CBSE Sample Papers for Class 10 Maths Paper 4 is part of CBSE Sample Papers for Class 10 Maths . Here we have given CBSE Sample Papers for Class 10 Maths Paper 4. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.
CBSE Sample Papers for Class 10 Maths Paper 4
|Sample Paper Set||Paper 4|
|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 4 of Solved CBSE Sample Papers for Class 10 Maths is given below with free PDF download solutions.
Time: 3 Hours
Maximum Marks: 80
- 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.
Question numbers 1 to 6 carry 1 mark each.
The decimal expansion of the rational number will terminate after how many places of decimal?
Is x = -2 a solution of the equation x² – 2x + 8 = 0?
In the adjoining figure, P and Q are points on the sides of AB and AC respectively of ∆ABC such that AP = 3.5 cm, PB = 7 cm, QC = 6 cm. Prove that PQ || BC.a
If tan θ = cot (30° + θ), find the value of θ.
A cylinder and a cone are of same base radius and c the ratio of volume of cylinder to that of the cone.
Two coins are tossed simultaneously. Find the probability of getting exactly one head.
Question numbers 7 to 12 carry 2 marks each.
Complete the following factor tree and find the numbers x, y and z.
If α and β are zeroes of the polynomial f(x) = x² – x – k such that α – β = 9, find k.
The coordinates of the vertices of ∆ABC are A(4, 1), B(-3, 2) and C(0, k). Given that the area of ∆ABC is 12 unit², find the value of k. .
Without using trigonometric tables, evaluate the following:
Area of a sector of a circle of radius 36 cm is 54π cm². Find the length of the corresponding arc of the sector.
Question numbers 13 to 22 carry 3 marks each.
The HCF of 65 and 117 is expressible in the form 65m – 117. Find the value of m. Also find the LCM of 65 and 117 using prime factorization
Solve the following pair of linear equations: px + qy = p – q;qx – py = p + q.
In the given figure, ABCDE is a pentagon with BE || CD and BC || ED. BC is perpendicular to CD. If the perimeter of pentagon ABCDE is 21 cm, find the values of x and y.
Find the roots of the following equation
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work for 30 days?
Jaspal Singh repays his total loan of Rs 118000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, what amount will be paid by him in the 30th instalment? What amount of loan does he still to pay after the 30th instalment?
Prove that : sin θ(1 + tan θ) + cos θ(1 + cot θ) = sec θ + cosec θ.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².
In the adjoining figure, DB ⊥ BC, DE ⊥ AB
and AC ⊥ BC. Prove that
In the adjoining figure, 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.
In the adjoining figure, APB and CQD are semicircles of diameter 7 cm each, while ARC and BSD are semicircles of diameter 14 cm each. Find the perimeter of the shaded region.
Two different dice are rolled together. Find the probability of getting:
(i) the sum of numbers on two dice as 5.
(ii) even numbers on both dice.
The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is . The probability of selecting a blue ball at random from the same jar is . If the jar contains 10 orange balls, find the total number of balls in the jar.
Find the value of p for the following distribution whose mean is 10:
Question numbers 23 to 30 carry 4 marks each.
If pth, qth and rth terms of an AP are a, b and c respectively, then show that (a – b) r + (b – c) p + (c – a) q = 0.
An icecream seller sells his icecreams in two ways:
(A) In a cone of r = 5 cm, h – 8 cm with hemispherical top.
(B) In a cup in shape of cylinder with r – 5 cm, h – 8 cm.
He charges the same price both but prefers to sell his icecream in a cone.
(a) Find the volume of the cone and the cup.
(b) Which out of the two has more capacity?
(c) By choosing a cone, which value is not followed by the icecream seller?
Solve the following pair of linear equations graphically:
x + 3y = 6; 2x – 3y = 12
Also find the area of the triangle formed by the lines representing the given equations with y-axis.
If the centroid of ∆ABC, in which A(a, b), B(b, c), C(c, a) is at the origin, then calculate the value of (a³ + b³ + c³).
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 parallelogram ABCD.
From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the cars. (Use √3 = 1.73)
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O’P are tangents to the two circles. Find the length of the common chord PQ.
Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ∆ABC in which PQ = 8 cm. Also justify the construction.
Size of agriculture holding in a survey of 200 families is given in the following table:
Compute the median and mode size of holding.
Hence, the decimal expansion of given rational number will terminate after 3 places of decimal.
(-2)² – 2 x (-2) + 8 = 4 + 4 + 8 = 16≠0
∴ x = -2 is not a solution of the equation x² – 2x + 8 = 0.
∴ By the converse of B.P.T., PQ || BC.
tan θ = cot (30° + θ)
= tan [90° – (30° + θ)]
= tan (60° – θ)
θ = 60° – θ => 2θ = 60° => θ = 30°.
Let r be the base radius of cylinder and cone and h be their heights.
Hence, the ratio of volume of cylinder to that of cone = 3:1.
When two coins are tossed simultaneously, then the outcomes are HH, HT, TH, TT.
So total number of outcomes = 4.
The outcomes favourable to the event ‘getting exactly one head’ are HT and TH.
∴The number of outcomes favourable to the event = 2.
∴ P(exactly one head) =
x = 3381 x 2 = 6762
y = 161 x 7 = 1127
and z = = 23.
Given f(x) = x² – x – k
Area of ∆ABC = | 14(2 – k) – 3(k – 1) + 0(1 – 2) |
=> 12 = |8 – 4k – 3k + 3|
=> 24 = |11 – 7k|
=> 11 – 7k = ±24
=> 11 – 7k = 24 or 11 – 7k = -24
=> 7k = -13 or 7k = 35
=> k = or k = 5
Hence, k = or k = 5
Let the central angle (in degrees) be θ, then
Hence, the length of the corresponding arc of the sector = 3π cm.
By Euclid’s division algorithm, we have
117 = 65 x 1 + 52; 65 = 52 x 1 + 13; 52 = 13 x 4 + 0
∴ HCF of 65 and 117 = 13
∴ 65m – 117 = 13 => 65m = 130 => m = 2.
Prime factorisation of 65 and 117 are as follows:
65 = 5 x 13 and 117 = 3 x 3 x 13
∴ LCM of 65 and 117 = 3 x 3 x 5 x 13 = 585.
Given equations are
px + qy = p – q …(i)
qx – py = p + q …(ii)
Multiplying (i) by p and (ii) by q, we get
p²x + pqy = p² – pq …(iii)
and q²x – pqy – pq + q² …(iv)
On adding (iii) and (iv), we have
Putting x = 1 in equation (i), we have
p x 1 + qy = p – q => qy = -q => y = -1
∴ x = 1 ,y = -1.
Given BE || CD and BC || ED, also BC ⊥ CD
=> BCDE is a rectangle
CD = BE
=> x + y = 5
Again, given perimeter of pentagon ABCDE = 21 cm
=> AB + BC + CD + DE + EA = 21 cm
=> 3 + x – y + x + y + BC + 3 = 21cm
=> 2x + x – y = 21 – 6
=> 3x – y = 15
Adding (i) and (ii), we get
4x = 20 => x = => x = 5
Putting x = 5 in (i), we get
5 + y = 5 =>y = 0
Hence, x = 5, y = 0.
(x + 4) (x – 7) = -30 => x² – 7x + 4x – 28 = -30
=> x² – 3x + 2 = 0 => x² – 2x – x + 2 = 0
=> x(x – 2) – 1(x – 2) = 0 => (x – 2) (x – 1) = 0
=> x – 2 = 0 or x – 1 = 0 => x = 2 or x = 1.
Hence, the roots of the given equation are 2 and 1.
200, 250, 300, … is in AP.
whose first term = 200 and common difference = 50.
Let the number of days for which work is delayed be n, then
=> n[200 + 25 (n- 1)] = 27750
=> 25n² + 175n = 27750
=> n² + 7n – 1110 = 0
=> n² + 37n – 30n – 1110 = 0
=> n(n + 37) – 30(n + 37) = 0
=> (n + 37) (n – 30) = 0
=> n = -37 or n = 30
But n cannot be negative
∴ n = 30
Hence, the work was delayed by 30 days.
Jaspal Singh starts repaying loan with first instalment of Rs 1000 and increases the instalment every month by Rs 100, so the numbers involved in instalments are 1000, 1100, 1200, …
These numbers form an AP with a = 1000 and d = 100.
30th term = a + 29d = 1000 + 29 x 100 = 3900.
Hence, his 30th instalment = Rs 3900
Sum of 30 terms = (100 + 3900)
= 15 x 4900 = 73500.
∴ The total amount of loan repaid = Rs 73500.
∴ Amount of loan still to be paid = Amount of total loan – amount of loan repaid
= Rs 118000 – Rs 73500 = Rs 44500.
In ∆ABC, ∠C = 90°
∴ AB² = AC² + BC²
In AECD, ∠ECD = 90°
∴ DE² = CD² + EC²
In ∆AEC, ∠ACE = 90°
∴ AE² = AC² + EC²
In ∆BCD, ∠BCD = 90°
∴ BD² = BC² + CD²
Adding (iii) and (iv), we get
AE² + BD² = (AC² + BC²) + (CD² + EC²)
= AB² + DE²
Given, DB ⊥ BC, AC ⊥ BC
=> ∠DBC = ∠ACB = 90°
∴ AC || DB(sum of interior ∠s on the same side of BC is 180°)
∠ABD = ∠BAC (alt ∠s)
Now, in ABED and AACB,
∠EBD = ∠ABD = ∠BAC
∠C = ∠E (each = 90°)
∴ ∆BED ~ ∆ACB (AA similarity)
Since tangents drawn from an external point to a circle are equal.
∴BF = BD = 4 cm (given)
and CE = CD = 3 cm (given)
Let AF = AE = x cm
Now, area (∆ABC) = area (∆AOB) + area (∆BOC) + area (∆AOC)
21 = AB x OF + BC x OD + AC x OE
=> 21 = (AF + BF) x 2 + (BD + CD) x 2 + (AE + CE) x 2
=> 21 = (x + 4) + (4 + 3) + (x + 3)
=> 21 = 2x + 14 => 2x = 7 => x = =>x = 3.5
AB = x + 4 = (3.5 + 4) cm = 7.5 cm
and AC = x + 3 = (3.5 + 3) cm = 6.5 cm.
Given BC = R = 7 cm
and AE = CF = cm = r (say)
The perimeter of the shaded region
= perimeter of semicircle APB + perimeter of semicircle BSD + perimeter of semicircle DQC + perimeter of semicircle CRA
Total possible outcomes = 6 x 6 = 36.
(i) The possible outcomes favourable to the event ‘sum of numbers on two dice is 5’ are (2, 3), (3, 2), (1, 4), (4, 1) i.e. 4 in number.
∴ Required probability =
(ii) The possible outcomes favourable to the event ‘even numbers on both dice’ are (2, 2), (2, 4), – (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) i.e. 9 in number.
∴ Required probability =
We know that the sum of probabilities of all elementary events = 1
=> P(red ball) + P(blue ball) + P(orange ball) = 1
=> x = 24.
Hence, the total number of balls in the jar = 24.
Let A be the first term and D be the common difference of the AP.
a = pth term => a = A + (p – 1)D ….(i)
b = qth term => b = A +(q – 1)D ….(ii)
c = rth term => c = A + (r – 1)D ….(iii)
Subtracting (ii) from (i), we get
a – b = (p – q) D ….(iv)
Similarly, b – c = (q – r) D ….(v)
and c – a = (r – p) D ….(vi)
∴ (a – b) r + (b – c) p + (c – a) q = [(p – q) r + (q – r) p + (r – p)q] D
= (pr – qr + pq + qr – pq) D
= 0 x D = 0.
Volume of type A = Volume of cone + Volume of hemisphere
The given equations can be written as
Select the coordinate axes. Take 1 cm = 1 unit on both axes.
Plot the points (0, 2) and (3, 1) on the graph paper and join these points by a straight line. Plot the points (3, -2) and (6, 0) on the same graph paper and join these points by a straight line.
The given lines intersect at the point P(6, 0). Therefore, the solution of the given pair of linear equations is x = 6, y = 0.
The area bounded by the given lines and y-axis has been shaded.
From the figure AB = 6 units and OP = 6 units
The area of shaded region = area of ∆ABP
= x AB x OP = x 6 x 6
= 18 sq. units
Given vertices of ∆ABC as A (a, b), B (b, c), C (c, a) and the centroid of ∆ABC is G(0, 0).
As we know that centroid of A whose vertices are (x1 y1), (x2, y2) and (x3, y3)
Let AB be a tower 100 m high. P and Q be the position of two cars.
Angles of depressions are shown in adjoining figure.
∴∠APB = 90° and ∠AQB = 45°.
In ∆APB, ∠ABP = 90°
Given: P is an external point to a circle with centre O. PA and PB are two tangents drawn from P to the circle, A and B being points of contact.
To prove: PA = PB
Construction: Join OA, OB and OP.
Proof: Since the tangent at any point of a circle is perpendicular to the radius through the point of contact.
∠OAP = 90° and ∠OBP = 90°
Now, in ∆OAP and ∆OBP,
OA = OB (radii of same circle)
OP = OP (common)
∠OAP = ∠OBP (each 90°)
As ∆PQR ~ ∆ABC, so the side PQ of ∆PQR is the corresponding side AB of ∆ABC.
Given AB = 6 cm and we need PQ to be 8 cm, so
Thus, we are required to construct ∆PQR similar to ∆ABC such that sides of ∆PQR are times of the corresponding sides of ∆ABC.
Steps of construction:
1. Draw AB = 6 cm.
2. With A as centre and radius 6 cm, draw an arc.
3. With B as centre and radius 5 cm, draw an arc to meet the previous arc at C.
4. Join AC and BC, then ABC is an isosceles triangle with AB = AC = 6 cm and BC = 5 cm.
5. Take points A and P same.
6. Draw any ray AX making an acute angle with AB on the side opposite to the vertex C.
7. Locate 4 (the greater of 4 and 3) points A1, A2, A3 and A4 on AX such that AA1 = A1A2 = A2A3 = A3A4.
8. Join A3B.
9. Through A4, draw a line parallel to A3B (making an angle equal to ∠AA3B) to intersect the extended line segment AB at Q.
10. Through Q, draw a line parallel to BC (making an angle equal to ∠ABC) to intersect the extended line segment AC at R.
Then AQR i.e. PQR is the required triangle.
Construct the cumulative frequency distribution table as under:
We hope the CBSE Sample Papers for Class 10 Maths Paper 4 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 4, drop a comment below and we will get back to you at the earliest.