CBSE Sample Papers for Class 10 Maths Paper 7 is part of CBSE Sample Papers for Class 10 Maths . Here we have given CBSE Sample Papers for Class 10 Maths Paper 7. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

## CBSE Sample Papers for Class 10 Maths Paper 7

 Board CBSE Class X Subject Maths Sample Paper Set Paper 7 Category CBSE Sample Papers

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 7 of Solved CBSE Sample Papers for Class 10 Maths is given below with free PDF download solutions.

Time: 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 comprises of 6 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 8 questions 1 of 4 marks each.
• There is no overall choice. However, internal choice has been provided in one question of 2 marks, 1 three questions of 3 marks each and two questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• In question of construction, drawings shall be neat and exactly as per the given measurements.
• Use of calculators is not permitted. However, you may ask for mathematical tables.

SECTION A

Question numbers 1 to 6 carry 1 mark each.

Question 1.
If the sum of zeroes of the quadratic polynomial 3x² – kx + 6 is 3, then find the value of k.

Question 2.
If 1 is a root of both the equations ay² + ay + 3 = 0 and y² + y + b = 0, then find the value of ab.

Question 3.
Consider the following distribution, find the frequency of class 30-40. Question 4.
Cards marked with number 3, 4, 5, ..50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number.

Question 5.
In which quadrant the point P that divides the line segment joining the points A(2, -5) and B(5, 2) in the ratio 2 : 3 lies?

Question 6.
If sec 2A = cosec (A – 27°), where 2A is an acute angle, find the measure of ∠A.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
Find whether the following pair of linear equations is consistent or inconsistent:
2x – 3y = 8; 4x – by = 9.

Question 8.
The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, -5) and R(-3, 6), find the coordinates of P.

Question 9.
In the adjoining figure, AP and BP are tangents to a circle with centre O, such that AP = 5 cm and ∠APB = 60°, find the length of chord AB. Question 10.
How many terms of the AP 18, 16, 14, … be taken so that their sum is zero?

Question 11.
Show that the mode of the sequences obtained by combining the two sequences S1 and S2 given below is different from that of S1 and S2 taken separately:
S1 : 3, 5, 8, 8, 9, 12, 13, 9, 9
S2 : 7, 4, 7, 8, 7, 8, 13.

Question 12.
In the adjoining figure, ABC and DBC are two right triangles. Prove that AP x PC = BP x PD. SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
Solve the following pair of equations
49x + 51y = 499
51x + 49y = 501.
OR
Sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number.

Question 14.
Divide 56 in four parts in AP such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5 : 6.

Question 15.
D and E are points on the sides AB and AC respectively of ∆ ABC such that DE is parallel to BC, and AD : DB = 4:5. CD and BE intersect each other at F. Find the ratio of the areas of ∆DEF and ∆CBF.

Question 16.
Find the area of ∆PQR with Q(3, 2) and the mid-points of the sides through Q being (2,-1) and (1, 2).
OR
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

Question 17.
Prove the following identity: Question 18.
Three alarm clocks ring at intervals of 4, 12 and 20 minutes respectively. If they start ringing together, after how much time will they next ring together?

Question 19.
Prove that 3 + 2√3 is an irrational number.

Question 20.
In the adjoining figure, ABCD is a square of side 14 cm. Semicircles are drawn with side of square as diameter. Find the area of the shaded region. Question 21.
In the adjoining figure shows two arcs PAQ and PBQ. Arc PAQ is a part of a circle with centre O and radius OP while arc PBQ is a semicircle drawn on PQ as diameter. If OP = PQ = 10 cm, show that the area of the shaded region is 25 (√3 – $$\frac { \pi }{ 6 }$$)cm². OR
Sides of a triangular field are 15 m, 16 m, 17 m. With the three comers of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.

Question 22.
Two different dice are tossed together. Find the probability that the product of the two numbers on the top of dice is
(i) 6
(ii) a perfect square number.
OR
A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Ramesh, a trader, will only accept the shirts which are good, but Kewal, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that
(i) it is acceptable to Ramesh?
(ii) it is acceptable to Kewal?

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
If α and β are the zeroes of the polynomial p(x) = 3x² + 2x + 1, find the polynomial whose zeroes are Question 24.
If sec θ – tan θ = x, show that sec θ + tan θ = $$\frac { 1 }{ x }$$ and hence, find the values of cos θ and sin θ.
OR
If tan (A + B) = √3 , tan (A – B) = $$\frac { 1 }{ \sqrt { 3 } }$$, 0° < A + B < 90°, A > B, find A and B. Also calculate tan A sin (A + B) + cos A tan (A – B).

Question 25.
From the top of a tower h metres high, the angles of depression of two objects, which are in line with the foot of tower are α and β (β>α). Find the distance between the two objects.

Question 26.
Sushant has a vessel of the form of an inverted cone, open at the top, of height 11 cm and radius of top as 2.5 cm and is full of water. Metallic spherical balls each of diameter 0.5 cm are put in the vessel due to which $$\frac { 2 }{ 5 }$$ th of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds.
What value has been shown by Sushant?

Question 27.
A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m away from the bottom of a pillar, a snake is coming to its hole at the base of a pillar, seeing the snake, the peacock pounces on it. If their speeds are equal, at what distance from the hole is the snake caught?
OR
Find the value(s) of p for which the quadratic equation
(2p + 1) x² – (7p + 2) x + (7p – 3) = 0 has equal roots. Also find these roots.

Question 28.
Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on the outer circle, construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculations.

Question 29.
In the adjoining figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS. OR
In the adjoining figure, O is the centre of the circle. Determine ∠ACB, if PA and PB are tangents and ∠APB = 50°. Question 30.
The following table gives the daily income of 50 workers of a factory. Draw both types (less than type and greater than type) ogives. Hence, obtain the median income. Given sum of zeroes of the polynomial 3x² – kx + 6 is 3
⇒ 3 = $$\frac { -(-k) }{ 3 }$$ ⇒ k =9

Given 1 is a root of both the equations => ay² + ay + 3 = 0 and y² + y + b =0
=> a.1² + a.1 + 3 = 0 and 1² + 1 + b = 0
=> a + a + 3 = 0 and 1 + 1 + b = 0
=> 2a + 3 = 0 and 2 + b = 0 Hence the values of ab is 3 Hence, the frequency of class 30-40 is 3

Total number of cards = 48
Perfect square number cards from 3 to 50 are 4, 9, 16, 25, 36 and 49. These are 6 in number.
∴ P(Perfect square number) = $$\frac { 6 }{ 48 }$$ = $$\frac { 1 }{ 8 }$$ sec 2A = cosec (A – 27°) => cosec (90° – 2A) = cosec(A – 27°)
=> 90° – 2A = A – 27° => 3A = 117°
=> A = $$\frac { { 117 }^{ o } }{ 3 }$$ => A = 39°
Hence, ∠A = 39°.

The given pair of linear equations can be written as 2x – 3y – 8 = 0 and 4x – 6y – 9 = 0. Given the coordinates of Q(2, -5) and R(-3, 6).
Let the coordinates of P be (2y, y).
According to given, PQ = PR
=> (PQ)² = (PR)²
=> (2 – 2y)² + (- 5 – y)² = (- 3 – 2y)² + (6 – y)²
=> 4 – 8y + 4y² + 25 + 10y + y² – 9 + 12y + 4y² + 36 – 12y + y²
=> 2y + 29 = 45 => 2y = 16 => y = 8
Hence, coordinates of P are (16, 8).

Given ∠APB = 60° and AP = 5 cm
In ∆APB, PA = PB (tangents drawn from an external point are equal)
∴∠PAB = ∠PBA (angles opposite to equal sides are equal)
Now, ∠PAB + ∠PBA + ∠APB = 180° (angle sum property of a triangle)
=> ∠PAB + ∠PAB + 60° = 180°
=> 2∠PAB = 120° => ∠PAB = 60°
=> ∠PAB = ∠PBA = ∠APB = 60°
∴∆APB is an equilateral triangle.
So, AB = AP = 5 cm. Given AP is 18, 16, 14, …
Here, a – 18, d = -2
If the sum of n terms of the given AP is zero, their Sn = 0 But n cannot be zero.
Hence, the required number of terms is 19.

Given S1 : 3, 5, 8, 8, 9, 12, 13, 9, 9
S2 : 7, 4, 7, 8, 7, 8, 13
In S1 : Number 9 occurs maximum number of times i.e. 3 times
Hence, the mode of sequence S1 = 9
In S2 : Number 7 occurs maximum number of times i.e. 3 times
Hence, the mode of sequence S2 = 7
After combining sequences S1 and S2, we have
3, 4, 5, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 12, 13, 13
Here number 8 occurs maximum number of times i.e. 4 times.
Hence, the mode of combined sequence = 8
Hence, mode of S1 and S2 taken combined is different from that of S1 and S2 taken separately.

In ∆APB and ∆DPC,
∠BAP = ∠CDP (each 90°)
∠APB = ∠DPC (vert. opp. ∠s)
∴∆APB ~ ∆DPC (by AA similarity criterion)  Given 49x + 51 y = 499
and 51x + 49y = 501
Adding (i) and (ii), we get
100x + 100y – 1000
=> x + y = 10
Subtracting (ii) from (i), we get
-2x + 2y = -2
=> -x + y = -1
Now, adding (iii) and (iv), we get
2y = 9 => y = $$\frac { 9 }{ 2 }$$
Putting y = $$\frac { 9 }{ 2 }$$ in (iii), we get OR
Let the two digit number be 10x + y, then
according to given, x + y = 8
and | 10x + y – (10y + x) | = 18
=> |9x – 9y| = 18 => |x – y| = 2 =>x – y = ±2
When x + y = 8 and x – y = 2, then
on adding these equations, we get 2x = 10 => x = 5.
Putting x = 5 in x + y = 8, we get y = 3
∴ The original number is 10 x 5 + 3 i.e. 53
When x + y = 8 and x – y = -2, then
on adding these equations, we get 2x – 6 => x = 3
Putting x = 3 in x + y = 8, we get y = 5
∴ The original number is 10 x 3 + 5 i.e. 35
Hence, the original number is 53 or 35.

Let the four numbers in AP be
a – 3d, a – d, a + d and a + 3d, then
according to given,
a – 3d + a – d + a + d + a + 3d = 56
4a = 56 => a = 14  Given AD : DB = 4 : 5 Let D and E are the mid-points of the sides through Q.
Let the coordinates of P and R be (x, y) and (u, v) respectively
∵D is the mid-point of QP    To find the time when the clocks will next ring together, we have to find the LCM of 4, 12 and 20.
Prime factorisation 4, 12 and 20
4 = 2 x 2
12 = 2 x 2 x 3
20 = 2 x 2 x 5
∴ LCM of 4, 12 and 20 = 2 x 2 x 3 x 5 = 60
Hence, the alarm clocks will ring together again after 60 minutes i.e. one hour.

Let us assume that 3 + 2√3 is a rational number, say r.  Radius of each semicircle = $$\frac { 14 }{ 2 }$$ cm = 7 cm
Mark the shaded regions I, II, III and IV (as shown in adjoining figure)
Area of region I + area of region III
= area of square ABCD – area of two semicircles each of radius 7 cm = (14 x 14 – 2.$$\frac { 1 }{ 2 }$$ π x 72) cm²
= (196 – $$\frac { 22 }{ 7 }$$ x 49) cm²
= (196 – 154) cm² = 42 cm²
Similarly, area of region II + area of region IV = 42 cm²
Required area = area of shaded regions I, II, III and IV
= (2 x 42) cm² = 84 cm²

OQ = OP (radii of same circle)
Given OP = PQ = 10 cm => OP = OQ = PQ = 10 cm.
Thus, OPQ is an equilateral triangle of side 10 cm.
Also ∠POQ = 60° (angle of an equilateral triangle)
Area of shaded region = area of a semicircle with PQ as diameter i.e. radius 5 cm + area of equilateral AOPQ of side 10 cm – area of sector of a circle of radius 10 cm and central angle 60° The region of the field left ungrazed by the three animals is shown shaded. Sum of areas of three sectors of radius 7 m (each) and having central angles as ∠A, ∠B and ∠C
= area of a sector of radius 7 m and having central angle as sum of ∠A, ∠B and ∠C
= area of a sector of radius 7 m and having central angle as ∠A + ∠B + ∠C i.e. an angle of 180° When two different dice are tossed together, the total number of outcomes is 36 and all the outcomes are equally likely.
(i) The outcomes favourable to the event ‘the product of two numbers on the top of two dice is 6’ are (1, 6), (6, 1), (2, 3) and (3, 2). These are 4 in number.
∴ P (product of two numbers is 6) = $$\frac { 4 }{ 36 }$$ = $$\frac { 1 }{ 9 }$$.
(ii) The outcomes favourable to the event ‘the product of two numbers on the top of dice is a perfect square number’ are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 4), (4, 1).
These are 8 in number.
∴ P (product of two numbers is a perfect square number) = $$\frac { 8 }{ 36 }$$ = $$\frac { 2 }{ 9 }$$.
OR
One shirt is drawn at random from a carton containing 100 shirts means all shirts are equally likely to be drawn. So, the sample space of the experiment has 100 equally likely outcomes.
(i) Since Ramesh accepts only good shirts and there are 88 good shirts in the carton. Therefore, the number of favourable (i.e. acceptable) outcomes to Ramesh = 88.
∴ P (acceptable to Ramesh) = $$\frac { 88 }{ 100 }$$ = 0.88
(ii) As Kewal rejects the shirts which have major defects and the number of such shirts is 4, so the number of shirts acceptable to Kewal = 100 – 4 = 96.
Therefore, the number of favourable (acceptable) outcomes to Kewal = 96.
∴ P (acceptable to Kewal) = $$\frac { 96 }{ 100 }$$ = 0.96

Given a and (3 are the zeroes of the polynomial p(x) = 3x² + 2x + 1  Given sec θ – tan θ = x
∵ sec² θ – tan² θ = 1
=> (sec θ tan θ) (sec θ + tan θ) = 1
x . (sec θ + tan θ) = 1
=> sec θ + tan θ = $$\frac { 1 }{ x }$$ …(ii)
Adding (i) and (ii), we get
sec θ – tan θ + sec θ + tan θ = x + $$\frac { 1 }{ x }$$   Let P be the top of the tower MP of height h metres, and A, B be the two objects. Let BM = x metres and AB = d metres. The angles of depression of the two objects are shown in the adjoining figure, then
∠MAP = α and ∠MBP = β.
From right angled ∆MBP, we get  Volume of water in the cone
= Volume of cone of height 11 cm and radius 2.5 cm Hence, the number of metallic balls put into the vessel = 440
Sushant’s love for plants, nature and environment has been depicted. Let CD be the pillar and initially the snake be at the point A and the peacock is sitting at the top of pillar i.e. at the point D. Then CD = 9 m and AC = 27 m.
Let the snake be caught at the point B. As the snake and peacock have equal speeds, they cover equal distance in same time.
Let AB = x metres, then BD = AB = x metres and BC = AC – AB = (27 – x) metres.
In ∆BCD, ∠C = 90°.
By Pythagoras theorem, we get
BD² = BC² + CD²
=> x² = (27 – x)² + 9² => x² = 729 – 54x + x² + 81
=> 54x = 810 => x = 15.
∴ BC = (27 – x) metres = (27 – 15) metres = 12 metres.
Hence, the snake is caught at a distance of 12 metres from its hole.
OR
The given equation is (2p + 1) x² – (7p + 2) x + (7p – 3) = 0 …(i)
Comparing it with ax² + bx + c = 0, we get
a = 2p + 1, b = – (7p + 2), c = 7p – 3.
Discriminant = b² – 4ac = (-(7p + 2))² – 4 (2p + 1) (7p – 3)
= 49p² + 28p + 4 – 4 (14p² – 6p + 7p – 3)
= 49p² + 28p + 4 – 56p² – 4p + 12
= -7p² + 24p + 16.
For equal roots, discriminant = 0
=> -7p² + 24p + 16 = 0 => 7p² – 24p – 16 = 0
=> 7p² – 28p + 4p – 16 = 0 => 7p (p – 4) + 4 (p – 4) = 0
=> (p – 4) (7p + 4) = 0 => p – 4 = 0 or 7p + 4 = 0 Steps of construction:
1. Draw two concentric circles of radii 3 cm and 5 cm with point O as their centre.
2. Let P be a point on the outer circle. Join OP and draw its perpendicular bisector to meet OP at M.
3. Taking M as centre and OM (or MP) as radius, draw a circle. Let this circle intersect the smaller circle i.e. circle of radius 3 cm at points A and B.
4. Join PA and PB. Then PA and PB are the required tangents on measuring PA (or PB), we find that PA = 4 cm. Calculation of length PA:
Join OA.
In ∆OAP, ∠OAP = 90° (angle in a semicircle)
By Pythagoras theorem, we get
PA² = OP² – OA² = 5² – 3² = 25 – 9 = 16
=> PA = 4 cm. Join QO and produce it to meet SR at M.
In ∆PRQ, PR = PQ (lengths of tangents)
=> ∠PRQ = ∠PQR …(i)
∠PQR + ∠PRQ + ∠RPQ = 180°
=> ∠PQR + ∠PQR + 30° = 180° (using (i))
=> 2∠PQR = 150° => ∠PQR = 75°.
As RS || PQ and RQ is transversal,
∠SRQ = ∠PQR
=> ∠SRQ = 75°
Since PQ is tangent to the circle with centre O at the point Q and OQ is radius, OQ ⊥ PQ.
Also RS || PQ (given), so QM ⊥ RS i.e. OM ⊥ RS
=> MR = MS (∵ perpendicular from centre to a chord bisects it)
In ∆QRM and ∆QSM,
MR = MS (proved above)
∠QMR = ∠QMS (each = 90°, as QM ⊥ RS)
QM = QM
∴∆QRM ≅ ∆QSM
∴ ∠MSQ = ∠MRQ (c.p.c.t.)
=> ∠RSQ = ∠SRQ.
In ∆QRS, ∠RQS + ∠SRQ + ∠RSQ = 180° => ∠RSQ + ∠SRQ + ∠SRQ = 180°
=> ∠RQS + 2 x 75° = 180° => ∠RQS = 30°.
OR As PA is tangent to the circle at A and OA is radius, OA ⊥ AP
i.e. ∠OAP = 90°
Similarly, ∠OBP = 90°
∠AOB + ∠OAP + ∠APB + ∠OBP = 360°
=> ∠AOB + 90° + 50° + 90° = 360°
=> ∠AOB = 130°
Reflex ∠AOB = 360° – 130° = 230°
∠ACB = $$\frac { 1 }{ 2 }$$ of reflex ∠AOB
(∵ angle at the centre = double the angle at the remaining part of circle)
=> ∠ACB = $$\frac { 1 }{ 2 }$$ x 230° = 115°   