These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 10. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.
CBSE Sample Papers for Class 10 Maths Paper 10
Board | CBSE |
Class | X |
Subject | Maths |
Sample Paper Set | Paper 10 |
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 10 of Solved CBSE Sample Paper for Class 10 Maths is given below with free PDF download solutions.
Time Allowed: 3 hours
Maximum Marks: 80
General Instructions:
- All questions are compulsory.
- The 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. Section D contains 8 questions of 4 marks each.
- There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
- Use of calculators is not permitted.
SECTION-A
Without actually performing long division, write down the decimal expansion of
\(\cfrac { 23 }{ 200 } \)
2.
In the following AP, find the missing term in the box: 2, □, 26.
3.
If θ is an acute angle and sin θ = cos θ, find the value of 2 tan2 θ + sin2 θ-1.
4.
In the given figure, DE || BC. If AB = 6 cm, AD = 4 cm, EC = 3 cm, find AC.
If \(\cfrac { 1 }{ 2 } \) is a root of the equation x2 + kx -< \(\cfrac { 5 }{ 4 } \) = 0, then find the value of k.
What is the area of the triangle formed by the points O< (0, 0), A< (-3, 0) and B< (5, 0) ?
<SECTION-B
7.
If A and B are (-2, – 2) and (2, – 4) respectively, find the coordinates of P such that AP = <\(\cfrac { 3 }{ 7 } \) AB and P lies on the line segment AB.
8.
Three unbiased coins are tossed together, find the probability of getting
(1) at least two heads
(2) at most two heads
9.
Without actually performing the long division, find if <\(\cfrac { 987 }{ 10500 } \) will have terminating or non-terminating repeating decimal expansion. Give reason for your answer.
10.
Find out whether the line representing the following pair of linear equations intersect at a point, are parallel or coincident. <\(\cfrac { 4 }{ 3 } \) x + 2y = 8; 2x + 3y = 12
11.
The sum of the first n terms of an AP is 3n2 + 6n. Find the nth term of this AP.
12.
A letter is chosen at random from the letters of the word ‘UNIVERSAL’. Find the probability the letter chosen is not a vowel.
13.
Prove that 3 + 2 \( \sqrt { 5 } \) is an irrational number.
OR
Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
14.
Check graphically whether the pair of equations x + 3y = 6 and 2x-3y = 12 is consistent. If so, solve them graphically.
15.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:(-1,-2), (1, 0), (-1, 2), (-3, 0)
OR
A median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, -6), B (3, -2) and C (5, 2).
16.
In an equilateral triangle ABC, D is a point on side BC such that BD = <\(\cfrac { 1 }{ 3 } \) BC. Prove that 9AD2 = 7AB2.
17.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D into lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.
18.
In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
19.
On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x-2 and – 2x + 4 respectively. Find g(x).
OR
If one zero of the polynomial 2x2 + 3x + λ is <\(\cfrac { 1 }{ 2 } \), find the value of A and other zero.
20.
Without using trigonometric tables, evaluate:
\(\cfrac { { Sec }^{ 2 }{ 36 }^{ o }-{ cot }^{ 2 }{ 54 }^{ o } }{ { cosec }^{ 2 }{ 57 }^{ o }-{ tan }^{ 2 }{ 33 }^{ o } } +2{ sin }^{ 2 }{ 22 }^{ o }{ 45 }^{ o }{ sec }^{ 2 }{ 68 }^{ o } \)
OR
If sin (A + B) = sin A cos B + cos A sin B, then find the value of sin 75°.
21.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of
र 500 per m2. (Note that the base of the tent will not be covered with canvas).
22.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students
SECTION—D
23.
The angles of elevation and depression of the top and bottom of a lighthouse from the top of a building, 60 m high, are 30° and 60° respectively. Find
(1) the difference between the heights of the lighthouse and the building.
(2) distance between the lighthouse and the building.
24.
Prove that ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Using the above result do the following:
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB – 2 CD, find the ratio of the areas of triangles AOB and COD.
25.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
OR
A 20 cm high metallic right circular cone whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter <\(\cfrac { 1 }{ 16 } \) cm, find the length of the wire.
26.
During the medical check-up of 35 students of a class, their weights were recorded as
Weight (in kg) | Number of students | Weight (in kg) | Number of students |
Less than 38 | 0 | Less than 46 | 14 |
Less than 40 | 3 | Less than 48 | 28 |
Less than 42 | 5 | Less than 50 | 32 |
Less than 44 | 9 | Less than 52 | 35 |
Draw a less than type give for the given data. Find out the median weight from the graph and verify the result by using the formula.
27.
One fourth of a group of people claim that they are creative, twice the square root of the group claim to be caring and the remaining 15 claim that they are optimistic. Find the total number of people in the group.
(a) How many persons in the group are creative?
(b) According to you, which one of the above three values is more important for development of a society?
28.
If the mth term of an AP is <\(\cfrac { 1 }{ n } \) and nth term is <\(\cfrac { 1 }{ m } \), then show that its (mn)th term is 1.
OR
In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the AP.
29.
Draw a circle of radius 6 cm from a point 10 cm away from its centre. Construct the pair of tangents to the circle and measure their lengths.
30.
Prove that: (cosec A – sin A) (sec A – cos A) = <\(\cfrac { 1 }{ tan A + cot A } \)
OR
Find the value of sin 45° geometrically.
Answers
SECTION-A
Answer 1.
Answer 2.
Answer 3.
Answer 4.
Answer 5.
Answer 6.
Area of ΔOAB = <\(\cfrac { 1 }{ 2 } \) [0(0 – 0) – 3(0 – 0) + 5(0 – 0)] = 0
⇒ Given points are collinear
Answer 7.
Answer 8.
The possible outcomes are:
(H,H,H), (H,H,T), (H,T,H), (T,H,H), (H,T,T), (T,H,T), (T,T,H) and (T,T,T) i.e., 8
(1) The outcomes favourable to the event A, at least two heads’ are (H,H,T), (H,T,H), (T,H,H), (H,H,H) i.e., 4
<Therefore,p(A) = \(\cfrac { Number of outcomes favourables to A }{ Number of all possible outcomes } \) = \(\cfrac { 4 }{ 8 } \) = \(\cfrac { 1 }{ 2 } \).
(2) The outcomes favourable to the event B, at most two heads are (H,H,T) (H,T,H), (T,H,H), (H,T,T), (T,H,T), (T,T,H), (T,T,T) i.e., 7
<Therefore,p(B) = \(\cfrac { Number of outcomes favourables to B }{ Number of all possible outcomes } \) = \(\cfrac { 7 }{ 8 } \)
Answer 9.
Answer 10.
Answer 11.
Given Sn=3n2+6n
Sn-1= 3(n – 1)2 + 6(n – 1) = 3(n2 + 1 – 2n) + 6n – 6
= 3n2 + 3-6n + 6n-6 = 3n2 – 3
The nth term will be an
Sn =Sn-1+ an
an = Sn– Sn-1 = 3n2 + 6n -3n2 + 3 = 6n + 3
Answer 12.
P (not a vowel) = 1 – P (vowel)
=1- <\(\cfrac { 4 }{ 9 } \) = \(\cfrac { 5 }{ 9 } \)
SECTION—C
Answer 13.
Let us assume, to the contrary, that 3 + 2 \( \sqrt { 5 } \) is a rational number. Such that
3 + 2 \( \sqrt { 5 } \) = <\(\cfrac { a }{ b } \) , where a and b are co-prime and b ≠ 0
Therefore <\(\cfrac { a }{ b } \) -3=2 \( \sqrt { 5 } \)
⇒ \(\cfrac { a }{ 2b } \) – \(\cfrac { 3 }{ 2 } \) < is a rational number.
But this contradicts the fact that \( \sqrt { 5 } \) is an irrational number. This contradiction has arisen because of our incorrect assumption that 3 + 2\( \sqrt { 5 } \) is a rational number. <So, we conclude that 3 + 2\( \sqrt { 5 } \) is an irrational number.
OR
Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q > 0 and 0 < r < 6
i.e., the possible remainders are 0,1, 2, 3,4, 5.
Thus, a can be of the form 6q, or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5, where q is some quotient.
Since a is odd integer, so a can not be of the form 6q, or 6q + 2, or 6q + 4, (since they are even).Thus, a is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Hence, any odd positive integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
Answer 14.
As the lines representing the pair of equations intersect each other at the point C(6, 0). Therefore, the given pair of equations is consistent. Solution: x – 6, y = 0
Answer 15.
Let A (-1, -2), B (1, 0), C (-1,2) and D (-3, 0) be the four given points.
Then, using distance formula, we have
Hence, four sides of quadrilateral are equal and diagonals AC and BD are also equal.
∴ Quadrilateral ABCD is a square.
OR
Since AD is the median of ΔABC therefore,
D is the mid-point of BC.
Here, area (Δ = area (ΔADC)
Hence, the median divides a triangle into two triangles of equal areas.
Answer 16.
Let ABC be an equilateral triangle and let D be a point on BC such that BD = <\(\cfrac { 1 }{ 3 } \) BC
To prove : 9AD2 = 7AB2
Constrution: Drew AE ⊥ BC.Join AD
Proof: ABC is an equilateral triangle and AE ⊥ DC.
Answer 17.
Let ΔABC be drawn to circumscribe a circle with centre O and radius 4 cm and circle touches the sides BC, CA and AB at D, E and F
We have given that
CD = 6 cm and BD = 8 cm
∴ BF = BD = 8cm {Length of two tangents drawn from an external point of circle are equal}
CE – CD = 6 cm
Now, let AF = AE = x cm
Now AB = (x + 8) cm; BC = 14 cm
CA = (x + 6) cm
2s = (x + 8) + 14 + (x + 6)
⇒ 2s = 2x + 28
⇒ s = x + 14
Answer 18.
Answer 19.
Answer 20.
Answer 21.
We have,
Radius of cylindrical base = <\(\cfrac { 4 }{ 2 } \) = 2 m
Height of cylindrical portion = 2.1 m
∴ Curved surface area of cylindrical portion = 2πrh
= 2 x <\(\cfrac { 22 }{ 7 } \) x 2 x 2.1 = 26.4 m2
Radius of conical base = 2 m
Slant height of conical portion = 2.8 m
.’. Curved surface area of conical portion = πrl
= <\(\cfrac { 22 }{ 7 } \) x 2 x 2.8 = 17.6 m2
Now, total area of the canvas = (26.4 + 17.6)m2 = 44 m2
.’. Total cost of the canvas used = र 500 x 44 = र 22,000
Answer 22.
SECTION—D
Answer 23.
Answer 24.
Answer 25.
Answer 26.
To represent the data in the table graphically, we mark the upper limits on the class interval on x-axis and their corresponding cumulative frequency on y-axis choosing a convenient scale.
Now, let us plot the points corresponding to the ordered pair given by (38,0), (40,3), (42,5), (44,9), (46,14), (48,28), (50,32) and (52,35) on a graph paper and join them by a freehand smooth curve.
Thus, the curve obtained is the less than type ogive.
Now, to locate y = <\(\cfrac { n }{ 2 } \) = <\(\cfrac { 35 }{ 2 } \) 17.5 on the y-axis,
we draw a line from this point parallel to x-axis cutting the curve at a point. From this point, draw a perpendicular line to the x-axis. The point of intersection of this perpendicular with the x-axis gives the median of the data. Here it is 46.5.
Let us make the following table in order to find median by using formula
Answer 27.
Hence, the number of people in the group = 36 (a) 9 persons
Value: All of these values have their own importance. A person having these values will certainly contribute to the development of society. However, the level of importance given to each of them depends upon a person’s own attitude. Hence, any value with justification is correct.
Answer 28.
Let a and d be the first term and common difference respectively of the given AP. Then
Answer 29.
Steps of Construction:
Step I: Take a point O and draw a circle of radius 6 cm.
Step II: Take a point P at a distance of 10 cm from Pf the centre O
Step III: Join OP and bisect it. Let M be the mid-point.
Step IV: With M as centre and MP as radius draw a circle to intersect the circle at Q and R
Step V: Join PQ and PR. Then, PQ and PR are the required tangents.
On measuring, we find, PQ = PR = 8 cm
Justification: On joining OQ we find that ∠PQO = 90°, as ∠PQO is the angle in the Semi-circle.
PQ⊥ OQ
Since OQ is the radius of the given circle, so PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle.
Answer 30.
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