Mathematics | Class 10th

CBSE Board Paper 2019

2

1. Find the coordinates of a point A, where AB is a diameter of a circle whose center is (2, -3) and B is the point (1, 4).

CBSE Board Paper 2019 Set - 1

General Instructions:

1. All Questions are compulsory.

2. This question paper consists of 30 questions divided into four

sections – A, B, C and D.

3. 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.

4. There is no overall choice. However, an internal choice has been

provided in two questions of 1 mark, two questions of 2 marks,

four questions of 3 marks each and three questions of 4 marks

each. You have to attempt only one of the alternatives in all such

questions.

5. Use of calculator is not permitted.

Time allowed: 3 Hours Max Marks: 80

Section A

1

3

2. For what values of k, the roots of equation x2 + 4x + k = 0 are real?

1

Find the value of k for which the roots of equation 3x2 – 10x + k = 0 are reciprocal of each other?

OR

3. Find A if tan 2A = cot (A - 24⁰)

1

OR

Find the value of (sin233⁰ + sin257⁰)

4. How many two digits numbers are divisible by 3?

1

5. In Fig. 1, DE ll BC, AD = 1 cm and BD = 2 cm. What is the ratio of the ar (∆ ABC) to the ar (∆ ADE)?

1

6. Find a rational number between √2 and √3.

1

2

Section B

7. Find the HCF of 1260 and 7344 using Euclid’s algorithm.

4

Show that every positive odd integer is of the form (4q + 1) or (4q + 3), where q is some integer.

OR

8. Which term of the AP 3, 15, 27, 39, ……… will be 120 more than its 21st term?

2

9. Find the ratio in which the segment joining the points (1, -3) and (4, 5) is divided by x-axis? Also find the coordinates of this point on x-axis.

2

10. A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all three tosses is a success, find the probability of losing the game.

2

11. A die is thrown once. Find the probability of getting a number which (i) is a prime number (ii) lies between 2 and 6.

2

12. Find c if the system of equations cx + 3y + (3 – c) = 0 has 12x + cy – c = 0 has infinitely many solutions?

2

5

Section C

13. Prove that √2 is an irrational number. 3

14. Find the value of k such that the polynomial x2 – (k + 6) x + 2(2k – 1) has sum of its zeroes equal to half of their product.

3

15. A father’s age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father.

3

OR

A fraction becomes 1/3 when 2 is subtracted from the numerator and it becomes 1/2 when 1 is subtracted from the denominator. Find the fraction.

16. Find the point on y-axis which is equidistant from the points (5, -2) and (-3, 2).

3

OR

The line segment joining the points A (2, 1) and B (5, -8) is trisected at the points P and Q such that P is nearer to A. If P lies on the line given by 2x – y + k = 0, find the values of k.

17. Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2θ + cot2θ.

3

OR

Prove that (1 + cot A – cosec A) (1 + tan A + sec A) = 2.

6

18. In Fig. 2, PQ is a chord of length 8 cm of a circle of radius 5 cm and center O. The tangents at P and Q intersect at point T. Find the length of TP.

3

19. In Fig. 3, ∠ACB = 90⁰ and CD ⊥ AB, prove that CD2 = BD × AD.

3

OR

If P and Q are the points on side CA and CB respectively of ∆ ABC, right angled at C, prove that (AQ2 + BP2) = (AB2 + PQ2)

20. Find the area of shaded region in Fig. 4, if ABCD is a

rectangle with sides 8 cm and 6 cm and O is the center of circle. (Take π = 3.14)

3

7

Section D

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

3

22. Find the mode of the following frequency distribution.

23. Two water taps together can fill a tank in

hours. The

tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.

OR

24. If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

3

4

4

8

25. Prove that

4

26. A man in a boat rowing away from a light house 100 m high takes 2 minutes to change the angle of elevation of the top of the light house from 60⁰ to 30⁰. Find the speed of the boat in meters per minute. [Use √3 = 1.732]

4

OR

Two poles of equal heights are standing opposite to each other on either side of the road, which is 80 m wide. From a point between them on the road, the angle of elevation of top of the poles are 60⁰ and 30⁰ respectively. Find the height of the poles and the distances of the point from the poles.

27. Construct a ∆ ABC in which CA = 6 cm, AB = 5 cm

and BAC = 45⁰. Then construct a triangle whose sides

are of the corresponding sides of a ∆ ABC.

4

28. A bucket open at the top is in the form of a frustum of a coin with a capacity of 12308.8 cm3. The radii of the top and bottom of circular ends of the bucket are 20 cm and 12 cm respectively. Find the height of the bucket and also the area of the metal sheet used in making it. (Use π = 3.14)

4

9

29. Prove that in a right-angle triangle the square of the hypotenuse is equal the sum of squares of the other two sides.

4

30. If the median of the following frequency distribution is 32.5. Find the values of f1 and f2.

4

OR

The marks obtained by 100 students of a class in an examination are given below.

Marks 0 -5

5-10

10-15

15-20

20-25

25-30

30-35

35-40

40-45

45-50

No. of Students

2 5 6 8 10 25 20 18 4 2

Draw ‘a less than’ type cumulative frequency curves (ogive). Hence find median.

10

Set - 2

1. Find the coordinates of a point A, where AB is a diameter of a circle with center (-2,2) and B is the point (3, 4).

1

7. find the value of k for which the following pair of linear equations have infinitely many solutions.

2x + 3y = 7

(k+1) x + (2k-1) y = 4k + 1

2

13. The arithmetic mean of the following frequency distribution is 53. Find the value of k.

3

14. Find the area of the segment shown in Fig. 2, if radius of the circle is 21 cm and ∠AOB = 120°. (use π = 22/7)

3

11

16. In Fig. 4, a circle is inscribed in a ΔABC having sides BC = 8 cm, AB = 10 cm and AC = 12 cm. Find the lengths BL, CM and AN.

3

23. Prove that

4

24. The first term of an AP is 3, the last term is 83 and the sum of all its terms is 903. Find the numbers of terms and the common difference of the AP.

4

25. Construct a triangle ABC with side BC = 6 cm, ∠B = 45°, ∠ A = 105°. Then construct another triangle whose sides are 3/4 times the corresponding sides of the Δ ABC.

4

12

Set - 3

1. Two positive integers a and b can be written as a = x3y2 and b = xy3. x, y are prime numbers. Find Lcm (a,b).

1

7. Find how many two-digit numbers are divisible by 7?

2

If the sum of first n terms of an AP is n2, then find its 10th term.

OR

13. Find all the zeroes of the polynomial 3x3 + 10x2 – 9x – 4 if one of its zero is 1.

3

15. Prove that is an irrational number, given that √3 is

an irrational number.

3

23. If 1

sec x , x 0,

4x

= + find (sec θ + tan θ).

4

24. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

4

13

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

4

OR

The table below shows the daily expenditure on food of 25 households in a locality

14

1. We know that center is the midpoint of diameter of circle. By section formula which states that if a point P (x, y) divides the line with end points A (x1, y1) and B (x2, y2) in the ration m: n, the coordinates of x and y are:

As midpoint divides the line in the ratio 1:1,

Then, A (x, y) is

⇒ 4 = x + 1 ⇒ x = 3

and

⇒ - 6 = y + 4 ⇒ y = -10

∴ coordinates of A are (3, -10).

2. For roots to be real D ≥ 0

⇒ b2 − 4ac ≥ 0 For x2 + 4x + k = 0,

⇒ 42 − 4(1)k ≥ 0 ⇒ 16 – 4k ≥ 0

⇒ – 4k ≥ − 16 ⇒ 4k ≤ 16

⇒ 𝐤 ≤ 𝟒

Solutions (Set-1)

Section A (Solutions)

15

Consider 3x2 – 10x + k = 0,

Compare the given equation with the general equation ax2 + bx + c = 0 So, a = 3, b = -10, c = 3

Let, the roots of the equation be x and 1

𝑥.

Product of the roots =𝑐

𝑎

⇒ k = 3

3. tan 2A = cot (A – 24⁰)

We know that tan θ = cot (90⁰ - θ)

cot (90⁰ - 2A) = cot (A – 24⁰)

⇒ 90⁰ - 2A = A – 24⁰

⇒ 90⁰ + 24⁰ = A + 2A

⇒ 90⁰ + 24⁰ = 3A

⇒ 114⁰ = 3A

⇒ A = 38⁰

Consider (sin233⁰ + sin257⁰),

We know that sin θ = cos (90⁰ - θ)

sin233⁰ + sin257⁰ = sin233⁰ + cos2(90⁰ - 57⁰)

= sin233⁰ + cos233⁰ = 1

∴ (sin233⁰ + sin257⁰) = 1

OR

OR

16

4. 1st two-digit number which is divisible by 3 is 12.

2nd two-digit number which is divisible by 3 is 15.

3rd two-digit number which is divisible by 3 is 18.

Last two-digit number which is divisible by 3 is 99.

12, 15, 18, ……, 99

This is an AP as the difference between the consecutive numbers is the same.

15 − 12 = 18 − 15 = 3

We know the nth term of AP is given by:

an = a + (n – 1) d

⇒ 99 = 12 + (n – 1)3 ⇒ 99 = 12 + 3n − 3

⇒ 99 = 9 + 3n ⇒ 99 − 9 = 3n

⇒ 90 = 3n

n = 30

So, there are 30 two-digit numbers which are divisible by 3.

5. In ∆ ADE and ∆ ABC

∠ ADE = ∠ ABC (alternate angles)

∠ DAE = ∠ BAC (same angle)

{By AA similarity criteria; which states that if two triangles have one pair of congruent angles, then the triangles are similar.}

∆ ADE ~ ∆ ABC

We know that if two triangles are similar then the ratio of their areas is equal to the square of alternating sides.

The ratio of the ar (∆ ABC) to the ar(∆ ADE)

Ratio of the ar (∆ ABC) to the ar (∆ ADE) = 9: 1

17

6. We know that

√2 = 1.414

… and √3 = 1.732…

There are infinite rational numbers between two irrational numbers.

They are terminating or not terminating but repeating.

So, numbers between √2 and √3 are just greater than 1.41 and less than 1.73.

They can be 1.45, 1.61, 1.6363…, 1.54 and so on.

18

7. By Euclid's algorithm

We know that

Dividend = Divisor × Quotient + Remainder

⇒ 7344 = 1260 × 5 + 1044

⇒ 1260 = 1044 × 1 + 216

⇒ 1044 = 216 × 4 + 180

⇒ 216 = 180 × 1 + 36 ⇒ 180 = 36 × 5 + 0

So, HCF of 1260 and 7344 is 36 because when the divisor is 36 remainder is 0.

We know by Euclid's algorithm if a and b are two positive integers, there exist unique integers q and r satisfying,

a = bq + r where 0 ≤ r < b

Let, b = 4

a = 4q + r, 0 ≤ r < 4

So, r can be 0,1,2 and 3.

If r = 0,

a = 4q

= 2(2q)

= 2n (where n = 2q)

which is an even integer.

If r = 1,

a = 4q + 1

= 2(2q) + 1

= 2n + 1 (where n = 2q)

which is an odd integer.

If r = 2,

Section B (Solutions)

OR

19

a = 4q + 2

= 2(2q + 1 )

= 2n (where n = 2q + 1)

which is an even integer

If r = 3,

a = 4q + 3

= 4q + 2 + 1 = 2(2q + 1) + 1

= 2n + 1 (where n = 2q + 1)

which is an odd integer

Thus "a" can be 4q, 4q + 1, 4q + 2, 4q + 3.

4q or 4q + 2 is not possible because they are even integers.

So, 4q + 1 or 4q + 3 are representing odd integers.

Hence, every positive odd integer is of the form (4q + 1) or (4q + 3).

8. Consider the AP 3, 15, 27, 39, ……

We know that the nth term of an AP is given by:

𝑎𝑛 = 𝑎 + (𝑛 – 1)𝑑

The 21st term will be:

𝑎21 = 3 + (21 – 1)12

⇒ 𝑎21 = 3 + (20)12 ⇒ 𝑎21 = 3 + 240

⇒ 𝑎21 = 243

According to the question,

𝑎𝑛 = 120 + 𝑎21

⇒ 𝑎 + (𝑛 – 1)𝑑 = 120 + 243

⇒ 3 + (𝑛 – 1)12 = 363

⇒ 3 + 12𝑛 − 12 = 363 ⇒ 12𝑛 − 9 = 363

⇒ 12𝑛 = 372 ⇒ 𝑛 = 31

So, 31st term of the AP is 120 more than its 21st term i.e. 363.

20

Sn = 3n2 – 4n

At n = 1, S1 = -1

At n = 2, S2 = 4

At n = 3, S3 = 15

We know that an AP is also represented by sum of nth terms.

AP ⇒ S1, (S2 – S1), (S3 – S2), … …

AP ⇒ −1, (4 + 1), (15 − 4), . . . . . ..

AP ⇒ −1, 5, 11, … …

an = a + (n – 1)d

an = −1 + (n – 1)6

= −1 + 6n − 6

= 6n – 7

So, nth term of an AP is given by 6n – 7.

9.

By section formula which states that if a point P (x, y) divides

the line with endpoints A (x1, y1) and B (x2, y2) in the ratio m:n, the coordinates of x and y are:

OR

21

If the segment is divided by the x-axis then the point at x-axis

is (x, 0).

Let, the ratio of segment AB be m: 1.

So, the ratio is 3: 5.

From the ratio we can find the co-ordinate

So, the coordinate of point is

.

10. If we toss a coin three times, we will get the following outcomes.

TTT, TTH, THT, THH, HTT, HTH, HHT, HHH

Now success is getting the same result in 3 tosses (TTT, HHH) i.e. 2 outcomes.

The outcomes where the game will be lost are (TTH, THT, THH, HTT, HTH, HHT) i.e. 6 outcomes.

As we know,

probablity =no.of favourable outcomes

total no.of possible outcomes

So,

Probability of losing the game is:

11. If a die is thrown once then the outcomes are -

1, 2, 3, 4, 5, 6

We know,

22

probablity =no.of favourable outcomes

total no.of possible outcomes

(i) getting a prime number

As prime numbers are 2,3,5.

Probability of getting a prime number is:

(ii) Number lies between 2 and 6

Numbers between 2 and 6 are 3,4,5.

Probability of getting a number between 2 and 6 is:

12. Consider cx + 3y + (3 – c) = 0 and 12x + cy – c = 0,

We know that if the system of linear equation

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has infinitely many solutions then

On comparing with given equations

Now,

⇒ c2 = 36 and 3(−c) = c(3 – c)

⇒ c = √36 and −3c = 3c − c2

⇒ c = ±6 and −3c − 3c = − c2

⇒ c = ±6 and −6c = − c2

⇒ c = ±6 and c2 − 6c = 0

⇒ c = ±6 and c(c − 6) = 0

23

⇒ c = −6,6 and c = 0,6

Hence, at c = 6, the system of the equations has infinitely many solutions.

13. Let us suppose that √2 is a rational number and represent it in

the form of where p and q are relatively prime and q≠0.

So,

On squaring both sides we get,

p2 = 2q2

As 2q2 is even then p2 is also even and p is also even.

Let, p = 2a

so p2 = 4a2

4a2 = 2q2

⇒ q2 = 2𝑎2

As 2a2 is even then q2 is also even and q is also even.

Section C (Solutions)

24

If p and q both are even, then they are further calculated which is not possible because of we assumed p and q are relatively prime number.

It means our assumption is wrong. So, by contradiction we can say that √2 is irrational number.

14. Consider the polynomial x2 – (k + 6) x + 2(2k – 1),

Compare it with general equation ax2 + bx + c,

Here, a = 1, b = -(k + 6) and c = 2(2k - 1)

We know,

⇒ Sum of zeroes = −b

a

= −[−(𝑘+6)]

1

⇒ Product of zeroes =c

a

According to the question,

Sum of zeroes = 1/2 Product of zeroes

⇒ 𝑘 + 6 = 2𝑘 − 1

⇒ 6 + 1 = 2𝑘 − 𝑘

⇒ 7 = 𝑘

15. Let, the present age of the father is x and the sum of age of two children be x1 and x2.

25

According to the question, the father’s age is three times the sum of the ages of his two children.

⇒ x = 3(x1 + x2) … (1)

According to the question after 5 years the father's age will be two times the sum of their ages.

x + 5 = 2(x1 + 5 + x2 + 5)

⇒ x + 5 = 2(x1 + 5 + x2 + 5)

⇒ x + 5 = 2(x1 + x2 + 10)

⇒ x + 5 = 2(x1 + x2 ) + 20

⇒ x = 2(x1 + x2) + 15 … (2)

Equate eq (1) and (2) to get,

⇒ 3(x1 + x2) = 2(x1 + x2) + 15

⇒ (x1 + x2) = 15

Put this value in (1) to get,

⇒ x = 3(15) ⇒ x = 45

So, present age of father is 45 years.

Let,

be the fraction.

According to the question, a fraction becomes 1/3 when 2 is subtracted from the numerator.

⇒ 3(𝑝 – 2) = 𝑞 … (1)

According to the question, a fraction becomes 1/2 when 1 is subtracted from the denominator.

⇒ 2𝑝 = 𝑞 – 1

⇒ 2𝑝 + 1 = 𝑞 … (2)

OR

26

equate (1) and (2),

⇒ 3(𝑝 – 2) = 2𝑝 + 1

⇒ 3𝑝 − 6 = 2𝑝 + 1

⇒ 𝑝 = 7

Put the value of p in (2) to get,

⇒ 2(7) + 1 = 𝑞

⇒ 14 + 1 = 𝑞

⇒ 15 = 𝑞

So, p = 7 and q = 15

Hence, the fraction is .

16.

Let, the equidistant point is (0, y)

If it is equidistant from (5, -2) and (-3, 2) then AC = BC

&

27

As, AC = BC

Squaring both sides, we get,

⇒ y2 + 4y + 29 = y2 − 4y + 13

⇒ 4y + 4y + 29 − 13 = 0

⇒ 8y + 16 = 0

⇒ 8y = −16

⇒ y = −2

So, the equidistant point is (0, -2)

Consider the points A (2, 1) and B (5, -8).

If P is one of the trisected points then the ratio of AB at

P = 1: 2

Let, coordinates of P are (x, y).

By section formula which states that if a point P (x, y) divides the line with endpoints A (x1, y1) and B (x2, y2) in the ration m: n, the coordinates of x and y are:

The coordinates of P are:

OR

28

x = 3 and y = -2

So, the coordinates of P are (3, -2).

P satisfies the equation of the given line 2x – y + k = 0.

⇒ 2(3) – (−2) + k = 0

⇒ 6 + 2 + k = 0

⇒ 8 + k = 0

⇒ k = − 8

17. LHS = (sin θ + cosec θ)2 + (cos θ + sec θ)2

As we know,

(a + b)2 = a2 + b2 + 2ab

= sin2θ + cosec2θ + 2 sin θ cosec θ + cos2θ + sec2θ + 2 cos θ sec θ

We know,

sin2θ + cos2θ = 1, cosec2θ = 1 + cot2θ and sec2θ = 1 + tan2θ

sin θ = 1/ cosec θ and sec θ = 1/cos θ

= 1 + 1 + cot2θ + 2 + 1 + tan2θ + 2

= 7 + tan2θ + cot2θ

= RHS

Hence proved.

LHS ⇒ (1 + cot A – cosec A)(1 + tan A + sec A)

= 1(1 + tan A + sec A) + cot A (1 + tan A + sec A) − cosec A (1 + tan A + sec A)

= 1 + tan A + sec A + cot A + cot A tan A + cot A sec A – cosec A – cosec A tan A – cosec A sec A

OR

29

= 2

= RHS

Hence proved.

18.

Join the point T with O.

By theorem, we know that the lengths of tangents from the external point are equal.

So, TP = TQ

In ΔTPQ,

TP = TQ

As two sides are equal, so the ΔTPQ is an isosceles triangle.

Here, OT is the bisector of ∠PTQ,

So, OT ⊥ PQ

∴ PR = RQ (perpendicular from center to the chord bisects the chord)

So,

PR = QR =1

2 PQ

30

=1

2(8)

= 4 cm

In ∆ OPR

OP2 = OR2 + PR2

52 = OR2 + 42

⇒ 25 = OR2 + 16

⇒ 9 = OR2

⇒ OR = 3 cm

In ∆ OPT

OT2 = OP2 + PT2

(TR + 3)2 = 52 + PT2 … (1)

In ∆ PRT

PT2 = TR2 + RP2

PT2 = TR2 + 42 … (2)

Put the value of PT2 in (1),

⇒ (TR + 3)2 = 52 + TR2 + 42

⇒ TR2 + 9 + 6TR = 25 + TR2 + 16

⇒ 6TR + 9 = 25 + 16 ⇒ 6TR = 25 + 16 − 9

⇒ 6TR = 32

From (2),

PT2 = (16

3)

2 + 16

31

So, TP = 20/3 cm

19. By Pythagoras theorem

In ∆ ABC

AB2 = AC2 + BC2 … (1)

In ∆ ACD

AC2 = AD2 + DC2 … (2)

In ∆ BCD

BC2 = BD2 + CD2 … (3)

Add eq (2) and eq (3)

AC2 + BC2 = AD2 + DC2 + BD2 + CD2

AB2 = AD2 + 2CD2 + BD2

AB2 − AD2 − BD2 = 2CD2

(AB – AD)(AB + AD) – BD2 = 2CD2

BD(AB + AD – BD) = 2CD2

BD(2AD) = 2CD2

CD2 = BD × AD

OR

32

Construction:

Join P to Q and A to Q.

In Δ ACQ,

AQ2 = AC2 + CQ2

In Δ PCB,

BP2 = PC2 + BC2

In ΔACB,

AB2 = AC2 + BC2

In ΔPCQ,

PQ2 = PC2 + CQ2

LHS:

AQ2 + BP2 = (AC2 + CQ2) + (PC2 + BC2)

RHS:

AB2 + PQ2 = (AC2 + BC2) + (PC2 + CQ2)

= (AC2 + CQ2) + (PC2 + BC2)

LHS = RHS

Hence proved.

20. In ∆ ADC

AC2 = AD2 + DC2

AC2 = 62 + 82

AC2 = 36 + 64

AC2 = 100 ⇒ AC = √100

AC = 10 cm

Radius of circle = AC/2 = 5 cm

Area of shaded region = area of circle – area of rectangle

= π(AO)2 – AD × DC = 3.14 × 25 – 6 × 8

= 30.5 cm2

33

21. We have the diagram of the canal as,

Given: Height of the canal = 1.5 m

Breadth of canal = 6 m

Speed of the water in the canal = 10 km/h

If speed of water = 10 km/h

⇒ In 1 hr., distance covered by water = 10 km

⇒ In 60 minutes, distance covered by water = 10 km

⇒ In 1 minute, distance covered by water = 10/60 km = 1/6 km

⇒ In 30 minutes, distance covered by water = 30 × 1/6 = 5 km = 5000 m [∵ 1 km = 1000m]

Volume of canal is given by,

Volume of canal = length × breadth × height

⇒ Volume of canal = 5000 × 6 × 1.5 [∵ length = 5000 m, breadth = 6 m & height = 1.5 m] …(i)

Now, we can understand that volume of the canal will itself be the volume of the area that needs to be irrigated. So,

Volume of the canal = Volume of the area of irrigation

⇒ 5000 × 6 × 1.5 = Area irrigated × height of irrigation

⇒ 5000 × 6 × 1.5 = Area irrigated × 8 cm

⇒ 5000 × 6 × 1.5 = Area irrigated × 8/100 m

⇒ Area irrigated =

⇒ Area irrigated = 562500 m2

Hence, it will irrigate 562500 m2 area in 30 minutes, if 8 cm of standing water is needed.

34

22.

Modal class = class which has maximum frequency = 30-40

l = lower limit of modal class = 30

h = class size = 10

f1 = frequency of modal class = 16

f0 = frequency of class preceding the modal class = 10

f2 = frequency of class succeeding the modal class = 12

⇒

= 30 + 6 = 36

23. Let, the time taken by smaller tap to fill the tank completely = x hours

The volume of the tank filled by a smaller tap in 1 hour =

Section D (Solutions)

35

The time taken by larger tap to fill the tank completely = x – 2 hours

The volume of the tank filled by a larger tap in 1 hour

Time taken by the tank to fill hours

hours

The volume of the tank filled by a smaller tap in

hours

The volume of the tank filled by a larger tap in hours

⇒ 15(x - 2) + 15x = 8x(x-2)

⇒ 15(x – 2 + x) = 8x (x – 2)

⇒ 15(2x - 2) = 8x2 - 16

⇒ 30x - 30 = 8x2 - 16

⇒ 8x2 – 46x + 30 = 0

⇒ 2(4x2 – 23x + 15) = 0

⇒ 4x2 – 23x + 15 = 0

⇒ 4x2 – 20x - 3x + 15 = 0

⇒ 4x (x -5) - 3(x - 5) = 0

⇒ (4x-3) (x -5) = 0

At x = 5

Time taken by smaller tap to fill the tank completely = 5 hours

Time taken by larger tap to fill tank completely = 5 – 2 = 3 hours

At

Time taken by smaller tap to fill the tank completely hours

Time taken by larger tap to fill the tank completely

36

hours

Time is not possible to be negative.

So, is not possible.

Hence the taps can fill the tank in 5hrs and 3 hrs.

Let, speed of the stream = x km/h

And the speed of the boat = y km/h

So, the speed of the upstream = (y – x) km/h

And the speed of the downstream = (y + x) km/h

… (1)

… (2)

Let,

The equations will be:

30u + 44v - 10 = 0

40u + 55v - 13 = 0

By cross multiplication method we know,

Here a1 = 30, b1 = 44, c1 = -10

a2 = 40, b2 = 55, c2 = -13

So,

OR

37

Now,

⇒ 10 = 2(y - x)

⇒ 10 = 2y - 2x

⇒ 5 = y - x .... (3)

⇒ 5 + x = y

And

⇒ y + x = 11 .... (4)

Put the value of y in eq (4),

⇒ 5 + x + x = 11

⇒ 5 + 2x = 11

⇒ 2x = 6 ⇒ x = 3

Put the value of x in (3) to get,

⇒ 5 = y – 3 ⇒ y = 8

So, speed of the stream = 3 km/h

And the speed of the boat = 8 km/h

24. Given,

S4 = 40

S14 = 280

We know that,

38

⇒ 2(2a + 3d) = 40

⇒ 4a + 12d = 40

⇒ 2a + 3d = 20 … (1)

⇒ 7(2a + 13d) = 280

⇒ 14a + 91 = 280

⇒ 2a + 13d = 40 … (2)

Subtract eq 1 from eq 2 to get,

⇒ 2a + 13d - (2a + 3d) = 40 - 20

⇒ 2a + 13d - 2a - 3d = 40 - 20

⇒ 10d = 20

⇒ d = 2

Put the value of d in (1) to get,

⇒ 2a + 3(2) = 20

⇒ 2a + 6 = 20

⇒ 2a = 20 – 6

⇒ 2a = 14 ⇒ a = 7

= n2 + 6n

25. LHS

Multiply and divide by (sin A + cos A + 1) to get,

39

As we know,

(a-b) (a + b) = a2 - b2

As we know,

(a + b)2 = a2 + b2 + 2ab

As we know,

1 - cos2A = sin2A

RHS

LHS = RHS

Hence proved.

26.

40

ACB = 60⁰ and ADB = 30⁰

Let, the length of CD = x m

And the length of CB = y m

⇒ y = 57.73 m

⇒ x + y = 173 m ⇒ x = 115.27 m

Speed of the boat = Distance/Time

= 57.635 m/min

AEB = 60⁰ and DEC = 30⁰

Let, length of BE = x m

And length of two equal poles = h m

Then, length of EC = 80 – x m

OR

41

… (1)

… (2)

From eq (1) and eq (2)

80 – x = 3x ⇒ x = 20 m

h = 20√3 m

So, the length of BE = 20 m

length of EC = 80 – 20 m = 60 m

length of two equal poles = 20√3 m

27. For ∆ ABC

Step 1: Draw the line of 5 cm with the help of scale and name it AB. Also draw the arc with the help of compass from point A.

Step 2: Open the compass at angle of 45⁰ with the help of

Protector. Mark the angle 0f 45⁰ on arc with the help of compass from point E. Joint the arc with point A and mark 6 cm on it.

42

Step 3: Joint Point C with Point B.

For corresponding triangle

Step 1: Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

Step 2: Cut the BD in 5 equal parts starting from B.

Step 3: Meet the point S with C and draw the angle equal to

BSC from the O with the help of compass i.e., draw line parallel to SC.

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Step 4: Draw the line parallel to AC from point C1 with help of compass.

∆ A1BC1 is our new triangle whose corresponding side is of the given triangle.

28. Volume of the bucket = 12308.8 cm3

Radius of top r1 = 20 cm

Radius of bottom r2 = 12 cm

Volume of Bucket

⇒

h = 5 × 3

h = 15 cm

Surface area of the metal sheet used = πr22 + π (r1 + r2) l

= √289 = 17 cm

Surface area of the metal sheet = π (12)2 + π (20 + 12)17

44

= 3.14×144 + 3.14×32×17

= 3.14(144 + 32×17) = 3.14 (144 + 544)

= 3.14 (688) = 2160.32 cm2

29. Draw a BD AC

In ∆ ADB and ∆ ABC

∠ BAD = ∠ BAC (same angle)

∠ ADB = ∠ ABC (90⁰)

By the AA criterion for triangle, similarity which states that if two triangles have two pairs of congruent angles, then the triangles are similar.

∆ ADB ≈ ∆ ABC

⇒ AB2 = AC×AD … (1)

In ∆ BDC and ∆ ABC

∠ DCB = ∠ ACB (same angle)

BDC = ABC (90⁰)

By AA theorem

∆ BDC ≈ ∆ ABC

⇒ BC2 = AC×CD … (2)

On adding eq (1) and eq (2)

AC×AD + AC×CD = AB2 + BC2

AC (AD + CD) = AB2 + BC2

AC×AC = AB2 + BC2

AC2 = AB2 + BC2

Hence proved.

45

30.

Sum of frequency = 40

f1 + 5 + 9 + 12 + f2 + 3 + 2 = 40

f1 + f2 = 9

l = lower limit of median class (30 – 40) = 30

n = number of observations = 31 + f1 + f2

cf = cumulative frequency of class preceding to the median

class = 14 + f1

f = frequency of the median class = 12

h = class size = 10

Median = l + (n

2−cf

f) × h

46

2.5 × 1.2 = 20 – 14 - f1

f1 = 3

f2 = 9 – 3

f2 = 6

n = number of observations = 100

n/2 = 50

median class = 56

l = lower limit of median class (25 – 30) = 25

cf = cumulative frequency of class preceding to the median

class = 31

f = frequency of the median class = 25

h = class size = 5

Median = 28.8

OR

47

48

1.

Let O be the center of AB.

Since AB is the diameter, so O is the midpoint of AB.

By section formula which states that if a point P (x, y) divides the line with end points A (x1, y1) and B (x2, y2) in the ration m: n, the coordinates of x and y are:

As midpoint divides the line in the ratio 1:1,

Then, the coordinates of A (x, y) can be calculated as:

⇒ a + 3 = 2(-2)

⇒ a + 3 = -4

⇒ a = -7

And,

⇒ b + 4 = 2(2) ⇒ b + 4 = 4

⇒ b = 0

Hence, coordinate of point A is (-7,0).

Solutions (Set-2)

49

7. We have,

2x + 3y = 7

⇒ 2x + 3y – 7 = 0 …. (1)

(k+1) x + (2k-1) y = 4k + 1

⇒ (k +1) x + (2k – 1) y – (4k+1) = 0 …. (2)

For the equations of the form:

a1x+ b1 y + c1 = 0

a2 x+ b2y + c2 = 0

The condition for having infinitely many solutions is:

For the equations (1) and (2),

Now,

2(2k-1) = 3(k+1)

⇒ 4k – 2 = 3k + 3

⇒ k = 5

or

2(4k+1) = 7(k+1)

⇒ 8k + 2 = 7k + 7

⇒ k = 5

Hence, the value k =5 will give infinitely many solutions for the given set of equations.

50

13.

Use the shortcut method to find the mean of given data.

For that,

Let the assumed mean be (A) = 2,

The deviation of values xi from assumed mean be di = xi – A. Now to find the mean:

51

First multiply the frequencies in column (ii) with the value of deviations in column (iv) as fidi.

Now add the sum of all entries in column (iv) to obtain and the sum of all frequencies in the column (ii) to obtain

So,

=

where, N = total number of observations

⇒ 3(72 + k) = -260 + 20k

⇒ 216 + 3k = -260 + 20k

⇒ 476 = 17k

⇒ k = 28

14. Given: Angle of sector = 120°

Radius of circle = 21 cm

To find: area of segment of a circle

Explanation:

We know,

For the given values,

52

Hence the area of segment of circle is .

16. Let AN = x cm, CM = z cm and BL = y cm

As we know that,

Tangents from an external point to a circle are equal,

In given Figure we have

AN= AM = x [Tangents from point A]

BN = BL = y [Tangents from point B]

CL = CM = z [Tangents from point C]

Now, Given: AC = 12 cm

AM + MC = 12

x + z = 12

z = 12 – x…. [1]

and BC = 8 cm

BL + LC = 8

y + z = 8

z = 8 – y …… [2]

and

AB = 10 cm

AN + BN = 10

x + y = 10 …. [3]

equate [1] and [2] to get,

12 – x = 8 – y

53

4 = x - y … [4]

Add [3] and [4] to get,

x + x + y – y = 10 + 4

2x = 14

x = 7 cm

put value of x in [4] to get,

4 = 7 – y

y = 3 cm

put value of y in [2] to get,

z = 8 – 3

= 5 cm

Hence,

AN = 7 cm

CM = 5 cm

BL = 3 cm

23. Taking LHS,

As we know,

54

As sin2θ = 1 – cos2θ

Hence proved

24. Now, a = 3

l = 83

Sn = 903

We know,

⇒ 43n = 903

⇒ n = 21

As we know an = a +(n-1) d and an = l.

⇒ 83 = 3 + (21-1) d

⇒ 80 = 20d ⇒ d = 4

∴ Number of terms is 21 and the common difference is 4.

25. Given,

• Side BC is 6 cm.

• ∠ B is 45°.

55

• ∠ A = 105°.

As the sum of angles in a triangle is 180°.

∠ A + ∠ B + ∠ C = 180°.

105° + 45° + ∠ C = 180°

150° + ∠ C = 180°

∠ C = 30°

Steps of construction:

1. Draw a line BC of length 6 cm.

2. With B as center construct an angle of 45°.

a. With B as center draw an arc of any convenient radius which cuts the line BC at D.

b. With D as center and with the same radius as above (step a)

draw an arc which cuts the previous arc at point E.

c. With E as center and with the same radius as above (step a)

draw an arc which cuts the previous arc at point F.

d. With E and F as centers and with radius more than half the

length of EF, draw two arcs which intersect at G.

56

e. The line BG makes 90° angle with the line BC.

f. With B as center and with some convenient radius draw an

arc which cuts BG and BC at H and I.

g. With H and I as centers and with radius more than half the

length of HI, draw two arcs which intersect at J.

h. Join BJ, which is the line which makes a 45° angle with line

BC.

3. With C as center construct an angle of 30°.

a. With C as center draw an arc with some convenient radius which cuts BC at K.

57

b. With K as center and with the same radius as above (step a)

draw an arc which cuts the previous arc at point L.

c. With K and L as centers and with radius more than half the

length of KL, draw two arcs which intersect at M.

d. Join CM which makes an angle of 30° with the line BC. Extend

BJ and CM to join at point A. This is the required triangle Δ ABC.

4. Now with B as center draw a ray BX making an acute angle

with BC on the opposite side of vertex A. (as the scale factor is

58

greater than 1). Mark 4 points B1, B2, B3 and B4 on BX which are equidistant. i.e. BB1 = B1B2 = B2B3 = B3B4.

5. Join B3C. Then draw a line B4C’ which is parallel to B3C and

meets the extended line BC at C’.

6. Now draw a line C’A’ which is parallel to CA and meets the

extended line AC at A’. The Δ A’BC’ is the required triangle of scale factor 3/4.

Justification:

59

Consider,

---- (1)

Also, as AC ∥ A’C’,

We can say that ∠ A’C’B = ∠ ACB (corresponding angles) – (2)

Now, consider the triangles Δ ABC and Δ A’BC’

∠ B = ∠ B (Common angle)

∠ A’C’B = ∠ ACB (from (2))

From the Angle Similarity of triangles, we can clearly say that

Δ ABC ≅ Δ A’BC’

By CPCT (Corresponding Parts of Congruent Triangles), we can say that,

From (1) we have

Hence justified.

60

1. Given,

a = x3y2 and b = xy3

a = x3y2

⇒ a = x × x × x × y × y

And

b = x × y × y × y

So, LCM of a and b will be x3y3.

7. The two-digit numbers that are divisible by 7 are 14, 21,….. ,98

Here, First term = a = 14

Common difference d = 21 – 14 = 7,

Last term = an = 98

Since, nth term is given by:

an = a + (n - 1) d

⇒ 98 = 14 + (n – 1)7

⇒ 98 = 14 + 7n – 7

⇒ 98 = 7 + 7n

⇒ 91 = 7n

⇒ n = 13

There are 13 two-digit numbers divisible by 7.

We know sum of n terms of an AP is:

Sn = n2

Solutions (Set-3)

OR

61

For n = 1,

S1 = 12

= 1

For n = 2,

S2 = 22

= 4

For n = 3,

S3 = 32

= 9

Now S1 = a1

a1 = 1

S2 – S1 = a2

4 – 1 = a2

3 = a2

Now, d = a2 – a1 = 3 -1

= 2

We know, an = a + (n – 1)d

For n = 10,

a10 = 1 + (10– 1)2

= 1 + 9(2) = 1 + 18

= 19

So, 10th term of the AP is 19.

62

13. Consider 3x3 + 10x2 – 9x – 4,

Since 1 is the zero of the given polynomial.

So, (x-1) is the factor of the polynomial.

So, 3x3 + 10x2 – 9x – 4 = (x-1) (3x2 + 13x +4)

Now, 3x2 + 13x +4 = 3x2 + 12x + x +4

= 3x (x + 4) + 1 (x+4)

= (3x+1) (x+4)

So, zeroes will be

Thus, the zeroes will be .

15. Let us assume

is a rational number.

So,

, b ≠ 0 and also a and b are integers.

⇒ b (2+√3) = 5a ⇒ 2b+ √3b = 5a

Since, a and b are integers.

5a – 2b will also be an integer.

So, will be rational number which means √3 is also

rational.

63

But √3 is irrational as given which is contradiction.

So, is an irrational number.

23. Consider ,

Squaring both sides, we get,

We know, tan2θ = sec2θ – 1

When,

= 2x

When

24. We have

64

Given: ∆ABC ∼ ∆PQR

To prove: Ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

i.e.,

Construction: Draw AM perpendicular to BC and PN perpendicular to QR.

Proof: We know area of the triangle is given by (1/2 × base × height).

So, in ∆ABC

Area (∆ABC) = 1/2 × BC × AM [∵ base = BC & height = AM] …(i)

Similarly, in ∆PQR

Area (∆PQR) = 1/2 × QR × PN [∵ base = QR & height = PN] …(ii)

Dividing equations (i) by (ii), we get

⇒ …(iii)

In ∆ABM & ∆PQN,

∠B = ∠Q [∵ ∆ABC ∼ ∆PQR; and corresponding angles are equal of similar triangles]

∠AMB = ∠PNQ [∵ they are 90°]

So, by AA-similarity property of triangle, ∆ABM ∼ ∆PQN.

⇒ [∵ corresponding sides of similar triangles are

proportional] …(iv)

Substituting equation (iv) in equation (iii), we have

⇒ …(v)

We know that,

65

∆ABC ∼ ∆PQR

Using property which says that corresponding sides of similar triangles are proportional, we can write as

Using this equality, re-write equation (v)

⇒

⇒ …(A)

Using this equality again, re-writing equation (v)

⇒

⇒ …(B)

Similarly, …(C)

Collecting equations (A), (B) & (C), we get

Hence, proved.

25.

66

The ogive formed is:

Solving by short-cut method: The above given data can be represented in the form of table as below:

OR

67

Formula of mean is given by

where, a = assumed mean fi = frequency of the ith class h = class width

= Rs. 211 So, the mean of the data is Rs. 211

68