# CBSE Previous Year Question Papers Class 10 Maths 2017 Delhi Term 2

## CBSE Previous Year Question Papers Class 10 Maths 2017 Delhi Term 2

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

• All questions are compulsory.
• This question paper consists of 30 questions divided into four sections- A, B, C and D.
• Section A contains 6 questions of 1 mark each, Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in two questions of 1 mark each, two questions of 2 marks each, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions.
• Use of calculators is not permitted.

### CBSE Previous Year Question Papers Class 10 Maths 2017 Delhi Term 2 Set I

Section – A

Question 1.
The ratio of the height of a tower and the length of its shadow on the ground is √3 : 1. What is the angle of elevation of the sun? 
Solution: Question 2.
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere? 
Solution:
Let radius of hemisphere be r units
Volume of hemisphere = S.A. of hemisphere $\frac { 2 }{ 3 }$πr3 = 3πr2
⇒ r = $\frac { 9 }{ 2 }$or diameter = 9 units

Question 3.
A number is chosen at random from the numbers -3, -2, -1,0, 1, 2, 3.
What will be the probability that square of this number is less then or equal to 1? 
Solution:
Possible outcomes {-3, -2, -1, 0, 1, 2, 3},
n = 7 and only three numbers -1,0, 1 fall under given condition so,
Required probability = $\frac { 3 }{ 7 }$

Question 4.
If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k? 
Solution:
Distance between (4, k) and (1, 0) = 5 k2 = 25 – 9 = 16
k = ±4

Section – B

Question 5.
Find the roots of the quadratic equation √2 x2 + 7x + 5√2 = 0. 
Solution: Question 6.
Find how many integers between 200 and 500 are divisible by 8. 
Solution:
Smallest divisible no. (by 8) in given range = 208
Last divisible no. (by 8) in range = 496
So, a = 208, d = 8, n = ?, an = 496
an = a + (n – 1 )d = 208 + (n – 1) 8 = 496
⇒ 8n + 208 – 8 = 496
⇒ 8n = 496 – 200 = 296
⇒ n = 37
So number of terms between 200 and 500 divisible by 8 are 37.

Question 7.
Prove that tangents drawn at the ends of a diameter of a circle are parallel to each other. 
Solution:
Given, PQ is a diameter of a circle with centre O.
The lines AB and CD are tangents at P and Q respectively.
To Prove: AB || CD Proof: AB is a tangent to the circle at P and OP is the radius through the point of contact
∠OPA = 90°
Similarly, CD is a tangent to circle at Q and
OQ is radius through the point of contact
∠OQD = 90°
⇒ ∠OPA = ∠OQD
But both form pair of alternate angles
AB || CD
Hence Proved.

Question 8.
Find the value of k for which the equation x2 + k(2x + k – 1) + 2 = 0 has real and equal roots. 
Solution:
Given equation is,
x2 + k(2x + k – 1) + 2 = 0
⇒ x2 + 2kx + k(k – 1) + 2 = 0
Here a = 1, b = 2k and c = k(k – 1) + 2
For real and equal roots
b2 – 4ac = 0
⇒ (2k)2 – 4 × 1 × (k (k – 1) + 2) = 0
⇒ 4k2 – 4(k2 – k + 2) = 0
⇒ 4k2 – 4k2 + 4k – 8 = 0
⇒ 4k = 8
⇒ k = 2

Question 9.
Draw a line segment of length 8 cm and divide it internally in the ratio 4 : 5. 
Solution:
Steps of construction:

1. Draw AB = 8 cm.
2. Draw any ray AX making an acute angle with AB.
3. Draw 9(4 + 5) points on ray AX namely A1, A2, A3, A4, A5, A6, A7, A8, A9 at equal distance.
4. Join BA9.
5. Through point A, draw a line parallel to A9B intersecting AB at the point C.
Then AC : CB = 4 : 5 Question 10.
In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD. Solution:
Given, PA = PB = 12 cm [Tangent from external point]
AC = CQ = 3 cm
BD = QD = 3 cm [Tangent from external point] So, PC + PD = (PA – AC) + (PB – BD)
= (12 – 3) + (12 – 3)
= 9 + 9 = 18 cm

Section – C

Question 11.
If mth term of an A.P. is $\frac { 1 }{ n }$and n term is $\frac { 1 }{ m }$, then find the sum of its first mn terms. 
Solution:
Let a and d be the first term and common difference respectively of the given A.P.  Question 12.
Find the sum of n terms of the series Solution:  Question 13.
If the equation (1 + m2)x2 + 2mcx + c2 – a2 = 0 has equal roots then show that c2 = a2( 1 + m2). 
Solution:
The given equation (1 + m2) x2 + 2mcx + c2 – a2 = 0 has equal roots
Here, A = 1 + m2, B = 2mc, C = c2 – a2
For equal roots, D = 0 = B2 – 4AC
⇒ (2mc)2 – 4(1 + m2) (c2 – a2) = 0
⇒ 4m2c2 – 4(c2 – a2 + m2c2 – m2a2) = 0
⇒ m2c2 – c2 + a2 – m2c2 + m2a2 = 0
⇒ -c2 + a2 (1 + m2) = 0
⇒ c2 = a2 (1 + m2)
Hence Proved.

Question 14.
The $\frac { 3 }{ 4 }$th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with an internal radius of 10 cm. Find the height of water in a cylindrical vessel. 
Solution:
According to the question, $\frac { 3 }{ 4 }$Volume of water in conical vessel = Volume of cylindrical vessel Hence the height of water in a cylindrical vessel is 1.5 cm.

Question 15.
In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region. Solution:
Area of shaded region = Area of quadrant OACB – Area of ΔDOB Hence, area of the shaded region is 6.125 cm2

Question 16.
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2 ∠OPQ. 
Solution:
Given, a circle with centre O, an external point T and two tangents TP and TQ.
Let ∠PTQ = θ To Prove: ∠PTQ = 2∠OPQ
Proof: TP = TQ [Tangent from an external point]
So ΔTPQ is an isosceles triangle
∠TPQ = ∠TQP [Angle opposite to equal sides of a Δ]
So, ∠TPQ = ∠TQP = $\frac { 1 }{ 2 }$(180° – θ) = 90° – $\frac { \theta }{ 2 }$
But, ∠TPO = 90° [Angle between tangent and radius]
∠OPQ = ∠OPT – ∠TPQ = 90° – (90° – $\frac { \theta }{ 2 }$)
= $\frac { \theta }{ 2 }$= $\frac { 1 }{ 2 }$∠PTQ
Or ∠PTQ = 2∠OPQ
Hence Proved.

Question 17.
Show that ∆ABC, where A(-2, 0), B(2, 0), C(0, 2) and ∆PQR where P(-4, 0), Q(4, 0), R(0, 4) are similar triangles. 
Solution:
Coordinates of vertices are
A(-2, 0), B(2, 0), C(0, 2)
P(-4, 0), Q(4, 0), R(0, 4) We see that sides of ∆PQR are twice the sides of ∆ABC.
Hence, both triangles are similar.
Hence Proved.

Question 18.
The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is ( $\frac { 7 }{ 2 }$, y), find the value of y. 
Solution:
Given, A(2, 1), B(3, -2) and C( $\frac { 7 }{ 2 }$, y)  Question 19.
Two different dice are thrown together. Find the probability that the numbers obtained
(i) have a sum less than 7
(ii) have a product less than 16
(iii) is a doublet of odd numbers. 
Solution:
Total possible outcomes in each case = 6 × 6 = 36
(i) Have a sum less than 7, Possible outcomes are,
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5), (2, 1) (2, 2) (2, 3) (2, 4),
(3, 1) (3, 2) (3, 3) (4, 1) (4, 2) (5, 1)
n(E) = 15
So, probability = $\frac { 25 }{ 36 }$= $\frac { 5 }{ 12 }$
(ii) Have a product less than 16, Possible outcomes are,
(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)
(4, 1), (4, 2) (4, 3)
(5, 1), (5, 2) (5, 3)
(6, 1), (6, 2)
n(E) = 25
So, probability = $\frac { 25 }{ 36 }$
(iii) Is a doublet of odd no.,
Possible outcomes are
(1, 1) , (3, 3), (5, 5)
n(E) = 3
P(doublet of odd no.) = $\frac { 3 }{ 36 }$= $\frac { 1 }{ 36 }$

Question 20.
A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/h. 
Solution:
From ∆ABC, $\frac { AB }{ BC }$= tan 60°  Section – D

Question 21.
Construct an isosceles triangle with base 8 cm and altitude 4 cm. Construct another triangle whose sides are $\frac { 2 }{ 3 }$times the corresponding sides of the isosceles triangle. 
Solution:
Steps of construction:

1. Draw BC = 8 cm.
2. Construct XY, the perpendicular bisector of line segment BC, meeting BC at M.
3. Cut MA = 4 cm on XM. Join BA & CA, ∆ABC is obtained.
4. At B, draw an acute angle in a downward direction. Draw 3 arcs B1, B2 and B3 on it.
5. Join B3C and at B2 draw line parallel to B3C, cutting BC at C’.
6. At C’, draw A’C’ parallel to AC.
Thus, ∆A’C’ B is the required triangle. Question 23.
The ratio of the sums of the first m and first n terms of an A. P. is m2 : n2. Show that the ratio of its mth and nth and terms is (2m – 1): (2n – 1). 
Solution:
Let a be first term and d is a common difference. Question 24.
Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream. 
Solution:
Let speed of the stream be x.
According to question, Question 25.
If a ≠ b ≠ c, prove that the points (a, a2), (b, b2) (c, c2) will not be collinear. 
Solution: This can never be zero as a ≠ b ≠ c
Hence, these points can never be collinear.
Hence Proved.

Question 26.
The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts. 
Solution:
Let BC = r cm & DE = R cm
Since B is mid-point of AD & BC || DE
C is mid-point of AE or AC = CE  Question 27.
Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25. 
Solution:
Total possible events in case of peter is 36 favourable outcome is (5, 5)
n(E) = 1
So, P(getting 25 as product) = $\frac { 1 }{ 36 }$
While total possible event in case of Rina is 6
Favourable outcome is 5
n(E) = 1
So, P(square is 25) = $\frac { 1 }{ 6 }$
As $\frac { 1 }{ 6 }$> $\frac { 1 }{ 36 }$, so Rina has better chance.

Question 28.
A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of circle. Find the area of major and minor segments of the circle. 
Solution:
r = 10 cm, i = 60°  Question 29.
The angle of elevation of a cloud from a point 60 m above the surface of the water of a lake is 30° and the angle of depression of its shadow in water of lake is 60°. Find the height of the cloud from the surface of water. 
Solution:
In ∆CMP, tan 30° = $\frac { CM }{ PM }$  ⇒ 3h = h + 120
⇒ 2h = 120
⇒ h = 60 m
Height of cloud from surface of water = h + 60 = 60 + 60 = 120 m.

Question 30.
In the given figure, the side of the square is 28 cm and the radius of each circle is half of the length of the side of the square where O and O’ are centres of the circles. Find the area of the shaded region. Solution:
r = $\frac { 1 }{ 2 }$(side) = 14 cm
side = 28 cm
Area of shaded region = 2 × (area of circle) + area of square – 2 × (area of quadrant) Question 31.
In a hospital, used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park of a hospital whose length is 25 m and breadth is 20 m. If tank is filled completely then what will be height of standing water used for irrigating the park.
Write your views on recycling of water. 
Solution:
Given, diameter of cylinder (d) = 2 m
Radius of cylinder (r) = 1 m
Heigth of cylinder (h1) = 5 m
Length of park (l) = 25 m
Breadth of park (b) = 20 m,
Let height of standing water in the park = h
Volume of water used to irrigate the park = Volume stored in a cylindrical tank ### CBSE Previous Year Question Papers Class 10 Maths 2017 Delhi Term 2 Set II

Note: Except for the following questions, all the remaining questions have been asked in the previous set.

Section – B

Question 10.
Draw a line segment of length 7 cm and divide it internally in the ratio 2 : 3. 
Solution: Steps of construction:

1. Draw AB = 7 cm.
2. At A draw an acute angle with 5 equidistant marks A1, A2, A3, A4, A5.
3. Join A5B.
4. Draw A2C || A5B to get point C on AB.
Thus, AC : CB = 2 : 3

Section – C

Question 19.
A metallic solid sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm. How many cones will be made? 
Solution:
Volumeofmetalincones = Volume of a solid sphere
Let n = number of cones
n × volume of each cone = volume of a solid sphere  Question 20.
From the top of a 7 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower. 
Solution:
Let C be top of a 7 m building CD and AB be the tower.
From C, draw CE ⊥ AB, so EBDC is a rectangle. From ∆CBD, tan 45° = $\frac { CD }{ BD }$
or BD = CD = 7 m
From ∆AEC,
tan 60° = $\frac { AE }{ EC }$
⇒ AE = EC tan 60° = 7√3 [∵EC = BD]
Height of tower is AB = AE + EB
= AE + DC
= 7√3 + 7
= 7(√3 + 1) m.

Section – D

Question 28.
Draw a right triangle in which the sides (other than the hypotenuse) are of lengths 4 cm and 3 cm. Now construct another triangle whose sides are $\frac { 3 }{ 5 }$times the corresponding sides of the given triangle. 
Solution: Steps of construction:

1. Draw AB = 4 cm.
2. Draw AC ⊥ AB of 3 cm.
3. Join BC.
4. Draw an acute angle at A with 5 equidistant marks.
5. Join A5B.
6. Draw A3B’ || A5B.
7. Draw B’C’ || BC.
Thus, AB’C’ is the required triangle.

Question 29.
If the sum of first m terms of an A.P. is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero. 
Solution:
Let a be first term and d is common difference of given A.P. then,  Question 30.
Two points A and B are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are 60° and 45° respectively. If the height of the tower is 15 m, then find the distance between these points. 
Solution:
Let PT be the tower Question 31.
The height of a cone is 30 cm. From its topside, a small cone is cut by a plane parallel to its base. If the volume of a smaller cone is $\frac { 1 }{ 27 }$of the given cone, then at what height it is cut from its base? 
Solution:
Volume of original cone OAB   ### CBSE Previous Year Question Papers Class 10 Maths 2017 Delhi Term 2 Set III

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Section – B

Question 10.
In the figure, AB and CD are common tangents to two circles of unequal radii. Solution:
Construction: Extend AB and CD to meet at a point P Now, PA and PC are tangents of circle with centre O
So, PA = PC …..(i)
PB and PD are tangent on circle with centre O’
So, PB = PD ….(ii)
On subtracting equation (ii) from equation (i),
PA – PB = PC – PD
AB = CD
Hence Proved.

Section – C

Question 18.
If the pth term of an A.P. is q and qth term is p, prove that its nth term is (p + q – n). 
Solution:
Let a be first term and d be common difference.
Then, pth term = q ⇒ a + (p – 1)d = q …(i)
qth term = p ⇒ a + (q – 1 )d = p …(ii)
On subtracting eq. (ii) from eq. (i)
(p – 1)d – (q – 1)d = q – p
⇒ pd – d – qd + d = q – p
⇒ (p – q) d = q – p
⇒ d = $\frac { q-p }{ p-q }$= -1
Putting value of d in eq. (i)
a + (p – 1) (-1) = 9
a = q + p – 1
⇒ nth term = a + (n – 1)d
= q + p – 1 + (n – 1)(-1)
= q + p – 1 + 1 – n
= q + p – n
⇒ Tn = q + p – n
Hence Proved.

Question 19.
A solid metallic sphere of diameter 16 cm is melted and recast into smaller solid cones, each of radius 4 cm and height 8 cm. Find the number of cones so formed. 
Solution:
No. of cones formed Question 20.
The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the height of the tower is 50 m, find the height of the hill. 
Solution:
Let AB be hill and DC be tower. Section – D

Question 29.
If the pth term of an A.P is $\frac { 1 }{ q }$and qth term is $\frac { 1 }{ p }$, prove that the sum of first pq terms of the A.P. is ( $\frac { pq+1 }{ 2 }$). 
Solution:
Let a be first term and d is common difference.  Question 30.
An observer finds the angle of elevation of the top of the tower from a certain point on the ground as 30°. If the observer moves 20 m towards the base of the tower, the angle of elevation of the top increases by 15°, find the height of the tower. 
Solution:
Let AB be the tower of height h.
From ∆ABC, $\frac { AB }{ BC }$= tan 45°   error: Content is protected !! 