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

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Question numbers 1 to 10 are multiple choice questions of 1 mark each.

Select the correct option.

Q 1. The point P on x-axis equidistant from the points A(–1, 0) and B(5, 0) is

A. (2, 0)

B. (0, 2)

C. (3, 0)

D. (2, 2)

 

Q 2. The co-ordinates of the point which is reflection of point (–3, 5) in x-axis are

A. (3, 5)

B. (3, -5)

C. (-3, -5)

D. (-3, 5)

 

Q 3. If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3:1, then the value of y is

A. 4

B. 3

C. 2

D. 1

 

Q 4. The sum of exponents of prime factors in the prime-factorisation of 196 is

A. 3

B. 4

C. 5

D. 2

 

Q 5. Euclid’s division Lemma states that for two positive integers a and b, there exists unique integer q and r satisfying a = bq + r, and

A. 0 < r < b

B. 0 < r < b

C. 0 ≤ r < b

D. 0 ≤ r ≤ b

 

Q 6. The zeroes of the polynomial x2 – 3x – m(m+3) are

A. m, m+3

B. –m, m+3

C. m, –(m+3)

D. –m, –(m+3)

 

Q 7. The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is

A. begin mathsize 12px style negative 14 over 3 end style

B. begin mathsize 12px style 2 over 5 end style

C. 5

D. 10

 

Q 8. The roots of the quadratic equation x2 – 0.04 = 0 are

A. ±0.2

B. ±0.02

C. 0.4

D. 2

 

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

begin mathsize 12px style 1 over straight p comma fraction numerator 1 minus straight p over denominator straight p end fraction comma fraction numerator 1 minus 2 straight p over denominator straight p end fraction comma...... end style

A. 1

B. begin mathsize 12px style 1 over straight p end style

C. -1

D. 2

 

Q 10. The nth term of the A.P. a, 3a, 5a, …. Is

A. na

B. (2n – 1)a

C. (2n + 1)a

D. 2na

 

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 ______, ______. 

 

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 ______.

 

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




Q 14. In fig. 3, MN || BC and AM : MB =1 : 2, then begin mathsize 12px style fraction numerator ar open parentheses increment AMN close parentheses over denominator ar open parentheses increment ABC close parentheses end fraction equals __________. end style

 

 

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


OR

 

The value of begin mathsize 12px style fraction numerator tan space 35 degree over denominator cot space 55 degree end fraction plus fraction numerator cot space 78 degree over denominator tan space 12 degree end fraction end style  is ______.

 

 

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?

 

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 x< 4. 


OR

 

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

 

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

 

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

 

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

  

 

Q.Nos.21 to 26 carry 2 marks each.

 

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, x3begin mathsize 12px style square root of 3 straight x end root end style  + 7, 7x + begin mathsize 12px style square root of 7 end style , 5x3 – 7x + 2, 2x2 + 3 – begin mathsize 12px style 5 over straight x end style , 5x – begin mathsize 12px style 1 half end style , ax+ bx2 + cx + d, x + begin mathsize 12px style 1 over straight x end style .

 

Answer the following questions:

 

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

 

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?

 

 

Q 23. In fig. 4, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that begin mathsize 12px style fraction numerator ar open parentheses increment ABC close parentheses over denominator ar open parentheses increment DBC close parentheses end fraction equals AO over DO end style.

 

 

OR

 

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

 

 

 

Q 24. Prove that begin mathsize 12px style 1 plus fraction numerator cot squared straight alpha over denominator 1 plus cosec space straight alpha end fraction equals cosec space straight alpha end style

  

OR

 

Show that tan4θ + tan2θ = sec4θ - sec2θ

 

 

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

 

 

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.

 

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 begin mathsize 12px style AQ equals 1 half open parentheses BC plus CA plus AB close parentheses end style.

 

                       

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. 


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).


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.


 

 

Q 31. If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x) and begin mathsize 12px style straight y over straight x minus 2 end style.

OR


Solve for  begin mathsize 12px style straight x space colon space fraction numerator 1 over denominator straight x space plus space 4 end fraction minus fraction numerator 1 over denominator straight x space minus space 7 end fraction equals 11 over 30 comma space straight x space not equal to space minus 4 comma space 7. end style

 

 

Q 32. Which term of A.P. 20, 19begin mathsize 12px style 1 fourth end style, 18begin mathsize 12px style 1 half end style, 17begin mathsize 12px style 3 over 4 end style,....  is the first negative term.

 

OR

 

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

 

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?

 

 

Q 34. Show that: begin mathsize 12px style fraction numerator cos squared space open parentheses 45 to the power of degree space plus space straight theta close parentheses space plus space cos squared open parentheses 45 degree minus straight theta close parentheses over denominator tan space open parentheses 60 degree space plus space straight theta close parentheses space tan space open parentheses 30 degree space minus space straight theta close parentheses end fraction equals 1 end style

 

Q. Nos. 35 to 40 carry 4 marks each. 

 

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.

 

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.

 

Q 38. Prove that begin mathsize 12px style square root of 5 end style 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 begin mathsize 12px style 2 over 3 end style  times the corresponding sides of ΔABC.


Q 40. A solid is in the shape of hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7cm and height of cone is 3.5cm, find the volume of the solid. (Take π = begin mathsize 12px style 22 over 7 end style).

Q 1. The point P on x-axis equidistant from the points A(–1, 0) and B(5, 0) is

A. (2, 0)

B. (0, 2)

C. (3, 0)

D. (2, 2)

Solution: Correct option : A

Let P(x, 0) … (Since, the point P is on x – axis)

The points A(–1, 0) and B(5, 0) are also lying on x – axis and P is equidistant from

the points A and B.

⇒ P is a midpoint of AB.

By using mid – point formula, we get

begin mathsize 12px style open parentheses fraction numerator negative 1 space plus space 5 over denominator 2 end fraction comma space fraction numerator 0 space plus space 0 over denominator 2 end fraction close parentheses equals open parentheses x comma space 0 close parentheses
rightwards double arrow 4 over 2 equals x rightwards double arrow x equals 2 end style

Therefore, the coordinates of the point P is (2, 0).

 

Q 2. The co-ordinates of the point which is reflection of point (–3, 5) in x-axis are

A. (3, 5)

B. (3, -5)

C. (-3, -5)

D. (-3, 5)

Solution: Correct option : C

The coordinates of the point which is the reflection of point (–3, 5) is (–3, –5).

 

 

Q 3. If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3:1, then the value of y is

A. 4

B. 3

C. 2

D. 1

Solution: Correct option : D

The point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3:1.

By using section formula, we get

begin mathsize 12px style open parentheses fraction numerator 12 space plus space 6 over denominator 4 end fraction comma fraction numerator 3 straight y space plus space 5 over denominator 4 end fraction close parentheses equals open parentheses 6 comma 2 close parentheses end style

Q 4. The sum of exponents of prime factors in the prime-factorisation of 196 is

A. 3

B. 4

C. 5

D. 2


Solution: Correct option : B

Prime factorisation of 196 = 2 × 2 × 7 × 7 = 22 × 72

The sum of the exponents of the prime factors in the prime factorisation of 196 is 2 + 2 = 4

 

Q 5. Euclid’s division Lemma states that for two positive integers a and b, there exists unique integer q and r satisfying a = bq + r, and

A. 0 < r < b

B. 0 < r < b

C. 0 ≤ r < b

D. 0 ≤ r ≤ b

Solution: Correct option : C

Euclid’s Division Lemma states that for two positive integers a and b, there exists unique integer q and r satisfying a = bq + r, and 0 ≤ r < b.

 

Q 6. The zeroes of the polynomial x2 – 3x – m(m+3) are

A. m, m+3

B. –m, m+3

C. m, –(m+3)

D. –m, –(m+3)

Solution: Correct option: B

p(x) = x2 – 3x – m(m + 3)

Substitute x = –m

p(–m) = (–m)2 – 3(–m) – m(m + 3)= m2 + 3m – m2 – 3m = 0  ...(i)

Substitute x = m + 3

p(m + 3) = (m + 3)2 – 3(m + 3) – m(m + 3)

= (m + 3)(m + 3 – 3 – m) ...(ii)

From (i) and (ii),

–m and m + 3 are the zeroes of the given polynomial.

 

Q 7. The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is

A. begin mathsize 12px style negative 14 over 3 end style

B. begin mathsize 12px style 2 over 5 end style

C. 5

D. 10

Solution: Correct option: D

 
 begin mathsize 11px style Here comma space straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 not equal to straight c subscript 1 over straight c subscript 2 space open parentheses Since comma space the space given space linear space equations space are space inconsistent close parentheses
rightwards double arrow straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2
rightwards double arrow 1 fifth equals 2 over straight k
rightwards double arrow straight k space equals space 10 end style
 

Q 8. The roots of the quadratic equation x2 – 0.04 = 0 are

A. ±0.2

B. ±0.02

C. 0.4

D. 2

Solution: Correct option: A

begin mathsize 11px style straight X squared space minus space 0.04 equals 0
rightwards double arrow straight X squared space equals space 0.04
rightwards double arrow straight X squared space equals 4 over 100 end style

 
Taking square root on both the side, we get
 
begin mathsize 11px style straight X equals plus-or-minus 2 over 10 equals plus-or-minus 0.2 end style
 

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

begin mathsize 12px style 1 over straight p comma fraction numerator 1 minus straight p over denominator straight p end fraction comma fraction numerator 1 minus 2 straight p over denominator straight p end fraction comma...... end style

A. 1

B. begin mathsize 12px style 1 over straight p end style

C. -1

D. 2

Solution: Correct option: C

begin mathsize 11px style common space difference equals fraction numerator 1 minus straight p over denominator straight p end fraction minus 1 over straight p equals fraction numerator 1 minus straight p minus 1 over denominator straight p end fraction equals fraction numerator negative straight p over denominator straight p end fraction equals negative 1 end style

 

Q 10. The nth 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,

an = a1 + (n – 1)d

= a + (n – 1)2a

= a + 2an – 2a

= 2an – 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 = begin mathsize 11px style 6 square root of 3 cm end style, AC = 12 cm and BC = 6 cm, then ∠B = 90°.

AB2 = begin mathsize 11px style open parentheses 6 square root of 3 close parentheses squared end style= 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 begin mathsize 12px style fraction numerator ar open parentheses increment AMN close parentheses over denominator ar open parentheses increment ABC close parentheses end fraction equals __________. end style

Solution: In fig. 3 MN || BC and AM: MB = 1: 2, then begin mathsize 11px style fraction numerator ar open parentheses increment AMN close parentheses over denominator ar open parentheses increment ABC close parentheses end fraction equals box enclose 1 over 9 end enclose end style.

AM: MB = 1: 2

Let AM = x and MB = 2x

⇒ AB = AM + MB = 3x

Here, ∆AMN ~ ∆ABC ... (AA test)

begin mathsize 11px style rightwards double arrow fraction numerator ar open parentheses increment AMN close parentheses over denominator ar open parentheses increment ABC close parentheses end fraction equals AM squared over AB squared equals straight X squared over open parentheses 3 straight X close parentheses squared equals 1 over 9 end style

 

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


OR

The value of begin mathsize 12px style fraction numerator tan space 35 degree over denominator cot space 55 degree end fraction plus fraction numerator cot space 78 degree over denominator tan space 12 degree end fraction end style  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 begin mathsize 12px style fraction numerator tan space 35 degree over denominator cot space 55 degree end fraction plus fraction numerator cot space 78 degree over denominator tan space 12 degree end fraction end style is 2.

Explanation:

begin mathsize 12px style fraction numerator tan space 35 degree over denominator cot space 55 degree end fraction plus fraction numerator cot space 78 degree over denominator tan space 12 degree end fraction
equals fraction numerator tan space 35 degree over denominator cot open parentheses 90 degree degree space minus space 35 degree close parentheses end fraction plus fraction numerator cot open parentheses 90 degree minus 12 degree close parentheses over denominator tan space 12 degree end fraction
equals fraction numerator tan space 35 degree over denominator tan space 35 degree end fraction plus fraction numerator tan space 12 degree over denominator tan space 12 degree end fraction
equals space 1 space plus thin space 1
equals 2 end style

 

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) =  begin mathsize 12px style fraction numerator straight n open parentheses straight A close parentheses over denominator straight n open parentheses straight S close parentheses end fraction equals 3 over 6 equals 1 half end style.

 

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 x< 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) =  begin mathsize 12px style fraction numerator straight n open parentheses straight A close parentheses over denominator straight n open parentheses straight S close parentheses end fraction equals 3 over 7 end style.

 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 = begin mathsize 12px style 5 over 7 end style

The probability that a randomly taken leap year has 52 Sundays is begin mathsize 12px style 5 over 7 end style.

 

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

 

= 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 = begin mathsize 12px style fraction numerator straight theta over denominator 360 degree end fraction cross times πr squared equals fraction numerator 30 degree over denominator 360 degree end fraction end style × 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 =  begin mathsize 12px style fraction numerator 20 plus 50 over denominator 2 end fraction equals 70 over 2 equals 35 end style


Class mark of the class 35 – 60 =  begin mathsize 12px style fraction numerator 35 plus 60 over denominator 2 end fraction equals 95 over 2 equals 47.5 end style

 

 

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 + begin mathsize 12px style square root of 3 straight x end root end style  + 7, 7x + begin mathsize 12px style square root of 7 end style , 5x3 – 7x + 2, 2x2 + 3 – begin mathsize 12px style 5 over straight x end style , 5x – begin mathsize 12px style 1 half end style , ax+ bx2 + cx + d, x + begin mathsize 12px style 1 over straight x end style .

 

Answer the following questions:

 

  1. How many of the above ten, are not polynomials?
  2. How many of the above ten, are quadratic polynomials?
 
Solution: 
 
  1. From the above ten, only three are not polynomials.

    They are begin mathsize 12px style straight x cubed space plus space square root of 3 straight x end root space plus space 7 comma space 2 straight x squared space plus thin space 3 space minus space 5 over straight x comma space straight x space plus space 1 over straight x end style  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  begin mathsize 12px style fraction numerator straight n open parentheses straight E close parentheses over denominator straight n open parentheses straight S close parentheses end fraction equals 2 over 6 equals 1 third. end style
  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  begin mathsize 12px style fraction numerator straight n open parentheses straight F close parentheses over denominator straight n open parentheses straight S close parentheses end fraction equals 1 over 6. end style
 

Q 23. In fig. 4, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that begin mathsize 12px style fraction numerator ar open parentheses increment ABC close parentheses over denominator ar open parentheses increment DBC close parentheses end fraction equals AO over DO end style.

 

 

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:  begin mathsize 12px style fraction numerator ar open parentheses increment ABC close parentheses over denominator ar open parentheses increment DBC close parentheses end fraction equals AO over DO end style
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
 
begin mathsize 12px style rightwards double arrow fraction numerator ar open parentheses increment AOM close parentheses over denominator ar open parentheses increment DON close parentheses end fraction equals AO squared over DO squared equals AM squared over DN squared
rightwards double arrow AO squared over DO squared equals AM squared over DN squared
Area space of space triangle space ABC space equals space 1 half space cross times space BC space cross times space AM
Area space of space triangle space DBC space equals space 1 half space cross times space BC space cross times space DN
rightwards double arrow fraction numerator ar open parentheses increment ABC close parentheses over denominator ar open parentheses increment DBC close parentheses end fraction equals fraction numerator begin display style 1 half end style cross times BC cross times AM over denominator begin display style 1 half end style cross times BC cross times DN end fraction equals AO over DO
Hence comma space fraction numerator ar open parentheses increment ABC close parentheses over denominator ar open parentheses increment DBC close parentheses end fraction equals AO over DO. end style
 
OR
 
In ΔADB,
AB2 = BD2 + AD2        Pythagoras theorem
⇒ AD2 = AB2 – BD2   … (i)
Also in ΔADC,
AC2 = CD2 + AD2         Pythagoras theorem
⇒ AD2 = AC2 – CD2   … (ii)
AB2 – BD2 = AC2 – CD2         from (i) and (ii)
⇒ AB2 + CD2 = BD2 + AC2
 
 

Q 24. Prove that begin mathsize 12px style 1 plus fraction numerator cot squared straight alpha over denominator 1 plus cosec space straight alpha end fraction equals cosec space straight alpha end style

  

OR

 

Show that tan4θ + tan2θ = sec4θ - sec2θ

 

Solution:

begin mathsize 12px style LHS space equals space 1 plus fraction numerator cot squared space straight alpha over denominator 1 plus cosec space straight alpha end fraction
equals 1 plus space fraction numerator cosec squared space straight alpha space minus 1 over denominator 1 plus cosec space straight alpha end fraction
equals 1 space plus space cosec space straight alpha space minus space 1
equals space cosec space straight alpha
Hence comma space 1 space plus fraction numerator cot squared space straight alpha over denominator 1 space plus space cosec space straight alpha end fraction equals cosec space straight alpha end style

 

 

OR


LHS =  tan4 θ +  tan2 θ 

= (sec2 θ - 1) 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 =  begin mathsize 12px style straight l plus open parentheses fraction numerator straight f subscript 1 minus straight f subscript 0 over denominator 2 straight f subscript 1 minus straight f subscript 0 minus straight f subscript 2 end fraction close parentheses cross times straight h equals 30 plus fraction numerator 10 minus 9 over denominator 20 minus 9 minus 3 end fraction cross times 5 space equals space 30.625 end style

 

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 - begin mathsize 12px style 1 third end styleπr2h

= begin mathsize 12px style 2 over 3 end styleπr2h

  

=  begin mathsize 12px style equals 2 over 3 cross times 22 over 7 cross times 6 cross times 6 cross times 14 end style

= 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 begin mathsize 12px style AQ equals 1 half open parentheses BC plus CA plus AB close parentheses end style.

 

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

begin mathsize 12px style 1 half end style (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

⇒ begin mathsize 12px style 22 over 7 end styler2 = 22176

⇒  r2 = 22176 × begin mathsize 12px style 7 over 22 end style

⇒ r = 84 cm

Circumference of a playground = 2πr2 = 2 × begin mathsize 12px style 22 over 7 end style × 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 


begin mathsize 12px style open parentheses fraction numerator 3 plus straight k over denominator 2 end fraction comma fraction numerator 4 plus 6 over denominator 2 end fraction close parentheses equals open parentheses fraction numerator 3 plus straight k over denominator 2 end fraction comma 5 close parentheses
Then space open parentheses fraction numerator 3 plus straight k over denominator 2 end fraction comma 5 close parentheses equals open parentheses straight x comma straight y close parentheses
rightwards double arrow fraction numerator 3 plus straight k over denominator 2 end fraction equals straight x space and space straight y space equals space 5 end style

      

It is given that x + y – 10 = 0 

   

 begin mathsize 12px style rightwards double arrow fraction numerator 3 plus straight k over denominator 2 end fraction plus 5 minus 10 equals 0 end style

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 = begin mathsize 12px style open parentheses fraction numerator 1 plus straight a over denominator 2 end fraction comma fraction numerator negative 4 plus straight b over denominator 2 end fraction close parentheses equals open parentheses 2 comma negative 1 close parentheses
end style

⇒ a = 3, b = 2

Midpoint of AC = begin mathsize 12px style open parentheses fraction numerator 1 plus x over denominator 2 end fraction comma fraction numerator negative 4 plus y over denominator 2 end fraction close parentheses equals open parentheses 0 comma negative 1 close parentheses
end style

⇒ x = -1, y = 2


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


Area of Δ ABC = begin mathsize 12px style 1 half end style x1(y2 - y3)+ x2(y3 - y1) + x3(y1 - y2)


begin mathsize 12px style 1 half end style 1(2 - 2 ) + 3(2 + 4) - 1(-4 - 2)

begin mathsize 12px style 1 half end style 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

 

begin mathsize 12px style rightwards double arrow AB over DE equals BC over EF equals AC over DF
rightwards double arrow fraction numerator 2 straight x minus 1 over denominator 18 end fraction equals fraction numerator 2 straight x plus 2 over denominator 3 straight x plus 9 end fraction equals fraction numerator 3 straight x over denominator 6 straight x end fraction equals 1 half
Consider comma
rightwards double arrow fraction numerator 2 straight x minus 1 over denominator 18 end fraction equals 1 half end style

 

⇒ 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 begin mathsize 12px style straight y over straight x minus 2 end style.

OR


Solve for  begin mathsize 12px style straight x space colon space fraction numerator 1 over denominator straight x space plus space 4 end fraction minus fraction numerator 1 over denominator straight x space minus space 7 end fraction equals 11 over 30 comma space straight x space not equal to space minus 4 comma space 7. end style

 

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

 

begin mathsize 12px style straight y over straight x minus 2 equals 9 over 7 minus 2 equals fraction numerator 9 minus 14 over denominator 7 end fraction equals fraction numerator negative 5 over denominator 7 end fraction end style

 

OR


According to the paper, condition given is x ≠ -4, 7 which implies that the equation should be  begin mathsize 12px style fraction numerator 1 over denominator straight x space plus space 4 end fraction minus fraction numerator 1 over denominator straight x space minus space 7 end fraction equals 11 over 30 end style.

begin mathsize 12px style fraction numerator 1 over denominator straight x space plus space 4 end fraction minus fraction numerator 1 over denominator straight x space minus space 7 end fraction equals 11 over 30 space where space straight x space not equal to space minus 4 comma space 7
rightwards double arrow space fraction numerator straight x space minus space 7 space minus space straight x space minus space 4 over denominator left parenthesis straight x space plus space 4 right parenthesis left parenthesis straight x space plus thin space 7 right parenthesis end fraction equals 11 over 30
rightwards double arrow fraction numerator negative 11 over denominator open parentheses straight x space plus space 4 close parentheses open parentheses straight x space minus space 7 close parentheses end fraction equals 11 over 30 end style


⇒ -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, 19begin mathsize 12px style 1 fourth end style, 18begin mathsize 12px style 1 half end style, 17begin mathsize 12px style 3 over 4 end style,....  is the first negative term.

 

OR

 

 

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

 

Solution: 20, 19begin mathsize 12px style 1 fourth end style, 18begin mathsize 12px style 1 half end style, 17begin mathsize 12px style 3 over 4 end style,...

The above sequence is an A.P.

⇒ a1 = 20 and d = begin mathsize 12px style 19 1 fourth minus 20 equals 77 over 4 minus 20 equals fraction numerator negative 3 over denominator 4 end fraction end style

Consider, an < 0 

⇒ a1 + (n – 1)d < 0

⇒  20 + (n - 1) × begin mathsize 12px style open parentheses fraction numerator negative 3 over denominator 4 end fraction close parentheses end style < 0

⇒ 80 - 3(n - 1) < 0

⇒ 80 – 3n + 3 < 0 

⇒ 83 < 3n 

⇒ n > begin mathsize 12px style open parentheses fraction numerator negative 3 over denominator 4 end fraction close parentheses end style  

Hence, 28th 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 an = 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: begin mathsize 12px style fraction numerator cos squared space open parentheses 45 to the power of degree space plus space straight theta close parentheses space plus space cos squared open parentheses 45 degree minus straight theta close parentheses over denominator tan space open parentheses 60 degree space plus space straight theta close parentheses space tan space open parentheses 30 degree space minus space straight theta close parentheses end fraction equals 1 end style

 

Solution: 

begin mathsize 12px style LHS space equals space fraction numerator cos squared open parentheses 45 degree space plus thin space straight theta close parentheses space plus space cos squared open parentheses 45 degree space minus space straight theta close parentheses over denominator tan open parentheses 60 degree space plus space straight theta close parentheses space tan open parentheses 30 degree space minus space straight theta close parentheses end fraction
equals fraction numerator sin squared open parentheses 90 degree minus 45 degree minus straight theta close parentheses space plus space cos squared open parentheses 45 degree minus straight theta close parentheses over denominator cot open parentheses 90 degree minus 60 degree minus straight theta close parentheses space tan open parentheses 30 degree minus straight theta close parentheses end fraction
equals fraction numerator sin squared open parentheses 45 degree minus straight theta close parentheses plus cos squared open parentheses 45 degree minus straight theta close parentheses over denominator cot open parentheses 30 degree minus straight theta close parentheses space tan open parentheses 30 degree minus straight theta close parentheses end fraction
equals 1
Hence comma space fraction numerator cos squared open parentheses 45 degree plus straight theta close parentheses plus cos squared open parentheses 45 degree minus straight theta close parentheses over denominator tan open parentheses 60 degree plus straight theta close parentheses space tan open parentheses 30 degree minus straight theta close parentheses end fraction equals 1 end style

 

 

 

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: 

 


 
 

 begin mathsize 12px style Mean equals fraction numerator sum fx over denominator sum straight f end fraction equals 18
rightwards double arrow fraction numerator 704 plus 20 straight f over denominator 40 plus straight f end fraction equals 18 end style

 

⇒ 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,

 

begin mathsize 12px style BC over CD equals tan space 45 degree
rightwards double arrow 20 over CD equals 1
end style 

CD = 20 m …(i)

In ΔACD,

 

begin mathsize 12px style rightwards double arrow AC over CD equals tan space 60 degree
rightwards double arrow fraction numerator AB plus BC over denominator CD end fraction equals square root of 3
rightwards double arrow fraction numerator AB plus 20 over denominator 20 end fraction equals square root of 3
rightwards double arrow AB equals 20 square root of 3 minus 20 equals 20 open parentheses square root of 3 minus 1 close parentheses end style 

 

 

Hence, the height of the tower is begin mathsize 12px style 20 open parentheses square root of 3 minus 1 close parentheses straight m. end style


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 begin mathsize 12px style 1 over straight x end style  and by the pipe is begin mathsize 12px style 1 over straight y end style .

In 1 hour the quantity of water filled by both the pipes is begin mathsize 12px style 1 over straight x plus 1 over straight y end style

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


begin mathsize 12px style 1 over straight x plus 1 over straight y equals 1 over 12.... comma comma left parenthesis straight i right parenthesis end style…(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  begin mathsize 12px style 4 over straight x end style.


In 9 hours the quantity of water filled by pipe B is  begin mathsize 12px style 9 over straight y end style.

According to the question,

 

 begin mathsize 12px style 4 over straight x plus 9 over straight y equals 1 half.... open parentheses ii close parentheses
Put space 1 over straight x equals straight m space and space 1 over straight y equals straight n
straight m space plus space straight n equals 1 over 12... left parenthesis iii right parenthesis
4 straight m space plus thin space 9 straight n equals 1 half.... left parenthesis iv right parenthesis end style 

 

Multiplying (i) by 4 we get

 

4m + 4n = begin mathsize 12px style 1 third end style  …(v)

Subtracting (v) from (iv) we get

 

n = begin mathsize 12px style 1 over 30 end style

 

Put it in (iii) we get

 

begin mathsize 12px style straight m space plus thin space 1 over 30 equals 1 over 12 rightwards double arrow straight m equals 1 over 20 end style

 

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 begin mathsize 12px style square root of 5 end style is an irrational number.

Solution: Let us assume that begin mathsize 12px style square root of 5 end style is a rational number

We can write it in the form of begin mathsize 12px style straight p over straight q end style where q ≠ 0.

Also p and q are co-prime numbers.

 

begin mathsize 12px style straight p over straight q equals square root of 5
rightwards double arrow straight p squared over straight q squared equals 5
rightwards double arrow straight P squared equals 5 straight q squared end style 

 

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, begin mathsize 12px style square root of 5 end style  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 begin mathsize 12px style 2 over 3 end style  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.
 
OR
 
Steps of construction:
Step 1: Draw a line segment AB = 6 cm
Step 2: Draw an angle of 60° at B so that ∠XBA = 60°
Step 3: With centre B and radius 5 cm, draw an arc intersecting XB at C
Step 4: Join AC, Δ ABC is the required triangle.
Step 5: Draw a line BY below BC which makes an acute angle ABY.
Step 6: Mark B1, B2 and B3 at equal distances from B such that BB= B1B2  = B2B3
Step 7: Join AB3
Step 8: Draw B2A parallel to AB3
Step 9: Draw C' A'  parallel to CA through A’ intersecting BC at C’
 
ΔA’BC’ is the required similar triangle.
 
 
Q 40. A solid is in the shape of hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7cm and height of cone is 3.5cm, find the volume of the solid. (Take π = begin mathsize 12px style 22 over 7 end style).

Solution: Hemisphere: r = 7 cm

 h = 3.5 cm, r = 7 cm

Volume of cone = begin mathsize 12px style 1 third πr squared straight h equals 1 third cross times 22 over 7 cross times 7 cross times 7 cross times 3.5 equals 179.6 space cm cubed end style
Volume of hemisphere = begin mathsize 12px style 2 over 3 straight pi cubed equals 2 over 3 cross times 22 over 7 cross times 7 cross times 7 cross times 7 equals 718.6 space cm cubed end style
Volume of solid = 718.6 - 179.6 = 898.2 cm3

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