are 6, 7, 8. OR Let altitude of the triangle be x cm, then base is (x + 4) cm.
35 Class| –10
⇒
Area of triangle = 48 =
⇒
= ½
it’s ½
× base × altitude ½ × (x + 4) × x
=
½
½
Hence, Area of ∆DEF : Area of ∆ ABC = 1 : 4. 1 OR EF || CD In ∆ ADE, By B.P.T.,
=
In ∆ ABC,
DE || BC
By B.P.T.,
=
…(i) 1
…(ii) 1
96 = x2 + 4x ⇒ x2 + 4x – 96 = 0 2 ⇒ x + 12x – 8x – 96 = 0 ½ ⇒x(x + 12) – 8(x + 12) = 0 ⇒ (x + 12) (x – 8) = 0 ½ x = 8 or x = – 12 (not possible) ⇒ ∴ Altitude = 8 cm. ½ 34. Proof : In ∆ ABC, D, E and F are the mid-points of AB, BC and AC respectively. ⇒
From equations (i) and (ii), we have = ⇒
AD2 = AF × AB.
1
35. Let the distance of the cloud P from the observation S is y.
FE joins the mid-points of AC and BC. ∴ FE || AB and FE =
AB
(∴ Mid-point theorem) Similarly,
DF || BC and DF =
BC
and
DE || AC and DE =
AC
1
In ∆ ABC and ∆EFD, ∠ ABC = ∠EFD
(·.· opp. angles of paralellogram BEFD) ∠BCA = ∠FDE
(·.· opp. angles of parallelogram ECFD) ∴ by AA-criteria, ∆ ABC ~ ∆EFD
1
36 ||
OSWAAL OSWAAL Karnataka CBSE(SSLC), (CCE), Mathematics Class –10
Height of cloud above lake = PT = H R be the reflection of P, then TR = H
½
In ∆PQS, tan α = H – h = x tan α
⇒
…(i) ½
In ∆SQR, tan β = H + h = x tan β
⇒
…(ii) ½
Apply [(ii) – (i)], we get 2h = x tan β – x tan α
½
x =
∴
Now in ∆PQS, sec α =
½
y = x sec α =
⇒
·
½
OR
Given,
sin θ – cos θ = (sin θ – cos θ)2 =
⇒ 2
2
⇒ sin θ + cos θ – 2 sin θ cos θ = ⇒ ⇒
½
1 – 2 sin θ cos θ =
½
2 sin θ cos θ =
½
Now, (sin θ + cos θ)2 = sin2 θ + cos2 θ + 2 sin θ cos θ = 1 + 2 sin θ cos θ =1+ 36. Let
In ∆PBQ, ⇒
3
=
4
7 4
½ ½ ½
∠BPQ = x°
BP = BQ (Radii of a same circle) ∠BQP = ∠BPQ (Angles opposite
to equal sides of a triangle) ∴
From (i) and (ii), we have ∠BQP = ∠ ARP ⇒ Corresponding angles are equal. ½ AR || BQ. ∴ Hence proved. ½ OR AB || CD Given, ∴ ∠ ABC = ∠BCD (Alternate angles) ½ But, again ∠ ABC = 55° ∴ ∠BCD = 55° ½ Again ∠BOD = 2∠BCD (central angle is twice the inscribed angle) ∴ ∠BOD = 2 × 55° = 110° 1
∠BQP = x°
∴
∠ ARP = x°
∠BOD + ∠BPD = 180°
⇒
110° + ∠BPD = 180°
∴
∠BPD = 180° – 110° = 70°.
½ ½
V. Solutions of four marks questions 37. Let the common ratio of G.P. is r and first term is a, then G.P. is
a, ar, ar2, ar3, ar4, …
According to first condition, a × ar × ar2 × ar3 × ar4 = 1
AP = AR ∠ ARP = ∠ APR
∴
…(i) 1
Similarly, in ∆PAR ⇒
Since the tangents at B and D intersect each other at point P.
…(ii) 1
⇒
a5r10 = 1
⇒
ar2 = 1
…(i) 1
Sample Question Papers 37 |
OSWAAL CBSE (CCE), Mathematics
37 Class| –10
According to second condition, a + ar + ar2 = a(1 + r + r2) =
⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒
½
1 r
2
(1 + r + r2) =
4(1 + r + r2) 4 + 4r + 4r2 3r2 – 4r – 4 3r2 – 6r + 2r – 4 3r(r – 2) + 2(r – 2) (r – 2) (3r + 2)
= 7r2 = 7r2 =0 =0 =0 =0
2 1 ½
r = 2 or r =
Hence, common ratio = 2 and
·
1
OR Let the common ratio of G.P. be r and first term be a, then G.P. is a, ar, ar2, ar3, … T 4 = 125 ⇒ ar3 = 125 …(i) 1
Sn =
When x =
then y =
From graph, when y = 5, we get x = + 3·1 or – 3·1 = ± 3·1
∴
1
39. In right ∆LMA, by Pythagoras theorem
AL2 = AM2 + LM2
½
…(i) ½
In right ∆LMB, by Pythagoras theorem LB2 = LM2 + MB2 ∴
=
⇒
=
1 + r3 = 126 r3 = 125 = (5)3 ⇒ ∴ r =5 From (i),a × 53 = 125 ⇒ 125a = 125 ⇒ a =1 ∴ Required G.P. is 1, 1 × 5, 1 × 5 2, 1 × 53, … i.e., 1, 5, 25, 125, …
…(ii) ½
⇒
38.
y =
1
In right ∆ ALB, by Pythagoras theorem AB2 = AL2 + LB2 2
2
½ 2
2
2
⇒ ( AM + BM) = AM + LM + LM + MB ½
½
2
2
⇒ AM + BM + 2 AM.BM
= 2LM2 + AM2 + BM2 1
x2
⇒
2LM2 = 2 AM.BM
⇒
LM2 = AM.BM
…(iii) ½
Dividing equation (i) by (ii), we get The values of y for corresponding values of x are given in the following table : 1
½
= =
AM 2 + AM.BM BM 2 + AM.BM
, using (iii)
½
