CBSE Class 10 Mathematics Previous Year Question Paper 2020 Delhi Set - 3

Q 9. The common difference of the A.P. 

A. 1

B. 

C. -1

D. 2

Solution: Correct option: C

Q 10. The n^th term of the A.P. a, 3a, 5a, …. Is

A. na

B. (2n – 1)a

C. (2n + 1)a

D. 2na

Solution: Correct option: B

Here a₁ = a, d = 3a – a = 2a

We know that,

a_n = a₁ + (n – 1)d

= a + (n – 1)2a

= a + 2a_n – 2a

= 2a_n – a = (2n – 1)a

In Q. Nos. 11 to 15, fill in the blanks. Each question carries 1 mark: 

Q 11. In fig. 1, the angles of depressions from the observing positions O₁ and O₂ respectively of the object A are ______, ______. 

Solution: In fig. 1, the angles of depressions from the observing positions O₁ and O₂ respectively of the object A are 30° and 45°.

In ΔO₁CA,

⇒ ∠O₁CA + ∠CAO₁ + ∠CO₁A = 180°

⇒ 90° + ∠CAO₁ + 60° = 180°

⇒ ∠CAO₁ = 30°

So, the angle of depression from the observing position O₁ of the object A is 30°

And the angle of depression from the observing position O₂ of the object A is 45°.

Q 12. In ΔABC, AB=6  cm, AC = 12 cm and BC = 6 cm, then ∠B=______.

OR

Two triangles are similar if their corresponding sides are ______.

Solution: In ∆ABC, AB = , AC = 12 cm and BC = 6 cm, then ∠B = 90°.

AB² = = 108, BC² = (6)² = 36 and AC² = (12)² = 144

Here, AC² = AB² + BC²

By the converse of Pythagoras theorem, we get

m∠B = 90°.

OR

Two triangles are similar if their corresponding sides are in the same ratio.

Q 13.  In given Fig. 2, the length PB=______cm.

Solution: In given Fig 2, the length PB = 4 cm.

AB is a tangent to the internal circle at P and OP is its radius.

⇒ OP ⊥ AB

In ∆APO,

OA² = OP² + AP² ... (OP ⊥ AB)

⇒ AP² = 5² – 3² = 16

⇒ AP = 4 cm

We know that,

In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

⇒ AP = PB = 4 cm

Q 14.  In fig. 3, MN || BC and AM : MB =1 : 2, then 

Solution: In fig. 3 MN || BC and AM: MB = 1: 2, then .

AM: MB = 1: 2

Let AM = x and MB = 2x

⇒ AB = AM + MB = 3x

Here, ∆AMN ~ ∆ABC ... (AA test)

Q 15. The value of sin 32° cos 58° +cos 32° sin 58° is ______.

OR

The value of   is ______.

Solution: The value of sin 32° cos 58° + cos 32° sin 58° is 1.

sin 32° cos 58° + cos 32° sin 58°

= sin 32° cos(90° – 32°) + cos 32° sin(90° – 32°)

= sin 32° sin 32°  + cos 32° cos 32°

= sin²32° + cos²32°

= 1 …. (sin²θ + cos²θ = 1)

OR

The value of  is 2.

Explanation:

Q Nos. 16 to 20 are short answer type questions of 1 mark each.

Q 16. A die is thrown once. What is the probability of getting a prime number?

Solution: A die is thrown once.

S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6

Let A be the event of getting a prime number.

A = {2, 3, 5} ⇒ n(A) = 3

⇒ P(A) =  .

Q 17. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3, then find the probability of x² < 4. 

OR

What is the probability that a randomly taken leap year has 52 Sundays?

Solution: Here, S = {–3, –2, –1, 0, 1, 2, 3} ⇒ n(S) = 7

Let A be the event that the square of the chosen number is less than 4.

A = {–1, 0, 1}

⇒ P(A) =  .

 OR

In the leap year, total number of days = 366 = 52 × 7 + 2

Total 52 weeks ⇒ 52 Sundays, now the 2 remaining days will be as follows.

S = {(Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday)}

⇒ n(S) = 7

So, the probability of not getting Sunday in the remaining 2 days = 

⇒ The probability that a randomly taken leap year has 52 Sundays is .

Q 18. If sin A + sin² A = 1, then find the value of the expression (cos² A + cos⁴ A).

Solution: sinA + sin²A = 1

⇒ sinA = 1 – sin²A = cos²A … (i)

⇒ sin²A = cos⁴A … (ii)

cos²A + cos⁴ A 

= sinA + sin²A   … from (i) and (ii)

= 1      (sinA + sin²A = 1)

Q 19. Find the area of the sector of a circle of radius 6 cm whose central angle is 30°. (Take π =3.14).

Solution: Here, r = 6cm, θ = 30° 

Area of the sector of a circle =  × 3.14 × 6 × 6 = 9.42 cm².

Q 20. Find the class marks of the classes 20 – 50 and 35 – 60.

Solution: Class mark of the class 20 – 50 =  

Class mark of the class 35 – 60 =  

Q 21. A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper.

The following were the answers given by the students:

2x + 3, 3x² + 7x + 2, 4x³ + 3x² + 2, x³ +   + 7, 7x +  , 5x³ – 7x + 2, 2x² + 3 –  , 5x –  , ax³ + bx² + cx + d, x +  .

Answer the following questions:

1. How many of the above ten, are not polynomials?
2. How many of the above ten, are quadratic polynomials?

1. From the above ten, only three are not polynomials.
   
   They are   because in all three polynomials, the degree of x is not a positive integer.
2. From the above ten, only one is a quadratic polynomial which is 3x² + 7x + 2 because highest degree of the variable x is 2.

Q 22 .A child has a die whose six faces show the letters as shown below: 

The die is thrown once. What is the probability of getting (i) A, (ii) D?

Solution: From the given question, n(S) = 6

1. Let E be the event of getting A on the die.
   
   As there are 2 A’s, n(E) = 2
   
   Hence, the required probability is  
2. Let F be the event of getting D on the die.
   
   As there is only one D, n(F) = 1
   
   Hence, the required probability is  

Q 23. In fig. 4, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that .

 

 

OR

 

In fig.5, if AD ⊥ BC, then prove that AB² + CD² = BD² + AC².

 

Solution: Given: Triangle ABC and triangle DBC are on the same base BC.

 

To prove:  

Construction:

Draw AM perpendicular to BC from point A and draw DN perpendicular to BC from point D.

 

 

Prove: 

In Δ AMO and Δ DNO

∠AOM = ∠DON                   vertically opposite angles

∠AMO = ∠DNO                   90˚ angles

⇒ ΔAMO ≈ ΔDNO                AA test of similarity  

Ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides

OR

LHS =  tan⁴ θ +  tan² θ 

= (sec² θ - 1)² +  sec² θ - 1

= sec⁴ θ - 2sec² θ + 1 + sec² θ - 1

= sec⁴ θ -  sec² θ

Hence,  tan⁴ θ +  tan² θ =  sec⁴ θ -  sec² θ.

Q 25. Find the mode of the following frequency distribution:

Solution: Here, the maximum frequency is 10, and the class corresponding to this is 30 – 35.

So the modal class is 30 – 35.

Therefore, l = 30, h = 5, f₀ = 9, f₁ = 10 and f₂ = 3

Mode =  

Hence, the mode of the above data is 30.625.

Q 26. From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of same height and same base radius is removed. Find the volume of the remaining solid.

Solution: h = 14 cm and r = 6 cm 

As the circular cone of same height and same radius is removed from the cylinder, therefore we have

Volume of the remaining solid 

= Volume of cylinder – volume of cone

=  πr²h - πr²h

= πr²h

=  

= 1056 cm³

Q. Nos. 27 to 34 carry 3 marks each.

Q 27. If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, respectively, prove that .

Solution:

Lengths of the tangents from an external point are equal.

Therefore, AQ = AR, BQ = BP and CP = CR

Perimeter of ΔABC = AB + BC + CA

= AB + BP + PC + AR – CR

= AB + BQ + PC + AQ – PC 

= AQ + AQ

⇒ AB + BC + CA = 2AQ

⇒  (AB + BC + CA) = AQ 

Q 28. The area of a circular play ground is 22176 cm². Find the cost of fencing this ground at the rate of Rs. 50 per metre.

Solution: Area of a circular playground = 22176 cm²

⇒ πr² = 22176

⇒ r² = 22176

⇒  r² = 22176 × 

⇒ r = 84 cm

Circumference of a playground = 2πr² = 2 ×  × 84 = 528 cm = 5.28 m

The cost of fencing the playground at Rs. 50 per metre 

= 5.28 × 50 = Rs. 264 

Q 29. If the mid – point of the line segment joining the points A (3, 4) and B (k, 6) is P(x, y) and x + y – 10 = 0, find the value of k.

OR

Find the area of triangle ABC with A (1, –4) and the mid-points of sides through A being (2, –1) and (0, –1).

Solution: The coordinates of the midpoint of the line segment joining A(3, 4) and B(k, 6) are 

It is given that x + y – 10 = 0 

⇒ 3 + k = 10

⇒ k = 7

Hence, the value of k is 7.

OR

In ∆ABC, coordinates of A are (1, –4)

Let (a, b) and (x, y) be the coordinates of points B and C respectively.

Midpoint of AB is (2, –1) and of AC is (0, –1)

Midpoint of AB = 

⇒ a = 3, b = 2

Midpoint of AC = 

⇒ x = -1, y = 2

Therefore, B(3, 2) and C(-1, 2)

Area of Δ ABC =  ∣x₁(y₂ - y₃)+ x₂(y₃ - y₁) + x₃(y₁ - y₂)∣

=  ∣1(2 - 2 ) + 3(2 + 4) - 1(-4 - 2)∣

=  ∣0 + 18 + 6∣ = 12

Hence the area of ΔABC is 12 square units.

Q 30. If Fig.6, if ΔABC ~ ΔDEF and their sides of lengths (in cm) are marked along them, then find the lengths of sides of each triangle.

Solution: Given that ΔABC ~ ΔDEF

⇒

⇒ 4x – 2 = 18

⇒ 4x = 20

⇒ x = 5

Then the sides of triangles are

AB = 2x – 1 = 9, BC = 2x + 2 = 12, AC = 3x = 15

DE = 18, EF = 3x + 9 = 24, DF = 6x = 30

Q 31. If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x) and .

OR

Solve for  

Solution: 2x + y = 23 …(i)

4x – y = 19 …(ii)

Adding (i) and (ii) we get

⇒ 6x = 42

⇒ x = 7

Put it in (i) we get

⇒ 14 + y = 23

⇒ y = 9

5y – 2x = 45 – 14 = 31

OR

According to the paper, condition given is x ≠ -4, 7 which implies that the equation should be  .

⇒ -30 = x² - 3x -28

⇒ x² - 3x + 2 = 0

⇒ (x - 2)(x - 1) = 0

⇒ x = 2 or x = 1x

Q 32. Which term of A.P. 20, 19, 18, 17,....  is the first negative term.

OR

Find the middle term of the A.P. 7, 13, 19, …., 247.

Solution: 20, 19, 18, 17,...

The above sequence is an A.P.

⇒ a₁ = 20 and d = 

Consider, an < 0 

⇒ a₁ + (n – 1)d < 0

⇒  20 + (n - 1) ×  < 0

⇒ 80 - 3(n - 1) < 0

⇒ 80 – 3n + 3 < 0 

⇒ 83 < 3n 

⇒ n >   

Hence, 28^th term of the given sequence is first negative term.

OR

7, 13, 19, …, 247

Given sequence is in A.P.

a₁ = 7, d = 6 and a_n = 247

⇒ a₁ + (n – 1)d = 247

⇒ 7 + (n – 1)6 = 247

⇒ 6(n – 1) = 240

⇒ n – 1 = 40

⇒ n = 41

The middle term should be a₂₁.

a₂₁ = a₁ + 20d = 7 + 20 × 6 = 127

The middle term of the given sequence is 127.

Q 33. Water in a canal, 6m wide and 1.5m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8cm standing water is required?

Solution: Area of the face of canal = 6 × 1.5 m² 

Speed of water flow = 10 km/h = 10000 m/h

So the water flow through the canal in 1 hour 

= 6 × 1.5 × 10000

= 90000 m³ 

Water flow through the canal in 30 min = 90000/2 = 45000 m³ 

The height of standing water = 8 cm = 0.08 m

Area can irrigate = 45000 ÷ 0.08 = 562500 m²

Q 34. Show that: 

Solution: 

Q 35. The mean of the following frequency distribution is 18. The frequency f in the class interval 19 – 21 is missing. Determine f.

OR

The following table gives production yield per hectare of wheat of 100 farms of a village:

Change the distribution to a ‘more than’ type distribution and draw its ogive.

Solution: 

⇒ 704 + 20f = 720 + 18f

⇒ 2f = 16

⇒ f = 8

OR

Q 36. From a point on the ground, the angles of elevation of the bottom and the top of a tower fixed at the top of a 20m high building are 45° and 60° respectively. Find the height of the tower. 

Solution: 

Let BC be the building, AB be the tower and D be the point on ground from where elevation angles are to be measured.

In ΔBCD,

CD = 20 m …(i)

In ΔACD,

Hence, the height of the tower is

Q 37. It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for four hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately?

Solution: Let the time taken to fill the pool by pipe A be x hours and by pipe B be y hours.

Thus, the quantity of water filled by the pipe A in one hour is   and by the pipe is  .

In 1 hour the quantity of water filled by both the pipes is 

It takes 12 hours to fill the pool for both the pipes.

…(i)

Let A be the pipe with larger diameter and B be the pipe with smaller diameter.

In 4 hours the quantity of water filled by pipe A is  .

In 9 hours the quantity of water filled by pipe B is  .

According to the question,

Multiplying (i) by 4 we get

4m + 4n =   …(v)

Subtracting (v) from (iv) we get

n = 

Put it in (iii) we get

Hence, x = 20 and y = 30

Hence, the pipe A and pipe B fills the pool in 20 hours and 30 hours respectively.

Q 38. Prove that  is an irrational number.

Solution: Let us assume that  is a rational number

We can write it in the form of  where q ≠ 0.

Also p and q are co-prime numbers.

Here p is divisible by 5

p = 5k where k is a positive integer

Hence, p² = 25k²

Substituting 5q² = p²

⇒ 5q² = 25k²

⇒ q² = 5k² 

q is divisible by 5

From this we can say that p and q have a common factor 5

It contradicts to our assumption p and q are co-prime.

Hence,   is an irrational number.

Q 39. Draw a circle of radius 3.5cm. From a point P, 6cm from its centre, draw two tangents to the circle.

OR

Construct a ΔABC with AB=6cm, BC=5cm and ∠B=60°. Now construct another triangle whose sides are   times the corresponding sides of ΔABC.

Solution: Steps of construction:

1. Draw a line segment OP = 6 cm
2. With centre O and radius 3.5 cm, draw a circle.
3. Draw the perpendicular bisector of OP intersecting it at M.
4. With centre M and radius OM, draw a circle which intersects the circle at T and S.
5. Join PT and PS which are the required tangents.