Download Maths Practice Paper Class-X

CBSE – CLASS X MATHS - 2012
General Instructions:
1. All questions are compulsory.
2. The question paper consists of 34 questions divided into four sections A, B, C and D.
3. Section A contains 10 questions of 1 mark each, which are multiple choices type
Questions, Section B contains 8 questions of 2 marks each, Section C contains 10 questions of 3
marks each, Section D contains 6 questions of 4 marks each.
4. There is no overall choice in the paper. However, internal choice is provided in one question of 2
marks, 3 questions of 3 marks each and two questions of 4 marks each.
5. Use of calculators is not permitted.
Q1. The length of shadow of a tower on the plane ground is √3 times the height of the tower. The angle
of elevation of sun is :
A.
B.
C.
D.
450
300
600
900
Q2.If the area of a circle is equal to sum of the areas of two circles of diameters 10 cm and 24 cm, then
the diameter of the larger circle (in cm) is :
A.
B.
C.
D.
34
26
17
14
Q3. If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the
ratio of the volume of the cylinder thus obtained to the volume of original cylinder is:
A.
B.
C.
D.
1:2
2:1
1:4
4:1
Q4. Two dice are thrown together. The probability of getting the same number on both dice is:
A.
B.
C.
D.
1/2
1/3
1/6
1/12
Q5. The coordinates of the point P dividing the line segment joining the points A (1, 3) and B (4, 6) in the
ratio 2 : 1
A.
B.
C.
D.
(2,4)
(3,5)
(4,2)
(5,3)
Q6.If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are
(-2, 5), then the coordinates of the other end of the diameter are :
A.
B.
C.
D.
(-6,7)
(6,-7)
(6,7)
(-6,-7)
Q7.The sum of first 20 odd natural numbers is:
A.
B.
C.
D.
100
210
400
420
Q8. If 1 is a root of the equations ay2 + ay + 3 = 0 and y2 + y + b = 0 then ab equals :
A. 3
B. −
C. 6
D. -3
7
2
Q9.In Fig. 1, the sides AB, BC and CA of a triangle ABC, touch a circle at P, Q and R respectively. If PA = 4
cm, BP = 3 cm and AC = 11 cm, then the length of BC (in cm) is :
A.
B.
C.
D.
11
10
14
15
Q10. In Fig 2, a circle touches the side DF of EDF at H and touches ED and EF produced at K and M
respectively. If EK = 9 cm, then the perimeter of
EDF (in cm) is:
A.
B.
C.
D.
18
13.5
12
9
Q11. If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value of p.
Q12. A number is selected at random from first 50 natural numbers. Find the probability that it is a
multiple of 3 and 4.
1
Q13. The volume of a hemisphere is 2425 cm3 . Find its curved surface area. [Use =
2
22
7
]
Q14 .Tangents PA and PB are drawn from an external point P to two concentric circles with centre O and
radii 8 cm and 5 cm respectively, as shown in Fig.3. If AP = 15 cm, then find the length of BP.
Q15. In Fig.4, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of
contact P bisects the base BC.
OR
In Fig.5, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle
at C. Prove that AC = CB.
Q16. In Fig.6, OABC is a square of side 7 cm. IF OAPC is a quadrant of a circle with centre O, then find the
area of the shaded regions. [Use =
22
7
]
Q17. Find the sum of all three digit natural numbers, which are multiples of 7.
Q18. Find the value(s) of k so that the quadratic equation 3x2 – 2kx + 12 = 0 has equal roots.
Q19. A point P divides the line segment joining the points A (3,-5) and B (-4, 8) such that
lies on the line x + y = 0, then fine the value of K.
AP
PB
=
K
1
. If P
Q20. If the vertices of a triangle are (1,-3), (4, p) and (-9, 7) and its area is 15 sq.units, find the value(s) of
p.
Q21. Prove that the parallelogram circumscribing a circle is a rhombus.
OR
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the
centre of the circle.
Q22. From a solid cylinder of height 7n cm and base diameter 12 cm, a conical cavity of same height and
same base diameter is hollowed out. Find the total surface area of the remaining solid. [Use =
OR
22
7
]
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied
on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, then find
the radius and slant height of the heap.
Q23. In Fig. 7, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm and 3.5 cm and
centre O. If ∠ POQ = 300, then find the area of the shaded region. [Use =
Q24.Solve for
OR
x : 4x2 – 4ax + (a2 – b2) = 0
Solve for x : 3x2− 2√6x + 2 = 0
22
7
]
Q25. A kite is flying at a height of 45 m above the ground. The string attached to the kite is temporarily
tide to a point on the ground. The inclination of the string with the ground is 600. Find the length of the
string assuming that there is no slack in the string.
Q26. Draw a triangle ABC with side BC = 6 cm, ∠ C = 300 and ∠ A = 1050. Then construct another
2
triangle whose sides are times the corresponding side of
3
ABC.
Q27.The 16th term of an AP is 1 more than twice its 8th term. If the 12th term of the AP is 47, then find its
nth term.
Q28. A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting (i) a king of red
colour (ii) a face and (iii) the queen of diamonds.
Q29. A bucket is in the form of a frustum of a cone and it can hold 28.49 litres of water. If the radii of its
circular ends are 28 cm and 21 cm, find the height of the bucket. [Use =
22
7
]
Q30. The angle of elevation of the top of a hill at the foot of a tower is 600 and the angle of depression
from the top of the tower to the foot of the hill is 300. If the tower is 50 m high, find the height of the
hill.
Q31. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of
contact.
OR
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Q32.A shopkeeper buys some books for Rs. 80. If he had bought 4 more books for the same amount,
each book would have cost Rs 1 less. Find the number of books he bought.
OR
1
The sum of two numbers is 9 and the sum of their reciprocals is . Find the numbers
2
Q33. Sum of first 20 terms of an AP is−240, and its first term is 7. Find its 24th term.
Q34. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 7 cm
and the height of the cone is equal to its diameter. Find the volume of the solid. [Use =
22
7
]
CBSE – CLASS X MATHS - 2012
General Instructions:
1. All questions are compulsory.
2. The question paper consists of 34 questions divided into four sections A, B, C and D.
3. Section A contains 10 questions of 1 mark each, which are multiple choices type
Questions, Section B contains 8 questions of 2 marks each, Section C contains 10 questions of 3
marks each, Section D contains 6 questions of 4 marks each.
4. There is no overall choice in the paper. However, internal choice is provided in one question of 2
marks, 3 questions of 3 marks each and two questions of 4 marks each.
5. Use of calculators is not permitted.
Q1
Let AB be the tower and BC be the length of the shadow of the tower.
Here, 𝜃𝜃 is the angle of elevation of the sun.
Given, length of shadow of tower = √3
BC = √3 AB …. (1)
In right
tan 𝜃𝜃 =
ABC,
AB
BC
�tan 𝜃𝜃 =
Opposite sid e
Adjacent side
�
x Height of the tower.
tan 𝜃𝜃 =
AB
√3AB
⇒ tan 𝜃𝜃 =
[ Using (1)]
1
√3
⇒ tan 𝜃𝜃 = tan 300
⇒ 𝜃𝜃 = 300
[ ∴tan 300 =
1
√3
]
Thus, the angle of elevation of the sun is 300.
Hence, the correct answer is B.
Q2
Let r 1 and r 2 be the radii of the two given circles.
Given, 2r 1 = 10 cm
∴ r 1 = 5 cm
Also, 2r 2 = 24 cm
∴ r 2 = 12 cm
Let R be the radius of the larger circle.
Given, area of larger circle = Sum of areas of two given circles
∴ 𝜋𝜋R2 = 𝜋𝜋r12 + 𝜋𝜋r22
⇒ R2 = (5 cm)2 + (12 cm)2
⇒ R2 = 25 cm2 + 144 cm2
⇒ R2 = 169 cm2
⇒ R = √169 cm
⇒ R = 13 cm
Thus, the diameter of the larger circle is ( 2 x 13 ) cm = 26 cm
Hence, the correct answer is B.
Q3
Let the radius and height of the original cylinder be r and h respectively.
∴ Volume of the original cylinder = 𝜋𝜋 r2h
According to the question, radius of the new cylinder is halved keeping the height same.
r
⇒ Radius of the new cylinder = 2
Also, height of the new cylinder = h
r 2
∴ Volume of the new cylinder = 𝜋𝜋 � � h =
∴
Volume of the new cylinder
Volume of the original cylinder
2
𝜋𝜋r 2 h
4
= (πr2h/4)/(πr2h) = (1/4) = 1 : 4
Hence, the correct answer is C.
Q4
Elementary events associated with random experiment of the given two dice are:
∴ Total number of outcomes = 36
Let A be the event of getting same number on both dice.
Elementary events favourable to event A are (1,1),(2,2)(3,3),(4,4),(5,5) and (6,6).
⇒ Favourable number of outcomes = 6
∴ P(A) =
6
36
=
1
6
So, required probability is
1
6
Hence, the correct answer is C.
Q5
Let P(x,y) divides line segment joining A(1,3) and B(4,6) in the ratio 2 : 1.
We know that , the coordinates of a point(x,y) dividing the line segment joining the points
(x 1 , y 1 ) and (x 2 ,y 2 ) in the ratio m 1 : m 2 are given by
x=
𝑚𝑚 1 𝑥𝑥 2 +𝑚𝑚 2 𝑥𝑥 1
𝑚𝑚 1 +𝑚𝑚 2
∴ Here x =
9
and y =
2(4)+ 1(1)
2+1
⇒ x = 3 and y =
𝑚𝑚 1 𝑦𝑦2 +𝑚𝑚 2 𝑦𝑦1
𝑚𝑚 1 +𝑚𝑚 2
and y =
2(6)+ 1(3)
15
2+1
3
⇒ x = 3 and y = 5
Thus, (3, 5) divides the line segment AB in the ratio 2: 1
Hence, the correct answer is B.
Q6
Let AB be the diameter and O be the centre of the circle.
We are given co-ordinates of one end point of circle and co-ordinates of its centre.
So, co-ordinates of A are (2, 3) and centre O are (-2, 5).
Let co-ordinates of point B be (x, y).
We know that centre of a circle is the midpoint of the diameter.
∴ By midpoint formula,
-2 =
2+x
2
and 5 =
3+y
2
⇒ −4 = 2 + x and 10 = 3 + y
⇒ x = −6 and y = 7
So, other end of the diameter is (-6, 7).
Hence, the correct answer is A.
Q7
Odd natural numbers are in the pattern 1,3,5,7,9…..
These numbers form an A.P. where a = 1, d = 3 – 1 = 2
n
We know that, S n = [ 2a + (n – 1)d ]
∴ 𝑠𝑠20 =
=
20
2
2
[ 2 x 1 + (20 – 1) x 2 ]
20
2
[ 2 + 19 x 2 ]
=10 [ 2 + 38 ]
=10 x 40
=400
Thus, the sum of first 20 odd natural numbers is 400.
Hence, the correct answer is C.
Q8
The given equations are ay2 + ay +3 = 0 and y2 + y + b = 0
Given, 1 is the root of both the equations
Therefore, y = 1 will satisfy both these equations.
Putting y = 1 in ay2 + ay +3 = 0, we get
a(1)2 + ax1 + 3 = 0
∴a+a+3=0
⇒ 2a + 3 = 0
⇒ 2a = −3
3
⇒a = −2
……..(1)
Putting y = 1 in y2 + y + b = 0, we get
(1)2 + 1 + b = 0
∴ 1+1+b = 0
⇒2 + b = 0
⇒ b = −2
……(2)
3
∴ ab = − ∗ (−2) = 3 [Using (1) & (2)]
2
Thus, the value of ab is 3.
Hence, the correct answer is A.
Q9
Given, AP = 4 cm, BP = 3 cm and AC = 11 cm.
The lengths of tangents drawn from an external point to the cirlce are equal.
AP = AR, BP = BQ, CQ = CR ……. (1)
AC = 11 cm
⇒ AR + RC = 11 cm
⇒ AP + CQ = 11 cm [ From equation (1) ]
⇒ 4 cm + CQ = 11 cm
⇒ CQ = (11 −4) cm
⇒ CQ = 7 cm
BP = BQ = 3 cm
Now, BC = BQ + QC
⇒ BC = (3+7) cm
⇒ BC = 10 cm
Hence, the correct option is B.
Q10
EK = 9 cm
As length of tangents drawn from an external point to the circle are equal.
∴ EK = EM = 9 cm
Also, DH = DK and FH = FM ……(1)
EK = EM = 9 cm
⇒ ED + DK = 9 cm and EF + FM = 9 cm
⇒ ED + DH = 9 cm and EF + FM = 9 cm [From equation (i)] ….(ii)
Perimeter of
EDF = ED + DF + EF
= ED + DH + HF + EF
= (9 + 9) cm
[From equation (ii)]
= 18 cm
Hence, the correct option is A.
Q11
The given points are A (0, 2), B (3, p) and C (p, 5).
According to the question, A is equidistant from point B and C.
∴ AB = AC
⇒ �(3 − 0)2 + (p − 2)2 = �(p − 0)2 + (5 − 2)2
⇒ �(3)2 + (p − 2)2 = �(p)2 + (3)2
⇒ �9 + p2 + 4 − 4p = �p2 + 9
⇒ �p2 − 4p + 13 = �p2 + 9
On squaring both sides, we obtain:
⇒ p2 – 4p + 13 = p2 + 9
⇒ −4𝑝𝑝 = −4
⇒ p=1
Q12
Total number of outcomes = 50
Multiples of 3 and 4 which are less than or equal to 50 are:
12,24,36,48
Favourable number of outcomes = 4
Probability of the number being a multiple of 3 and 4
=
=
=
Number of favourable outcomes
Total number of outcomes
4
50
2
25
Q13
1
Given, volume of hemisphere = 2425 cm3 =
2
Let the radius of the hemisphere be ‘r’ cm.
Volume of hemisphere =
2
⇒ 3 πr 3 =
P
2
⇒3 x
22
7
⇒ r3 =
4851
2
x r3 =
P
2
3
πr 3
P
4851
2
4851 x 3 x 7
2 x 2 x 22
P
441 x 21
⇒ r3 = 2 x 2 x 2
P
⇒ r 3=
P
21 x 21 x 21
2x2x2
21 x 21 x 21
3
⇒r = �
⇒r =
21
2
2x2x2
cm
…..(1)
∴ Curved surface area of hemisphere = 2 𝜋𝜋r2
=2x
22
=2x
22
21 2
7
x� �
7
2x2
x
2
21 x 21
[ Using (1)]
4851
2
cm3
= 693 cm2
Q14
Given that : OA = 8 cm, OB = 5 cm and AP = 15 cm
To find : BP
Construction : Join OP.
Now, OA
AP and OB
BP Tangent to a circle is perpendicular to the
Radius through the point of contact
⇒ ∠ OAP = ∠ OBP = 900
On applying Pythagoras theorem in
OAP, we obtain:
(OP)2 = (OA)2 + (AP)2
⇒ (OP)2 = (8)2 + (15)2
⇒ (OP)2 = 64 + 225
⇒ OP = √289
⇒ (OP)2 = 289
⇒ OP = 17
Thus, the length of OP is 17 cm.
On appling Pythagoras theorem in
(OP)2 = (OB)2 + (BP)2
⇒ (17)2 = (5)2 + (BP)2
⇒ 289 = 25 + (BP)2
OBP, we obtain :
⇒ (BP)2= 289 - 25
⇒ (BP)2= 264
⇒ BP = 16.25 cm (approx.)
Hence, the length of BP is 16.25 cm.
Q15
Given : An isoceles
ABC with AB = AC, circumscribing a circle.
To prove : P bisects BC
Proof : AR and AQ are the tangents drawn from an external point A to the circle.
∴AR = AQ (Tangents drawn from an external point to the circle are equal)
Similarly, BR = BP and CP = CQ.
It is given that in
ABC, AB = AC.
⇒ AR + RB = AQ + QC.
⇒ BR = QC (As AR = AQ)
⇒ BP = CP (As BR = BP and CP = CQ)
⇒ P bisects BC.
Hence, the result is proved.
OR
Given : Two concentric circles C 1 and C 2 with centre O, and AB is the chord of C 1 touching C 2 at C.
To prove : AC = CB
Construction : Join OC.
Proof : AB is the chord of C 1 touching C 2 at C, then AB is the tangent to C 2 at C with OC as its radius.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of
contact.
∴ OC
AB
Considering, AB as the chord of the circle𝐶𝐶1 . So, OC
AB.
∴ OC is the bisector of the chord AB.
Hence, AC = CB (Perpendicular from the centre to the chord bisects the chord).
Q16
It is given that OABC is a square of side 7 cm.
∴ Area of square OABC = (7)2 cm2 = 49 cm2
Also, it is given that OAPC is a quadrant of circle with centre O.
∴ Radius of the quadrant of the circle = OA = 7 cm
∴Area of the quadrant of circle =
=
1
=
49𝜋𝜋
(𝜋𝜋 ∗ 72) cm2
4
4
=
49
=
77
4
2
1
4
(𝜋𝜋r2)
cm2
x
22
7
cm2
cm2
∴ Area of the shaded region = Area of Square − Area of Quadrant of circle.
= [ 49 −
=[
=
77
98−77
21
2
2
2
] cm2
] cm2
cm2
= 10.5 cm2
Thus, the area of the shaded region is 10.5 cm2.
Q17
Three digit natural numbers which are multiples of 7 are 105, 112, 119,…..,994.
105, 112, 119,……..994 are in A.P.
First term (a) = 105
Commom difference (d) = 7
Let 994 be the nth term of A.P.
∴ a n = 994
⇒ 105 + (n-1) x 7 = 994 [ a n = a + (n-1)d ]
⇒ 7(n-1) = 994 - 105
⇒ 7(n-1) = 889
⇒ n-1 = 127
⇒ n = 128
Sum of all the terms of A.P. =
128
2
(105 + 994)
= 64 x 1099
n
[ Sn = (a + l), l being last term]
2
= 70336
Thus, the sum of all three digit natural numbers which are multiples of 7 is 70336.
Q18
The given quadratic equation is 3x2 – 2kx + 12 = 0
On comparing it with the general quadratic equation ax2 + bx + c = 0, we obtain
a = 3, b = -2k and c = 12
discriminant, ‘D’ of the given quadratic equation is given by
D = b2 – 4ac
= (-2k)2– 4 ∗ 3 ∗ 12
= 4k2– 144
For equal roots of the given quadratic equations, Discriminant will be equal to 0.
i.e., D = 0
⇒ 4k2 – 144 = 0
⇒ 4 (k2 – 36) = 0
⇒ k2 = 36
⇒k = ± 6
Thus, the values of k for which the quadratic equation 3x2 – 2kx + 12 = 0 will have equal roots are 6 and
−6.
Q19
The given points are A (3,-5) and B (-4, 8).
Here, x 1 = 3, y 1 = -5, x 2 = -4 and y 2 = 8.
Since
AP
PB
K
= , the point P divides the line segment joining the points A and B in the ratio K : 1. The
1
coordinates of P can be found using the section formula
𝑚𝑚𝑚𝑚 2 +𝑛𝑛𝑛𝑛 1 𝑚𝑚𝑚𝑚 2 +𝑛𝑛𝑛𝑛 1
here, m = K and n = 1
Co-ordinates of P = �
K x (−4)+ 1 x 3
K+1
,
K x 8+1 x (−5)
K+1
It is given that, P lies on the line x+y = 0.
∴
−4K+3
⇒
K+1
+
8K−5
K+1
−4K+3+8K−5
K+1
=0
=0
⇒ 4K – 2 = 0
⇒ 4K = 2
1
⇒K = 2
1
Thus, the required value of K is .
2
� =�
−4K+3
K+1
,
𝑚𝑚 +𝑛𝑛
,
8K−5
K+1
�
𝑚𝑚 +𝑛𝑛
n
Q20
Given, vertices of a triangle are (1,-3), (4, p) and (-9, 7).
x 1 = 1 , y 1 = -3
x2 = 4 , y2 = p
x 3 = -9 , y 3 = 7
Area of given triangle
1
= [ x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 )]
2
1
= [ 1(p-7) + 4(7+3) + (-9) (-3-p)]
2
1
= [p-7+40+27+9p]
2
1
= [10p + 60]
2
= 5 (p+6)
Here, the obtained expression may be positive or negative.
∴ 5(p+6) = 15 or 5(p+6) = -15
⇒ p + 6 = 3 or p + 6 = −3
⇒ p = −3 or p = −9
Q21
Since ABCD is a parallelogram,
AB = CD …(1)
BC = AD …(2)
It can be observed that
DR = DS (Tangents on the circle from point D)
CR = CQ (Tangents on the circle from point C)
BP = BQ (Tangents on the circle from point B)
AP = AS (Tangents on the circle from point A)
Adding all these questions, we obtain
DR + CR +BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (1) and (2) in this equation, we obtain
2AB = 2BC
AB = BC …(3)
Comparing equations (1),(2) and (3), we obtain
AB = BC = CD = DA
Hence, ABCD is a rhombus.
OR
Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point
P, Q, R, S. let us join the vertices of the quadrilateral ABCD to the center of the circle.
Consider
OAP and
OAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the same circle)
OA = OA (Common side)
OAP ≅
Therefore, A
OAS (SSS congruence criterion)
A, P
S, O
O
And thus, ∠ POA = ∠ AOS
∠1= ∠8
Similarly,
∠2= ∠3
∠4= ∠5
∠6= ∠7
∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 3600
( ∠ 1 + ∠ 8) + ( ∠ 2 + ∠ 3) + ( ∠ 4 + ∠ 5) + ( ∠ 6 + ∠ 7) = 3600
2 ∠ 1 + 2 ∠ 2 + 2 ∠ 5 + 2 ∠ 6 = 3600
2( ∠ 1 + ∠ 2) + 2( ∠ 5 + ∠ 6) = 3600
( ∠ 1 + ∠ 2) + ( ∠ 5 + ∠ 6) = 1800
∠ AOB + ∠ COD = 1800
Similarly, we can prove that ∠ BOC + ∠ DOA = 1800
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the
centre of the circle.
Q22
It is given that, height (h) of cylindrical part = height (h) of the conical part = 7 cm
Diameter of the cylindrical part = 12 cm
∴ Radius (r) of the cylindrical part =
∴ Radius of conical part = 6 cm
12
2
cm = 6 cm
Slant height (l) of conical part = √r 2 + h2 cm
= √62 + 72 cm
= √36 + 49 cm
= √85 cm
= 9.22 cm (approx.)
Total surface area of the remaining solid
= CSA of cylindrical part + CSA of conical part + Base area of the circular part
= 2 𝜋𝜋𝜋𝜋ℎ + 𝜋𝜋𝜋𝜋𝜋𝜋 + 𝜋𝜋r 2
=2x
22
7
x 6 x 7 cm2 +
22
7
x 6 x 9.22 cm2 +
22
= 264 cm2 + 173.86 cm2 + 113.14 cm2
7
x 6 x 6 cm2
= 551 cm2
OR
Height (h 1 ) of cylindrical bucket = 32 cm
Radius (r 1 ) of circular end of bucket = 18 cm
Height (h 2 ) of conical heap = 24 cm
Let the radius of the circular end of conical heap be r 2 .
The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.
Volume of sand in the cylindrical bucket = Volume of sand in conical heap
1
𝜋𝜋 ∗ 𝑟𝑟12 ∗ ℎ1 = 𝜋𝜋 ∗ 𝑟𝑟22 ∗ ℎ2
3
1
𝜋𝜋 ∗ 182 ∗ 32 𝜋𝜋 ∗ 24
3
𝜋𝜋 ∗ 182 ∗ 32 =
𝑟𝑟22 =
3∗ 18 2 ∗32
24
1
𝜋𝜋 ∗ 𝑟𝑟22 ∗ 24
3
= 182 x 4
r 2 = 18 x 2 = 36 cm
slant height = √362 + 242 = 122 + (32 + 22 ) = 12 √13 cm
therefore, the radius and slant height of the conical heap are 36 cm and 12 √13 cm
respectively.
Q23
PQ and AB are the arcs of two concentric circles of radii 7 cm and 3.5 cm respectively.
Let r 1 and r 2 be the radii of the outer and the inner circle respectively.
Suppose 𝜃𝜃 be the angle subtended by the arcs at the centre O.
Then r 1 = 7 cm, r 2 = 3.5 cm and 𝜃𝜃 = 300
Area of the shaded region
= Area of sector OPQ − Area of sector OAB
=
=
𝜃𝜃
360 0
𝜃𝜃
360 0
=
=
=
30 0
𝜋𝜋𝑟𝑟12 −
1
1
12
x
x
𝜋𝜋𝑟𝑟22
𝜋𝜋 (𝑟𝑟12 − 𝑟𝑟22 )
360 0
12
𝜃𝜃
360 0
x
22
7
22
7
22
7
[ (7 cm)2– (3.5 cm)2 ]
x (49 – 12.25) cm2
x 36.75 cm2
= 9.625 cm2
thus, the area of the shaded region is 9.625 cm2.
Q24
The given quadratic equation is 4x2 – 4ax + (a2 – b2) = 0
4x2 – 4ax + (a2 – b2) = 0
∴ 4x2 – 4ax +(a-b)(a+b) = 0
4x2 + [-2a -2a + 2b – 2b]x + (a-b)(a+b) = 0
4x2 + (2b-2a)x – (2a+2b)x + (a-b)(a+b) = 0
4x2 + 2(b-a)x – 2(a+b)x + (a-b)(a+b) = 0
2x [2x – (a-b)] – (a+b) [2x – (a-b)] = 0
[2x – (a – b)] or [ 2x – (a + b) = 0
2x = a – b or 2x = a + b
x=
a−b
2
or x =
a+b
2
Thus, the solution of the given quadratic equation is given by x =
OR
The given quadratic equation is 3x2 - 2√6x + 2 = 0
Comparing with the quadratic equation ax2 + bx + c = 0, we have
a = 3, b = -2√6 and c = 2
discriminant of the given quadratic equation
D = b2 – 4 ac = (2√6)2 -4×3×2 = 24 – 24 = 0
∴x=
−(−2√6)±√0
2x3
 x=
 x=
2√6
6
√6
3
x =
−b ±√D
2a
Thus, the solution of the given quadratic equation is x =
Q25
√6
3
.
a−b
2
or x =
a+b
2
Let C be the position of kite above the ground such that it subtends an angle of 600 at point A on the
ground.
Suppose the length of the string. AC be l m.
Given, BC = 45 m and ∠ BAC = 600.
In
ABC :
sin 600 =
BC
AC
therefore,
 l=
 sin 𝜃𝜃 =
√3
2
=
45 x 2
√3
45
l
=
90
√3
Perpendicular
Hypotensue
= 30√3
Thus, the length of the string is 30√3 m.
Q26
Given:
BC = 6 cm, ∠ C = 300 and ∠ A = 1050
Therefore ∠ B = 1800– ( ∠ A + ∠ C) (In
= 1800 – (1050 + 300)
= 450
Steps of construction :
1. draw a line BC = 6 cm.
ABC, ∠ A + ∠ B + ∠ C = 1800)
draw a ray CN making an angle of 300 at C.
draw a ray BM making an angle of 450 at B.
locate the point of intersection of rays CN and BM and name it as A.
ABC is the triangle whose similar triangle is to be drawn.
Draw any ray BX making anacute angle with BC on the side opposite to the vertex A.
Locate 3 (Greater of 2 and 3 in 2/3 ) points B 1 ,B 2 and B 3 on BX so that BB 1 = B 1 B 2 = B 2 B 3 .
Join B 3 C and draw a line through B 2 (smaller of 2 and 3 in 2/3) parallel to B 3 C to intersect BC at
C’.
9. Draw a line through C’ parallel to the line CA to intersect BA at A’.
10. A ‘BC’ is the required similar triangle whose sides are 2/3 times the corresponding sides of
ABC.
2.
3.
4.
5.
6.
7.
8.
Q27
Let a be the first term and d be the common difference of the given A.P.
According to the given question,
16th term of the AP = 2 x 8th term of the AP + 1
i.e., a 16 = 2𝑎𝑎8 + 1
(because a n = a+(n-1)d)
a+ (16-1)d = 2[a+(8-1)d] +5
 a+15d = 2[a+7d]+5
 a+15d = 2a+14d+5
 d = a+1
…..(1)
Also, 12th term, a 12 = 47






a+(12-1)d = 47
a+11d = 47
a+11(a+1) = 47
a+11a+11 = 47
12 a = 36
a = 36
[Using (1)]
On putting the value of a in (1), we get d = 3+1 = 4
Thus, nth term of the AP, a n = a + (n-1)d
On putting the respective values of a and d, we get
a n = 3+(n-1) 4 = 3 + 4n- 4 = 4n – 1
hence, nth term of the given AP is 4n – 1.
Q28
Total number of cards in a deck of cards = 52
Therefore, Total number of outcomes = 52
(i)
let A denote the event of getting a king of red colour.
There are two cards in favour of getting a king of red colour i.e., king of heart and king of diamond.
Number of outcomes in favour of event A =2
∴ P(A) =
Outcomes in favour of event A
Total number of outcomes
(ii)
=
2
52
=
1
26
let B denote the event of getting a face card.
There are 12 cards in favour of getting a face card i.e., 4 King, 4 Queen and 4 Jack cards.
Number of outcomes in favour of event B = 12
P(B) =
Outcomes in favour of event B
(iii)
Total number of outcomes
=
12
52
=
3
13
let C denote the event of getting a queen of diamond.
There is one queen of diamond in the deck of cards.
P(C) =
Q29
Outcomes in favour of event C
Total number of outcomes
=
1
52
Let the height of the bucket be h cm.
Suppose r 1 and r 2 be the radii of the circular ends of the bucket.
Given, r 1 = 28 cm and r 2 = 21 cm
Capacity of bucket = 28.49 litres
∴Volume of the bucket = 28.49 ×1000 cm3 [1 litre = 1000 cm3]



1
3
1
𝜋𝜋ℎ(𝑟𝑟12 + 𝑟𝑟22 + 𝑟𝑟1 𝑟𝑟2 ) = 28.49 ×1000 cm3
×
3
22
21
22
7
× h [(28)2 + (21)2 + (28 × 21)]cm2 = 28490 cm3
x h x 1813 = 28490 cm
 h=
28940 ∗ 21
22 ∗ 1813
 h = 15 cm
Thus, the height of the bucket is 15 cm.
Q30
Let AB be the hill and CD be the tower.
Angle of elevation of the hill at the foot of the tower is 600. i.e., ∠ ADB = 600 and the angle of
depression of the foot of hill from the top of the tower is 300, i.e., ∠ CBD = 300.
Now in right angled
tan 300 =
CBD :
CD
BD
 BD =
 BD =
CD
tan 30 0
50
[
1
]
√3
 BD = 50√3m
In right
tan 600 =
ABD :
AB
CD
 AD = BD X tan 600
= 50√3 ×√3 m
= 50 ×3 m
= 150 m
Hence, the height of the hill is 150 m.
Q31
Given : A circle (O,r) and a tangent l at point A.
To prove : OA
l
Construction : take any point B other than A on the tangent l. Join OB.
Suppose OB meets the cirlce at C.
Proof : Among all line segments joining the centre O to any point on l, the perpendicular is the shortest
to l.
So, in order to prove OA l we need to prove that OA is shorter than OB.
OA = OC (Radius of same circle)
Now, OB = OC + BC
 OB > OC
 OB > OA
 OA > OB
B is an arbitrary point on the tangent l. thus, OA is shorter than any other line segment joining O to
any point on l.
Hence OA is perpendicular to l.
OR
Let the sides of the quadrilateral ABCD touch the circle at points P, Q, Rand S as shown in the figure.
We know that, tangents drawn from an external point to the circle are equal in length.
Therefore,
AP = AS
BP = BQ
CQ = CR
DR = DS
…..(1)
Therefore,
AB + CD = (AP+BP) + (CR+DR)
= (AS+BQ) + (CQ+DS) [Using (1)]
= (AS+DS) + (BQ+CQ)
= AD + BC
Hence, AB + CD = AD + BC
Q32
Let the number of books purchased by the shopkeeper be x.
Cost price of x books = Rs 80
∴ Original cost price of one book = Rs
80
x
If the shopkeeper had pruchased 4 more books, then the number of books purchased by him would be
(x+4).
∴ New cost price of one book = Rs
80
x+4
Given, Original cost price of one book – New cost price of one book = Rs 1









∴
80
x
80
−
x+4
=1
80(x+4)− 80 x
x(x+4)
=1
80x + 320 – 80x = x (x+4)
x2 + 4x = 320
x2 + 4x – 320 = 0
x2 + 20x – 16x – 320 = 0
x(x+20) – 16 (x+20) = 0
(x-16) (x+20) = 0
x– 16 = 0 or x + 20 = 0
x = 16 or x = - 20
∴ x = 16
(because number of books cannot be negative)
𝑂𝑂𝑂𝑂
Let the first number be x.
Given : First number + Second number = 9
∴ x + Second number = 9
 Second number = 9 – x
Given,








1
First number
9−x+x
x(9−x)
=
+
1
Second number
1
2
9 × 2 = 9x – x2
x2 – 9x + 18 = 0
x2 – 6x -3x + 18 = 0
x(x-6) -3(x-6) = 0
(x-3)(x-6) = 0
x - 3 = 0 or x – 6 = 0
x = 3 or x = 6
when x = 3, we have
9-x = 9-3 = 6
When x=6, we have
9-x = 9-6 = 3
=
1
2
∴
1
1
1
+
=
x 9−x
2
Thus, the two numbers are 3 and 6.
Q33
First term of the A.P (a) = 7; sum of first 20 terms = −240.
n
The sum of first in terms of an AP, Sn = [ 2a + (n − 1)d], where a is the first term and d is the common
2
difference.
∴ 𝑆𝑆20 =





20
[ 2 × 7 + (20-1)d] = -240
10 [ 14 + 19d ] = -240
14 + 19 d = - 24
19d = - 24 – 14
19d = - 38
d = -2
2
now, 24th term of the AP, a 24 = a+(24-1)d
on putting respective values of a and d, we get
a 24 + 7 + 23 × (-2) = 7 – 46 = - 39
hence, 24th term of the given AP is -39.
Q34
Let r and h be radius and height of the cone respectively.
Radius of cone (r) = 7 cm (Given)
Diameter of cone = 2 × r = (2 × 7)cm = 14 cm
According to the question, height of the cone is equal to its diameter.
∴ Height of cone (h) = 14 cm
Radius of hemisphere = Radius of cone = 7cm
∴ Volume of solid = Volume of cone + Volume of hemisphere
=
=
2
1
πr 2 h + πr 3
3
3
πr 2
=
3
1 22
×
× 7 × 7 × [14 + (2 × 7)]cm3
7
3
22
=
=
[h + 2r]
3
× 7 × 28cm3
4312
2
cm3
= 1437.33 cm3
Thus, the volume of the solid is 1437.33 cm3 .
CBSE – CLASS X MATHS -2013
General Instructions:
(i)
(ii)
(iii)
All questions are compulsory.
The question paper consists of 34 questions divided into four sections A,B,C and D.
Sections A contains 8 questions of one mark each, which are multiple choice type questions,
section B contains 6 questions of two marks each, section C contains 10 questions of three
marks each, and section D contains 10 questions of four marks each.
Use of calculations is not permitted.
(iv)
Section A
Q1 The common difference of the AP
(A)
(B)
(C)
(D)
P
–P
-1
1
1
p
,
1−p
p
,
1−2p
p
, …… is :
Q2 In Fig.1, PA and PB are two tangents drawn from an external point P to a circle with centre C
radius 4 cm. if PA PB, then the length of each tangent is:
(A)
(B)
(C)
(D)
3 cm
4 cm
5 cm
6 cm
Q3 In Fig.2, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches the sides BC,
AB, AD and CD at points P,Q,R and S respectively, If AB = 29 cm, AD = 23 cm, ∠B = 900 and DS = 5 cm,
then the radius of the circle (in cm.) is :
(A)
(B)
(C)
(D)
11
18
6
15
Q4 The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 300.
The distance of the car from the base of the tower (in m.)is :
(A) 25√3
(B) 50√3
(C) 75√3
(D) 150
Q5 The probability of getting an ever number, when a die is thrown once, is:
(A)
(B)
(C)
(D)
1
2
1
3
1
6
5
6
Q6 A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the
probability that it bears a prime-number less than 23, is :
(A)
(B)
7
90
10
90
(C)
(D)
4
45
9
89
Q7 In Fig.3, the area of triangle ABC (in sq.units) is:
(A)
(B)
(C)
(D)
15
10
7.5
2.5
Q8 If the difference between the circumference and the radius of a circle is 37 cm, then us 𝜋𝜋 =
circumference (in cm) of the circle is:
(A)
(B)
(C)
(D)
154
44
14
7
Section B
Q9 Solve the following quadratic equation for x :
22
7
, the
4√3x 2 + 5x -2√3 = 0
Q10 How many three – digit natural numbers are divisible by 7?
Q11 In Fig.4, a circle inscribed in triangle ABC touches its sides AB,BC and AC at points D,E and F
respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF.
Q12 Prove that the parallelogram circumscribing a circle is a rhombus.
Q13 A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that
the drawn card is neither a kind nor a queen.
Q14 Two circular pieces of equal radii and maximum area, touching each other are cut out from a
rectangular card board of dimensions 14 cm. 7 cm. find the area of the remaining card board.
�Use π =
Section C
22
�
7
Q15 For what value of k, are the roots of the quadratic equation kx (x – 2) + 6 = 0 equal?
1
1
Q16 Find the number of terms of the AP 18, 15 , 13,…., -49 and find the sum of all its terms.
2
2
Q17 Construct triangle with sides 5 cm, 4 cm, and 6 cm. then construct another triangle whose sides are
2
3
times the corresponding sides of triangle.
Q18 The horizontal distance between two poles is 15 m. the angle of depression of the top of first pole
as seen from the top of second pole is 300. If the height of the second pole is 24 m, find the height of the
first pole. [ Use √3 = 1.732]
Q19 Prove that the points (7,10), (-2,5) and (3,-4) are the vertices of an isosceles right triangle.
Q20 Find the ratio in which the y-axis divides the line segment joining the points (-4,-6) and (10,12). Also
find the coordinates of the point of division.
Q21 In Fig.5, AB and CD are two diameters of a circle with centre O, which are perpendicular to each
other. OB is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
[Use 𝜋𝜋 =
22
7
]
Q22 A vessel is in the form of hemispherical bowl surmounted by a hollow cylinder of same diameter.
The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. find the
total surface are of the vessel. [Use 𝜋𝜋 =
22
7
]
Q23 A wooden toy was made by scooping out a hemisphere of same radius from each end of a solid
cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the volume of wood
in the toy. [Use 𝜋𝜋 =
22
7
]
Q24 In a circle of radius 21 cm, an arc subtends an angle of 600 at the centre. Find : (i) the length of the
arc (ii) area of the sector formed by the arc. [Use 𝜋𝜋 =
Section D
Q25 Solve the following for x :
1
2a+b+2x
=
1
2a
1
+ +
b
1
2x
22
7
]
Q26 Sum of the areas of two squares is 400 cm2. If the difference of their perimeters is 16 cm, find the
sides of the two squares.
Q27 If the sum of first 7 terms of an AP is 49 and that of last 17 terms is 289, find the sum of its first n
terms.
Q28 Prove that the tangent at any point of a circle is perpendicular to the radius through the points of
contact.
Q29 In fig.6, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B
respectively. Another tangent at C intersects the line l at D and m at E. prove that ∠DOE = 900.
Q30 The angle of elevation of the top of a building from the foot of the tower is 300 and the angle of
elevation of the top of the tower from the foot of the building is 600. If the tower is 60 m high, find the
height of the building.
Q31 A Group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and
rest are extremely kind. A person from the group is selected at random. Assuming that each person is
equally likely to be selected, find the probability of selecting a person who is (i) extremely patient (ii)
extremely kind or honest. Which of the above values you prefer more?
Q32 The three vertices of a parallelogram ABCD are A (3,-4), B (-1,-3) and C (-6, 2). Find the coordinates
of vertex D and find the area of ABCD.
Q33 Water is flowing through a cylindrical pipe, of internal diameter 2 cm, into a cylindrical tank of base
radius 40 cm, at the rate of 0.4 m/s. determine the rise in level of water in the tank in half an hour.
Q34 A bucket open at the top, and made up of a metal sheet is in the form of a frustum of a cone. The
depth of the bucket is 24 cm and the diameters of its upper and lower circular ends are 30 cm and 10 cm
respectively. Find the cost of metal sheet used in it at the rate of Rs 10 per 100 cm2. [Use 𝜋𝜋 = 3.14]
CBSE – CLASS X MATHS 2013
General Instructions:
(i)
(ii)
(iii)
(iv)
All questions are compulsory.
The question paper consists of 34 questions divided into four sections A, B, C and D.
Sections A contains 8 questions of one mark each, which are multiple choice type questions,
section B contains 6 questions of two marks each, section C contains 10 questions of three
marks each, and section D
Use of calculations is not permitted.
Q1
1
The common difference of the AP ,
the second term and first term i.e.
1−p
p
p
1−p 1
p
,
1−2p
- =
p
,….. can be found out by finding the difference between
p
1−p−1
p
=
The correct answer is -1 which is given by option C.
−p
p
= -1
Q2
Since, AP
PB, CA
AP, CB
having side equal to 4 cm.
BP and AC = CB = radius of the circle, therefore APBC is a square
Therefore, length of each tangent is 4 cm.
The correct answer is 4 cm which is given by option B.
Q3
Given that AB, BC, CD and AD are tangents to the circle with centre O at Q,P,S and R respectively. AB =
29 cm, AD = 23, DS = 5 cm and ∠B = 900.
Join PQ .
We know that, the lengths of the tangents drawn from an external point to a circle are equal.
DS = DR = 5 cm
∴ AR = AD – DR = 23 cm – 5 cm = 18 cm
AQ = AR = 18 cm
∴ QB = AB – AQ = 29 cm – 18 cm = 11 cm
QB = BP = 11 cm
In right
PQB, PQ2 = QB2 + BP2 = (11 cm)2 + (11 cm)2 = 2 x (11 cm)2
PQ = 11√2 cm…..(1)
In right
PQB,
PQ2 = OQ2 + OP2 + r2 + r2 = 2 r2
PQ = √2r …..(2)
From (1) and (2), we get
r = 11 cm
thus, the radius of the circle is 11 cm.
The correct answer is 11 which is given by option A.
Q4
Let AB be the tower of height 75 m.
∠CBD =∠ ACB = 300
Suppose C be the position of the car from the base of the tower.
In right
Cot 300 =
ABC,
AC
AB
 AC = AB cot300
 AC = 75 m x √3
 AC = 75√3m
Thus, the distance of the car from the base of the tower is 75 √3 m.
The correct answer is 7 5√3 which is given by option C.
Q5
When a die is thrown once, the sample space is given by, S = {1,2,3,4,5, and 6}
Then, the event, E of getting an ever number is given by, E = {2,4, and 6}
∴ Probability of getting an even number = P (E) ∴ P(E) =
1
The correct answer is which is given by option A.
Number of favourable outcomes
Number of all possible outcom es
3
= =
6
1
2
2
Q6
If the given that the box contains 90 discs, numbered from 1 to 90.
As one disc is drawn at random from the box, the sample space is given by, S = {1,2,3,…90}
The prime number less than 23 are 2,3,5,7,11,13,17, and 19.
Then, the event, E of getting a prime number is given by, E = {2,3,5,7,11,13,17,19}
∴ P(E) =
Number of favourable outcomes
Number of all possible outcomes
The correct answer is
Q7
4
45
=
8
90
=
4
45
which is given by option C.
Construction : Draw AM
BC.
It can be observed from the given figure that BC = 5 unit and AM = 3 unit.
It
ABC, BC is the base and AM is the height.
1
Area of triangle ABC = x base x height
1
= x BC x AM
2
2
1
= x 5 x 3 sq.units
2
= 7.5 sq.units
The correct answer is 7.5 which is given by option C.
Q8
Difference between circumference and radius of the circle = 37 cm
Let r be the radius of the circle.
∴ 2𝝅𝝅r – r = 37 cm
 r (2𝝅𝝅 – 1) = 37 cm
 r �2 x
 rx
37
7
22
7
− 1� = 37 cm
= 37 cm
 r = 7 cm
∴ Circumference of the circle = 2 𝝅𝝅r = 2 x
22
7
x 7 cm = 44 cm
The correct answer is 44 which is given by option B.
Q9
4√3x 2 + 5x – 2 √3 = 0
 4√3 x 2 +8x-3x - 2 √3 = 0
 4x(√3x + 2) - √3 (√3 + 2) = 0
 (4x - √3 ) (√3x + 2) = 0
∴x=
√3
4
or x = -
2
√3
Q10
All the three-digit natural numbers that are divisible by 7 will be of the form 7n.
2
Therefore, 100≤7n≤999=> 14 ≤ n ≤142
7
5
7
Since, n is an integer, therefore, there will be 142 – 14 = 128 three-digit natural numbers that will be
divisible by 7.
Therefore, there will be 128 three – digit natural numbers that will be divisible by 7.
Q11
Given that AB = 12 cm, BC = 8 cm and AC = 10 cm.
Let, AD = AF = p cm, BD = BE = q cm and CE = CF = r cm
(Tangents drawn from an external point to the circle are equal in length)
 2(p+ q +r) = AB+BC+AC = AD+DB+BE+EC+AF+FC = 30 cm
 p +q +r = 15
AB = AD + DB = p + q = 12 cm
Therefore, r = CF = 15 – 12 = 3 cm.
AC = AF + FC = p + r = 10 cm
Therefore, q = BE = 15 – 10 = 5 cm.
Therefore, p = AD = p + q + r – r – q = 15 – 3 – 5 = 7 cm.
Q12.
GIVEN : ABCD be a parallelogram circumscribing a circle with centre O.
TO PROVE : ABCD is a rhombus.
We know that the tangents drawn to a circle from an exterior point are equal in length.
Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS.
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
Therefore, AB + CD = AD + BC or 2AB = 2BC (Since, AB = DC and AD = BC)
Therefore, AB = BC = DC = AD.
Therefore, ABCD is a rhombus.
Hence, proved.
Q13
Let E denote the event that the drawn card is neither a king nor a queen.
Total number of possible cases = 52.
Total number of cards that are king and those that are queen in the pack of playing cards = 4 + 4 = 8.
Therefore, there are 52-8 = 44 cards that are neither a king nor a queen.
Total number of favorable cases = 44.
∴ Required probability = P (E) =
Favourable number of cases
Total number of cases
=
44
52
=
11
13
Thus, the probability that the drawn card is neither a king nor a queen is
Q14
11
13
.
Dimension of the rectangular card board = 14 cm x 7 cm
Since, two circular pieces of equal radii and maximum area touching each other are cut from the
rectangular card board, therefore, the diameter of each of each circular piece is 14/2 = 7 cm.
7
Radius of each circular piece = cm.
2
7 2
∴ Sum of area of two circular pieces = 2 x 𝜋𝜋 � � = 2 x
2
22
7
x
49
4
= 77 cm2
Area of the remaining card board = Area of the card board – Area of two circular pieces
= 14 cm x 7 cm – 77 cm2 = 98 cm2 – 77 cm2 = 21 cm2
Thus, the area of the remaining card board is 21 cm2 .
Q15
The given quadratic equation is k x (x – 2) + 6 = 0.
This equation can be rewritten as kx2 – 2kx + 6 = 0.
For equal roots, it discriminate, D = 0.




b2 – 4ac = 0, where a = k, b = -2k and c = 6
4k2 – 24k = 0
4k (k – 6) = 0
K = 0 or k = 6
But k cannot be 0, so the value of k is 6.
Q16
1
1
The AP is given as 18, 15 , 13, …, -49 .
2
2
1
1
1
First term a = 18, common difference d = 15 − 18 = −2 and the last term of the AP = -49 .
2
Let the AP has n terms.
2
2
a n = a + (n-1) d
-(99/2) = 18 – (5/2)(n-1)
5(n-1) =135
n =27 +1
n =28
∴ n = 27 + 1 = 28
Thus, the given AP has 28 terms.
Now, the sum of all the terms (Sn ) is given by,
Sn =
n
2
[2a + (n-1)d =
28
2
5
2
Thus, the sum of all the terms of the AP is – 441.
Q17
5
[ 2 x 18 + (28 – 1) x �− �] = 14 [ 36 – 27 x ] = - 441
2
Step 1
Draw a line segment AB = 4 cm. taking point A as centre, draw an arc of 5 cm radius.
Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at
point C. Now, AC = 5 cm and BC = 6 cm and ABC is the required triangle.
Step 2
Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.
Step 3
Locate 3 points A 1 , A 2 , A 3 (as 3 is greater between 2 and 3) on line AX such that
AA 1 = A 1 A 2 = A 2 A 3 .
Step 4
Join BA 3 and draw a line through A 2 parallel to BA 3 to intersect AB at point B’.
Step 5
Draw a line through B’ parallel to the line BC to intersect AC at C’.
AB’C’ is the required triangle.
Q18
Let AB and CD be two poles, where CD = 24 m.
It is given that angle of depression of the top of the pole AB as seen from the top of the pole CD is 300
and horizontal distance between the two poles is 15 m.
∴ ∠ CAL = 300 and BD = 15 m.
To find: height of pole AB
Let the height of pole AB be h m.
AL = BD = 15 m and AB = LD = h
Therefore, CL = CD – LD = 24 – h
Consider right
tan ∠ CAL =
ACL:
Perpendicular
 tan 300 =

1
√3
=
24−h
15
 24 – h =
Base
24−h
=
CL
AL
15
15
√3
 24 – h = 5√3
 h = 24 - 5√3
 h = 24 – 5 x 1.732 [ Taking √3 = 1.732 ]
 h = 15.34
Therefore, height of the pole AB = h m = 15.34 m.
Q19
Let the given points be A (7,10) B(-2,5) and C(3,-4).
Using distance formula, we have
AB = �(7 + 2)2 + (10 − 5)2 = √81 + 25 = √106
BC = �(−2 − 3)2 + (5 + 4)2 = √25 + 81 = √106
CA = �(7 − 3)2 + (10 + 4)2 = √16 + 196 = √212
Since AB = BC, therefore,
ABC is an isosceles triangle.
Also, AB2+BC2 = 106 + 106 = 212 = AC2
So,
ABC is a right triangle right angled at ∠B.
So,
ABC is an isosceles triangle as well as a right triangle.
Thus, the points (7, 10), (-2, 5) and (3,-4) are the vertices of an isosceles right triangle.
Q20
Let the y-axis divide the line segment joining the points (-4,-6) and (10,12) in the ratio λ :1 and the
Point of the intersection be (0,y).
So, by section formula, we have:
10
�
∴
λ +(−4) 12 λ +(−6)
,
� = (0,y)
λ +1
λ +1
λ −4
= 0 => 10 λ − 4 = 0
λ +1
10
=
2
Therefore, y =
12
=> λ =
4
10
5
λ +(−6) 12 x 25 − 6
= 2
=
+1
λ +1
5
24−30
)
5
2+5
)
(
5
(
= -
6
7
Thus, the y –axis divides the line segment joining the given points in the ratio 2 : 5 and the point of
6
division is ( 0 , - ).
7
Q21
AB and CD are the diameters of a circle with centre O.
Therefore, OA = OB = OC = OD = 7 cm (Radius of the circle)
Area of the shaded region
= Area of the circle with diameter OB + (Area of the semi – circle ACDA – Area of
7 2
1
= 𝜋𝜋 � � + � ∗ 𝜋𝜋 ∗ (7)2 −
=
22
=
77
7
2
2
x
49
4
2
1
cm2 + ∗
2
22
7
1
2
ACD)
∗ 14 𝑐𝑐𝑐𝑐 ∗ 7 𝑐𝑐𝑐𝑐�
1
∗ 49 cm2 - * 14 cm * 7 cm
cm2 + 77 cm2 - 49 cm2
2
= 66.5 cm2
Thus, the area of the shaded region is 66.5 cm2.
Q22
Let the radius and height of cylinder is r cm and h cm respectively.
Diameter of the hemisphere bowl = 14 cm
∴ Radius of the hemispherical bowl = Radius of the cylinder = r =
Total height of the vessel = 13 cm
14
2
cm = 7 cm
∴ Height of the cylinder, h = Total height of the vessel – Radius of the hemispherical bowl
= 13 𝑐𝑐𝑐𝑐 − 7 𝑐𝑐𝑐𝑐 = 6 𝑐𝑐𝑐𝑐
Total surface area of the vessel = 2 (curved surface area of the cylinder + curved surface area of the
hemisphere) (Twice because the vessel is hollow)
= 2(2𝜋𝜋rh + 2𝜋𝜋r2) = 4𝜋𝜋r(h+r) = 4 x
= 1144 cm2
22
7
x 7 x (6+7)cm2
Thus, the total surface area of the vessel is 1144 cm2.
Q23
Height of the cylinder, h = 10 cm
Radius of the cylinder = Radius of each hemisphere = r = 3.5 cm
Volume of wood in the toy = Volume of the cylinder – 2 x Volume of each hemisphere
= 𝜋𝜋r2h – 2 x
=
22
7
616
3
3
𝜋𝜋r3
4
x (3.5 cm)2 x 10 cm - x
= 385 cm3 =
2
cm3
539
3
cm3
3
22
7
x (3.5 cm)3
= 205.33 cm3 (Approx)
Thus, the volume of the wood in the toy is approximately 205.33 cm3.
Q24
It is given that, radius = 21 cm.
The Arc subtends an angle of 600.
length (l) of the arc is given by : 3600
(i)
θ
l=
=
x 2𝜋𝜋r , where r = 21 cm and 𝜃𝜃 = 600
360 0
60
360
22
∗2∗
7
= 22 cm
(ii)
A=
=
∗ 21 cm
Area, A of the sector formed by the arc is given by
θ
x 𝜋𝜋r2 , where r = 21 cm and 𝜃𝜃 = 600
360 0
60 0
360 0
22
x
x 21 x 21 cm2
7
= 231 cm2
Q25
The given equation is
1
2a+b+2x









=
1
1
2a
1
1
+ +
-
b
1
2a+b+2x
2x
2x−2a−b−2x
2x (2a+b+2x )
−2a−b
2x (2a+b+2x )
−(2a+b)
1
=
=
=
2x (2a+b+2x )
−1
x(2a+b+2x )
2x
=
=
1
2a+b+2x
+
=
1
2a
1
+ +
b
1
2x
1
2a
b
b+2a
2 ab
b+2a
1
2 ab
b+2a
ab
2 ab
2x2 + 2ax + bx +ab = 0
2x(x +a) + b (x+a) = 0
(x+a) (2x+b) = 0
x+a = 0 or 2x+b = 0
 x = -a, or x =
−b
2
Q26 let the sides of the two squares be x cm and y cm where x>y.
Then, their areas are x2 and y2 and their perimeters are 4x and 4y.
By the given condition, x2 + y2 = 400 and 4x – 4y = 16
4x – 4y = 16 => 4(x – y) = 16 => x – y = 4
 x = y + 4 ….(1)
Substituting the value of y from (1) in x2 + y2 = 400, we get that (y+4)2 + y2 = 400






y2 + 16 + 8y + y2 = 400
y2 + 4y – 192 = 0
y2 + 16y – 12y – 192 = 0
y(y+16) – 12(y+16) = 0
(y+16) (y-12) = 0
y = -16 or y = 12
since, the value of y cannot be negative, the value of y = 12.
So, x = y+4 = 12+4 = 16
Thus, the sides of the two squares are 16 cm and 12 cm.
Q27
Given that, S 7 = 49 and S 17 = 289
n
S n = [ 2a + (n-1)d ]
2
7
7
∴ S 7 = 49 = [ 2a + (7-1)d ] => 49 = = (2a + 6d)
2
 (a + 3d) = 7 ……(i)
Similarly, S 17 =
 289 =
17
2
17
2
2
[ 2a + (17 – 1)d ]
[ 2a + 16d ]
 (a + 8d) = 17 …..(ii)
Subtracting equation (i) from equation (ii), we get that 5d = 10.
Therefore, the value of d = 2 and a = 7 – 3d = 1
n
n
∴ S n = [2a + (n-1)d ] = [ 2(1) + (n-1)(2) ]
n
2
n
= (2+2n-2) = (2n) = n2
2
2
2
Therefore, the sum of n terms of the AP is n2.
Q28
Given : A circle C (0,r) and a tangent / at point A.
To prove : OA
l
Construction : take a point B, other than A, on the tangent l. Join OB. Suppose OB meets the circle in C.
Proof : we know that, among all line segment joining the point O to a point on l, the perpendicular is
shortest to l.
OA = OC (Radius of the same circle)
Now, OB = OC + BC.
Therefore, OB > OC
 OB > OA
 OA > OB
B is an arbitrary point on the tangent l. thus, OA is shorter than any other line segment joining O to
any point on l.
Hence, proved.
Q29
Given : l and m at are two parallel tangents to the circle with centre touching the circle at A and B
respectively. DE is a tangent at the point C, which intersects l at D and m at E.
To prove : ∠ DOE = 900
Construction : Join OC.
Proof :
In
ODA and
ODC,
OA = OC (Radii of the same circle)
AD = DC (Length of tangents drawn from an external point to a circle are equal)
DO = OD (Common side)
ODA ≅
ODC (SSS congruence criterion)
∴ ∠DOA = ∠COD …(1) (C.P.C.T)
OEB ≅
Similarly,
OEC
∠EOB = ∠ COE …(2)
AOB is a diameter of the circle. Hence, it is a straight line.
∴ ∠DOA + ∠COD + ∠COE + ∠EOB = 1800
From (1) and (2), we have
2 ∠COD + 2 ∠COE = 1800
 ∠COD + ∠COE = 900
 ∠DOE = 900
Hence, proved.
Q30
Let AB be the building and CD be the tower. Suppose the height of the building be h m.
Given, ∠ACB = 300 , ∠CBD = 600 and CD = 60 m
In right
cot 600 =
BCD :
BC
CD
=> BC = CD cot 600
 BC = 60 m *
In right
ACB :
tan 300 =

1
√3
1
√3
=> BC =
60
√3
m=
60√3
3
m = 20√3 m …..(1)
AB
=
BC
h
20√3
(Using (1))
 h = 20 m
Thus, the height of the building is 20 m.
Q31
Since the group consists of 12 persons, sample space consists of 12 persons.
∴ Total number of possible outcomes = 12
Let A denote event of selecting persons which are extremely patient
∴ Number of outcomes favorable to A is 3.
Let B denote event of selecting persons which are extremely kind or honest.
Number of persons which are extremely honest is 6.
Number of persons which are extremely kind is 12 – (6+3) = 3
∴ Number of outcomes favorable to B = 6+3 = 9.
(i)
Probability of selecting a person who is extremely patient is given by P(A).
P(A) =
(ii)
Number of outcomes favourable to A
Total number of possible outcomes
=
3
12
1
= .
4
Probability of selecting a person who is extremely kind or honest is given by P(B)
P(B) =
Number of outcomes favourable to B
Total number of possible outcomes
=
9
12
=
3
4
Q32
The three vertices of the parallelogram ABCD are A(3,-4), B(-1,-3) and C(-6,2).
Let the coordinates of the vertex D be (x,y)
It is known that in a parallelogram, the diagonals bisect each other.
∴ Mid point of AC = Mid point of BD
 �
3−6
2
3
,
−4+2
2
2
� =�
 �− , − � = �
 −
2
3
2
=
2
−1+x
2
 x = -2, y = 1
−1+𝑥𝑥
2
−1+𝑥𝑥
2
2
,− =
2
,
,
−3+𝑦𝑦
2
−3+𝑦𝑦
2
−3+y
2
�
�
So, the coordinates of the vertex D is (-2,1).
Now, area of parallelogram ABCD
= area of triangle ABC + area of triangle ACD
= 2 x area of triangle ABC [ Diagonal divides the parallelogram into two triangles of equal area ]
The area of a triangle whose vertices are (x 1 ,y 1 ), (x 2 ,y 2 ) and (x 3 ,y 3 ) is given by the numerical value of the
1
expression [ x 1 (y 2 -y 3 ) + x 2 (y 3 -y 1 ) + x 3 (y 1 -y 2 )]
2
1
Area of triangle ABC = [ 3(-3-2) + (-1) {2-(-4)} + (-6) {-4-(-3)} ]
1
2
= [ 3 * (-5) + (-1) *6 + (-6)* (-1) ] =
2
∴ Area of triangle ABC =
15
2
1
[ -15-6+6 ] = −
2
15
2
sq units ( Area of the triangle cannot be negative)
Thus, the area of parallelogram ABCD = 2 *
Q33
𝟏𝟏𝟏𝟏
𝟐𝟐
= 15 sq units.
Diameter of circular end of pipe = 2 cm
∴ Radius (r 1 ) of circular end of pipe =
2
200
m = 0.01 m.
Area of cross –section = 𝜋𝜋 * r12 = 𝜋𝜋 x (0.01)2 = 0.0001 𝜋𝜋 m2
Speed of water = 0.4 m/s = 0.4 * 60 = 24 metre/min
Volume of water that flows in 1 minute from pipe = 24 x 0.0001 𝜋𝜋 m3 = 0.0024 𝜋𝜋 m3
Volume of water that flows in 30 minute from pipe = 30 x 0.0024 𝜋𝜋 m3 = 0.072 𝜋𝜋 m3
Radius (r 2 ) of base of cylindrical tank = 40 cm = 0.4 m
Let the cylindrical tank be filled up to h m in 30 minutes.
Volume of water filled in tank in 30 minutes is equal to the volume of water flowed in 30 minutes from
the pipe.
∴ 𝜋𝜋 * (r 2 )2 x h = 0.072 𝜋𝜋
 (0.4)2 * h = 0.072
 0.16 * h = 0.072
h=
0.072
0.16
 h = 0.45 m = 45 cm
Therefore, the rise in level of water in the tank in half an hour is 45 cm.
Q34
Diameter of upper end of bucket = 30 cm
∴ Radius (r 1 ) of upper end of bucket = 15 cm
Diameter of lower end of bucket = 10 cm
∴ Radius (r 2 ) of lower end of bucket = 5 cm
Height (h) of bucket = 24 cm
Slant height (l) of frustum = �(r1 − r2 )2 + h2
=�(15 − 5)2 + (24)2 = �(10)2 + (24)2 = √100 + 576
=√676 = 26 cm
Area of metal sheet used to make the bucket = 𝜋𝜋 (r1 + r2 )l + 𝜋𝜋𝜋𝜋22 = 𝜋𝜋 (15 + 5)26 + 𝜋𝜋 (5)2
= 520𝜋𝜋 + 25𝜋𝜋 = 545 𝜋𝜋 cm2
Cost of 100 cm2 metal sheet = Rs 10
Cost of 545 𝜋𝜋 cm2 metal sheet = Rs
545 x 3.14 x 10
100
= Rs. 171.13
Therefore, cost of metal sheet used to make the bucket is Rs 171.13.
CBSE -MATHS-SET 1-2014
Q1. The first three terms of an AP are 3y-1, 3y+5 and 5y+1, respectively.
We need to find the value of y.
We know that if a, b and c are in AP, then:
b − a = c − b ⇒ 2b = a + c
∴ 2 (3y+5) = 3y – 1 +5y +1
⇒6y +10 = 8y
⇒10 = 8y -6y
⇒ 2y =10
⇒y=5
Hence the correct option is C.
Q2.
It is known that the length of the tangents drawn from an external point to a circle is equal.
∴ QP = PT = 3.8 cm
PR = PT = 3.8 cm
... (1)
... (2)
From equations (1) and (2), we get:
QP = PR = 3.8 cm
Now, QR = QP + PR
= 3.8 cm + 3.8 cm
= 7.6 cm
Hence, the correct option is B.
Q3. Given: ∠QPR = 46°
PQ and PR are tangents.
Therefore, the radius drawn to these tangents will be perpendicular to the tangents.
So, we have OQ ⊥ PQ and OR ⊥ RP.
⇒ ∠OQP = ∠ORP = 90∘
So, in quadrilateral PQOR, we have
∠OQP +∠QPR + ∠PRO + ∠ROQ = 360∘
⇒ 90° + 46° + 90° + ∠ROQ = 360∘
⇒ ∠ROQ = 360∘ − 226∘ = 134∘
Hence, the correct option is B.
Q4.
In the figure, MN is the length of the ladder, which is placed against the wall AB and makes an
angle of 60° with the ground.
The foot of the ladder is at N, which is 2 m away from the wall.
∴ BN = 2 m
In right-angled triangle MNB:
cos 60° =
1
⇒ =
2
BN
MN
2m
MN
=
2m
MN
⇒MN = 4 m
Therefore, the length of the ladder is 4 m.
Hence, the correct option is D
Q5. Possible outcomes on rolling the two dice are given below:
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Total number of outcomes = 36
Favourable outcomes are given below:
{(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
Total number of favourable outcomes = 9
∴ Probability of getting an even number on both dice =
Total number of favourable outcomes
=
Total number of outcomes
Hence, the correct option is D.
9
36
=
1
4
Q6. Total number of possible outcomes = 30
Prime numbers between 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.
Total number of favourable outcomes = 10
∴ Probability of selecting a prime number from 1 to 30
=
Total number of favourable outcomes
Total number of outcomes
=
10
30
=
1
3
Hence, the correct option is C.
Q7. It is given that the three points A(x, 2), B(−3, −4) and C(7, −5) are collinear.
∴ Area of ∆ABC = 0
1
⇒ [x 1 (y 2 - y 3 ) + x 2 (y 3 - y 1 ) + x 3 (y 1 - y 2 ) ] = 0
2
Here, x 1 =x, y 1 =2, x 2 = -3, y 2 = -4, and x 3 = 7, y 3 =-5
⇒ x[-4 - (-5)] - 3(-5 - 2) + 7[2 -(-4) ] = 0
⇒ x(-4 + 5) - 3(-5 -2) + 7(2 + 4) = 0
⇒ x - 3 × (-7) + 7 × 6 = 0
⇒ x + 21 + 42 = 0
⇒ x + 63 = 0
⇒ x = -63
Thus, the value of x is − 63.
Hence, the correct option is A.
Q8. Let r and h be the radius and the height of the cylinder, respectively.
Given: Diameter of the cylinder = 4 cm
∴ Radius of the cylinder, r = 2 cm
Height of the cylinder, h = 45 cm
Volume of the solid cylinder = πr2h =π × (2)2 × 45 cm3 = 180π cm3
Suppose the radius of each sphere be R cm.
Diameter of the sphere = 6 cm
∴ Radius of the sphere, R = 3 cm
Let n be the number of solids formed by melting the solid metallic cylinder.
∴ n × Volume of the solid spheres = Volume of the solid cylinder
⇒n×
⇒n×
⇒n =
4
3
4
3
πR3 = 180π
π33 = 180π
180×3
4×27
=5
Thus, the number of solid spheres that can be formed is 5.
Hence, the correct option is B.
Q9. We have: 2x2 + ax –a2 = 0
Comparing the given equation with the standard quadratic equation (ax2 +bx + c = 0), we get
a =2, b = a and c =-a2
Using the quadratic formula,𝑥𝑥
−𝑎𝑎±�𝑎𝑎 2 −4×2×(−𝑎𝑎)2
𝑥𝑥 =
=
=
−𝑎𝑎±√9𝑎𝑎 2
4
−𝑎𝑎±3𝑎𝑎
⇒x=
2×2
4
−𝑎𝑎+3𝑎𝑎
= 𝑎𝑎2
4
=
−𝑏𝑏±√𝑏𝑏 2 −4𝑎𝑎𝑎𝑎
2𝑎𝑎
𝑜𝑜𝑜𝑜 𝑥𝑥 = 𝑎𝑎−3𝑎𝑎
= −𝑎𝑎
4
, we get:
So, the solutions of the given quadratic equation are x =
𝑎𝑎
2
or x = -a.
Q10. Let a be the first term and d be the common difference.
Given: a = 5
T n = 45
S n = 400
We know:
T n = a + (n - 1) d
⇒ 45 = 5 + (n - 1)d
⇒ 40 = (n - 1) d
………….(1)
n
And S n = a + T n
⇒ 400 =
⇒
n
2
=
2
n
2
400
(5+45)
50
⇒n = 2 × 8 = 16
On substituting n = 16 in (1), we get:
40 =(16-1)d
⇒ 40 = (15)d
⇒d =
40
15
=
8
3
8
Thus, the common difference is .
3
Q11. Let XBY and PCQ be two parallel tangents to a circle with centre O.
Construction: Join OB and OC.
Now, XB ||AO
⇒ ∠XBO +∠AOB = 180°
Now, ∠XBO = 90°
(sum of adjacent interior angles is 180°)
(A tangent to a circle is perpendicular to the radius through the point of contact)
⇒90° +∠AOB =180°
⇒∠AOB = 180° - 90° =180°
Similarly , ∠AOC = 90⁰
∠AOB + ∠AOC = 90⁰ + 90⁰= 180°
Hence, BOC is a straight line passing through O.
Thus, the line segment joining the points of contact of two parallel tangents of a circle passes through its
centre.
Q12. Let us draw the circle with extent point P and two tangents PQ and PR.
We know that the radius is perpendicular to the tangent at the point of contact.
∴∠OQP = 90°
We also know that the tangents drawn to a circle from an external point are equally inclined to the
joining the centre to that point.
∴∠QPO = 60°
Now, in ∆QPO:
Cos 60°=
1
⇒ =
2
PQ
PQ
PO
PO
⇒2PQ = PO
Q13. Rahim tosses two coins simultaneously. The sample space of the experiment is {HH, HT, TH, and
TT}.
Total number of outcomes = 4
Outcomes in favour of getting at least one tail on tossing the two coins = {HT, TH, TT}
Number of outcomes in favour of getting at least one tail = 3
∴ Probability of getting at least one tail on tossing the two coins
=
Number of favourable outcomes
Total number of outcomes
Q14.
=
Let us join OB.
3
4
In ∆OAB:
OB2 = OA2 +AB2 = (20)2 + (20)2 = 2 × (20)2
⇒OB = 20 √2
Radius of the circle, r = 20 √2 cm
Area of quadrant OPBQ =
=
90°
360°
1
θ
360°
× πr2
×3.14 × (20√2 )2cm2
= × 3.14 ×800 cm2
4
= 628 cm2
Area of square OABC = (Side) 2 = (20)2 cm2 = 400 cm2
∴ Area of the shaded region = Area of quadrant OPBQ – Area of square OABC
= (628 - 400) cm2
= 228 cm2
Q 15. Given equation:
4
-3=
4
-3 =
x
⇒
x
4−3x
x
5
2x+3
; x≠ 0, -
5
3
2
2x+3
=
5
2x+3
⇒ (4 -3x) (2x+3) = 5x
⇒-6x2 +8x-9x+12 =5x
⇒6x2 +6x -12 =0
⇒ x2 +x -2 =0
⇒ x2 +2x –x-2 =0
⇒(x+2) (x-1) =0
⇒(x+2) =0 (x–1)=0
⇒ x = -2 or x = 1
Thus, the solution of the given equation is -2 or 1.
Q16. Let a be the first term and d be the common difference of the given A.P.
Given:
a7 =
a9 =
1
9
1
7
a 7 = a + (7-1) d = 19
⇒ a + 6d =
1
9
……… (1)
a 9 = a + (9 -1) d =
⇒a + 8d =
1
7
1
7
………. (2)
Subtracting equation (1) from (2), we get:
2d =
2
63
⇒d=
1
63
Putting d =
a + (6 ×
⇒a=
1
63
1
63
1
in equation (1), we get:
)=
1
63
9
∴ a 63 =a + (63 - 1) d =
1
63
+62 (
1
63
)=
63
63
=1
Thus, the 63rd term of the given A.P. is 1.
Q17. Follow the given steps to construct the figure.
Step 1
Draw a line BC of 8 cm length.
Step 2
Draw BX perpendicular to BC.
Step 3
Mark an arc at the distance of 6 cm on BX. Mark it as A.
Step 4
Join A and C. Thus, ∆ABC is the required triangle.
Step 5
With B as the centre, draw an arc on AC.
Step 6
Draw the bisector of this arc and join it with B. Thus, BD is perpendicular to AC.
Step 7
Now, draw the perpendicular bisector of BD and CD. Take the point of intersection as O.
Step 8
With O as the centre and OB as the radius, draw a circle passing through points B, C and D.
Step 9
Join A and O and bisect it Let P be the midpoint of AO.
Step 10
Taking P as the centre and PO as its radius, draw a circle which will intersect the circle at point
B and G. Join A and G.
Here, AB and AG are the required tangents to the circle from A.
Q 18. The given points are A (0, 2), B (3, p) and C (p, 5).
It is given that A is equidistant from B and C.
∴ AB = AC
⇒ AB2 = AC2
⇒ (3-0)2 + (p-2)2 = (p-0)2 + (5-2)2
⇒ 9+p2+ 4- 4p=p2+9
⇒ 4 - 4p =0
⇒ 4p =4
⇒ p =1
Thus, the value of p is 1.
Length of AB = �(3 − 0)2 + (1 − 2)2 = �32 + (−1)2 = √9 + 1 = √10units.
Q19. Let d be the distance between the two ships. Suppose the distance of one of the ships from the
light house is X meters, then the distance of the other ship from the light house is (d-x) meter.
In right –angled ∆ADO, we have.
tan 45° =
⇒1=
200
OD
AD
200
=
X
X
⇒ x =200 … (1)
In right-angled ∆ BDO, we have
tan 60°=
⇒ √3 =
OD
BD
=
200
200
d−x
d−x
⇒ d− x =
200
√3
Putting x=200. We have:
d - 200=
d=
200
√3
200
√3
+ 200
⇒ d = 200(
√3+1
√3
)
⇒ d = 200 × 1.58
⇒d = 316
(approx.)
Thus, the distance between two ships is approximately 316 m.
Q20. The given points are A (-2, 1), B (a, b) and C (4-1).
Since the given points are collinear, the area of the triangle ABC is 0.
1
⇒ [x 1 (y 2 -y 3 )+x 2 (y 3 -y 1 ) +x 3 (y1 - y2)] =0
2
Here, x 1 =-2,y 1 =1,x 2 =a,y 2 =b and x 3 =4, y 3 =-1
∴
1
2
[− 2(b+1)+a(-1-1) +4 (1-b) =0
⇒-2b -2-2a+4-4b=0
⇒2a+6b=2
⇒a+3b=1
…….(1)
Given :
a-b= 1
………(2)
Subtracting equation (1) from (2) we get:
4b =0
⇒b=0
Subtracting b= 0 in (2), we get:
a= 1
Thus, the values of a and b are 1 and 0, respectively.
Q21. It is given that ABC is an equilateral triangle of side 12 cm.
Construction:
Join OA, OB and OC.
Draw.
OP ⊥ BC
OQ ⊥ AC
OR ⊥ AB
Let the radius of the circle be r cm.
Area of ∆AOB +Area of ∆BOC + Area of ∆AOC = Area of ∆ABC
⇒
1
2
1
1
1
× AB× OR + × BC × OP + × AC × OQ =
1
2
⇒ × 12 × r + × 12 ×r +
2
⇒3×
1
2
2
√3
4
× 12 × r =
1
2
2
× 12 × r =
× 12 × 12
√3
4
√3
4
× (12)2
× (Side)2
⇒ r = 2√3 =2 × 1.73 = 3.46
Therefore the radius of the inscribed circle is 3.46 cm.
Now, area of the shaded region = Area of ∆ABC – Area of the inscribed circle
=[
√3
×
4
(12)2 – π (2√3)2] cm2
= [36√3 -12π] cm2
= [36 ×1.73 – 12 × 3.14] cm2
= [62.28 – 37.68] cm2
= 24.6 cm2
Therefore, the area of the shaded region is 24.6 cm2.
1
Q22. Radius of Semicircle PSR = × 10 cm =5 cm
2
1
Radius of Semicircle RTQ = × 3 = 1.5 cm
2
1
Radius of semicircle PAQ = ×7 cm = 3.5 cm
2
Perimeter of the shaded region = Circumference of semicircle PSR + Circumference of
semicircle RTQ + Circumference of semicircle PAQ
1
1
1
= [ × 2π(5) + × 2π (1.5)+ × 2π(3.5)] cm
2
2
2
= π (5+1.5 +3.5) cm
=3.14 × 10 cm
= 31.4 cm
Q23. For the given tank.
Diameter = 10 m
Radius, R = 5m
Depth, H =2m
20
Internal radius of the pipe = r =
2
cm =10 cm =
Rate of flow of water = v = 4 km/h = 4000 m/h
Let t be the time taken to fill the tank.
1
10
m
So, the water flown through the pipe in t hours will equal to the volume of the tank
∴ πr2×v × t =ΠR2H
1
⇒ ( ) 2 × 4000 × t = (5)2× 2
10
⇒t=
25×2×100
4000
=1
1
4
1
Hence, the time taken is 1 hours.
4
Q 24.
Let ACB be the cone whose vertical angle ∠ACB = 60°. Let R and x be the radii of the lower and
upper end of the frustum.
Here, height of the cone, OC = 20 cm = H
Height CP = h = 10 cm
Let us consider P as the mid-Point of OC.
After cutting the cone into two parts through P.
OP =
20
2
= 10 cm
Also, ∠ACO and ∠OCB =
1
2
× 60° = 30°
After cutting cone CQS from cone CBA, the remaining solid obtained is a frustum.
Now, in triangle CPQ:
tan 30° =
⇒
1
√3
=
⇒x=
x
10
10
√3
x
10
cm
In triangle COB:
Tan 30° =
R
CO
⇒
1
R
=
√3
⇒R=
20
20
cm
√3
1
Volume of the frustum, V = π (R2H – x2h)
20 2
1
10
3
2
⇒𝑉𝑉 = 3 𝜋𝜋(� � . 20 − � � . 10)
1
8000
1
7000
= π(
3
= π(
3
√3
−
3
)
3
1
1000
3
√3
)
= π ×7000
=
9
7000
9
π
The volumes of the frustum and the wire formed are equal.
π×(
1
24
⇒l =
)2 ×l =
7000
9
7000
9
π [ Volume of wire = πr2h]
× 24 × 24
⇒ l = 448000 cm = 4480 m
Hence, the length of the wire is 4480 m.
Q25. Let the two natural numbers be X and Y such that x > y.
Given:
Difference between the natural numbers = 5
∴ X - Y =5
…..(1)
Difference of their reciprocals =
1
y
-
⇒
1
=
x−y
=
x
xy
1
10
1
10
1
10
(given)
⇒
5
xy
=
1
10
⇒ xy = 50
… (ii)
Putting the value of x from equation (i) in equation (ii), we get
(y+5) y =50
⇒ y2 +5y - 50 = 0
⇒ y2 + 10y -5y - 50 = 0
⇒ y (y+10) – 5 (y+10) = 0
⇒ (y -5) (y + 10) = 0
⇒ y = 5 or -10
As y is a natural number, therefore y = 5
Other natural number = y + 5 = 5 + 5 = 10
Thus, the two natural numbers are 5 and 10.
Q26. Let AP and BP be the two tangents to the circle with centre O.
To Prove : AP = BP
Proof:
In ∆AOP and ∆BOP
OA =OB
(radii of the same circle)
∠OAP = ∠OBP = 90° (since tangent at any point of a circle is perpendicular to the radius through
the point of contact)
OP = OP
(common)
∴ ∆AOP ≅ ∆ OBP
(by R.H.S. congruence criterion)
∴ AP = BP
(corresponding parts of congruent triangles)
Hence the length of the tangents drawn from an external point to a circle are equal.
Q27. Let AB be the building and CD be the tower.
In right ∆ABD.
AB
BD
⇒
= tan 60°
60
BD
= √3
60
⇒ BD =
√3
⇒ BD = 20√3
In right ∆ACE:
CE
AE
⇒
= tan 30°
CE
BD
⇒ CE =
=
1
√3
20√3
√3
(∴ AE = BD)
= 20
Height of the tower = CE + ED = CE + AB = 20 m + 60 m = 80 m
Difference between the heights of the tower and the building = 80 m – 60m = 20 m
Distance between the tower and the building = BD = 20√3 m.
Q28. Total number of cards = 49
(1)
Total number of outcomes = 49
The odd numbers form 1 to 49 are 1, 3,5,7,9,11,13,15,17, 19,21, 23, 2527,29,31,33,35, 37,
39, 41, 43, 45, 47 and 49.
Total number of favourable outcomes = 25
∴ Required probability =
Total number of favourable outcomes
=
Total number of outcomes
(ii)
25
49
Total number of outcomes = 49
The number 5,10,15,20,25,30,35,40 and 45 multiples of 5.
The number of favourable outcomes = 9
∴ Required probability =
Total number of favourable outcomes
(iii) Total number of outcomes = 49
=
Total number of outcomes
The number 1, 4, 9, 16, 25, 36 and 49 are perfect squares.
Total number of favourable outcomes = 7
∴ Required probability =
Total number of favourable outcomes
(iv)
Total number of outcomes
=
7
49
=
Total number of outcomes = 49
We know that there is only one even prime number which is 2
Total number of favourable outcomes = 1
∴ Required probability =
Total nu mber of favourable outcomes
Total number of outcomes
=
1
49
1
7
9
49
Q29. Let the Point P (x, 2) divide the line segment joining the points A (12, 5) and B (4, -3) in the ratio
k: 1
Then, the coordinates of P are (
4k+12
k+1
Now, the coordinates of P are (x,2)
∴
4k+12
k+1
−3k+5
= x and
−3k+5
=2
k+1
k+1
,
−3k+5
k+1
)
=2
⇒− 3k+ 5 = 2k+2
⇒ 5k = 3
3
⇒k =
5
Substituting k =
X=
3
5
4× +12
⇒x=
3
5
in
4k+12
k+1
= x, we get
3
+1
5
12+60
⇒x=
3+5
72
8
⇒x=9
Thus, the value of x is 9
3
Also, the point P divides the line segment joining the points A (12, 5) and (4, -3 ) in the ratio : 1, i.e. 3:5.
Q30. Given quadratic equation:
( k + 4)x2 + (k + 1)x + 1=0
Since the given quadratic equation has equal roots, Its discriminant should be zero.
∴D=0
⇒(k+1)2- 4 × (k+4) × 1 = 0
⇒k2+2k+1- 4k - 16=0
⇒k2- 2k -15 =0
5
⇒k2 - 5k+ 3k -15 = 0
⇒(k-5) (k+3) =0
⇒k-5 =0 or k+3 =0
⇒k = 5 or -3
Thus, the values of k are 5 and -3
For k = 5
(k+4)x2 +(k+1)x +1= 0
⇒9x2 +6x +1 = 0
⇒(3x)2 +2(3x) +1 =0
⇒(3x +1)2 =0
1
⇒x = − , −
3
For k = -3
1
3
(k +4)x2 +(k+1)x + 1 = 0
⇒x2 – 2x +1 =0
⇒(x - 1)2 = 0
⇒ x = 1, 1
1
Thus, the equal root of the given quadratic equation is either 1 or − .
3
Q31 Let a and d be the first term and the common difference of an A. P. respectively.
nth term of an A. P, a n = a + ( n -1 )d
n
Sum of n terms of an A. P , S n = [2a+(n -1)d]
2
We have:
Sum of the first 10 terms =
10
⇒210 = 5[2a+9d ]
⇒ 42 = 2a+9d
………. (1)
2
[2a + 9d]
15th term from the last = ( 50−15 + 1) th= 36th term from the beginning
Now,
a 36 = a + 35d
∴ Sum of the last 15 terms =
=
15
2
15
2
(2a 36 +(15 −1)d)
[2(a +35d) +14d]
=15[a +35d +7d]
⇒2565 = 15[a+42d]
⇒ 171= a+42d
………….(2)
From (1) and (2), we get,
d=4
a= 3
So, the A. P. formed is 3, 7, 11, 15 …. and 199.
Q32 Given ABCD be a parallelogram circumscribing a circle with centre O.
To Prove: ABCD is a rhombus.
We know that the tangents drawn to a circle from an exterior point are equal is length.
∴ AP = AS, BP = BQ, CR =CQ AND DR = DS.
AP +BP +CR +DR = AS +BQ +CQ +DS
(AP +BP) + (CR+DR) = (AS+DS) +BQ +CQ)
∴AB +CD =AD +BC OR 2AB = 2BC
∴ AB = BC = DC =AD
Therefore, ABCD is a rhombus.
(since AB =DC and AD = BC)
Q33.
Height (h) of the conical vessel = 11 cm
Radius (r 1 ) of the conical Vessel = 2.5 cm
Radius (r 2 ) of the metallic spherical balls =
0.5
2
= 0.25 cm
Let n be the number of spherical balls =that were dropped in the the vessel.
Volume of the water spilled = Volume of the spherical balls dropped
2
5
× Volume of cone = n × Volume of one spherical ball
⇒
2
5
2
×
1
3
2
4
πr h = n × πr
1
⇒r h= n × 10r
1
3
3
3
2
2
⇒ (2.5)2 ×11 = n × 10 × (0.25)3
⇒ 68.75 = 0.15625n
⇒ n = 440
Hence, the number of spherical balls that were dropped in the vessel is 440.
Sushant made the arrangement so that the water that flows out, irrigates the flower beds.
This shows the judicious usage of water.
Q34.
The following figure shows the required cylinder and the conical cavity.
Given
Height (b) of the conical Part = Height (h) of the cylindrical part = 2.8 cm
Diameter of the cylindrical part = Diameter of the conical part = 4.2 cm
∴ Radius ® of the cylindrical part = Radius ® of the conical part = 2.1 cm
Slant height (l) of the conical part =√r 2 + h2
=�(2.1)2 + (2.8)2 cm
=√4.41 + 7.81 cm
= √12.25 cm
= 3.5 cm
Total surface area of the remaining solid = Curved surface area of the cylindrical part +Curved surface
area of the conical part + Area of the cylindrical base
= 2πrh +πrl +πr2
=
(2 ×
22
7
22
× 2.1 ×2.8+
7
× 2.1 × 3.5 +
= (36.96 + 23.1 + 13.86) cm2
22
7
× 2.1× 2.1) cm2
=73.92 cm2
Thus, the total surface area of the remaining solid is 73.92 cm2
CLASS X : MATH SOLUTIONS
Q1. It is given that x = ∴ 3�
⇒
−12
3
4
k=
1
1
2
� + 2k (- ) - 3 =0
3
–3=
2
- k - 3=0
4
is the solution of the quadratic equation 3x2+2kx-3 = 0.
2
−9
4
Hence, the value of k is
−9
4
Q2. Let AB and CD be the two towers of heights x and y, respectively.
Suppose E is the centre of the line joining the feet of the two towers i.e. BD.
Now, in ∆ ABE,
𝐴𝐴𝐴𝐴
𝐵𝐵𝐵𝐵
⇒
= tan30°
𝑥𝑥
𝐵𝐵𝐵𝐵
= √3
⇒BE = √3x ……… 1
Also,
In ∆ CDE,
𝐶𝐶𝐶𝐶
𝐷𝐷𝐷𝐷
= tan60°
⇒DE =
𝑦𝑦
√3
………. (2)
NOW BE = DE …… (3)
(E is mid – point of BD)
So, from (1) , (2) and (3) we get
√3x =
𝑥𝑥
⇒ =
𝑦𝑦
𝑦𝑦
√3
1
3
Hence, the ratio of x and y is 1: 3.
Q3. There are 26 letters in English alphabets.
∴ Total number of outcomes = 26
We know that there are 5 vowels and 21 consonants in English alphabets.
∴ Total number of favourable outcomes = 21
∴ Probability that the chosen letter is a consonant =
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
Q4. PA and PB are tangents drawn from an external point P to the circle.
∴ PA = PB (Length of tangents drawn from an external point to the circle are equal)
In ∆ PAB,
PA = PB
⇒ ∠ PBA = ∠PAB
……… (1) (Angles opposite to equal sides are equal.)
Now,
∠APB + ∠PBA + ∠PAB = 180°
⇒ 50° + ∠PAB + ∠PAB = 180° [Using (1)]
⇒ 2∠PAB = 130°
⇒ ∠PAB =
130°
2
= 65°
=
21
26
We know that radius is perpendicular to the tangent at the point of contact
∴ ∠ OAP = 90°
(OA ⊥ PA)
⇒ ∠PAB + ∠OAB = 90°
⇒ 65° + ∠OAB = 90°
⇒ ∠OAB = 90° - 65° = 25°
Hence the measure of ∠OAB is 25°
SECTION – B
Q5. We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended
by it at any point on the remaining part of the circle.
∴ ∠AOQ = 2 ∠ ABQ
1
⇒ ∠ABQ = ∠ AOQ
2
1
⇒ ∠ABQ = × 58° = 29°
2
or ∠ABT = 29°
We know that the radius is perpendicular to the tangent at the point of contact
∴ ∠OAT = 90° (OA ⊥ AT)
Or ∠BAT = 90°
Now, in ∆BAT,
∠BAT + ABT + ∠ATB = 180°
⇒90° + 29°+∠ATB= 180°
⇒∠ATB = 61°
∠ATQ = 61°
Q6. We have
4x2 – 4a2x + (a4 – b4) =0
⇒ (4x2 -4a2x +a4) –b4 =0
⇒[(2x) 2 -2 (2x) (a) + (a2)2] – b4 = 0
⇒(2x – a2)2 – (b2)2 = 0
⇒ (2x – a2 +b2) (2x – a2 –b2) =0
⇒ 2x – a2 +b2 = 0 or 2x – a2 – b2 = 0
⇒ 2x = a2 – b2 or 2x = a2 + b2
⇒x=
𝑎𝑎 2 −𝑏𝑏 2
Q7.
2
or
𝑎𝑎 2 +𝑏𝑏 2
2
TP and TQ are tangents drawn from an external point T to the circle. O is the centre of the circle.
Suppose OT intersect PQ at point R.
In ∆OPT AND ∆OQT,
OP= OQ (Radii of the circle)
TP = TQ (Lengths of tangents drawn from an external point to a circle are equal )
OT = OT (Common sides)
∴ ∆ OPT ≅ ∆ OQT (BY SSS congruence rule)
So, ∠PTO = ∠QTO (BY CPCT) .……… (1)
Now, in ∆PRT and ∆QRT,
TP =TQ (Lengths of tangents drawn from an external point to a circle are equal)
∠PTO = ∠QTO [From (1)]
RT = RT
(Common sides)
∴ ∆PRT ≅ ∆QRT (By SAS congruence rule)
So, PR = QR ………. (2) (By CPCT)
And, ∠
PRT = ∠QRT (BY CPCT)
Now,
∠PRT +∠QRT = 180°
(Linear Pair)
⇒2∠PRT = 90°
⇒ 2 ∠PRT = 90°
∴ ∠PRT = ∠QRT = 90° ……(3)
From (2) and (3) we can conclude that
OT is the right bisector of the line segment PQ.
Q8. The given anthmetic progression is
6, 13, 20…., 216
Let 216 be the nth term of the given AP.
So,
a=6
d =7
an = 216
Now,
a n = a + (n - 1)d
⇒216 = 6 + (n - 1) × 7
⇒7(n-1) = 210
⇒n - 1 = 30
⇒n = 31, which is odd
∴ Middle term of the AP
31+1 th
)
2
=(
P
term of the AP
= 16th term of the AP
∴ a 16 =6+16-1×7=6+15×7=6+105=111
Thus, the middle term of the given AP is 111.
Q9. Using distance formula, we have
AB = �(2 − 5)2 + (−2 − 2)2
= √9 + 16 = 5
AC = �(−2 − 5)2 + (𝑡𝑡 − 2)2
= = √𝑡𝑡 2 − 4𝑡𝑡 + 53
BC =�(−2 − 2)2 + (𝑡𝑡 + 2)2 =√𝑡𝑡 2 + 4𝑡𝑡 + 20
……. (1)
……… (2)
…. (3)
Now, it is given that ∆ABC is right angled at B.
Using the Pythagorean theorem, we have
AB2 +BC2 = AC2
∴ 25 + t2 + 4t + 20 = t2 - 4t + 53
[From (1) (2) and (3)]
⇒45 + 4t = -4t + 53
⇒ 8t = 8
⇒t=1
Hence, the value of t is 1.
3
Q10. Let P ( ,
5
) divide AB in the ratio m 1 :m 2.
4 12
Using the section formula, we have
3
5
( ,
4 12
3
4
5
=
12
1
2
2𝑚𝑚 1 + 𝑚𝑚 2
𝑚𝑚 1 +𝑚𝑚 2
=
= ��
3
2
−5𝑚𝑚 1 + 𝑚𝑚 2
𝑚𝑚 1 +𝑚𝑚 2
4
=
1
2
𝑚𝑚 1 +𝑚𝑚 2
…….. (1)
…… (2)
From (1) we have
3
1
2𝑚𝑚 1+ 𝑚𝑚 2
2
)
2𝑚𝑚 1 + 𝑚𝑚 2
𝑚𝑚 1 +𝑚𝑚 2
3
−5𝑚𝑚 1 + 𝑚𝑚 2
2
�,�
𝑚𝑚 1 +𝑚𝑚 2
��
1
3(𝑚𝑚1 + 𝑚𝑚2 ) = 4(2𝑚𝑚1 + 𝑚𝑚2 )
3m 1 +3m 2 =8m 1 + 2m 2
2
-5 m 1 =- m 2
m 1: m 2 = 1: 5
SECTION - C
Q11. Let the coordinates of points B and C of the ∆ABC be (a 1 , b 1 ) and (a 2 , b 2 ), respectively.
Q is the midpoint of AB.
Using midpoint formula, we have
(0, -1)=
⇒
𝑎𝑎 1 +1 𝑏𝑏1 −4
𝑎𝑎 1 +1
2
2
,
2
=0 and
𝑏𝑏1 −4
2
=-1
⇒ a 1 =-1 and b 1 =2
Therefore, the coordinates of B are (-1,2).
P is the midpoint of AC.
Now,
(2, -1) =
⇒
𝑎𝑎 2 +1
2
𝑎𝑎 2 +1 𝑏𝑏2 −4
2
,
2
= 2 and
𝑏𝑏2 −4
⇒ a 2 = 3 and b 2 =2
2
= -1
Therefore, the coordinates of C are (3,2).
Thus, the vertices of ∆ABC are A(1, -4), B (-1,2) and C(3,2).
Now,
Area of the triangle having vertices (x 1, y 1 ),(x 2 , y 2 ),(x 3 ,y 3 ) = 12x1y2-y3+2y3-y1+x3y1-y2
1
∴ Area of ∆ABC = [x 1 (y 2 - y 3 ) + x 2 (y 3 - y 1) + 3 (y 1 - y 2 )]
2
1
= × [1 (2 - 2) + -1(2 + 4) + 3(-4 - 2)]
2
=-12
Since area is a measure that cannot be negative, we will take the numerical value of – 12, that is 12.
Thus , the area of ∆ABC is 12 square units.
Q12
We have
kx2 + 1 - 2(k - 1) x + x2 = 0
This equation can be rearranged as
(k + 1) x2 - 2(k - 1)x + 1 = 0
Here, a = k+1, b = -2(k-1) and c =1
∴ D = b2 – 4ac
= [-2(k-1)2] – 4 × (k + 1) × 1
=4[(k-1)2 – (k+1)]
= 4[k2 – 3k]
= 4[k (k -3)]
The given equations will have equal roots, if D = 0.
⇒ 4[k (k-3)] = 0
⇒ k = 0 or k -3 = 0
⇒ k= 3
Putting k= 3 in the given equation , we get
4x2 - 4x + 1 =0
⇒(2x-1)2 = 0
⇒ 2x -1 =0
⇒ x =1/2
Hence, the roots of the given equation are 1/2 and 1/2.
Q13
Let AB the building and CD be the tower.
CD/BD=tan 45°
⇒30/BD =1
⇒ BD =30m
In ∆ABD,
𝐴𝐴𝐴𝐴
𝐵𝐵𝐵𝐵
= tan 30°
⇒ AB = BD ×
1
√3
⇒ AB= 10√3 m
Therefore, the height of the building is 10√3 m.
Q14.
When two dice are thrown simultaneously, the possible outcomes are
∴ Total number of possible outcomes = 36
We know
Probability of an event =
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
(1) The outcomes favorable of the event’ the sum of the numbers on two dice to be 5’ denoted by E
are (1, 4), (2, 3), (4, 1) and (3, 2).
Total number of favorable outcomes = 4
∴ PE =4/36 =1/9
(ii) The outcomes favourable to the event ‘even numbers on both dice’ denoted by F are
(2, 2), ( 2,4),(2,6), (4,2),(4,4),(4,6), (6,2), (6,4) and (6,6).
Total number of favouable outcomes = 9
∴PF =9/36 = 1/4
Q15
Let a and d be the first term and the common difference of AP, respectively,
We know
𝑛𝑛
Sn = [2a+ (n-1) d]
2
8
∴S 8 = [2a+ (8-1) d and S4 =
2
8
2
[2a+ (4-1) d]
⇒ S 8 = 4(2a + 7d) and S 4 = 2(2a + 3d)
⇒S 8 = 8a + 28d and S 4 =4a + 6d
Now,
3(S 8 – S 4 ) =3(8a+28d-4a-6d)
= 3(4a+22d)
=6(2a+11d)
=
12
2
[2a+ (12-1) d]
=S 12
∴ S 12 =3(S 8 -S 4 )
16.
Let the radius of the semi-circle APB be r.
⇒ The radius of the semi- circle AQO =
Now,
𝑟𝑟
2
Perimeter of the given figure = Length of arc AQO + Length of arc APB + OB
3
=r( ×
=r
2
80
22
7
+ 1)
14
40
40
⇒ r( ) = 40
= r( ) cm
7
7
⇒ r = 7cm
∴ Area of the shaded region = Area of semi – circle AQO + Area of semi- circle APB
=
7
𝜋𝜋(2 )2
2
+
1
𝜋𝜋72
2
1
= 49 𝜋𝜋 ( + )
8
2
= 96.25 cm2
17. The remaining solid is a frustum of the given cone.
Total surface area of the frustum
Where
h = Height of the frustum = 12 - 4 = 8 cm
r 1 = Larger radius of the frustum = 6 m
r 2 = Smaller radius of the frustum
l = Slant height of the frustum
In the given figure, ∆ABC ~∆ ADE by AA similarity criterion.
∴ BC/DE = AB/AD
𝑟𝑟2
6
=
4
12
⇒ r 2 = 2 cm
We know
⇒ l = 4√5 cm.
∴ Total surface area of the frustum = 𝜋𝜋𝜋𝜋(r 1 +r 2 ) +πr 1 2 + πr 2 2
= 350.592 cm2
Hence, the total surface area of the remaining solid is 350.592 cm2.
18. Let h be the height of the cone and r be the radius of the base of cone.
The volume of the wooden toy
𝜋𝜋𝑟𝑟 2 ℎ
=
3
2
+ πr3
3
According to the question,
77
6
(h + 7) =
1001
6
⇒ h= 6 cm
The height of the wooden toy = 6 cm + 3.5 cm
= 9.5 cm
Now,
Curved surface area of the hemispherical part = 2X(22/7)X3.5X3.5 = 2 ×
=77 cm2
22
7
× (3.5)2
Hence, the cost painting the hemispherical part of the toy = 77 X 10
= Rs 770
Q19. Surface area of the remaining block = Surface area of the cuboid + Curved surface area of
cylinder -2 × Area of base of cylinder
= 2( 15 × 10 + 10 × 5 +15 × 5) + (2 ×
= 583 cm2
22
7
×
7
2
× 5 ) –( 2 ×
22
7
×
Thus, the surface area of the remaining block is 583 cm2.
Q20. From the symmetry of the given figure, we have
7
2
×
7
2
)
∴ Area of the shaded region = Area of the square with side 14 cm – (Area of square with side 4
cm + 4 × Area of each semicircle with radius 2 cm )
= 14 × 14 –[ 4 × 4 + 4 ×
= 196 –( 16 + 25.12)
1
2
× 3.14 ×(2)2]
= 196 – 41.12
= 154.88 cm2
Section - D
21. Let the numerator and the denominator of the fraction be x + 3, respectively.
∴ Original fraction =
𝑥𝑥
𝑥𝑥+3
Now, 2 is added to both the numerator and the denominator.
∴ New fraction =
(𝑥𝑥+2)
(𝑥𝑥+3)
According to the question,
𝑥𝑥 + 2 29
𝑥𝑥
+
=
𝑥𝑥 + 3 𝑥𝑥 + 5 20
Solving we get
𝑥𝑥(𝑥𝑥 + 5) + (𝑥𝑥 + 2)(𝑥𝑥 + 3)
29
=
(𝑥𝑥 + 3)(𝑥𝑥 + 5)
20
𝑥𝑥 =
7
10
−45
Or𝑥𝑥 =
11
Now, 𝑥𝑥 ≠ (
−45
11
)as it is a fraction.
So, the original fraction becomes (7/10).
22. Since Ramkali increased her weekly savings uniformly every week by a fixed number, her
savings will form an AP.
Let S n be the sum of savings in all 12 weeks.
𝑛𝑛
∴ Sn= (2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑)
2
(Here a is the money saved in the first week and d is the fixed
increase in the weekly savings.)
12
⇒ Sn= (2 × 100 + (12 − 1)20)=Rs 2,520
2
Ramkali required Rs2,500 after 12 weeks, but she saved Rs 2,520. So, she will be able to send
her daughter to school after 12 weeks.
It shows that Ramkali is aware of the importance of girl child education.
23.
2
𝑥𝑥+1
+
3
2(𝑥𝑥−2)
=
23
5𝑥𝑥
4(𝑥𝑥 − 2) + 3(𝑥𝑥 + 1) 23
=
2(𝑥𝑥 + 1)(𝑥𝑥 − 2)
5𝑥𝑥
5𝑥𝑥�4(𝑥𝑥 − 2) + 3(𝑥𝑥 + 1)� = 46(𝑥𝑥 + 1)(𝑥𝑥 − 2)
11𝑥𝑥 2 − 21𝑥𝑥 − 92 = 0
11𝑥𝑥 2 − 44𝑥𝑥 + 23𝑥𝑥 − 92 = 0
11𝑥𝑥(𝑥𝑥 − 4) − 23(𝑥𝑥 − 4) = 0
Solving we get x =(-23/11) or 4
24.
Given: A circle C (o,r) and a tangent l at point A
To prove: OA ⊥ /
Construction:
Take a point B, other than A, on the tangent l. Join OB. Suppose OB meets the circle in C.
Proof: We know that among all line segments joining the point O to a point on /, the
perpendicular is shortest to /.
OA = OC (Radius of the same circle)
Now,
OB=OC+BC
⇒ OB > OC
⇒ OB > OA
⇒ OA < OB
B is an arbitrary point on the tangent l . Thus, OA is shorter than any other line segment joining
O to any point on l .
Here,
OA ⊥ l
Hence, the tangent at any point of a circle is perpendicular to the radius through the point of
contact.
25.
We know that tangents from an external point are equal in length.
∴ PQ = PR
In ∆ PQR,
PQ = PR
∴ ∠PQR = ∠PRQ (Angles opposite to equal sides are equal).
Now in ∆ PQR,
∠ PQR + ∠ PRQ + ∠ RQP =1800
2 ∠ RQP = 1800-300
∠ RQP = 750
Also, radius is perpendicular to the tangent at the point of contact.
∴ ∠OQP = ∠ORP = 90 0
Now, in PQOR,
∠ ROQ + ∠ OQP + ∠ QPR + ∠ PRO = 3600
900 + 900+300 + ∠ ROQ = 3600
∠ ROQ = 1500
Since, angle subtended by an arc at any point on the circle is half the angle subtended at the
centre by the same arc.
angle QSR = 750
Also, ∠QSR = ∠SQT (Alternate interior angles)
∴ ∠SQT =750
Now,
∠ SQT + ∠ PQR + ∠ SQR = 1800
750 + 750 + ∠ SQR = 1800
∠ SQR =300
26.
Given: A ∆ABC with BC = 7 cm, ∠B = 60 0 = and AB = 6 cm
Steps of Construction:
1. Draw a line segment AB = 6 cm.
2.
3.
4.
5.
6.
7.
8.
With B as centre, ∠B = 60 0 .
With B as centre and radius BC=7 cm, draw an arc.
Join AC to obtain ∆ ABC.
Below AB, make an acute angle ∠BAX .
Along AX, mark off four points A 1 , A 2 , A 3 and A 4 such that AA 1 =A 1 A 2 =A 2 A 3 =A 3 A 4 .
Join A 4 B.
From A 3 , draw A 3 B’//A 4 B
9. From B’ draw B’C’//BC
Thus, ∆ AB’C’ is the required triangle with each of its side (3/4) times the size of the
corresponding sides of ∆ ABC.
27.
Let RQ be the tower and SR be the flag staff.
In PQR
Tan300 = (RQ/PQ)
1
√3
ℎ
= ---------------------------(i)
𝑥𝑥
x = h√3
In PQS,
Tan600=
√3 =
𝑆𝑆𝑆𝑆
𝑃𝑃𝑃𝑃
ℎ+5
𝑥𝑥
----------------(ii)
From (i) and (ii), we get
3h = h+5
2h = 5
H = 2.5 m
Hence, the height of the tower is 2.5metres.
28. Total number of cards = 20
(i) Numbers divisible by 2 or 3 are 2,3,4,6,8,9,10,12,14,15,16,18and 20.
Total favorable number of cards =13
Probability that the number on the drawn card is divisible by 2 or 3=(Total favorable number of
cards /Total number of cards)
= (13/20)
(ii) Prime numbers are 2, 3, 5, 7, 11, 13,17and19
Total favorable number of cards =8
Probability that the number on the drawn card is a prime number=
(Total favorable number of cards/ Total number of cards)=(8/20)=(2/5)
29.
Let the vertices of the quadrilateral be A(-4,8), B(-3,-4), C(0,-5)and D(5,6).
Join AC to form two triangles, namely, ∆ ABC and ∆ ACD.
Area of quadrilateral ABCD = Area of ∆ ABC+ Area of ∆ ACD
We know
Area of triangle having vertices (x 1 ,y 1 ),(x 2 ,y 2 ),(x 3 ,y 3 ) = (1/2)(x 1 (y 2 -y 3 )+x 2 (y 3 -y 1 )+x 3 (y 1 -y 2 ))
Now,
Area of ∆ ABC = (1/2)((-4)(-4+5) + (-3)(-5-8) + 0 (8+4))
= (1/2)(-4+39)
= (35/2)
Area of ∆ ACD =(1/2)((-4)(-6-5) +5(-5-8) +0 (8-6))
= (1/2)(-44-65)
= (109/2)
∴ Area of quadrilateral ABCD=Area of ∆ ABC + Area of ∆ ACD = (35/2)+(109/2)=72 square units
Thus, the area of quadrilateral ABCD is 72 square units.
30.
Depth (h 1 ) of the well = 14 m
Radius (r 1 ) of the circular end of the well = (4/2) m =2 m
Height (h 2 ) of embankment = 40 cm = 0.4 m
Let the width of embankment be x.
From the figure, it can be observed that the embankment will be cylindrical in shape having
outer radius (r 2 ) as (2 +x) m and inner radius (r 1 ) as 2 m.
Volume of earth dug from the well = Volume of earth used to form embankment
Πr 1 2h 1 = π(r 2 2 – r 1 2)h 2
Π(2)214 = π((2+x)2 – 22)
X2+4x-40 =0
(x-10)(x+14) =0
X =10 m
Therefore, the width of the embankment will be 10 m.
31. Increase in the level of water in half an hour, h=3.15 m=315 cm
Radius of the water tank, r=40 cm
Volume of water that falls in the tank in half an hour
= πr2h
= π(40)2(315)
= 5,04,000π cm3
Rate of flow of water = 2.52 km/h
Length of water column in half an hour = 2.522 = 1.26km = 1,26,000 cm
Let the internal diameter of the cylindrical pipe be d.
Volume of the water that flows through the pipe in half an hour = π (d/2)2(126000) cm3
We know
Volume of the water that flows through the pipe in half an hour = Volume of water that falls in
the tank in half an hour
Π(d/2)2(126000) =5,04,000π
(d/2)2 = 4
d2 =16
d =4
Thus, the internal diameter of the pipe is 4 cm.
CBSE CLASS X MATH - 2011
Q.1) Which of the following cannot be the probability of an event?
A. 1.5
B.
3
5
C. 25%
D. 0.3
Q.2 The mid-point of segment AB is the point P (0, 4). If the Coordinates of B are (-2, 3) then the
coordinates of A are
A. (2, 5)
B. (-2.-5)
C. (2, 9)
D. (-2, 11)
Q3. The point P which divides the line segment joining the point A (2, -5) and B (5, 2) in the lies in the
ratio 2:3 quadrant.
A.
B.
C.
D.
1
Ii
Iii
Iv
Q4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away foot of
the tower, is 45°. The height of the tower (in metres) is
A. 15
B. 30
C. 30√3
D. 10√3
Q5. A sphere of diameter 18 cm is dropped into a cylindrical vessel of diameter 36 cm partly filled with
water. If the sphere is completely submerged, then the water level rises (in cm) by
A.
B.
C.
D.
3
4
5
6
Q 6 In figure 1, O is the centre of a circle, AB is a chord and AT is the tangent at A. If ∠AOB = 100°,
then ∠BAT is equal to
A.
B.
C.
D.
Q7.
100°
40°
50°
90°
In figure 2, PA and PB are tangents to the circle with centre O. If ∠APB = 60°, then ∠OAB is
A.
B.
C.
D.
30°
60°
90°
15°
Q8. The roots of the equation x2 + x – (p +1) = 0, where p is a constant are
A.
B.
C.
D.
p, p+1
–p, p+1
p, -(p+1)
–p, - (p +1)
Q9. In an AP, if a= 15, d = -3 and a n = 0, then the value of n is
A.
B.
C.
D.
5
6
19
4
Q10. The radii of two circles are 8 cm and 6 cm respectively. The diameter of the circle having area
equal to the sum of the areas of the two circles (in cm) is
A.
B.
C.
D.
10
14
20
28
Q11. A coin is tossed two times. Find the probability of getting both heads and both tails.
Q12. Find the value of m so that the quadratic equation mx (5x – 6) + 9 =0 has two equal roots.
Q13. Find the value (s) of x for which the distance between the points P(x, 4) and Q (9, 10) is 10 units.
Q14. Two cubes, each of side 4 cm are joined end to end. Find the surface area of the resulting cuboid.
Q15. Draw a line segment of length 6 cm. Using compasses and ruler, find a point P on it which divides it
in the ratio 3:4.
Q 16. Two concentric circles are of radii 7 cm and r cm respectively, where r > 7. A chord of the larger
Circle, of length 48 cm, touches the smaller circle. Find the value of r.
Q17. Find whether - 150 is a term of the AP 17, 12, 7, 2 …….?
Q18. In figure 3, APB and CQD are semi-circles of diameter 7 cm each, while ARC and BSD are semi22
circles of diameter 14 cm each. Find the perimeter of the shaded region. [Use π = ]
7
OR
Find the area of a quadrant of a circle, where the circumference of circle is 44 cm.
[Use π =
22
7
]
Q19. If two vertices of an equilateral triangle are (3, 0) and (6, 0), find the third vertex.
OR
Find the value of k, if the points P (5, 4), Q (7, k) and R (9, -2) are collinear.
Q20. From the top of a tower 100 m high, a man observes two cars on the opposite of the tower
with angles of depression 30° and 45° respectively. Find the distance between the cars.
Use [ √3 = 1.73]
Q21. Two dice are rolled once. Find the probability of getting such numbers on two dice, whose
product is a perfect square.
OR
A game consists of tossing a coin 3 times and noting its outcomes each time. Hanif wins if he gets three
heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Q22. The radii of the circles ends of a bucket of height 15 cm are 14 cm and r cm (r < 14 cm).
22
If the volume of bucket is 5390 cm3, then find the value of r. [Use π = ]
7
Q23. In fig. 4, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD
and DC into which BC is divided by the point of contact D are of lengths 4 cm and 3 cm respectively. If
area of ∆ABC = 21 cm2, then find the lengths of sides AB and AC.
Q24. Find the value of the middle term of the following AP:
-6,-2, 2 ………….58.
OR
Determine the AP whose fourth term is 18 and the differences of the ninth term from the fifteenth
term is 30.
Q25. Find the roots of the following quadratic equation:
2√3x2 – 5x + √3 = 0
Q26. Find the area of the major segment APB, in Fig 5, of a circle of radius 35 cm and ∠AOB = 90°
[Use π =
22
7
]
Q27. Draw a triangle PQR such that PQ = 5 cm, ∠P = 120° and PR = 6 cm. Construct another triangle
3
Whose sides are times the corresponding sides of ∆PQR?
4
Q28. Find the point of y-axis which is equidistant from the points (-5, -2) and (3, 2).
Q29. From a solid cylinder of height 20 cm and diameter 12 cm, a conical cavity of height 8 cm and
22
radius 6 cm is hollowed out. Find the total surface area of the remaining solid. [Use π = ]
7
Q30. The length and breadth of a rectangular piece of paper are 28 cm and 14 cm respectively. A semicircular portion is cut off from the breadth’s side and a semicircular portion is added on length’s side, as
shown in Fig. 6. Find the area of the shaded region. [Use π =
22
7
]
Q31. From the top of a 15 m high building, the angle of elevation of the top of a cable tower is 60° and
the angle of depression of its foot is 30° . Determine the height of the tower.
Q32. A motor boat whose speed is 20 km/h is still water, takes 1 hour more to go 48 km upstream than
to return downstream to the same spot. Find the speed of the steam.
OR
Find the roots of the equation
1
x+4
-
1
x−7
=
11
30
, x≠ -4, 7
Q33. Prove that the lengths of tangents drawn from an external point to a circle are equal.
Q34. If the sum of first 4 terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n
terms.
OR
Find the sum of the first 30 positive integers divisible by 6.
For more sample papers visit :
www.4ono.com
For more sample papers visit :
www.4ono.com
For more sample papers visit : www.4ono.com
1040105 - A1
Class - X
MATHEMATICS
Time : 3 to 3½ hours
¥çÏ·¤Ì× â×Ø : 3 âð 3½ ƒæ‡ÅUð
Maximum Marks : 80
¥çÏ·¤Ì× ¥´·¤ : 80
Total No. of Pages : 14
·é¤Ü ÂëcÆUæð´ ·¤è ⴁØæ : 14
General Instructions :
1.
All questions are compulsory.
2.
The question paper consists of 34 questions divided into four sections A, B, C and D.
Section - A comprises of 10 questions of 1 mark each. Section - B comprises of 8 questions of
2 marks each. Section - C comprises of 10 questions of 3 marks each and Section - D
comprises of 6 questions of 4 marks each.
3.
Question numbers 1 to 10 in Section - A are multiple choice questions where you are to select
one correct option out of the given four.
4.
There is no overall choice. However, internal choice has been provided in 1 question of two
marks, 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.
5.
Use of calculator is not permitted.
6.
An additional 15 minutes time has been allotted to read this question paper only.
âæ×æ‹Ø çÙÎðüàæ Ñ
1.
âÖè ÂýàÙ ¥çÙßæØü ãñ´Ð
2.
§â ÂýàÙ˜æ ×ð´ 34 ÂýàÙ ãñ´, Áæð ¿æÚU ¹‡ÇUæð´ ×ð´ ¥, Õ, â ß Î ×ð´ çßÖæçÁÌ ãñÐ ¹‡ÇU - ¥ ×ð´ 10 ÂýàÙ ãñ´ ¥æñÚU ÂýˆØð·¤
ÂýàÙ 1 ¥´·¤ ·¤æ ãñÐ ¹‡ÇU - Õ ×ð´ 8 ÂýàÙ ãñ´ ¥æñÚU ÂýˆØð·¤ ÂýàÙ 2 ¥´·¤æð´ ·ð¤ ãñ´Ð ¹‡ÇU - â ×ð´ 10 ÂýàÙ ãñ´ ¥æñÚU ÂýˆØð·¤
ÂýàÙ 3 ¥´·¤æð´ ·¤æ ãñÐ ¹‡ÇU - Î ×ð´ 6 ÂýàÙ ãñ´ ¥æñÚU ÂýˆØð·¤ ÂýàÙ 4 ¥´·¤æð´ ·¤æ ãñÐ
3.
Âýà٠ⴁØæ 1 âð 10 Õãéçß·¤ËÂèØ ÂýàÙ ãñ´Ð çΰ »° ¿æÚU çß·¤ËÂæð´ ×ð´ âð °·¤ âãè çß·¤Ë ¿éÙð´Ð
4.
§â×ð´ ·¤æð§ü Öè âßæðüÂçÚU çß·¤Ë Ùãè´ ãñ, Üðç·¤Ù ¥æ´ÌçÚU·¤ çß·¤Ë 1 ÂýàÙ 2 ¥´·¤æð´ ×ð´, 3 ÂýàÙ 3 ¥´·¤æð´ ×ð´ ¥æñÚU 2 ÂýàÙ
4 ¥´·¤æð´ ×ð´ çΰ »° ãñ´Ð ¥æ çΰ »° çß·¤ËÂæð´ ×ð´ âð °·¤ çß·¤Ë ·¤æ ¿ØÙ ·¤Úð´UÐ
5.
·ñ¤Ü·é¤ÜðÅUÚU ·¤æ ÂýØæð» ßçÁüÌ ãñÐ
6.
§â ÂýàÙ-Â˜æ ·¤æð ÂɸÙð ·ð¤ çÜ° 15 ç×ÙÅU ·¤æ â×Ø çÎØæ »Øæ ãñÐ §â ¥ßçÏ ·ð¤ ÎæñÚUæÙ ÀUæ˜æ ·ð¤ßÜ ÂýàÙ-Â˜æ ·¤æð Âɸð´»ð
¥æñÚU ß𠩞æÚU-ÂéçSÌ·¤æ ÂÚU ·¤æð§ü ©žæÚU Ùãè´ çܹð´»ðÐ
1
P.T.O.
SECTION - ‘A’
1.
Question numbers 1 to 10 carry one mark each :
Which of the following numbers has terminating decimal expansion ?
(A)
37
45
(B)
21
(C)
3 6
2 5
17
49
(D)
89
2 2
2 3
2.
The value of p for which the polynomial x314x22px18 is exactly divisible by (x22) is
(A) 0
(B) 3
(C) 5
(D) 16
3.
DABC and DPQR are similar triangles such that ‘ A=328 and ‘ R=658 then ‘ B is
(A) 838
(B) 328
(C) 658
(D) 978
4.
In fig. 1, the value of the median of the data using the graph of less than ogive and more than
ogive is
(A)
5.
(B)
40
(C)
80
(D)
15
(C)
1
2
(D)
2
(C)
0
(D)
1
If u5458, the value of cosec2u is
(A)
6.
5
1
2
(B)
1
sin (6081u)2cos (3082u) is equal to
(A) 2 cosu
(B) 2 sinu
1040105 - A1
2
7.
The [HCF3LCM] for the numbers 50 and 20 is
(A)
8.
100
(C)
1000
(D)
50
k53
(B)
k52
(C)
k54
(D)
k522
22
(D)
0
cosA
(D)
sinA
If sinA1sin2A51, then the value of cos2A1cos4A is
(A)
10.
(B)
The value of k for which the pair of linear equations 4x16y2150 and 2x1ky2750
represents parallel lines is
(A)
9.
10
2
(B)
1
(C)
The value of [(secA1tanA) (12sinA)] is equal to
(A)
tan 2 A
(B)
sin2A
(C)
SECTION - ‘B’
Question numbers 11 to 18 carry 2marks each.
11.
Find a quadratic polynomial with zeroes 31 2 and 32 2 .
12.
In figure 2, ABCD is a parallelogram. Find the values of x and y.
13.
If sec4A5cosec (A2208) where 4A is an acute angle, find the value of A.
OR
If 5 tanu=4, find the value of
1040105 - A1
5 sin M 3 cos M
.
5 sin M 2 cos M
3
P.T.O.
AQ
AR
.
QD
RB
14.
In figure 3, PQ??CD and PR??CB. Prove that
15.
In figure 4, two triangles ABC and DBC are on the same base BC in which ‘ A5 ‘ D5908.
If CA and BD meet each other at E, show that AE3CE5BE3DE.
16.
Check whether 6n can end with the digit 0 for any natural number n ?
17.
Find the mean of the following frequency distribution :
Class
Frequency
18.
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
8
12
10
11
9
Find the mode of the following data :
Class
Frequency
1040105 - A1
0 - 20
20 - 40
40 - 60
60 - 80
15
6
18
10
4
SECTION - ‘C’
Question number 19 to 28 carry 3 marks each.
19.
Prove that
7 is an irrational number.
OR
Prove that 3+ 5 is an irrational number.
20.
Use Euclid’s division algorithm to find the HCF of 10224 and 9648.
21.
If a and b are zeroes of the quadratic polynomial x226x1a; find the value of ‘a’ if 3a+2b=20.
22.
Solve for x and y.
y
8
3
3
3y
x
5
2
4
2
4x OR
The sum of the numerator and the denominator of a fraction is 8. If 3 is added to both the
numerator and the denominator, the fraction becoms
3
. Find the fraction.
4
tan M cot M
tan 2 M cot 2 M .
sin M cos M
23.
Prove that
24.
In figure 5, DABC is right angled at B, BC57cm and AC2AB51cm. Find the value of
cosA2sinA.
1040105 - A1
5
P.T.O.
25.
In figure 6, P and Q are the midpoints of the sides CA and CB respectively of DABC right
angled at C. Prove that 4 (AQ21BP2)55AB2.
26.
The diagonals of a trapezium ABCD with AB?? DC intersect each other at point O. If AB52CD,
find the ratio of the areas of triangles AOB and COD.
27.
The mean of the following frequency distribution is 50. Find the value of p.
Classes
Frequency
28.
0 - 20
20 - 40
40 - 60
60 - 80
80 - 100
17
28
32
p
19
Compute the median for the following cumulative frequency distribution :
Weight in
(kg)
Number of
students
less than less than less than less than less than less than less than less than
46
48
50
52
38
40
42
44
0
3
5
9
14
28
32
35
OR
Find the missing frequencies in the following frequency distribution table, if N=100 and median
is 32.
Marks obtained
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
50 - 60
Total
No. of Students
10
?
25
30
?
10
100
1040105 - A1
6
SECTION - ‘D’
29.
Question number 29 to 34 carry 4 marks each.
Divide 30x 4 111x 3 282x 2 212x148 by (3x 2 +2x24) and verify the result by division
algorithm.
30.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct
points, prove that the other two sides are divided in the same ratio.
OR
Prove that in a triangle if the square of one side is equal to the sum of the squares of the other
two side then the angle opposite to the first side is a right angle.
31.
Without using trigonometric tables, evaluate the following :
sec 37 12cot 158 cot258 cot458 cot758 cot658- 3 (sin2188+sin2728)
cosec 53
OR
Prove that:
tan M
cot M
511secu cosecu.
1 cot M
1 tan M
32.
If 2cosu2sinu5x and cosu23sinu5y. Prove that 2x21y222xy55.
33.
Check graphically whether the pair of linear equations 4x2y2850 and 2x23y1650 is
consistent. Also, find the vertices of the triangle formed by these lines with the x-axis.
34.
The following table shows the ages of 100 persons of a locality.
Age (yrs)
Number of persons
0 - 10
5
10 - 20
15
20 - 30
20
30 - 40
23
40 - 50
17
50 - 60
11
60 - 70
9
-oOo-
1040105 - A1
7
P.T.O.
¹‡ÇU-¥
1.
Âýà٠ⴁØæ 1 âð 10 ÂýˆØð·¤ ÂýàÙ 1 ¥´·¤ ·¤æ ãñÐ
çِÙçÜç¹Ì ×ð´ âð ·¤æñÙ âè ⴁØæ ·¤æ Îàæ×Üß ÂýâæÚU âæ´Ì ãæð»æ?
(A)
2.
p ·¤æ
(A)
3.
21
(C)
3 6
2 5
17
49
(D)
89
2 2
2 3
0
(B)
838
3
(C)
Âý·¤æÚU ·ð¤ â×M¤Â ç˜æÖéÁ ãñ´ ç·¤
(B)
328
5
(D)
‘ A=328 ÌÍæ ‘ R=658 ãñ,
(C)
658
16
Ìæð
‘ B ·¤æ
(D)
978
×æÙ ãñ Ñ
¥æ·ë¤çÌ 1 ×ð´, çÎ¹æ° »Øð ÒÒâð ·¤× Âý·¤æÚUÓÓ ·ð¤ ÌæðÚU‡æ ÌÍæ Òâð ¥çÏ·¤ Âý·¤æÚU ·ð¤ ÌæðÚU‡æÓ ·ð¤ ¥æÜð¹ âð ¥æ¡·¤Ç¸æð´ ·¤æ
×æŠØ·¤ ãñ Ñ
(A)
5.
(B)
ßã ×æÙ çÁâ·ð¤ çÜ° ÕãéÂÎ x314x22px18, Âê‡æüÌØæ çßÖæçÁÌ ãæð»æ (x22) âð, ãñ Ñ
DABC ÌÍæ DPQR §â
(A)
4.
37
45
5
(B)
40
(C)
80
(D)
15
(C)
1
2
(D)
2
ØçÎ u5458 ãñ, Ìæð cosec2u ·¤æ ×æÙ ãñ Ñ
(A)
1
2
1040105 - A1
(B)
1
8
6.
sin (6081u)2cos (3082u)
(A)
7.
ⴁØæ¥æð´
(A)
8.
k ·¤æ
(A)
9.
(B)
2sinu
50 ÌÍæ 60 ·¤æ [HCF3LCM] ãñ
10
(B)
(B)
0
(D)
1
(C)
1000
(D)
50
Ñ
100
ßã ×æÙ çÁâ·ð¤ çÜ° ÚñUç¹·¤ â×è·¤ÚU‡æ Øé‚×
k53
(C)
4x16y2150 ÌÍæ 2x1ky2750 â×æ‹ÌÚU
k52
(C)
ÚðU¹æ°¡ ÎàææüÌð ãñ;
k54
(D)
k522
(C)
22
(D)
0
(C)
cosA
(D)
sinA
ÌÍæ 32
2
ØçÎ sinA1sin2A51 ãñ, Ìæð cos2A1cos4A ·¤æ ×æÙ ãñ Ñ
(A)
10.
2cosu
ÕÚUæÕÚU ãñ Ñ
2
(B)
(secA1tanA) (12sinA) ·¤æ
(A)
tan 2 A
(B)
1
×æÙ ÕÚUæÕÚU ãñ Ñ
sin2A
¹‡ÇU-Õ
Âýà٠ⴁØæ 11 âð 18 Ì·¤ ÂýˆØð·¤ ÂýàÙ 2 ¥´·¤æð´ ·¤æ ãñÐ
11.
ßã çmƒææÌ ÕãéÂÎ ™ææÌ ·¤èçÁ° çÁâ·ð¤ àæê‹Ø·¤ 31
12.
¥æ·ë¤çÌ 2 ×ð´ ABCD °·¤ â×æ´ÌÚU ¿ÌéÖéüÁ ãñÐ
1040105 - A1
2
x ÌÍæ y ·ð¤
9
ãñÐ
×æÙ ™ææÌ ·¤èçÁ°Ð
P.T.O.
13.
ØçÎ sec4A5cosec (A2208), Áãæ¡ 4A °·¤ ‹ØêÙ ·¤æð‡æ ãñ, Ìæð
¥Íßæ
ØçÎ
5 tanu54 Ìæð
5 sin M 3 cos M
5 sin M 2 cos M
A
·¤æ ×æÙ ™ææÌ ·¤èçÁ°Ð
·¤æ ×æÙ ™ææÌ ·¤èçÁ°Ð
AQ
AR
.
QD
RB
14.
¥æ·ë¤çÌ 3 ×ð´, PQ??CD ÌÍæ PR??CB ãñÐ çâh ·¤èçÁ° ç·¤
15.
¥æ·ë¤çÌ 4 ×ð´, Îæð ç˜æÖéÁð´ ABC ÌÍæ DBC °·¤ ãè ¥æÏæÚU BC ÂÚU ãñ´ çÁÙ×𴠑 A5 ‘ D5908 ãñÐ ØçÎ CA ÌÍæ BD
çՋÎé D ÂÚU ÂÚUSÂÚU ç×ÜÌð ãñ´, Ìæð Îàææü§° ç·¤ AE3CE5BE3DE.
16.
Áæ¡¿ ·¤èçÁ° ç·¤ ç·¤âè Âýæ·ë¤Ì ⴁØæ n ·ð¤ çÜ° €Øæ 6n ·¤æ §·¤æ§ü ·¤æ ¥´·¤ àæê‹Ø ãæð â·¤Ìæ ãñ?
17.
çِÙçÜç¹Ì ÕæÚ´UÕæÚUÌæ Õ´ÅUÙ ·¤æ ×æŠØ ™ææÌ ·¤èçÁ°Ð
Æ¢ã
¼ÍÁ›<¼ÍÁ<³Í
1040105 - A1
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
8
12
10
11
9
10
18.
çِÙçÜç¹Ì ¥æ¡·¤Ç¸æð´ ·¤æ ÕãéÜ·¤ ™ææÌ ·¤èçÁ°Ð
Æ¢ã
0 - 20
20 - 40
40 - 60
60 - 80
15
6
18
10
¼ÍÁ›<¼ÍÁ<³Í
¹‡ÇU-â
Âýà٠ⴁØæ 19 âð 28 Ì·¤ ÂýˆØð·¤ ÂýàÙ 3 ¥´·¤æð´ ·¤æ ãñÐ
19.
çâh ·¤èçÁ° ç·¤
7
°·¤ ¥ÂçÚU×ðØ â´Øæ ãñÐ
¥Íßæ
çâh ·¤èçÁ° ç·¤ 3+
5
°·¤ ¥ÂçÚU×ðØ â´Øæ ãñÐ
20.
Øêç€ÜÇU ·ð¤ çßÖæÁÙ °Ü»æðçÚUÍ× ·¤æ ÂýØæð» ·¤ÚU·ð
21.
ØçÎ çmƒææÌ ÕãéÂÎ
22.
x ÌÍæ y ·ð¤
10224 ÌÍæ 9648¤·¤æ HCF ™ææÌ
x226x1a ·ð¤ a ÌÍæ b àæê‹Ø·¤
·¤èçÁ°Ð
ãñ, Ìæð ‘a’ ·¤æ ×æÙ ™ææÌ ·¤èçÁ° ØçÎ 3a+2b=20 ãñÐ
çÜØð ãÜ ·¤èçÁ° Ñ
y
8
3
3
3y
5
x
2
4
2
4x ¥Íßæ
°·¤ çÖóæ ¥´àæ ÌÍæ ãÚU ·¤æ Øæð» 8 ãñÐ ØçÎ ¥´àæ ÌÍæ ãÚU ÎæðÙæð´ ×ð´ ÌèÙ-ÌèÙ ÁæðǸ çÎØæ Áæ°, Ìæð çÖóæ
™ææÌ ·¤èçÁ°Ð
23.
çâh ·¤èçÁ° ç·¤
1040105 - A1
3
4
ãæð ÁæÌè ãñÐ çÖóæ
tan M cot M
tan 2 M cot 2 M .
sin M cos M
11
P.T.O.
24.
¥æ·ë¤çÌ 5 ×ð´, D ABC ×ð´ B ÂÚU â×·¤æð‡æ ãñ, BC57 âð.×è. ÌÍæ AC2AB51 âð.×è. ãñÐ
™ææÌ ·¤èçÁ°Ð
25.
¥æ·ë¤çÌ 6 ×ð´ DABC °·¤ â×·¤æð‡æ ç˜æÖéÁ ãñ çÁâ×ð´ C ÂÚU â×·¤æð‡æ ãñ P ÌÍæ Q ·ý¤×àæÑ ÖéÁæ¥æð´ CA ÌÍæ CB ·ð¤ ׊Ø
çՋÎé ãñ´Ð çâh ·¤èçÁ° ç·¤ Ñ 4 (AQ21BP2)55AB2.
26.
°·¤ â×ÜÕ ABCD ×ð´, çÁâ×ð´ AB??DC ãñ, ·ð¤ çß·¤‡æü °·¤ ÎêâÚðU ·¤æð çՋÎé
ØçÎ AB52 CD ãñ, Ìæð DAOB ÌÍæ DCOD ·ð¤ ÿæð˜æȤÜæð´ ×ð´ ¥ÙéÂæÌ ™ææÌ ·¤èçÁ°Ð
27.
çِÙçÜç¹Ì ÕæÚ´UÕæÚUÌæ Õ´ÅUÙ ·¤æ ×æŠØ 50 ãñ Ìæð p ·¤æ ×æÙ ™ææÌ ·¤èçÁ°Ð
Æ¢ã
¼ÍÁ›<¼ÍÁ<³Í
1040105 - A1
0 - 20
20 - 40
40 - 60
60 - 80
80 - 100
17
28
32
p
19
12
O
cosA2sinA
·¤æ ×æÙ
ÂÚU Âýç̑ÀðUÎ ·¤ÚUÌð ãñ´Ð
28.
çِÙçÜç¹Ì â´¿Øè ÕæÚ´UÕæÚUÌæ Õ´ÅUÙ ·¤æ ×æŠØ·¤ ™ææÌ ·¤èçÁ°Ð
½ÍÁ<Ξ‹¢ä;כ 38 Éמ‹¾
ÎÆSÍδã¿Íכž‹Ï
0
ɛh¿Í
40
Éמ‹¾
42
Éמ‹¾
3
44
5
Éמ‹¾
46
9
Éמ‹¾
48
14
Éמ‹¾
50
28
Éמ‹¾
52
32
Éמ‹¾
35
¥Íßæ
ØçÎ n=100 ÌÍæ ×æŠØ·¤ 32 ãñ Ìæð çِÙçÜç¹Ì ÕæÚ´UÕæÚUÌæ Õ´ÅUÙ ×ð´ ÜéŒÌ ÕæÚ´UÕæÚUÌæ°¡ ™ææÌ ·¤èçÁ°Ð
©äÍs³Í›ž‹
ÎÆSÍδã¿Íכž‹Ïɛh¿Í
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
50 - 60
žÐ‹Ã
10
?
25
30
?
10
100
¹‡ÇU-Î
Âýà٠ⴁØæ 29 âð 34 Ì·¤ ÂýˆØð·¤ ÂýàÙ 4 ¥´·¤ ·¤æ ãñÐ
29.
ÕãéÂÎ 30x4111x3282x2212x148 ·¤æð (3x212x24) âð Öæ» ÎèçÁ° ÌÍæ ÂçÚU‡ææ× ·¤æ âˆØæÂÙ çßÖæÁÙ
°Ë»æðçÚUÍ× âð ·¤èçÁ°Ð
30.
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1040105 - A1
14
CBSE CLASS X MATH -SOLUTION – 2011
Q1 The probability of an event is always greater than or equal to zero and less
than or equal to one.
Here,
3
5
= 0.6
25% =
25
100
= 0.25
Therefore, 0.6, 0.25 and 0.3 are greater than or equal to 0 and less than or
equal to 1.
But 1.5 is greater than 1.
Thus, 1.5 cannot be the probability of an event.
The correct answer is A.
Q2. Let the coordinates of point A be ( X, Y).
It is given that P (0, 4) is the mid-point of AB.
x−2 y+3
∴ (0, 4) = (
⇒
x−2
2
2
,
= 0 and
2
)
y+3
2
=4
⇒ x - 2 =0 and y+3 = 8
⇒x =2 and y =5
Thus, the coordinates of point A are (2, 5).
The correct answer is A.
Q3. The point P divides the line segment joining the point A (2, -5) and (5, 2) in
the ratio 2: 3.
∴ 𝑃𝑃 = (
=(
2×5+3×2
2+3
10+6 4−15
5
,
16 −11
=( ,
5
5
5
)
,
2×2+3×(−5)
2+3
)
)
16 −11
The point P ( ,
5
5
) lies in quadrant IV.
The correct answer is D.
Q4.
Let AB be the tower and P be the point on the ground.
It is given that BP = 30 m, ∠P = 45°
Now,
⇒
AB
BP
AB
30 m
= tan 45°
=1
⇒AB = 30m
Thus, the height of the tower is 30m.
The correct answer is B.
Q5. Radius of the sphere =
18
Radius of the cylinder =
2
cm = 9 cm
36
2
cm = 18 cm
Let the water level in the cylinder rises by h cm.
After the sphere is completely submerged.
Volume of the sphere = Volume of liquid raised in the cylinder
4
⇒ π (9cm)3 = π (18cm)2 ×h
3
⇒h=
4×9×9×9
3×18×18
cm
⇒ h = 3cm
Thus, the water level in the cylinder rises by 3 cm.
The correct answer is A.
Q6. It is given that ∠AOB = 100°
∆AOB is isosceles because
OA = OB = radius
∴ ∠OAB = ∠OBA
∠AOB +∠OAB +∠OBA = 180° [Angle sum property of triangle]
⇒ 100° + ∠OAB + ∠OAB = 180°
⇒2 ∠OAB = 80°
⇒∠OAB = 40°
Now, ∠OAT = 90° [AT is tangent and OA is radius]
Thus, ∠BAT = ∠OAT - ∠OAB = 90° - 40° = 50°
The correct answer is C.
Q 7. Since PA and PB are tangents to the circle from an external point O.
Therefore, PA = PB
∴∆PAB is an isosceles triangle where ∠PAB = ∠PBA
∠P + ∠PAB +∠PBA = 180° [angle sum property of triangle]
⇒ 60° +2∠PAB = 180°
⇒ 2 ∠PAB = 180° - 60° = 120°
⇒∠PAB =
120
2
= 60°
It is known that the radius is perpendicular to the tangent at the point of
contact.
∴ ∠OAP = 90°
⇒ ∠PAB +∠OAB = 90°
⇒ ∠OAB = 90° - 60° = 30°
The correct answer is A.
Q 8. The roots of the equation is x2 +x –p (p +1) = 0, where p is a constant.
Its solution can be solved by using quadratic formula x =
This can be done as
−b±√b 2 −4ac
2a
.
On comparing the given equation with ax2 +bx +c = 0
a = 1, b = 1, c = -p(p+1)
∴x=
−1±�12 −4×1×{−p(p+1)}
2×1
−1±�1−4(−p 2 −p)
X=
2
=
−1±�(2p+1)2
=
−1±(2p+1)
2
2
=
−1+(2p+1)
=
−1+(2p+1) 2p
2
2
or
=
−1−(2p +1)
2
=p
−1−(2p+1) −2−2p
=
2
=
2
2
= -1-p = - (p+1)
Therefore, the roots are given by x = p, - (p +1)
The correct answer is C.
Q9. We have, a= 15 and d = -3
Given a n = 0
⇒ a + (n-1) d = 0
⇒ 15 + (n-1) (-3) = 0
⇒ 15 - 3n +3 = 0
⇒-3n = -18
⇒n=6
The correct answer is B.
Q10. Let the radius of the required circle be r cm.
Area of required circle = area of circle of radius 8 cm + area of circle of radius 6 cm
⇒ πr2 = π (8 cm) 2 + π (6cm) 2
⇒ r2 = 64 cm2 +36 cm2
⇒ r2 = 100 cm2
⇒r = 10 cm
Thus, the diameter of the required circle is 2× 10 cm = 20 cm.
The correct answer is C.
Q11. Let E be the event of getting both heads or both tails.
The sample space for the given experiment is {(H, H), (H, T), (T, H), (T, T)}
Total number of outcomes = 4
Favorable outcomes = {(H, H), (T,T)}
Favorable number of outcomes = 2
Required probability, P (E) =
Favorable number of outcomes
=
Total number of outcomes
2
4
=
1
2
Q12. The given quadratic equation is mx (5x – 6) + 9 = 0
∴ 5mx2 -6mx + 9 = 0 …..(1)
For equation (1) to have equal roots, the discriminant of the equation D
should be 0.
⇒ (-6m)2 – 4 × 5 m × 9 = 0
⇒ 36m2 – 180m = 0
⇒ 36m (m -5) = 0
⇒ m = 0 or m – 5 = 0
⇒ m = 5 (If m = 0, then equation (1) will not be a quadratic equation)
Thus, the value of mi is 5.
Q 13. It is given that the distance between the points P (x, 4) and Q (9, 10) is 10
units.
Let x 1 = x, y 1 = 4, x 2 = 9, y 2 = 10
Applying distance formula, if is obtained.
d = �(x2 − x1 )2 + (y2 − y1 )2
10 =�(9 − x)2 + (10 − 4)2
10 = √81 + x 2 − 18x + 36
10= √x 2 − 18x + 117
On squaring both sides, it is obtained.
100 = x2 – 18x + 117
⇒ x2 – 18x + 17 = 0
⇒ x2 - 17x – x + 17 = 0
⇒ x (x-17) – 1 (x - 17 ) = 0
⇒ (x -1) (x – 17) = 0
⇒ x = 1, 17
Thus, the values of x are 1 and 17.
Q14. If two cubes of sides 4 cm are joined end to end, then the length (l), breadth
(b) and height (h) of the resulting cuboid are 8 cm, 4 cm, and 4 cm, respectively.
∴ Surface area of the resulting cuboid = 2 (lb +bh + lh )
= 2 (8 cm × 4 cm + 4 × 4 cm = 8 cm ×4 cm)
= 2 × (32 + 16 +32) cm2
= 160 cm2
Thus, the surface area of the resulting cuboid is 160 cm2.
Q15. A point P can be marked on a line segment of length 6 cm which divides the
line segment in the ratio of 3:4 as follows.
(1) Draw line segment AB of length 6 cm and draw a ray AX making an acute
angle with line segment AB.
(2) Locate 7 (3+4 ) points, A 1 . A 2 , A 3 , A 4 ………….A 7 , on AX such that AA 1 = A 1 A 2
= A 2 A 3 and so on.
(3) Join BA 7 .
(4) Through the point A 3 , draw a line parallel to BA 7 (by making an angle equal
to ∠AA 7 B at A 3 ),
intersecting AB at point P.
P is the point that divides line segment AB of length 6 cm in the ratio of 3:
4.
Q16. Let O be the centre of the two concentric circles. Let PQ be the chord of
larger circle touching the smaller circle at M. This can be represented
diagrammatically as:
We have PQ = 48 cm.
Radius of the smaller circle, OM = 7 cm
Let the radius of the larger circle be r, i.e. OP = r
Since PQ is a tangent to the inner circle, OM ⊥ PQ
Thus, OM bisects PQ.
⇒ PM = MQ =
48
2
cm = 24 cm
Now applying Pythagoras Theorem in ∆OPM
OP2 = OM2 + PM2
⇒ OP2 = (7 cm) 2 + (24cm) 2 = (49+576) cm2 = 625 cm2 = (25 cm)2
⇒ OP = 25 cm
∴ Radius of the larger circle is 25 cm.
Thus, the value of r is 25 cm.
Q17. The given A. P. is 17, 12, 7, 2, …….
First term, a = 17
Common difference, d = 12 - 17 = -5
If -150 is a term of the given A.P., then for a natural number n, a n = -150
⇒ a + (n-1) d = -150
⇒17+ (n-1) (-5) = -150
⇒(-5) (n -1) = -150 -17 = -167
⇒ n -1 =
⇒n =
167
167
5
5
+1 =
172
5
= 34.4
Now, 34.4 is not a natural number.
Thus, -150 is not a term of the A.P, 17, 12, 7, 2 ……….
Q18. Perimeter of the shaded region = Length of APB + Length of ARC + Length
CQD + Length of DSB
1
7
1
× 2π (7cm) =
Now, perimeter of APB = × 2π ( ) cm =
Perimeter of ARC=
2
2
2
1
7
22
7
22
7
Perimeter of CQD = ×2π ( cm) =
2
1
2
Perimeter of DSB = ×2π (7cm) =
2
7
2
cm = 11 cm
× 7cm = 22 cm
22 7
22
7
×
× cm = 11 cm
7 2
×7cm = 22 cm
Thus, perimeter of the shaded region = 11 cm + 22 cm +11 cm = 66 cm
OR
Let the radius of the circle be r.
It is given that perimeter of the circle is 44 cm.
∴ 2πr = 44 cm
⇒ 2×
22
7
× r = 44 cm
⇒ r = 7 cm
Area of a quadrant of a circle
=
1
4
1 22
× πr2 = ×
=
1
4
×
4
22
7
7
× (7cm) 2
× 49 cm2 = 38.5 cm2
Thus, the area of a quadrant of the given circle is 38.5 cm2.
Q19. Let the two given vertices be A (3, 0) and B (6, 0).
Let the coordinates of the third vertex be C (x, y).
It is given that the triangle ABC is equilateral.
Therefore, AB = BC = CA (Sides of an equilateral triangle)
⇒ �(6 − 3)2 + (0 − 0)2 =�(x − 6)2 + (y − 0)2 =�(x − 3)2 + (y − 0)2
⇒ 9= (x - 6)2 + y2= (x - 3)2 + y2
∴ (x- 6)2 + y2 = (x - 3)2 + y2
⇒ -12x + 36 = -6x + 9
⇒-6x = -27
⇒x=
9
2
Now, y2 +(x -6)2 = 9
9
9
⇒ y2 + ( − 6) 2 = 9 (∴x = )
2
⇒ y2 = 9 ⇒ y2 =
27
4
2
9
4
27
⇒y=±�
4
3√3
=±
2
9 3√3
Thus, the coordinates’ of the third vertex are ( ,
2
OR
2
9,
) or ( −
2
3√3
2
)
Let Q (7, k) divide the line segment joining P (5, 4) and (9,-2) in the ratio 𝝀𝝀: 1
∴ Coordinates of the Point Q = (
∴
9λ+5
λ+1
= 7 and k =
−2λ+4
λ+1
9λ+5 −2λ+4
λ+1
,
λ+1
)
⇒9𝝀𝝀 +5 = 7𝝀𝝀 +7
⇒2𝝀𝝀 =2
⇒𝝀𝝀 =1
Now, k =
−2λ+4
λ+1
⇒k=
−2×1+4
⇒k =
−2+4
1+1
⇒ k =1
2
Thus, the value of k is 1.
Q20.
The given information can be diagrammatically represented as,
Here, AB is the tower of height 100 m. The Points C and D are the position
of the two cars.
In right ∆ACB,
tan 45° =
⇒1=
AB
BC
100m
BC
⇒ BC = 100 m
In right ∆ABD
tan 30° =
⇒
1
√3
=
AB
BD
100m
BD
⇒BD = 100√3 m
Distance between the two cars = CD
= BC +CD
= 100m +100√3m
= 100m +100×1.73m
= 100m + 173 m
= 273 m
Thus, the distance between two cars is 273 m.
Q21 The sample space of the given experiment is:
∴ n (S)= 36
Let E be the event that the product of numbers obtained on the upper face is
a perfect square
∴ E = {(1,1), (1,4), (2,2), (3,3), (4,1), (4,4), (5,5), (6,6)}
∴ n (E) = 8
P(E) =
𝑛𝑛(𝐸𝐸)
OR
𝑛𝑛(𝑆𝑆)
=
8
36
=
2
9
The set of possible outcomes of the given experiment are:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Let E be the event of getting three heads or three tails.
∴E = {HHH, TTT}
∴ Probability of winning = P (E)
=
𝑛𝑛(𝐸𝐸)
𝑛𝑛(𝑆𝑆)
2
= =
8
1
4
∴ Probability of losing = P (E’)
= 1- P (E)
1
= 1- =
4
3
4
3
Therefore, the probability that Hanif will lose the game is .
Q22.
4
Height of the bucket (which is in the shape of a frustum of a cone), h = 15
cm
Radius of one end of bucket, R = 14 cm
Radius of the other end of the bucket is r.
It is given that the volume of the bucket is 5390 cm3.
1
⇒ π (R2 + r2 + Rr)h = 5390
⇒
3
1
3
×
22
7
× [142 + r2 + 14r} × 5390
⇒196 +r2 + 14r =
5390×7
22×5
= 343
⇒r2 +14r +196 - 343 =0
⇒ r2 +14r -147 =0
⇒ r2 +21r-7r -147 =0
⇒ r (r+21)-7(r+21)=0
⇒ (r+21) (r -7) =0
⇒ r -7 = 0 or r+21 = 0
⇒ r = 7 or r = -21
Since the radius cannot be a negative number, r= 7 cm
Thus, the value of r is 7 cm.
Q23. Join O with A, O with B, O with C, O with E, and O with F.
We have, OD =OF =OE = 2cm (radii)
BD = 4 cm and DC = 3 cm
∴ BC = BD + DC = 4 cm + 3 cm = 7 cm
Now, BF = BD = 4 cm [Tangents from the same point]
CE = DC = 3 cm
Let AF = AE = x cm
Then, AB = AF + BF = (4 + x) cm and AC = AE + CE = (3 + x) cm
It is given that
ar (∆OBC) +ar (∆OAB) +ar(∆OAC) = ar (∆ABC)
1
1
⇒ × BC × OD + × AB × OF +
2
1
1
2
1
1
2
× AC × OE = 21
⇒ × 7 × 2 + × (4 + x) × 2 + × (3 + x) × 2 = 21
2
1
2
2
⇒ × 2(7 + 4 + x + 3 + x ) = 21
2
⇒14+2x = 21
⇒2x = 7
⇒x = 3.5
Thus, AB = (4 + 3.5) cm = 7.5 cm and AC = (3 + 3.5) cm = 6.5 cm.
Q24. The given AP is -6, -2, 2, ….., 58
Here, first term a = - 6 and common difference d = -2 –(-6) = -2 + 6 = 4
Last term, l = 58
⇒ a+ (n -1)d = 58
⇒ -6 + (n -1)× 4 = 58
⇒(n-1) × 4 = 64
⇒ (n-1) =16
⇒ n = 17
Middle term of the A.P. (
𝑛𝑛 + 1
2
) th term = (
17+ 1 th
) term =
2
9th term
a 9 = a+ (9-1) d = -6 + 8 × 4 = -6 +32 = 26
Thus, the middle term of the given A.P. is 26.
OR
Let the first term of the given AP be ‘a’ and the common difference be ‘d’.
We have a 4 = 18
⇒ a+(4-1)d = 18
⇒ a+3d = 18
…………..(i)
Also, it is given that
a 15 - a 9 = 30
⇒ a + (15 - 1) d – {a + (9 - 1) d} = 30
⇒ a +14d – (a + 8d) = 30
⇒ 6d = 30
⇒d=5
Putting the value of d in (i):
a+ 3 × 5 = 18
⇒ a + 15 = 18
⇒ a = 18 -15
⇒ a =3
Therefore, the first term and the common difference of the AP are 3 and 5
respectively
Thus, the A.P. is 3, 3+5, 3+ (2×5), 3+ (3×5) ………
That is 3, 8, 13, 18…
Q25. The given quadratic equation is 2√3x2 - 5x + √3 = 0
Comparing with the standard from ax2 +bx +c = 0
The value of a, b and c are
a = 2√3, b = -5, c = √3.
√D = √b 2 − 4ac = �25 − 4 × 2√3 × √3
= √25 − 24 = √1
=1
∴ x=
=
=
=
=
−𝑏𝑏 ±√𝐷𝐷
2𝑎𝑎
5 ±1
2×2√3
6
or
3
or
4√3
2√3
√3
2
or
4
4√3
1
√3
√3
3
Therefore, the roots of the given quadratic equation are
Q26. Radius of the given circle = 35 cm
√3
2
and
√3
.
3
Area of the minor segment = Area of sector OAB – area of ∆AOB
Area of section OAB =
𝜃𝜃
360°
=
×πr2
90°
360°
1 22
= ×
4
7
22
× × (35cm) 2
7
×1225 cm2
= 962.5 cm2
1
Area of ∆AOB = ×OA×OB
2
1
= ×35cm×35cm2
2
= 612.5 cm2
∴ Area of the minor segment = 962.5 cm2 – 612.5 cm2 = 350 cm2
Area of the major segment = Area of the circle – area of the minor segment
Area of the circle = πr2 =
22
7
× (35 cm)2 = 3850 cm2
Thus, the area of the major segment APB = 3850 cm2 -350 cm2 = 3500 cm2
3
4
Q27. A ∆PQ’R’ whose sides are of the corresponding sides of ∆PQR can be
drawn as follows.
Step1. Draw a ∆PQR with side PQ = 5 cm, PR = 6 cm and ∠P = 120°
Step2. Draw a ray PX making an acute angle with PR on the opposite side
of vertex Q.
Step3. Locate 4 points (as 4 is greater in 3 and 4), P 1 , P 2 , P 3 , P 4 , on line
segment PX.
Step4. Join P 4 R and draw a line through P 3 , parallel to P 4 R intersecting PR
at R’.
Step5. Draw a line through R’ parallel to QR intersecting PQ at Q’. ∆PQ’R’
is the required triangle.
Q28. Let the coordinates of the point on y-axis be P (0, y).
Let the given points be A (-5,-2) and B (3, 2).
It is given that PA = PB
⇒�(0 − (−5))2 + y − (−2))2 = �(0 − 3)2 + y − 22
⇒ 25+(y + 2)2 = 9 + (y - 2)2
⇒ 25 + y2 + 4y + 4 = 9 + y2 - 4y + 4
⇒ 8y = -16
⇒ y = -2
Thus, the coordinates of the required point is (0,-2)
Q29. The remaining solid, after removing the conical cavity, can be drawn as,
Height of the cylinder, h 1 = 20 cm
∴ Radius of the cylinder, r =
12 cm
Height of the cone, h 2 = 8 cm
2
=6 cm
Radius of the cone, r = 6 cm
Total surface area of the remaining solid
⇒ Areas of the top face of the cylinder + curved surface area of the cylinder +
curved surface area of the cone
Slant height of the cone, l = �(8cm)2 + (6cm)2
= √64cm2 + 36cm2
= √100 cm
= 10 cm
Curved surface are of the cone = πrl =
22
7
× 6cm × 10cm =
Curved surface are of the cylinder = 2πrh = 2 ×
22
7
1320
7
cm2
× 6cm × 10cm =
5280
7
cm2
Area of the top face of the cylinder = πr2 =
22
7
∴Total surface area of the remaining solid = (
× (6cm) 2 =
1320
7
=
792
7
5280
+
7392
7
7
cm2
cm2
+
792
7
) cm2
= 1056 cm2
Q30. Length of the rectangular piece of paper = 28 cm
Breadth of the rectangular piece of paper = 14 cm
Area of the rectangular paper, A 1 = (28.14) cm2 = 392 cm2
Radius (r) of the removed semicircular portion
1
= πr2
2
1
= ×
=
2
11
7
22
7
× (7)2 cm2
× 49cm2
=77 cm2
Radius R of the semicircular portion added =
Area (A 3 ) of the added semicircular portion
1
= πR2
2
1
= ×
2
22
7
× (14)2 cm2
= 308 cm2
28
2
cm = 14cm
∴ Area of the shaded region = A 1 –A 2 +A 3 = (392-77+308) cm2 = 623 cm2
Q31 The given situation can be represented as:
Here, AB is the building of height 15 m and CE is the cable tower of height h m.
CD = AB = 15m, DE = CE –CD = (h -15) m
In right ∆ADE,
tan 60° =
⇒√3 =
DE
AD
h−15
⇒ AD =
AD
h−15
√3
In right ∆ACD,
tan 30° =
⇒
1
√3
=
CD
AD
15 m
AD
…….
(1)
⇒ AD = 15√3m
………
(2)
From (1) and (2)
⇒
h−15
=15√3
√3
⇒h -15 = 45
⇒ h = 60
Thus, the height of the cable tower is 60 m.
Q32. Let the speed of the stream be x km/h
Speed of the boat while going upstream = (20 -x) km/h
Speed of the boat while going downstream = (20 +x) km/h
Time taken for the upstream journey =
48𝑘𝑘𝑘𝑘
(20−𝑥𝑥)𝑘𝑘𝑘𝑘 /ℎ
Time taken for the downstream journey =
It is given that,
=
48𝑘𝑘𝑘𝑘
(20+𝑥𝑥)𝑘𝑘𝑘𝑘 /ℎ
=
48
h
48
h
20−𝑥𝑥
20+x
Time taken for the upstream Journey = Time taken for the downstream journey +
1 hour
⇒
⇒
⇒
48
20−x
-
48
20+x
=1
960+48x−960+48x
(20−x)(20+x)
96x
400−x 2
=1
=1
⇒ 400 -x 2 =96x
⇒ 𝑥𝑥 2 + 96𝑥𝑥 − 400 = 0
⇒ x 2 + 100x − 4x − 400 = 0
⇒ x (x+100)-4(x+100)=0
⇒ (x+100)(x-4) = 0
⇒ x+100 = 0 or x-4 =0
⇒x = -100 pr x = 1
∴x = 4
[Since speed cannot be negative]
Thus, the speed of the stream is 4 km/h
OR
1
x+4
⇒
⇒
−
1
x−7
=
11
30
(x−7)−(x+4)
(x+4)(x−7)
−11
x 2 −3x−28
=
=
11
11
30
30
⇒x 2 - 3x - 28 = -30
⇒x 2 - 3x +2 =0
⇒x 2 -2x –x +2 =0
⇒x(x-2)-1(x-2) =0
⇒ (x-1)(x-2) = 0
⇒x-1 = 0 or x-2 = 0
⇒ x = 1 or x = 2
Hence, the roots of the given equation are 1 and 2.
Q33.
Let O be the centre of a circle.
Let PA and PB are two tangents drawn from a point P, lying outside the circle .
Join OA, OB, and OP.
We have to prove that PA = PB
In ∆OAP and ∆OPB,
∠OAP = ∠OPB (Each equal to 90°)
(Since we know that a tangent at any point of a circle is perpendicular to the
radius through the point of contact and hence, OA ⊥ PA and OB ⊥ PB)
OA = OB (Radii of the circle)
OP = PO (Common side)
Therefore, by RHS congruency criterion,
∆OPA ≅ ∆OPB
∴ By CPCT,
PA = PB
Thus, the lengths of the two tangents drawn from an external point to a circle are
equal.
Q34. Let a and d respectively be the first term and the common difference of the
given A. P
The sum of first four terms
S4 = 40
4
⇒ {2a+ (4-1)d} = 40
2
⇒2a +3d = 20
…….
(1)
The sum of first 14 terms
S14 = 280
14
⇒ {2a + (14 -1)d} = 280
2
⇒ 2a +13d = 40
………
(2)
Subtracting equation (1) from equation (2)
(2a +13d)-(2a +3d) = 40-20
⇒ 10d = 20
⇒d=2
Substituting d = 2 in equation (1),
2a +3×2 = 20
⇒ 2a = 20 -6 = 14
14
⇒a = 7
2
n
∴Sum of first n terms, Sn = {2a+ (n-1) d}
n
= {2×7+ (n-1) × 2}
2
2
n
= {14+2n-2}
n
2
= (2n+12)
2
n
= ×2(n+6)
2
=n(n+6)
=n2 +6n
OR
The first 30 integers divisible by 6 are 6 , 12, 18 ……180
Sum of first 30 integers
= 6 + 12 + 18 +…. + 180
=
30
2
(6 + 180)
=15 × 186
= 2790
[Sn= n (a+1) ]
2