**Table Of Contents**

With our detailed notes, solved papers, and critical questions, you may improve your MATHEMATICS skills for CBSE 10th Class. These useful resources will help you improve your maths proficiency.

**Time allowed:3 Hours Maximum Marks:80 **

## SECTION – A

**1. If HCF (336, 54) = 6, find LCM (336, 54). [1]**

**Sol. **Given, HCF (336, 54) = 6

We know,

HCF x LCM = one number x other number

⇒ 6 x LCM = 336 x 54

⇒ LCM = 336 x 54 / 6

= 336 x 9

= 3024 **Ans.**

**2. Find the nature of roots of the quadratic equation [1]**

**2x**^{2}** – 4x + 3 = 0. **

**Sol. **Given, 2x^{2} – 4x + 3 = 0

Comparing it with quadratic equation

ax^{2} + bx + c = 0

Here, a = 2, b = -4 and c = 3

∴ D = b^{2} – 4ac

= (-4)^{2} – 4 x (2) (3)

= 16 – 24

= -8 < 0

D < 0 shows that roots will not be real. Hence, roots will be imaginary. **Ans. **

**3. Find the common difference of the Arithmetic Progression (A.P) [1]**

**Ans.** Given,

**4. Evaluate : sin**^{2}** 60° + 2 tan 45°- cos**^{2}** 30° [1]**

^{2}

^{2}

**Sol. **We know,

sin 60° = √3/2, tan 45° = 1 and cos 30° = √3/2

∴ sin^{2} 60° +2 tan 45° – cos^{2} 30°

= 2 **Ans. **

** OR**

**If sin A = 3/4, calculate sec A.**

**Sol. **Given, sin A = 3/4

**5. Write the coordinates of a point P on x-axis which is equidistant from the point A(-2, 0) and B(6, 0). [1]**

**Sol. **Let coordinates of P on x-axis is (x, 0)

Given, A(-2, 0) and B(6, 0)

Here, PA = PB

On squaring both sides, we get

(x + 2)^{2} = (x – 6)^{2}

x^{2} + 4 + 4x = x^{2} + 36 – 12x

⇒ 4 + 4x = 36 – 12

⇒ 16x = 32

⇒ x = 32/16 = 2

Co-ordinates of P are (2, 0) **Ans.**

**6. In Figure 1, ABC is an isosceles triangle right angled at C with AC = 4 cm. Find the length of AB. ** [1] **

**Sol. **Given, ∠C=90° and AC = 4 cm

AB = ?

∵∆ABC is an isosceles triangle so,

BC = AC = 4 cm

On applying Phythagoras theorem, we have

AB^{2} = AC^{2} + BC^{2}

= 4^{2} + 4^{2}

⇒ AB² = 16 + 16 = 32

⇒ AB = √32

= 4√2 cm

** OR**

**In Figure 2, DE || BC. Find the length of side AD, given that AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm. **

**Sol. **Given, DE || BC

On applying, Thales theorem, we have

AD/AB = AE/AC

4AD = AD +7.2

3AD = 7.2

AD = 2.4 cm **Ans. **

## SECTION – B

**7. Write the smallest number which is divisible by both 306 and 657. [2] **

**Sol.** Smallest number which is divisible by 306 and 657 is LCM (657, 306)

657 = 3 x 3 x 73

306 = 3 x 3 x 2 x 17

LCM = 3 x 3 x 73 x 2 x 17

= 22338 **Ans. **

**8. Find a relation between x and y if the points A(x, y), B(-4,6) and C(-2,3) are collinear.** ** [2]

** OR**

**Find the area of a triangle whose vertices are given as (1, – 1) (-4,6) and (-3,- 5).** **

**9. The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is 1/5 The probability of selecting a black marble at random from the same jar is 1/4. If the jar contains 11 green marbles, find the total number of marbles in the jar. [2]**

**Sol.** Let probability of selecting a blue marble, black marble and green marble are P(x), P(y), P(z) respectively.

P(x) = 1/5, P(y) = 1/4

We know,

P(x) + P(y) + P(z) = 1

1/5 + 1/4 + P(z) = 1

9/20 + P(z) = 1

(∵ No. of green marbles = 11)

⇒ Total no. of marbles = 20

∴ There are 20 marbles in the jar. **Ans. **

**10. Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique solution. [2]**

**Sol. **Given, x + 2y = 5

3x + ky + 15 = 0

Comparing above equations with

a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0,

We get,

a_{1} = 1, b_{1} = 2, c_{1} = -5

a_{2} = 3, b_{2} = k, c_{2} = 15

Condition for the pair of equations to have unique solution is

k ≠ 6

k can have any value except 6. **Ans. **

**11. The larger of two supplementary angles exceeds the smaller by 18°, Find the angles.** [2]

**Sol. **Let two angles A and B are supplementary.

∴ A + B = 180° …(i)

Given, A = B +18°

On putting A = B+ 18° in equation (i), we get

B + 18° + B = 180°

2B + 18° = 180°

⇒ 2B = 162°

⇒ B = 81°

⇒ A = B + 18°

⇒ A = 99° **Ans.**

** OR**

**Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present? **

**Sol. **Let age of Sumit be x years and age of his son be y years. Then, according to question we have,

x = 3y .…(1)

Five years later,

On putting x = 3y in equation (ii)

y = 15 years

Then, present age of sumit is

3 x y = 3 x 15

= 45 years **Ans. **

**12. Find the mode of the following frequency distribution: **

Class Interval: | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 | 50-55 |

Frequency: | 25 | 34 | 50 | 42 | 38 | 14 |

**Sol. **

Class Interval | Frequency |

25-30 | 25 |

30-35 | 34 |

35-40 | 50 |

40-45 | 42 |

45-50 | 38 |

50-55 | 14 |

Here, maximum frequency is 50.

So, 35-40 will be the modal class.

l = 35, f_{0} = 34, f_{1} = 50, f_{2} = 42 and h = 5

= 38.33 **Ans. **

## SECTION – C

**13. Prove that 2 + 5√3 is an irrational number, given that √3 is an irrational number. [3]**

**Sol. **Let 2 + 5**√**3 = r, where, r is rational.

∴ (2 + 5**√**3)^{2} = r^{2}

4 + 75 + 20 **√**3 = r^{2}

79 + 20 **√**3 = r^{2}

20**√**3 = r^{2} -79

**√**3 = r^{2} – 79 / 20

Now, r^{2} – 79 / 20 is a rational number. So, **√**3 must also be a rational number. But **√**3 is an irrational number (Given).

So, our assumption is wrong.

∴ 2 + 5**√**3 is an irrational number. **Hence Proved.**

** OR**

**Using Euclid’s Algorithm, find the HCF of 2048 and 960.****

**14. Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP x PC = BP x DP. [3] **

**Sol. **Given, ∆ABC, ∆DBC are right angle triangles, right angled at A and D, on same side of BC. AC & BD intersect at P.

In ∆APB and ∆PDC,

∠A = ∠D = 90°

∠APB = ∠DPC (Vertically opposite)

∴ ∆APB ~ ∆DPC (By AA Similarity)

⇒ AP x PC = BP x PD. **Hence Proved. **

** OR**

**Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3RS. Find the ratio of the areas of triangles POQ and ROS.** **

**15. In Figure 3, PQ and RS are two parallel tangents to a circle with centreO and another tangent AB with point of contact C intersecting PO at A and RS at B. Prove that ∠AOB = 90°. [3]**

**Sol.** Given, PQ || RS

To prove: ∠AOB = 90°

Construction: Join O and C, D and E

In ∆ODA and ∆OCA

OD = OC (radii of circle)

OA = OA (common)

AD = AC

(tangent drawn from same point)

By SSS congruency

∆ODA ≅ ∆OCA

Then, ∠DOA = ∠AOC ….(i)

Similarly, in ∆EOB and ∆BOC, we have

∆EOB ≅ ∆BOC

∠EOB = ∠BOC ….(ii)

EOD is a diameter of circle, therefore it is a straight line.

Hence,

∠DOA + ∠AOC + ∠EOB + ∠BOC = 180°

2(∠AOC) + 2(∠BOC) = 180°

∠AOC + ∠BOC = 90°

∠AOB = 90°. **Hence Proved. **

**16. Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (-2,-5) and (6, 3). Find the coordinates of the point of intersection. [3] **

**Sol. **Let required ratio be k : 1

By section formula, we have

Here, x_{1} = -2, x_{2} = 6, y_{1} = -5, y_{2} = 3

m = k, n = 1

6k – 2 – 9k + 15 = 0

-3k + 13 = 0

k = 13/3

Hence required ratio is

Here, intersection point are,

= 3/2

∴ intersection point is (9/2, 3/2) **Ans. **

**17. Evaluate :** [3]**

**18. In Figure 4, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of the shaded region. (Use 𝛑 = 3.14) **

**Sol. **Given, OABC is a square with OA = 15 cm

OB = radius = r

Let side of square be a then,

a^{2} + a^{2} = r^{2}

2a^{2} = r^{2}

r = **√**2a

r = 15**√**2 cm (∵ a = 15 cm)

Area of square = Side x Side

= 15 x 15

= 225 cm^{2}

Area of quadrant OPBQ

= 225 x 1.57

= 353.25 cm^{2}

Area of shaded region

= Area of quadrant OPBQ – Area of square OABC

=353.25 – 225

= 128.25 cm²

** OR**

**In Figure 5, ABCD is a square with side 2√2 cm and inscribed in a circle. Find the area of the shaded region. (Use 𝛑 = 3.14) **

**Sol. **Given, ABCD is a square with side 2√2 cm

∵ BD = 2r

In ∆BDC

BD^{2} = DC^{2} + BC^{2}

4r^{2} = 2(DC)^{2}

(∵ DC = CB = Side = 2√2 )

4r^{2} = 2 x 2√2 x 2√2

4r^{2} = 8 x 2

4r^{2} = 16

⇒ r^{2} = 4

r = 2 cm

Area of square BCDA = Side x Side

= DC x BC

= 2√2 x 2√2

= 8 cm^{2}

Area of circle = 𝛑r^{2}

= 3.14 x 2 x 2

= 12.56 cm^{2}

Area of shaded region

= Area of circle – Area of square.

= 12.56 – 8

= 4.56 cm^{2} **Ans. **

**19. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. (Use 𝛑 = 22/7) [3]**

**Sol. **ABCD is a cylinder and BFC and AED are two hemisphere which has radius (r) = 7/2 cm

= 1001/2

= 500.5 cm^{3}

Volume of two hemisphere

= 539/3

= 179.67 cm^{3}

Total volume of solid = Volume of two hemisphere + Volume of cylinder

= 179.67 + 500.5

= 680.17 cm^{3} **Ans. **

**20. The marks obtained by 100 students in an examination are given below: [3]**

Marks | Number of Students |

30 – 35 | 14 |

35 – 40 | 16 |

40 – 45 | 28 |

45 – 50 | 23 |

50 – 55 | 18 |

55 – 60 | 8 |

60 – 65 | 3 |

**Find the mean marks of the students.**

**Sol. **

Marks | Number of Students | x_{i} | f_{i}x_{i} |

30 – 35 | 14 | 32.5 | 455 |

35 – 40 | 16 | 37.5 | 600 |

40 – 45 | 28 | 42.5 | 1190 |

45 – 50 | 23 | 47.5 | 1092.5 |

50 – 55 | 18 | 52.5 | 945 |

55 – 60 | 8 | 57.5 | 460 |

60 – 65 | 3 | 62.5 | 187.5 |

𝚺f_{i} = 110 | 𝚺f_{i}x_{i} = 4930 |

= 44.82 **Ans. **

**21. For what value of k, is the polynomial ** [3]

**f(x) = 3x**^{4}** – 9x**^{3}** + x**^{2}** + 15x + k **

**completely divisible by 3x**^{2}** – 52** **

** OR**

**Find the zeroes of the quadratic polynomial 7y**^{2}** – 11/3y – 2/3 and verify the relationship between the zeroes and the coefficients.**

^{2}

**Sol. **The given polynomial is

P(y) = 7y^{2} – 11/3y – 2/3

∵ P(y) = 0

7y^{2} – 11/3y – 2/3 = 0

⇒ 21y^{2} – 11y – 2 = 0

⇒ 21y^{2} – 14y + 3y – 2 = 0

⇒ 7y (3y – 2) + 1 (3y – 2) = 0

⇒ (3y – 2) (7y + 1) = 0

∴ y = 2/3, – 1/7

So zeroes of P(y) are 2/3, – 1/7 **Ans. **

Verification : On comparing 7y – 11/3y – 2/3

with ax^{2} + bx + c, we get

a = 7, b = 11/3, c = 2/3

Sum of zeroes = -b/a

11/21 = 11/21

Product of zeroes = c/a

**Hence Verified. **

**22. Write all the values of p for which the quadratic equation x**^{2}** + px + 16 = 0 has equal roots. Find the roots of the equation so obtained. [3]**

^{2}

**Sol.** Given, equation is x^{2} + px + 16 = 0

This is of the form ax^{2} + bx + c = 0

where, a = 1, b = p and c = 16

∴ D = b^{2} – 4ac

= p^{2} – 4 x 1 x 16

= p^{2} – 64

for equal roots, we have

D = 0

p^{2} – 64 = 0

p^{2} = 64

p = ± 8

Putting p = 8 in given equation we have,

x^{2} + 8x + 16 = 0

(x – 4)^{2} = 0

x = 4, 4

∴ Required roots are -4 and -4 or 4 and 4. **Ans. **

## SECTION – D

**23. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio. [4] **

**Sol. **Given, A ∆ABC in which DE| |BC and DE intersect AB and AC at D and E respectively.

To prove: AD/DB = AE/EC

Construction: Join BE and CD

Draw EL ⊥ AB and DM ⊥ AC

**Proof:** we have

Now, ∆DBE and ∆ECD, being on same base DE and between the same parallels DE and BC, we have

area (∆DBE) = area (∆ECD) …..(iii)

from equations (i), (ii) and (iii), we have

AD/DB = AE/EC **Hence Proved. **

**24. Amit, standing on a horizontal plane, inds a bird Aying at a distance of 200 m from him at an elevation of 30°. Deepak standing on the roof of a 50 m high building, finds the angle of elevation of the same bird to be 45°. Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak. [4] **

**Sol.** Let Amit be at C point and bird is at A point. Such that ∠ACB = 30°. AB is the height of bird from point B on ground and Deepak is at D point, DE is the building of height 50 m.

Now, In right triangle ABC, we have

sin 30° = P/H = AB/AC

1/2 = AB/200

AB = 100 m

In right ∆AFD, we have

sin 45° = P/H = AF/AD

(∵ AB = AF + BF

100 = AF + 50

AF = 50)

AD = 50√2 m

Hence, the distance of bird from Deepak is 50√2 m. **Ans. **

**25. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm**^{3}** of iron has approximately 8 gm mass. (Use 𝛑 = 3.14) [4]**

^{3}

**Sol.** Let AB be the iron pole of height 220 cm with base radius 12 cm and there is an other cylinder CD of height 60 cm whose base radius is 8 cm. ** **

Volume of AB pole = 𝛑r_{1}^{2}h_{1}

= 3.14 x 12 x 12 x 220

= 99475.2 cm^{3}

Volume of CD pole = 𝛑r_{2}^{2}h_{2}

= 3.14 x 8 x 8 x 60

= 12057.6 cm^{3}

Total volume of the poles

= 99475.2 + 12057.6

= 111532.8 cm^{3}

It is given that,

Mass of 1 cm^{3} of iron = 8 gm

Then mass of 111532.8 cm^{3} of iron

= 111532.8 x 8 gm

Then total mass of the pole is = 111532.8 x 8 gm

= 892262.4 gm

= 892.26 kg **Ans. **

**26. Construct an equilateral ∆ABC with each side 5 cm. Then construct another triangle whose sides are 2/3 times the corresponding sides of ∆ABC.** [4]**

** OR**

**Draw two concentric circles of radii 2 cm and 5 cm. Take a point P on the outer circle and construct a pair of tangents PA and PB to the smaller circle. Measure PA.** **

**27. Change the following data into less than type’ distribution and draw its ogive : [4]**

Class Interval | Frequency |

30-40 | 7 |

40-50 | 5 |

50-60 | 8 |

60-70 | 10 |

70-80 | 6 |

80-90 | 6 |

90-100 | 8 |

**Sol. **

Class Interval | Frequency |

less than 40 | 7 |

less than 50 | 12 |

less than 60 | 20 |

less than 70 | 30 |

less than 80 | 36 |

less than 90 | 42 |

less than 100 | 50 |

On graph paper, we take the scale.

**28. Prove that: [4]**

**Sol. **

**Hence Proved. **

** OR**

**Prove that : **

**Sol. **LH.S.

R.H.S.

= 2 – (1 + cos 𝜃)

= 1 – cos 𝜃 …..(i)

From equation (i) and (ii), we get

L.H.S. = R.H.S. **Hence Proved. **

**29. Which term of the Arithmetic Progression -7, – 12, – 17, – 22, …. will be – 82? Is – 100 any term of the A.P. ? Give reason for your answer. [4]**

**Sol. **-7, -12, -17, -22, ….

Here a = -7, d = -12 – (-7)

= -12 + 7

= -5

Let T_{n} = -82

∴ T_{n} = a + (n – 1) d

– 82 = -7 + (n – 1)(-5)

– 82 = -7 – 5n + 5

– 80 = -5n

n = 16

Therefore, 16^{th} term will be – 82.

Let T_{n} = -100

Again, T_{n} = a + (n – 1) d

– 100 = -7 + (n – 1) (-5)

– 100 = -7 -5n + 5

– 98 = -5n

n = 98/5

But the number of terms can not be in fraction. So, – 100 can not be a term of this A.P. **Ans. **

** OR**

**How many terms of the Arithmetic Progression 45, 39,33, ….. must be taken so that their sum is 180? Explain the double answer. **

**Sol. **45, 39, 33, ……..

Here a = 45, d = 39 – 45 = -6

Let S_{n} = 180

n (96 – 6n) = 360

96n – 6n^{2} = 360

6n^{2} – 96n + 360 = 0

On dividing the above equation by 6

n^{2} – 16n + 60 = 0

n^{2} – 10n – 6n + 60 = 0

n(n – 10) – 6(n -10) = 0

(n – 10) (n – 6) = 0

n = 10, 6

∴ Sum of first 10 terms = Sum of first 6 terms

= 180

This means that the sum of all terms from 7^{th} to 10^{th} is zero. **Ans. **

**30. In a class test, the sum of Arun’s marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects. [4]**

**Sol. **Let Arun marks in Hindi be x and marks in English be y.

Then, according to question, we have

x+ y = 30 …..(i)

(x + 2) (y – 3) = 210 …..(ii)

from equation (i), put x = 30 – y in equation (ii)

(30 – y + 2) (y – 3) = 210

(32 – y) (y – 3) = 210

32y – 96 – y^{2} + 3y = 210

y^{2} – 35y + 306 = 0

y^{2} -18y – 17y + 306 = 0

y (y – 18) – 17 (y – 18) = 0

(y – 18) (y – 17) = 0

y = 18, 17

Put y = 18 and 17 in equation (i), we get

x = 12, 13

Hence his marks in Hindi can be 12 and 13 and in English his marks can be 18 and 17. **Ans. **