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 12.

## CBSE Sample Papers for Class 10 Maths Paper 12

Board | CBSE |

Class | X |

Subject | Maths |

Sample Paper Set | Paper 12 |

Time Allowed | 3 hours |

Maximum Marks | 80 |

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 12 of Solved CBSE Sample Paper for Class 10 Maths is given below with free PDF download solutions.

**General Instructions:**

- All questions are compulsory.
- This questions paper consists of 30 questions, distributed in four sections – A, B, C and D.
- Section A contains six questions of 1 mark each;

Section B contains six questions of 2 marks each;

Section C contains ten questions of 3 marks each;

and Section D contains eight 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 in three questions of 4 marks each. You have to attempt only one of the two alternatives given in these questions.
- Use of calculators is not permitted.

**SECTION – A**

**Question 1.**

Write whether \(\frac { 2\surd 45+3\surd 20 }{ 2\surd 5 }\) on simplification gives an irrational or a rational number.

**Solution:**

**Question 2.**

If x = a, y = b is the solution of the pair of equations x – y = 2 and x + y = 4, find the values of a and b.

**Solution:**

As x = a, y = b is a solution of x – y = 2 and x + y = 4

we have

a – b = 2 and a + b = 4

On solving these two equations in a and b, we get

a = 3 and b = 1

**Question 3.**

If one root of 5x^{2} + 13x + k = 0 is the reciprocal of the other root, then find value of k.

**Solution:**

**Question 4.**

**Solution:**

**Question 5.**

Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their surface areas.

**Solution:**

Let x and y be the lengths of the edges to the two cubes. Then,

x^{3} : y^{3} = 1 : 27

⇒ x : y = 1 : 3

So, 6x^{2} : 6y^{2} = 6 : 54 i.e. 1 : 9 [Surface areas ot cubes are 6x^{2} and 6y^{2}]

**Question 6.**

A (5, 1); B (1, 5) and C (-3, -1) are the vertices of ∆ABC. Find the length of median AD.

**Solution:**

**SECTION – B**

**Question 7.**

Given that √3 is an irrational number, prove that (2 + √3) is an irrational number.

**Solution:**

Let, on the contrary, (2 + √3) is a rational number.

Then, 2 + √3 = r, r is a rational number

⇒ √3 = r – 2

As (r – 2) is a rational number, we get √3 as a rational number.

This is in contradiction to the given fact that √3 is irrational.

Hence, our assumption is wrong.

So, (2 + √3) is irrational.

**Question 8.**

X is a point on the side BC of ∆ABC. XM and XN are drawn parallel to AB and AC respectively meeting AB in N and AC in M. MN produced meets CB produced at T. Prove that TX^{2} = TB x TC

**Solution:**

**Question 9.**

In Fig. (1), ABC is a triangle in which ∠B = 90°, BC = 48 cm and AB = 14 cm. A circle is inscribed in the triangle, whose centre is O. Find radius r of in-circle.

**Solution:**

**Question 10.**

Find the linear relation between x and y such that P(x, y) is equidistant from the points A(1, 4) and B(-1, 2).

**Solution:**

As P(x, y) is equidistant from A(1, 4) ed B (-1, 2),

AP² = BP²

or AP² = BP²

i.e. (x – 1)² + (y – 4)² = (x + 1)² + (y – 2)²

i.e. x² – 2x + 1 + y² – 8y + 16 = x² + 2x + 1 + y² – 4y + 4

i.e. 4x + 4y – 12 = 0

i.e. x + y – 3 = 0

This is the required linear equation between x and y

**Question 11.**

A, B, C are interior angles of ∆ABC. Prove that cosec (\(\frac { A+B }{ 2 }\)) = sec \(\frac { C }{ 2 }\)

**Solution:**

**Question 12.**

A right circular cylinder and a cone have equal bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the ratio between radius of their bases to their heights is 3 : 4.

**Solution:**

Let ‘r’ cm be the base radius and ‘h’ cm be the height.

Then Curved surface area of the cylinder = 2πrh

**SECTION – C**

**Question 13.**

Using Euclid’s division algorithm find the HCF of the numbers 867 and 225.

**Solution:**

Since 867 > 255, we apply the division lemma to 867 and 255, to get

867 = 255 x 3 + 102

Since the remainder 102 ≠ 0, we apply the division lemma to 255 and 102, to set

255 = 102 x 2 + 51

Since the remainder 51 ≠ 0, we apply the division lemma to 102 and 51, to set

102 = 51 x 2 + 0

The remainder has now become zero, so our procedure stops.

Since the division at this stage is 51, the HCF of 867 and 255 is 51.

**Question 14.**

Divide 27 into two parts such that the sum of their reciprocals is \(\frac { 3 }{ 20 }\)

**Solution:**

Let x and (27 – x) be two parts of 27. So,

**Question 15.**

In an A.P., if the sum of its first n terms is 3n² + 5n and its k^{th} term is 164, find the value of k.

**Solution:**

**Question 16.**

If the coordinates of two adjacent vertices of a parallelogram are (3, 2), (1, 0) and diagonals bisect each other at (2, -5), find the coordinates of the other two vertices.

**OR**

If the area of triangle with vertices (x, 3), (4, 4) and (3, 5) is 4 square units, then find x.

**Solution:**

**Question 17.**

In Fig. 3 AB is a chord of length 8 cm of a circle of radius 5 cm. The tangents to the circle at A and B intersect at P.

Find the length of AP.

**OR**

Prove that the lengths of tangents drawn from an external point to a circle are equal.

**Solution:**

**Proof:**

Let O be the centre of a circle, a point P lying outside the circle and two tangents PQ, PR on the circle from P (see figure).

We are required to prove PQ = PR.

For this, we join OP, OQ and OP. Then,

∠OQP and ∠ORP are right angles.

Now, in ∆s OQP and ORP,

OQ = OR (Radii of the same circle)

OP = OP (Common)

By RHS congruence criterian,

∆OQP = ∆ORP

⇒ PQ = PR

**Question 18.**

Construct a triangle with sides 6 cm, 8 cm and 10 cm. Construct another triangle whose sides are \(\frac { 3 }{ 5 }\) of the corresponding sides of original triangle.

**Solution:**

**Question 19.**

Prove that

**Solution:**

**Question 20.**

The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled by their tips in 48 hours.

**OR**

The side of a square is 10 cm. Find the area between inscribed and circumscribed circles of the square.

**Solution:**

We know that the long hand of a clock completes one round in 1 hour

distance covered in 1 round = 2π (6) cm = 12π cm

⇒ distance covered in 48 hours = 12π x 48 i.e. 576π cm

Also, short hand of a clock complete one round in 12 hours.

So, in 48 hours, this hand covers 4 rounds.

Thus, distance covered in 48 hours = 4 x 2π (4) = 32π cm

Thus, the total distance travelled by their tips = (576π + 32π) cm = 608π cm

**Question 21.**

If sin (A + 2B) = \(\frac { \surd 3 }{ 2 }\) and cos (A + 4B) = 0, A > B, and A + 4B ≤ 90°, then find A and B.

**Solution:**

As, sin (A + 2 B) = \(\frac { \surd 3 }{ 2 }\), and cos (A + 4B) = 0

A + 2B = 60° and A + 4B = 90°

Solving the two, we get

A = 30° and B = 15°

**Question 22.**

By changing the following frequency distribution to less than type distribution, draw its ogive.

**Solution:**

**SECTION – D**

**Question 23.**

For what values of m and n the following system of linear equations has infinitely many solutions.

3x + 4y = 12

(m + n) x + 2 (m – n) y – 5m – 1

**Solution:**

The given system of equations has infinitely many solutions, if

**Question 24.**

Obtain all zeroes of 3x^{4} – 15x^{3} + 13x^{2} + 25x – 30, if two of its zeroes are \(\surd \frac { 5 }{ 3 }\) and –\(\surd \frac { 5 }{ 3 }\)

**Solution:**

**Question 25.**

A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of slower train is 10 km/h less than that of faster train, find the speeds of two trains.

**OR**

Solve for x

**Solution:**

Let the speed of faster train be x km/h.

Then, speed of the slower train is (x – 10) km/h.

Time taken by the faster train in covering a journey of 200 km = \(\frac { 200 }{ x }\) hours

Time taken by the slower train in covering a journey of 200 km = \(\frac { 200 }{ x-10 }\)

According to the question,

**Question 26.**

Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

**Solution:**

**Question 27.**

The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of depression from the top of tower to the foot of hill is 30°. If tower is 50 metres high, find the height of the hill.

**OR**

Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point in between them on the road, the angles of elevation of the top of poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.

**Solution:**

**Question 28.**

A man donates 10 aluminum buckets to an orphanage. A bucket made of aluminum is of height 20 cm and has its upper and lowest ends of radius 36 cm and 21 cm respectively. Find the cost of preparing 10 buckets if the cost of aluminum sheet is ₹ 42 per 100 cm². Write your comments on the act of the man.

**Solution:**

**Question 29.**

Find the mean and mode for the following data:

**Solution:**

**Question 30.**

A box contains cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that number on the drawn card is

(i) a prime number.

(ii) a composite number.

(iii) a number divisible by 3.

**OR**

The king, Queen and Jack of clubs are removed from a pack of 52 cards and then the remaining cards are well shuffled. A card is selected from the remaining cards. Find the probability of getting a card

(i) of spade.

(ii) of black king.

(iii) of club.

(iv) of jack.

**Solution:**

(i) There are 8 prime numbers in between 1 to 20.

So, P (a prime number) = \(\frac { 8 }{ 20 }\) i.e. \(\frac { 2 }{ 5 }\)

(ii) There are 11 composite numbers in between 1 to 20.

So, P (a composite number) = \(\frac { 11 }{ 20 }\)

(iii) There are 6 multiples of 3 in between 1 to 20.

So, P (a multiple of 3) = P (a number divisible by 3) = \(\frac { 6 }{ 20 }\) i.e. \(\frac { 3 }{ 10 }\)

**OR**

Total number of cards in the experiment = 49

Of 49 cards, 13 cards are of spade, hearts and diamonds each; and 10 cards of club.

So,

(ii) P (a black king) = \(\frac { 13 }{ 49 }\)

(iii) P (a club) = \(\frac { 1 }{ 49 }\)

(iv) P (a jack) = \(\frac { 3 }{ 49 }\)

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