## CBSE Sample Papers for Class 10 Maths Paper 10

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.

**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 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.

❖ Use of calculator is not permitted.

**SECTION – A**

**Question 1.**

Find the distance of the point P(-6, 8) from the origin.

**Question 2.**

∆ABC is similar to ∆DEF. Area of ∆ABC is nine-sixteenth the area of ∆DEF. If perimeter of ∆DEF is 24 cm then find the perimeter of ∆ABC.

**Question 3.**

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

**Question 4.**

Find the sum and product of zeroes of the polynomial 7x² + 5x – 2.

**Question 5.**

Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.

**Question 6.**

∆ABC is right angled at A and ∆PQR is right angled at P. If cos B = cos Q, show ∠B = ∠Q.

**SECTION – B**

**Question 7.**

Find the fourth vertex D of a parallelogram ABCD whose three vertices are A(-2, 3), B(6, 7) and C(8, 3).

**Question 8.**

Show that 5 – √3 is irrational.

**Question 9.**

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to drawn. Find the probability that the card drawn is :

(i) A face card

(ii) A red face card

(iii) ‘2’ of spades

**Question 10.**

A jar contains 54 marbles each of which is blue, green or white. The probability of selecting a blue marble at random from the jar is \(\frac { 1 }{ 3 }\), and the probability of selecting a green marble at random is \(\frac { 4 }{ 9 }\) . How many white marbles does the jar contain ?

**Question 11.**

If the product of zeroes of the polynomial ax² – 6x – 6 is 4, find the value of ‘a’.

**Question 12.**

Solve for x and y : \(\frac { x }{ 2 }\) + \(\frac { y }{ 4 }\) = 3, 2x – y = 4.

**SECTION – C**

**Question 13.**

In what ratio does the X-axis divide the line segment joining the points (-4, -6) and (-1, 7) ? Also find the coordinates of the point of division.

**Question 14.**

Points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in 5 equal parts. Find the coordinates of the point P, Q and R.

**OR**

If the points A(-1, -4), B(b, c) and C(5, -1) are collinear and 2b + c = 4. Find the value of b and c.

**Question 15.**

In the given figure, a circle is inscribed in ∆PQR with PQ = 10 cm, QR = 8 cm and PR = 12 cm. Find the length QM, RN and PL.

**Question 16.**

Find the largest number which can divide 1002 and 1288 leaving remainder 1 in each case.

**Question 17.**

Find the area of the circle in which a square of area 64 cm^{2} is inscribed. (Use π = 3.14)

**OR**

A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, find the area of the road.

**Question 18.**

How many cubic centimetres of iron is required to construct an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm provided the thickness of the iron is 1.5 cm. If one cubic cm of iron weighs 7.5 g, find the weight of the box.

**OR**

The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

**Question 19.**

Solve for x and y : \(\frac { 44 }{ x + y }\) + \(\frac { 30 }{ x – y }\) = 10 and \(\frac { 55 }{ x + y }\) +\(\frac { 40 }{ x – y }\) = 13, where x ≠ y and x ≠ -y.

**OR**

Solve for x and y : 1001x + 999y = 2; 999x + 1001y = -2.

**Question 20.**

Determine the values of a and b for which the given system of equations has infinitely many solutions : (2a – 1) x + 3y – 5 = 0 and 3x + (b – 1) y – 2 = 0.

**Question 21.**

If cosec 3A = sec (A – 36°), where 3A is an acute angle, find the value of A.

**Question 22.**

Find the mode of the following data:

Classes |
Frequency |

140-145 | 3 |

145-150 | 7 |

150-155 | 10 |

155-160 | 7 |

160-165 | 6 |

165-170 | 2 |

170-175 | 3 |

**SECTION – D**

**Question 23.**

Draw a circle of radius 3 cm. Take a point P at a distance of 5.5 cm from the centre of the circle. From point P, draw tangents to the circle. Write steps of construction.

**Question 24.**

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

**Question 25.**

The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

**Question 26.**

A cylindrical bucket of height 32 cm and base radius 18 cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

**OR**

A wooden toy rocket is in the shape of a cone mounted on a cylinder. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Use π = 3.14)

**Question 27.**

An express train covers a distance of 240 km at a particular speed. Another train whose speed is 12 km/hr less takes one hour more to cover the same distance. Find the speed of the express train.

**OR**

A shopkeeper buys a number of books for ₹ 80. If he had bought 4 more books for the same amount, each book would have cost him ₹ 1 less. How many books did he buy ?

**Question 28.**

Mr. Singh repays his total loan of ₹ 118000 by paying every month starting with the first instalment of ₹ 1000. If he increases the instalment by ₹ 100 every month, what amount will be paid by him in the 30th instalment ? What amount of loan does he still have to pay after the 30th instalment ?

**OR**

A thief runs with a uniform speed of 100 m/minute. After one minute, a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief ?

**Question 29.**

The angle of elevation of the top of a chimney from the foot of a tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40 m, find the height of the chimney.

**Question 30.**

The mean of the following frequency distribution is 62.8 and the sum of all frequency is 50. Compute the missing frequencies f_{1} and f_{2}:

Classes |
Frequency |

0-20 | 5 |

20-40 | f_{1} |

40-60 | 10 |

60-80 | f_{2} |

80-100 | 7 |

100-120 | 8 |

Total | 50 |

**SOLUTIONS**

**SECTION – A**

**Solution 1.**

Given, point P(-6, 8) and origin O(0, 0)

Using distance formula

**Solution 2.**

**Solution 3.**

**Solution 4.**

Given polynomial is 7x² + 5x – 2 = 0

On comparing with ax² + bx + c = 0,

we get,

a = 7, b = 5, c = -2

Sum of zeroes = \(\frac { -b }{ a }\) = \(\frac { -5}{ 7 }\)

Product of zeroes = \(\frac { c }{ a }\) = \(\frac { -2}{ 7 }\)

**Solution 5.**

Given A.P. is 2x, x + 10, 3x + 2.

Common difference between two consecutive terms of A.P. is same.

b – a = c – b

⇒ 2b = a + c

⇒ 2(x + 10) = 2x + 3x + 2

⇒ 2x + 20 = 5x + 2

⇒ 20 – 2 = 5x – 2x

⇒ 18 = 3x

⇒ x = 6.

**Solution 6.**

Given : ∠A = ∠P = 90° ……(i)

and

cos B = cos Q

⇒ \(\frac { AB }{ BC }\) = \(\frac { PQ }{ QR }\) ……(ii)

From equation (i) and (ii),

∆ABC ~ ∆DEF [S.A.S.]

∠B = ∠Q.

Hence Proved.

**SECTION – B**

**Solution 7.**

Given, coordinates of parallelogram ABCD are A(-2, 3), B(6, 7) and C(8, 3)

Let the coordinates of D be (α, β).

We know that diagonals of parallelogram bisect each other.

So, mid-point of AC = mid-point of BD

Coordinates of vertex D = (0, -1).

**Solution 8.**

Let us assume on the contrary that 5 – √3 is rational number. Then there exist co-prime positive integers x and y such that

5 – √3 = \(\frac { x }{ y }\)

5 – \(\frac { x }{ y }\) = √3

\(\frac { 5y – x }{ y }\) = √3

Since x, y are integers, so \(\frac { 5y – x }{ y }\) is a rational number

√3 is rational.

This contradicts the fact that √3 is irrational. So, our assumption is incorrect. Hence, 5 – √3 is an irrational number.

Hence Proved.

**Solution 9.**

Given :

Total cards = 52

(i) We know that there are 12 face cards

P(a face cards) = \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\)

(ii) We know that there are 6 red face cards

P(Red face cards) = \(\frac { 6 }{ 52 }\) = \(\frac { 3 }{ 26 }\)

(iii) There is only one card of ‘2’ of spades

P(2 of spade) = \(\frac { 1 }{ 52 }\)

**Solution 10.**

We have,

Total marbles = 54

P(blue marble) = \(\frac { 1 }{ 3 }\)

P(green marble) = \(\frac { 4 }{ 9 }\)

Let number of white marbles be x.

**Solution 11.**

Given polynomial is,

ax² – 6x – 6

On comparing with Ax² + Bx + C = 0, we get

A = a, B = -6, C = -6

Product of roots = \(\frac { C }{ A }\)

4 = \(\frac { -6 }{ a }\)

a = \(\frac { -6 }{ 4 }\) = \(\frac { -3 }{ 2 }\)

Value of a = \(\frac { -3 }{ 2 }\)

**Solution 12.**

We have,

**SECTION – C**

**Solution 13.**

**Solution 14.**

We have, points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) into five equal parts.

**Solution 15.**

Given, PQ = 10 cm, PR = 12 cm and QR = 8 cm.

Let the length of QM be x cm

Thus,

RM = (8 – x) cm

and QL = x cm [LQ and QM are tangents from Q]

RN = (8 – x) cm [RN and RM are tangents from R]

Now, PN = PR – RN = [12 – (8 – x)] cm = (4 + x) cm

and PN = PL [PL and PN are tangents from P]

Again, QL = PQ – PL = [10 – (4 + x)] cm = (6 – x) cm

But QL = x cm [From above]

x = 6 – x

⇒ 2x = 6

⇒ x = 3 cm

QM = 3 cm

RN = (8 – 3) cm = 5 cm

and PL = (4 + 3) cm = 7 cm.

**Solution 16.**

Let x be the required number which divides 1002 and 1288 leaving remainder 1 each case.

This means x perfectly divide (1002 – 1) and (1288 – 1) i.e., 1001 and 1287.

So, x is the HCF of 1001, 1287

Using Euclid’s division alogrithm,

1287 = 1001 x 2 + 286

1001 = 286 x 3 + 143

286 = 143 x 2 + 0

Hence, HCF of 1001 and 1287 is 143.

**Solution 17.**

Given : Area of square = 64 cm²

Let side of square and radius of circle be a cm and r cm, respectively.

Now, we know

**Solution 18.**

Given:

External length of the box = 36 cm

External breadth of the box = 25 cm

External height of the box = 16.5 cm

Width of the box = 1.5 cm

Internal length of the box = 36 – 1.5 = 34.5 cm [Internal length = External length – Width]

Internal breadth of the box = 25 – 1.5 = 23.5 cm

Internal height of the box = 16.5 – 1.5 = 15 cm

Volume of iron in the box = External volume of the box – Internal volume of the box

= 36 x 25 x 16-5 – 34.5 x 23.5 x 15

= 14850 – 12161.25

= 2688.75 cm^{3}

1 cm^{3} iron = 7.5 gram

2688.75 cm^{3} iron = 7.5 x 2688.75 gram

= 20165.625 gram.

**OR**

Let the radii of circular ends of frustum be R cm and r cm and its slant height be 1 cm.

Then, 2πR = 18 cm ⇒ πR = 9 cm

and 2πr = 6 cm ⇒ πr = 3 cm

Also, l = 4 cm

Curved surface area of the frustum = πl(R + r) = l(πR + πr) = 4(9 + 3) = 48 cm^{2}.

**Solution 19.**

Given equations are,

**Solution 20.**

Given system of equation is,

(2a – 1)x + 3y – 5 = 0 …..(i)

and 3x + (b – 1)y – 2 = 0 …..(ii)

**Solution 21.**

We have,

cosec 3A = sec (A – 36°)

sec (90° – 3A) = sec (A – 36°) [cosec A = sec (90° – A)]

90° – 3A = A – 36°

90° + 36° = A + 3A

4A = 126°

A = 31.5°.

**Solution 22.**

Classes |
Frequency |

140-145 | 3 |

145-150 | 7 |

150-155 | 10 |

155-160 | 7 |

160-165 | 6 |

165-170 | 2 |

170-175 | 3 |

**SECTION – D**

**Solution 23.**

**Steps of construction :**

1. Draw a circle of radius 3 cm taking O as centre.

2. Draw a point P at a distance of 5.5 cm from the centre and join OP.

3. Draw the perpendicular bisector of OP, intersecting OP at point M.

4. Taking MP as radius and M as centre, draw a circle which cut the given circle at two points A and B.

5. Join A to P and B to P.

Thus, AP and BP are required tangents.

**Solution 24.**

Let P be a point outside a circle from which two tangents PA and PB are drawn.

To prove : PA = PB.

Proof : In ∆PBO and ∆AOP,

OP = OP [Common]

OB = OA [Radius of the circle]

∠OBP = ∠OAP

[Both 90°, any point joined from centre to point of contact of tangent is perpendicular]

By R.H.S. congruency rule,

∆PBO = ∆AOP

PA = PB [cpct]

So, tangents are equal when drawn from an exterior point to the circle.

Hence Proved.

**Solution 25.**

Let AB be the tower of height ‘h’ m, BC be the shadow of tower when sun’s altitude is 60° and BD be the shadow of tower when sun’s altitude is 30°.

**Solution 26.**

Given:

Height of cylindrical bucket, H = 32 cm

Radius of cylindrical bucket, R = 18 cm

and Height of cone, h = 24 cm

Let r be the radius of cone.

According to the question,

Volume of cone = Volume of cylinder

\(\frac { 1 }{ 3 }\) πr²h = πR²H

**Solution 27.**

Given:

Let Total distance to be covered = 240 km

Speed of Express train = x km/h

Speed of another train = (x – 12) km/h

**Solution 28.**

Let total installment paid be ‘n’.

Since, the amount of each installment increases by ₹ 100 every month, so installments are paid in A.P.

Given : Total loan, S_{n} = 118000

Amount of first installment, a = 1000

Increase in amount of each installment, d = 100

Amount to be paid in 30th installment = a_{30}

We know that

a_{n} = a + (n – 1 )d

a_{30} = 1000 + (30 – 1)100 = 1000 + 2900

a_{30} = 3900

So, amount to be paid in 30th installment is ₹ 3900.

Total amount paid in 30th installment = \(\frac { 30 }{ 2 }\) [2 x 1000 + 29 x 100]

= 15[2000 + 2900] = 15 x 4900 = 73500

Amount left to be paid = 118000 – 73500

= ₹ 44500.

**OR**

Let total time be n minutes.

Since, policeman runs after 1 minutes, so he will catch the thief in (n – 1) minutes.

Total distance covered by thief = 100 m/minute x n minute =(100n) m

Now, total distance covered by the policeman = (100) m + (100 + 10) m + (100 + 10 + 10) m +…+ (n – 1) terms

= 100 + 110 + 120 +…+ (n – 1) terms

⇒ n^{2} – 6n + 3n – 18 = 0

⇒ n(n – 6) + 3(n – 6) = 0

⇒ (n + 3) (n – 6) = 0

n = 6 or n = – 3 (Neglect)

Hence, policeman will catch the thief in (6 – 1) i.e., 5 minutes.

**Solution 29.**

Let AB be the tower and CD be the chimney of height h m.

AB = 40 m CD = h m

Let the distance between the foot of the tower and the foot of the chimney be x m.

**Solution 30.**

Let the assumed mean be A = 50 and class size (h) = 20.

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