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 11.

## CBSE Sample Papers for Class 10 Maths Paper 11

Board | CBSE |

Class | X |

Subject | Maths |

Sample Paper Set | Paper 11 |

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 11 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.**

If x = 3 is one root of the quadratic equation x² – 2kx – 6 = 0, then find the value of k.

**Solution:**

Given that x = 3 is one root of x² – 2kx – 6 = 0, we have

(3)² – 2k(3) – 6 = 0

This gives

9 – 6k – 6 = 0 or 6k = 3

⇒ k = \(\frac { 1 }{ 2 }\)

**Question 2.**

What is the HCF of smallest prime number and the smallest composite number?

**Solution:**

The smallest prime number = 2;

and The smallest composite number = 4.

So, the required HCF = HCF (2, 4) = 2

**Question 3.**

Find the distance of a point P(x, y) from the origin.

**Solution:**

The distance (d) of a point P(x, y) from the origin is given by

d = \(\sqrt { \left( { x-0 } \right) ^{ 2 }+\left( y-0 \right) ^{ 2 } }\)

i.e., \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } }\)

**Question 4.**

In an A.P., if the common difference (d) = – 4 and the seventh term (a_{7}) is 4, find the first term.

**Solution:**

Let the first term be V. Then

a_{7} = a + 6d

⇒ 4 = a + 6 (-4)

⇒ a = 4 + 24

⇒ a = 28

Thus, the first term is 28.

**Question 5.**

What is the value of (cos²67° – sin²23°) ?

**Solution:**

Here, cos²67° – sin²23°

= cos²(90° – 23°) – sin²23°

= sin²23° – sin²23°

= 0

**Question 6.**

Given ΔABC ~ ΔPQR, if \(\frac { AB }{ PQ }\) = \(\frac { 1 }{ 3 }\), then find \(\frac { ar\left( \Delta ABC \right) }{ ar\left( \Delta PQR \right)}\)

**Solution:**

We know that if ΔABC ~ ΔPQR, then

**SECTION – B**

**Question 7.**

Given that √2 is irrational, prove that (5 + 3√2) is an irrational number.

**Solution:**

Given that √2 is irrational, we have

3√2 is irrational.

and hence 5 + 3√2 is irrational.

**OR**

Let, on the contrary, (5 + 3√2) is an irrational number.

Then, 5 + 3√2 = \(\frac { a }{ b }\)

where a, b ∈ I and b ≠ 0.

Here, L.H.S is an irrational number and R.H.S a rational number. Thus, we arrives at a contradiction.

So, (5 + 3√2) is an irrational number.

**Question 8.**

In Fig. 1, ABCD is a rectangle. Find the value of x and y.

**Solution:**

As ABCD is a rectangle, we have

x + y = 30 and x – y = 14 cm

Solving the two equations simultaneously, we have

x + (x – 14) = 30

This gives

2x = 44 or x = 22

⇒ y = 8 (∴ x – y = 14)

**Question 9.**

Find the sum of first 8 multiples of 3.

**Solution:**

First 8 multiples of 3 are:

3, 6, 9, 12, 15, 18, 21, 24.

It is clearly an AP with first term (a) = 3; and common difference (d) = 3.

Hence,

sum of first 8 terms (S_{8}) = \(\frac { 8 }{ 2 }\) [2 x 3 + (8 – 1) (3)]

= 4 [6 + 21]

= 4 x 27, or 108

**Question 10.**

Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3). Hence, find m.

**Solution:**

**Question 11.**

Two different dice are tossed together. Find the probability:

i. of getting a doublet;

ii. of getting a sum 10, of the numbers on the two dice.

**Solution:**

When two different dice are tossed together, the sample space (S) consists of 36 elements,

i. of the 36 elements, 6 elements are doublets. Thus,

P(a doublet) = \(\frac { 6 }{ 36 }\) , i.e. \(\frac { 1 }{ 6 }\)

ii. of the 36 elements, following elements have sum 10

(4, 6), (5, 5), (6, 4)

Thus,

P(a sum of 10) = \(\frac { 3 }{ 36 }\) , i.e. \(\frac { 1 }{ 12 }\)

**Question 12.**

An integer is chosen from 1 and 100. Find the probability that it is:

i. divisible by 8;

ii. not divisible by 8

**Solution:**

Here, sample space S consists of 100 elements.

So, n(S) = 100

i. Let A be the event the number chosen is divisible by 8

A = {8, 16, 24, ………. , 96}

⇒ n(A) = 12

Now, P(a number divisible by 8) = \(\frac { n(A) }{ n(s) }\) = \(\frac { 12 }{ 100 }\) = \(\frac { 3 }{ 25 }\)

ii. P (a number not divisible by 8) = 1 – P(A) = 1 – \(\frac { 3 }{ 25 }\) = \(\frac { 22 }{ 25 }\)

**SECTION – C**

**Question 13.**

Find HCF and LCM of 404 and 96 and verify that

HCF x LCM = Product of the two given numbers.

**Solution:**

Here, the prime factorisation of 404 and 96 are:

404 = 2 x 2 x 101 or 2² x 101

96 = 2 x 2 x 2 x 2 x 2 x 3 or 2^{5} x 3.

Thus, HCF and LCM of 404 and 96 are

HCF = 2² i.e. 4; LCM = 25 x 3 x 101, i.e. 9696.

Now,

HCF x LCM = 4 x 9696, i.e. 38784

Product of 404 and 96 = 38784.

Thus,

HCF x LCM = Product of 404 and 96 is verified.

**Question 14.**

Find all zeroes of the polynomial (2x^{4} – 9x^{3} + 5x^{2} + 3x – 1) if two of its zeroes are (2 + √3 ) and (2 – √3).

**Solution:**

Since (2 + √3) and (2 – √3) are zeroes of the polynomial,

x – (2 + √3) and x – (2 – √3) are factors of the polynomial

⇒ [x – (2 + √3)][x – (2 – √3)] i.e. [(x – 2)^{2} – 3] i.e. (x^{2} – 4x + 1) is a factor of the polynomial.

Let us now divide the given polynomial by (x^{2} – 4x + 1) to get the other two zeroes from the quotient.

Further,

Quotient = 2x^{2} – x – 1 can be factorised set as.

2x^{2} – x – 1 = 2x^{2} – 2x + x – 1 = 2x (x – 1) + 1 (x – 1) = (2x + 1)(x – 1)

Thus, the other two zeroes are 1 and \(\frac { -1 }{ 2 }\).

**Question 15.**

If A (-2, 1), B (a, 0), C (4, b) and D(1, 2) are the vertices of a parallelogram ABCD. Find the values of a and b. Hence, find the lengths of the sides.

**OR**

If A (-5, 7), B (-4, -5), C (-1, -6) and D (4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral.

**Solution:**

**Question 16.**

A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

**Solution:**

Let the usual speed of the plane be ‘x’ km/h.

Then, the increased speed = (x + 100) km/h

Solving this quadratic equation, we have

3000 (x + 100) – x (x + 100) = 3000x

⇒ 3000x + 300000 – x^{2} – 100x = 3000x

⇒ x^{2} + 100x – 300000 = 0

⇒ (x + 600) (x – 500) = 0

⇒ x + 600 = 0 or x – 500 = 0

⇒ x = 500 (x ≠ -600)

Thus, the usual speed is 500 km/h

**Question 17.**

Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral described on one of its diagonal.

**OR**

If the area of two similar triangles are equal, prove that they are congruent.

**Solution:**

Let ‘a’ cm be the side of square ABCD. Then,

**Question 18.**

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

**Solution:**

Let PA and PB be two tangents drawn from P to the circle C (O, r).

Join OP, OA and OB

As OA ⊥ PA and OB ⊥ PB,

∠PAO = 90° = ∠PBO

OA = OB = r

OP = OP (Common)

Hence, by RHS congruence criterion,

∆PAO = ∆PBO

Consequently,

PA = PB.

**Question 19.**

If 4 tanθ = 3, evaluate \(\left( \frac { 4sin\theta -cos\theta +1 }{ 4sin\theta +cos\theta -1 } \right)\)

**OR**

If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

**Solution:**

Given 4 tanθ = 3, we have

sinθ = \(\frac { 3 }{ 5 }\) and cosθ = \(\frac { 4 }{ 5 }\).

OR

Given tan 2A = cot (A – 18°),

We have

tan 2A = tan [90° – (A – 18°)]

⇒ 2A = 90° – (A – 18°)

or 2A = 108° – A

or 3A = 108°

or A = 36°

**Question 20.**

Find the area of the shaded region in Fig. 7, where arcs drawn with centres A, B, C and D intersect in pairs at mid-point P, Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 12 cm (use π = 3.14).

**Solution:**

In Fig. 7,

quad. APS = quad. BPQ = quad. DSR = quad. CRQ

⇒ ar (quad. APS) = ar (quad. BPQ) = ar(quad. DSR) = ar(quad. CRQ)

= \(\frac { 1 }{ 4 }\) [π (6)^{2}] sq cm = 9π sq cm = 28.26 sq cm.

Thus, the area of the shaded region

= Area of square ABCD – 4 x ar (quad. APS)

= 12 x 12 – 4 x 28.26 = 144 – 113.04

= 30.96 sq cm.

**Question 21.**

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 8. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm. Find the total surface area of the article.

**OR**

A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

**Solution:**

From the question, we note that

Total surface area of the article = 2 x surface area of a hemisphere + curved surface area of the cylinder

= [2 x 2π (3.5)^{2} + 2π (3.5) (10)] sq cm.

= [49π + 70π] sq cm

= 119π sq cm, or 374 sq cm.

**OR**

**Question 22.**

The table below shows the salaries of 280 persons:

Salary(in ₹ 1000s) | No. of Persons |

5 – 10 | 49 |

10 – 15 | 133 |

15 – 20 | 63 |

20 – 25 | 15 |

25 – 30 | 6 |

30 – 35 | 7 |

35 – 40 | 4 |

40 – 45 | 2 |

45 – 50 | 1 |

Calculate the median salary of the data.

**Solution:**

**SECTION – D**

**Question 23.**

A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

**OR**

A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed?

**Solution:**

Let the speed of the stream be ‘x’ km/h

Given the motor boat speed in still water as 18km/h

upstream speed = (18 – x) km/h

and downstream speed = (18 + x) km/h

According to the question,

Time taken by boat in downstream = Time taken by boat in upstream – 1 hour

**Question 24.**

The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of two middle terms is 7 : 15. Find the numbers.

**Solution:**

Let a and d denote the first term and the common difference of an AP.

Then, following four terms are four consecutive numbers:

a – 3d, a – d, a + d, a + 3d

As per the question,

(a – 3d) (a + 3d) : (a – d)(a + d) = 7 : 15

**Question 25.**

In an equilateral triangle ∆ABC, D is a point on side BC such that BD = \(\frac { 1 }{ 3 }\) BC.

Prove that 9 (AD)^{2} = 7 (AB)^{2}.

**OR**

Prove that in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

**Solution:**

**Question 26.**

Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then, construct a triangle whose sides are \(\frac { 3 }{ 4 }\) of the corresponding sides of the ∆ABC.

**Solution:**

**Steps of construction:**

- Draw a line segment BC of length 6 cm.
- Draw a ray BX making 60° with BC.
- Draw an arc with radius 5 cm from B so that it cuts BX at A. Join AC to from ∆ABC
- Draw a ray BY making an acute angle with BC
- Locate B
_{1}, B_{2}, B_{3}, B_{4}on BY such that BB_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4} - Join B
_{4}C. Draw B_{3}C’ || B_{4}C - From C’, draw C’A’ || CA.

Then, ∆A’BC’ is the required triangle.

**Question 27.**

Prove that

**Solution:**

Here,

**Question 28.**

The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find:

i. The area of the metal sheet used to make the bucket

ii. Why should we avoid the bucket made by ordinary plastic?

**Solution:**

i. Areas of the metal sheet used to make the bucket (Use π = 3.14)

ii. Use of a plastic bucket is not suitable for carrying hot water.

**Question 29.**

As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exacdy behind the other on the same side of the light house, find the distance between the two ships [use √3 = 1.732]

**Solution:**

**Question 30.**

The mean of the following distribution is 18. Find the frequency ‘f’ of the class 19 – 21.

OR

The following distribution gives the daily income of 50 workers of a factory.

Convert the above distribution to less than type cumulative frequency distribution and draw its ogive.

**Solution:**

We hope the CBSE Sample Papers for Class 10 Maths Paper 11 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 11, drop a comment below and we will get back to you at the earliest.

## Leave a Reply