10 th CBSE - · PDF file · 2013-07-28Product of the zeroes of –2x2 –...
Transcript of 10 th CBSE - · PDF file · 2013-07-28Product of the zeroes of –2x2 –...
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General Instructions All questions are compulsory.
The question paper consists of 34 questions divided into four sections A, B, C and D. Section
A comprises of 8 questions of 1 mark each, Section B comprises of 6 questions of 2 marks
each. Section C comprises of 10 questions of 3 marks each and Section D comprises of 10
questions of 4 marks each.
Question numbers 1 to 8 in Section A are multiple choice questions where you are
to select one correct option out of the given four.
There is no overall choice. However, internal choice has been provided in 1 question of two
marks each, 3 questions of three marks each and 2 questions of four marks each. You have
to attempt only one of the alternatives in all such questions,
Use of calculators is not permitted
10th CBSE Mega Test - II
Time: 3hours Max. Marks: 80
NAME OF THE CANDIDATE CONTACT NUMBER
Solution Visits: www.pioneermathematics.com/latest_updates.com
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Section - A [Questions number 1 to 10 carry one mark each]
1. The product of two equal irrational numbers is always
(a) rational (b) irrational (c) rational or irrational (d) one
Ans. (a)
2. A terminating decimal when expressed in fractional form, always has denominator in
the form of
(a) 2m. 3n, m, n 0 (b) 3m5n, m, n 0 (c) 5m7n, m, n 0 (d) 2m . 5n, m, n 0
Ans. (d)
3. A polynomial of the form ax3 + bx2 + cx + d has atmost
(a) 2 zeroes (b) 3 zeroes (c) 4 zeroes (d) 5 zeroes
Ans. (b)
4. Product of the zeroes of –2x2 – kx + 6 is
(a) – 3 (b) 3 (c) k/2 (d) k /2
Ans. (a)
5. The graph of y = - 3 is a line parallel to the
(a) x-axis (b) y-axis (c) both x and y-axis (d) none of these
Ans. (a)
6. Geeta has only Re 1 and Rs 2 coins with her. If the total number of coins that she has is
50 and the amount of money with her is Rs. 75, then the number of Re 1 and Rs. 2 coins
are, respectively
(a) 35 and 15 (b) 30 and 20 (c) 10 and 40 (d) 25 and 25
Ans. (d)
7. 0 0ABC ~ DEF; A 47 , E 83 , then C is equal to
(a) 400 (b) 500 (c) 600 (d) 700
Ans. (b)
8. The area of two similar triangles is 16 cm2 and 25 cm2 respectively. The ratio of their
corresponding altitudes is
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(a) 4 : 5 (b) 4 : 3 (c) 3 : 4 (d) 2 : 3
Ans. (a)
9. If sin θ + sin2θ = 1, then cos2 θ + cos4 θ is equal to
(a) –1 (b) 0 (c) 1 (d) 2
Ans. (c)
10. While computing mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the classes.
(b) centred at the calls marks of the classes.
(c) centred at the upper limits of the classes.
(d) centred at the lower limits of the classes.
Ans. (b)
Section - B
[Questions number 11 to 18 carry 2 marks each.]
11. Use Euclid’s division algorithm to find the HCF of 198 and 315.
198 315 1
198
117 198 1
117
81 117 1
81
36 81 2
72
9 36 4
36
X
315 > 198
315 = 198 × 1 + 117
198 = 117 × 1 + 81
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117 = 81 × 1 + 36
81 = 36 × 2 + 9
36 = 9 × 4 + 0
H. C. F. = 9
12. In Fig. 1, the graph of some polynomial p(x) is given. Find the zeroes of the polynomial.
Sol :
The graph intersects the X-axis at x = – 3 and x = - 1
Zeroes of the polynomial are –3 and - 1
13. Find the value of k for which the following system of equations has no solutions:
2x + ky = 1; 3x – 5y = 7.
Sol :
Equations are 2x + ky = 1; 3x – 5y = 7
Since these equations have no solution we have
1 1 1
2 2 2
a b c
a b c
Here a1 = 2 b1 = k c1 = 1
a1= = 3 b2= = –5 c2 = 7
2 k 1
3 5 7
10k
3
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14. Solve for x and y:
2 313
x yx, y 0
5 42
x y
OR
Solve for x and y : ax + by – a + b = 0, bx – ay – a – b = 0.
Sol :
2 313
x y ..(i)
5 42
x y ..(ii)
Multiplying (i) by 4 and (ii) by 3, we get
8 1252
x y
15 126
x y23
46x
..(by adding)
46x = 23 1
x2
Putting the value of x in (i), we get
2 313
1 y
2
3
4 13y
3
y= 13 – 4 = 9
9y = 3 3 1
y9 3
1 1x , y
2 3
Or
ax + by – a + b = 0 ax + by = a – b ..(i)
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bx – ay – a – b = 0 bx – ay = a + b ..(ii)
Multiplying (i) by a and (ii) by b, we get
2 2
2 2
2 2 2 2
a x aby a ab
b x aby ab b
a b x a b
…(by adding)
2 2
2 2
a bx 1
a b
Putting the value of x in (i), we get
a(1) + by = a – b by = a – b – a
by = – b y = –1
x = 1, y = – 1
15. A right triangle has hypotenuse of length q cm and one side of length p cm. If (q – p) = 2,
express the length of third side of the right triangle in terms of q.
Sol :
Let the third side b x
A.T.Q. x2 + p2 = q2 ..(i)
Given that q – p = 2
p = q – 2
Putting value of p in (i), we get
x2 + (q – 2)2 = q2 x2 + q2 + 4 – 4q = q2
x2 = q2 – q2 – 4 + 4q
x2 = 4q – 4 x2 = 4 (q – 1)
x 2 q 1
16. If 3 cot =2, find the value of 2 sin 3 cos
.2sin 3 cos
Sol :
23 cot 2 cot
3
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Required expression = 2 sin 3 cos
2 sin 3 cos
=
2 sin 3 cos
2 3 cotsin sin2 sin 3cos 2 3 cotsin sin
=
22 3
2 2 03 02 2 2 42 33
17. If 3
sin5
, find the value of tan sec 2.
Sol :
3Sin
5
22 2 2 3
cos 1 sin cos 15
2 9cos 1
25
9 16
cos 125 25
4
cos5
1 5
seccos 4
3sin 3 5 35tan
4cos 5 4 45
2 2
2 3 5 8tan sec
4 4 4= (2)2 = 4
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18. If tan cot 2,find the value of 2 2tan cot .
Sol :
tan cot 2
Squaring both sides
2 2tan 2 tan cot cot 4
2 2tan 2 cot 4 … [ tan . cot 1 ]
2 2tan cot 4 2 2
Section - C
[Question numbers 19 to 28 carry 3 marks each.]
19. Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7
respectively.
Sol :
960 2048 2
1920
128 960 7
896
64 128 2
128
X
2053 – 5 = 2048
967 – 5 = 960
2048 > 960
2048 = 960 × 2 + 128
960 = 128 × 7 + 64
128 = 64 × 2 + 0
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H. C. F. = 64
Required number is 64
20. Prove that 7 6 5 is an irrational number.
Sol :
Let us assume, to the contray, that 7 6 5 is rational
That is, we can find coprimes a and b b 0 , such that a
7 6 5b
a
7 6 5b
7 a 7b a
56 6b 6b
Since a and b are integers , we get 7 a
6 6b is rational, and so 5 is rational
But this contradicts the fact that 5 is irrational This contradiction has arisen because
of our incorrect assumption that 7 6 5 is rational.
So, we conclude that 7 6 5 is irrational.
21. Obtain all other zeroes 3x4 – 15x3 + 13x2 + 25x – 30, if two of its zeroes are
5 5and .
3 3
OR
On dividing p(x) by a polynomial x – 1 – x2, the quotient and remainder were x – 2 and
3 respectively. Find p(x).
Sol :
Since two of the zeroes are 5 5
and3 3
25 5 5x x x
3 3 3is a factor of given polynomial
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2
2 4 3 2
4 2
3 2
3
2
2
3x 15x 18
x 5/3 3x 15x 13x 25x 30
3x 5x
15x 18x 25x 30
15x 25x
18x 30
18x 30
3x4 – 15x3 + 13x2 + 25x – 30
= 2 25x 3x 15x 18
3
= 2 5x .3 x 2 x 3
3
2
2
2
3x 15x 18
3 x 5x 6
3 x 3x 2x 6
3[x x 3 2 x 3 ]
3 x 2 x 3
Remaining two zeroes are
x – 2 = 0 or x – 3 = 0
x = 2 or x = 3
Remaining two zeroes are 2 and 3
Or
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Dividend = Divisor × Quotient + Remainder
or p(x) = g(x) × q(x) + r(x) ..(i)
Here p(x) = ? g(x) = x – 1 – x2
q(x) = x – 2, r(x) = 3
Putting these values in (i), we get
p(x) = (x – 1 – x2) (x – 2) + 3
= x2 – 2x – x + 2 – x3 + 2x2 + 3
= – x3 + 3x2 – 3x + 5
22. Draw the graph of x – y + 1 = 0 and 3x + 2y – 12 = 0. Calculate the area bounded by
these lines and X-axis.
OR
Determine the value of a and b so that the following linear equations have infinite
solutions :
(2a – 1) x + 3y – 5 = 0
3x + (b – 1) y – 2 = 0
Sol :
By plottig
In the graph, the two lines intersect at B(2, 3)
x = 2, y = 3 is the solution .
Area bounded by ABC
= 1 1
AC BD 5 32 2
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= 7.5 sq. units
Or
(2a – 1) x + 3y = 5
3x + (b – 1) y = 2
a1 = 2a – 1 b1 = 3 c1 = 5
a2 = 3 b2 = b – 1 c2 = 2
For linear equations to have infinite solutions
1 1 1
2 2 2
a b c
a b c
2a 1 3 5
3 b 1 2
I and III
2a 1 5
3 2 2a 1 2 3 5
4a – 2 = 15 4a = 15 + 2
4a = 17 17
a4
II and III
3 5
b 1 2 3 2 5 b 1
5b = 5 – 6 5b = 6 + 5
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5b = 11 b = 11
5
17 11
a and b4 5
23. In Fig, XY BC. Find the length of XY.
Sol:
Here AX = 1 cm, XB = 3 cm
Now AX + AB = 1 + 3 = 4 cm AB = 4 cm
In AXY and ABC
XAY = BAC ..(common)
AXY = BAC …(corresponding angles)
AXY ~ ABC (AA Similarly)
AX XY
AB BC …[sides are proportional]
1 XY
4 6
6 3XY
4 2 = 1.5 cm
24. In given figure ABC ~ PQR. Also ar ( ABC) = 4. ar ( PQR). If BC = 12 cm, find QR.
Sol :
ABC ~ PQR ..(Given)
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2
2
ar ABC BC
ar PQR QR [ Area of two similar s is equal to the
ratio of the squares of corresponding sides]
2
2
4 ar PQR 12
ar PQR QR …[ ar ABC 4ar PQR ]
2
2
124
1 QR
2 12
1 QR [taking square-root]
2 QR = 12 QR = 6 cm
25. Prove the identity : sin 1 cos 1 sin
cos 1 sin cos
Sol :
sin 1 cosL.H.S.
cos 1 sin
Dividing numerator and denominator by cos
sin 1 cos
tan sec 1cos cos coscos 1 sin 1 sec tancos cos cos
Using formula; 2 2sec tan 1
We get
2 2tan sec sec tan
1 sec tan
= tan sec sec tan sec tan
1 sec tan
= tan sec [1 sec tan ]
1 sec tan
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= tan sec 1 sec tan
1 sec tan
= tan sec
= sin 1 sin 1
cos cos cos = R.H.S.
26. Prove that : 2 2
2 2
2 2
sin A sin Btan A tan B
cos Acos B
OR
Find the value of :
0 0 2 0 2 0
0 0 0 0 0
tan .cot 90 sec .cosec 90 sin 35 sin 55
tan10 tan20 tan30 tan70 tan80
Sol :
L. H. S. = tan2A – tan2B
= 2 2
2 2
sin A sin B
cos A cos B
= 2 2 2 2
2 2
sin A.cos B sin B.cos A
cos A.cos B
=
2 2 2 2
2 2
sin A 1 sin B sin B 1 sin A
cos A.cos B
= 2 2 2 2 2 2
2 2
sin A sin A.sin B sin B sin A. sin B
cos A.cos B
= 2 2
2 2
sin A sin B
cos A.cos B = R. H. S.
Or
0 0 2 0 2 0
0 0 0 0 0
tan .cot 90 sec .cosec 90 sin 35 sin 55
tan10 tan20 tan30 tan70 tan80
=
2 0 2 0 0
0 0 0 0 0 0
tan .tan sec . sec sin 35 sin 90 35
1tan 10 tan20 tan 90 10 .tan 90 10
3
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= 2 2 2 0 2 0
0 0 0 0
tan sec sin 35 cos 35
1tan 10 tan20 cot 20 . cot 10
3
…
0
0
0
cot 90 tan
cosec 90 sec
1tan 30
3
= 0 0
0 0
1 1
1 1 1tan10 tan20
tan 20 tan 103
..
2 2
2 2
sec tan 1
sin cos 2
1cot
tan
= 2
2 31
3
27. Find the median from the following data :
Class interval 0–10 10–20 20–30 30–40 40-50
Frequency 3 4 6 4 3
Sol :
Median class = thnitem
2
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= 20
2= 10th item
Median lies in class 20 – 30
I = 20, c.f. = 7, f = 6, h = 10
Median – L +
nc.f.
2 hf
= 10 7
20 106
= 20 + 3 10
6 = 20 + 5 = 25
28. Find mode from the following data :
Class interval 1–3 3–5 5–7 7–9 9-11
Frequency 4 5 8 7 6
Sol :
Maximum frequency = 8
Modal Class is 5 – 7
l = 5, f0 = 5, f1 = 8, f2 = 7, h = 2
Mode = L + 1 0
1 0 2
f fh
2f f f
= 8 5
5 22 8 5 7
= 5 + 3
16 12 × 2
= 5 + 3 2
4
= 5 + 3
5 1.5 6.52
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Section - D
[Questions number 29 to 34 carry 4 marks each.]
29. Find the mean age in years using step deviation method from the frequency
distribution given below :
Class interval (age in years) Frequency
25-29
30–34
35–39
40–44
45–49
50–54
55–59
4
14
22
16
6
5
3
Total 70
Sol :
C.I.
Class interval
fi
Frequency
(xi)
Midpoint
ii
x au
h i if u
25–29
30–34
35–39
40–44
45–49
50–54
55–59
4
14
22
16
6
5
3
27
32
37
42
47
52
57
–3
–2
–1
0
1
2
3
12
28 62
22
0
6
10 25
9
Total in f 70 i if u 37
Here a = 42, in f 70, h = 5
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Mean i i
i
f ux a h
f
= 42 + 5 × 37
70
= 42 – 2.64 = 39.36
30. The following table gives production yield per hectare of wheat of 100 farms of a
village :
Production yield
(in kg/hec.)
No. of farms
50–55
55–60
60–65
65–70
70–75
75–80
2
8
12
24
38
16
Total 70
Convert the above distribution to more than type cumulative frequency distribution
and draw its ogive.
Sol :
Production
yield
(in kg/hec.)
f
Number
of farms
Production
yield
more than
Cumulative
frequency
50–55
55–60
60–65
65–70
70–75
75–80
2
8
12
24
38
16
50
55
60
65
70
75
100
98
90
78
54
16
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Now, plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) taking
production yield along X-axis and number of farms along Y-axis
31. If m = sin A cos A
, n ,sin B cos B
prove that (m2 – n2) sin2B = 1 – n2.
Sol :
L. H. S. = (m2 – n2) sin2B
= 2 2
2
2 2
sin A cos Asin B
sin B cos B
= 2 2 2 2
2
2 2
sin Acos B cos Asin Bsin B
sin B.cos B
=
2 2 2 2
2
1 cos A .cos B cos A 1 cos B
cos B
= 2 2 2 2 2 2
2
cos B cos Acos B cos A cos A cos B
cos B ..(i)
= 2 2
2
cos B cos A
cos B
R. H. S. = 1 – n2
= 2
2
cos A1
cos B
= 2 2
2
cos B cos A
cos B ..(ii)
L.H.S. = R.H.S.
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32. If 3 3x sin ycos sin cos , and x sin = y cos , then prove that x2 + y2 = 1.
OR
Prove that: 2 2
2
2 2 2
tan cos1 2 cos
tan 1 sin cos
Sol :
3 3x sin y cos sin cos
x sin y cos
xsiny ...(i)
cos
3 3sinx sin x cos sin cos
cos [From (i)]
3 2 2x sin xsin cos sin cos
2 2x sin sin cos sin cos
x sin .1 sin cos
sin cos
xsin
x cos
From (i), y = sin
cos .cos
y sin
L. H. S. = x2 + y2
= 2 2sin cos
= 1= R. H. S.
Or
L. H. S = 2 2
2 2 2
tan cos
tan 1 sin cos
=
2
22
2 2 2
2
sincoscos
sin 1 sin cos
1cos
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=
2
22
2 2 2 2
2
sincoscos
sin cos sin cos
cos
= 2 2
2 2 2 2
sin cos
sin cos sin cos
= 2 2
2 2 2 2
sin cos 1
sin cos sin cos
= 2 2
1
1 cos cos
= 2
1
1 2 cos = R. H. S.
33. The ratio of the areas of similar triangles is equal to the ratio squares on the
corresponding sides–Prove This.
Using the above theorem, prove that of the area of the equilateral triangle described on
the side of a square is half the area of the equilateral triangle described on its diagonal.
Sol :
Part I. See theorem 1, Page 96
Part II.
Given : ACP and BCQ are two similar s, drawn on the diagonal and side of a
square ABCD
To Prove : 2 BCQ = ACP
Proof : ACP ~ BCQ
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2
2
ar ACP AC
ar BCQ BC
But AC2 = BC2 + AB2
= 2BC2 [ AB = BC]
2
2
ar ACP 2BC2
ar BCQ BC
ar ACP 2 ar BCQ O.E.D.
34. Solve the system of equations:bx ay
a b + a + b = 0 and bx – ay + 2ab = 0.
OR
The sum of the digits of a two-digit number is 12. The number obtained by
interchanging the two digits exceeds the given number by 18. Find the number.
Sol:
bx aya b 0
a b
bx ay
a ba b
2 2b x a y
a bab
b2x – a2y = –a2b – ab2 ..(i)
Again, bx – ay = 2ab = 0
bx – ay = – 2ab ..(ii)
Multiplying (i) by 1 and (ii) by a, we get
2 2 2 2
2 2
2 2 2
b x a y a b ab
abx a y 2a b
b x abx a b ab ....(By subtracting)
bx (b – a) = ab(a – b)
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ab a b
x ab b a
Putting the value of x in (ii), we get
b(–a) – ay = –2ab
–ab – ay = – 2ab
–ay = –2ab + ab
ab
y ba
x =–a, y = b
Or
Let the unit’s place digit be x and the ten’s place digit be y
Original number = x + 10y and Interchanged number = y + 10x
Given, x + y = 12
y = 12 – x ..(i)
A.T.Q.,
(10x + y) – (x + 10y) = 18
10x + y –x –10y = 18
9x – 9y = 18
x – y = 2 …(Dividing by 9)
x – (12 – x) = 2 ..[From (i)]
x – 12 + x = 2
2x = 2 + 12 = 14 x = 7
Putting the value of x in (i), we get
y = 12 – 7 = 5
The original number = x + 10y
= 7 + 10(5) = 57