class x time allowed: 3 hrs max marks: 80 · class x mathematics sample question paper 2018-19 time...

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CLASS X

Mathematics

Sample Question Paper 2018-19

Time allowed: 3 hrs Max Marks: 80

General Instructions:

1. All the questions are compulsory.

2. The questions paper consists of 30 questions divided into 4 sections A, B, C

and D.

3. Section A comprises of 6 questions of 1 mark each. Section B comprises of 6 questions of 2 marks each. Section C comprises of 10 questions of 3 marks

each. Section D comprises of 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 each, two questions of 2 marks each, 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 calculators is not permitted.

SECTION – A

(Question numbers 1 to 6 carries 1 mark each)

1. Find the value of a, for which point a

P ,23

is the mid – point of the line

segment joining the points Q( – 5,4) and R( – 1,0).

2. Find the value of k, for which one root of the quadratic equation

kx2 – 14x + 8 = 0 is 2.

OR

Find the value(s) of k for which the equation x2 + 5kx + 16 = 0 has real

and equal roots.

3. Write the value of 2

2

1cot

sin −

.



3

OR

If sin θ = cos θ, then find the value of 2tan θ + cos2θ

4. If nth term of an A.P. is (2n + 1), what is the sum of its first three terms?

5. In figure if AD = 6 cm, DB = 9cm, AE = 8cm and EC = 12cm and ∠ADE = 48°. Find ∠ABC

6. After how many decimal places will the decimal expansion of 4 3

23

2 5

terminate?

SECTION – B

(Question numbers 7 to 12 carries 2 marks each)

7. The HCF and LCM of two numbers are 9 and 360 respectively. If one

number is 45, find the other number.

OR

Show that 7 – 5 is irrational, given that 5 is irrational.



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8. Find the 20th term from the last term of the AP 3,8, 13,….,253.

OR

If 7 times the 7th term of an A.P is equal to 11 times its 11th term, then find its 18th term.

9. Find the coordinates of the point P which divides the join of A(–2,5) and B(3, –5) in the ratio 2:3.

10. A card is drawn at random from a well shuffled deck of 52 cards. Find the probability of getting neither a red card nor a queen.

11. Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is a prime number.

12. For what value of p will the following pair of linear equations have infinitely many solutions:

(p – 3) x + 3y = p

px + py = 12

SECTION – C


13. Use Euclid’s Division Algorithm to find the HCF of 726 and 275.

14. Find the zeroes of the following polynomial:

25 5x 30x 8 5+ +

15. Places A and B are 80 km apart from each other on a highway. A car starts from A and another from B at the same time. If they move in same direction, they meet in 8 hours and if they move towards each other they meet in 1 hour 20 minutes. Find the speed of cars.



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16. The points A (1, – 2), B(2,3), C (k,2) and D( – 4, – 3) are the vertices of a parallelogram. Find the value of k.

OR

Find the value of k for which the points (3k – 1, k – 2), (k,k – 7) and (k – 1, – k – 2) are collinear.

17. Prove that

22cos θ 1cotθ tanθ

sinθcosθ

−− =

OR

Prove that: sin θ (1 + tan θ) + cos θ (1 + cot θ) = sec θ + cosec θ.

18. The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D and intersecting the larger circle at P on producing. Find the length of AP.

19. In figure ∠1 = ∠2 and ΔNSQ≅ΔMTR, then prove that ΔPTS~ΔPRQ.

OR

In ΔABC, if AD is the median, then show that AB2 + AC2 = 2(AD2 + BD2)



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20. Find the area of the minor segment of a circle of radius 42cm, if length of the corresponding arc is 44cm.

21. Water is flowing at the rate of 15 km per hour through a pipe of diameter 14cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tank will rise by 21cm.

OR

A solid sphere of radius 3 cm is melted and then recast into small spherical balls each of diameter 0.6cm. Find the number of balls.

22. The table shows the daily expenditure on grocery of 25 households in a locality. Find the modal daily expenditure on grocery by a suitable method.

SECTION – D


23. A train takes 2 hours less for a journey of 300km if its speed is increased by 5 km/h from its usual speed. Find the usual speed of the train.

OR

Solve for x: ( )1 1 1 1

, a 0, b 0, x 0, x a ba b x a b x

= + + − + + +

24. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106.

Find the 29th term.

25. Prove that in a right – angled triangle square of the hypotenuse is equal to sum of the squares of the other two sides.



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26. Draw a ΔABC with sides 6cm, 8cm and 9 cm and then construct a triangle

similar to ΔABC whose sides are 3

5 of the corresponding sides of ΔABC.

To construct: a ΔABC with sides 6cm, 8cm and 9 cm and then construct a

triangle similar to ΔABC whose sides are 3

5 of the corresponding sides of

ΔABC

27. A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how long will the car take to reach the observation tower from this point?

OR

The angle of elevation of a cloud from a point 60m above the surface of the water of a lake is 30° and the angle of depression of its shadow from the same point in water of lake is 60°. Find the height of the cloud from the surface of water.

28. The median of the following data is 525. Find the values of x and y if the

total frequency is 100.



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OR

The following data indicates the marks of 53 students in Mathematics.

Draw less than type ogive for the data above and hence find the median.

29. The radii of circular ends of a bucket of height 24 cm are 15 cm and 5 cm. Find the area of its curved surface.

30. If sec θ + tan θ = p, then find the value of cosec θ.

***



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SOLUTIONS

1. Given: line segment joining the points Q( – 5,4) and R( – 1,0), and also

given the mid – point of the line segment i.e., P (a

3, 2)

To find: the value of ‘a’.

Explanation: As it is given P is the midpoint of line QR, so the coordinates of P are

x – coordinate,

a

3 =

x1 + x2

2

Here x1 = – 5 and x2 = – 1, so the above equation becomes

a

3 =

(−5) + (−1)

2

⇒ 2a = 3( – 6)

⇒ 2a =–18

⇒ a =–9

Hence the value of a is–9.

2. Given: quadratic equation kx2 – 14x + 8 = 0, one of its root is 2

To find: the value of ‘k’

Explanation: Given 2 is the root of the given equation hence the 2 must satisfy the given equation.

So now putting x = 2 in the given equation, we get

kx2 – 14x + 8 = 0

⇒ k(2)2 – 14(2) + 8 = 0

⇒ 4k – 28 + 8 = 0

⇒ 4k – 20 = 0

⇒ 4k = 20

⇒ k = 5

Hence the value of k is 5.



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OR

Given: quadratic equation x2 + 5kx + 16 = 0, it has real and equal roots


Explanation: The given quadratic equation has real and equal roots, so its determinant will be equal, i.e.,

D = 0

But we know determinant is b2 – 4ac

Hence for real and equal roots, we get

b2 – 4ac = 0………..(i)

Now comparing the given quadratic equation x2 + 5kx + 16 = 0

With the standard quadratic equation ax2 – bx + c = 0, we get

a = 1, b = 5k, c = 16

Substituting these values in equation (i), we get

b2 – 4ac = 0

(5k)2 – 4(1)(16) = 0

⇒ 25k2 – 64 = 0

⇒ 25k2 = 64

⇒ k2 = 64

25

Taking square root on both sides, we get

⇒ k = ±√64

25

⇒ k = ±8

5

Hence the value of k is ±8

5

3. Given: cot2 θ −1

sin2 θ

To find: the value of given expression

Explanation: given cot2 θ −1

sin2 θ

But we know, cot θ = cos θ

sin θ



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Substituting this in given expression, we get

= (cos θ

sin θ)

2

−1

sin2 θ

= cos2 θ

sin2 θ−

1

sin2 θ

On combining we get

= cos2 θ − 1

sin2 θ

= −(1 − cos2 θ)

sin2 θ… . . (i)

But we know sin2θ + cos2θ = 1

⇒ sin2θ = 1 – cos2θ

Now substituting the above value in equation (i), we get

= −(sin2 θ)

sin2 θ

= – 1

Hence the given value of the expression is–1

OR

Given: sin θ = cos θ

To find: the value of 2 tan θ + cos2θ

Explanation: given sin θ = cos θ

⇒ sin θ

cos θ = 1

But we know this equal to tan θ

Hence tan θ = 1

But this is possible for the value

θ = 45°………. (i)

We will take the next expression,

2 tan θ + cos2θ

Substituting the value of θ from equation (i), we get

2tan (45°) + cos2(45°) ………. (ii)



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We know tan (45°) = 1 and cos (45°) = 1

√2

Substituting these values in equation (ii), we get

= 2(1) + (1

√2)

2

= 2 + 1

2

= 4 + 1

2

= 5

2

Hence the value of 2 tan θ + cos2θ = 5

2

4. Given: nth term of an A.P. is (2n + 1),

To find: the sum of its first three terms

Explanation: given nth term of an A.P. is (2n + 1),

∴ an = 2n + 1……….(i)

Now the first term of the AP can be found by substituting n = 1, so we get

a1 = 2 × 1 + 1

= 2 + 1

a1 = 3………(ii)

Now the second term of the AP can be found by substituting n = 2, so we get

a2 = 2 × 2 + 1

= 4 + 1

a2 = 5………(iii)

Now the third term of the AP can be found by substituting n = 3, so we get

a3 = 2 × 3 + 1

= 6 + 1

a3 = 7………(iv)

So, the sum of the first three terms of the AP is

a1 + a2 + a3



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Now substituting values from equation (ii), (iii) and (iv) we get

a1 + a2 + a3 = 3 + 5 + 7

= 15

Hence the sum of its first three terms is 15.

5. Given: AD = 6cm, DB = 9cm, AE = 8cm and EC = 12cm and ∠ADE = 48°,

in ΔABC in the above figure

To find: ∠ABC

Explanation:

From given figure we see that line DE divides the two sides of the triangle.

Now we will find the ratio of the sides, i.e., we will find

AD

DB and

AE

EC

Now substituting the given values in the above equation, we get

AD

DB =

6

9 =

2

3… … (i)

And

AE

EC =

8

12 =

2

3… . . (ii)

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

AD

DB =

AE

EC… … (iii)

We know that

If line divides two sides of a triangle in same ratio then the line is parallel to the third side.

Hence from equation (iii),

DE||BC……(iv)

Now taking AB as the transversal line, ∠ADE and ∠ABC form corresponding

angles, and we know corresponding angles are equal, hence

∠ABC = ∠ADE

Now given ∠ADE = 48°, substituting this in above equation, we get

∠ABC = 48°

This is the required value.



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6. Consider 23

24 × 53,

Now to make the denominator of same power we will multiply and divide by 5, we get

= 23

24 × 53 ×

5

5

= 23 × 5

24 × 53 × 5

= 115

24 × 54

= 115

(2 × 5)4

= 115

(10)4

= 115

10000

= 0.0115

Hence so after 4 decimal places, the decimal expansion of 23

24 × 53 terminates.

7. Let LCM and HCF of the two numbers be X and Y respectively and let a and b be the two numbers.

According to the given criteria, X = 360 and Y = 9

There is an important formula to remember,

LCM (two numbers) × HCF (two numbers) = Multiplication of two numbers.

Now substituting the corresponding values, we get

X × Y = a × b

⇒ 360 × 9 = a × b

⇒ a × b = 3240

But given one number is 45, so let a = 45, we get

⇒ 45 × b = 3240

⇒ b = 3240

45



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⇒ b = 72

Hence the other number is 72.

OR

Given: √5 is irrational

To show: 7 − √5 is irrational

Explanation: we will prove this by contradiction method.

So, let us assume 7 − √5 is rational.

And we know a rational number can be written in a

b form, where a and b are

co prime (i.e., a and b have no common factor other than 1) and also b≠0,

so

7 − √5 = a

b

⇒ −√5 = a

b− 7

⇒ √5 = 7 −a

b

⇒ √5 = 7b − a

b… . . (i)

Hence 7b−a

b is rational from our assumption.

But given √5 is irrational

So, from equation (i), irrational = rational

But this is not possible, hence this is a contradiction.

So, our assumption is incorrect.

Hence 7 – √5 is irrational.

Hence proved

8. Given: AP 3,8, 13,…,253

To find: the 20th term from the last term of the AP

Explanation: The given AP is 3,8, 13,….,253

So, the first term is 3

∴ a = 3……(i)

Now finding the difference of second and term, we get



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d = 8 – 3 = 5………(ii)

So last term can be written as

an = 253

But we know an = a + (n – 1)d (standard form)

⇒ 253 = a + (n – 1)d

Substituting the values from equation (i) and (ii), we get

⇒ 253 = 3 + (n – 1)(5)

⇒ 253 = 3 + 5n – 5

⇒ 253 = 5n – 2

⇒ 5n = 253 + 2

⇒ 5n = 255

⇒ n = 51

Hence 253 is the 51st term of the given AP

Now to the 20th term from last means 51 – 19 = 32nd term, as the last term is 51 – 0, second last term is 51 – 1, third last term is 51 – 2 and so on.

So 32nd term can be found from using the standard form formula, i.e., an = a + (n – 1)d, so

a32 = a + (32 – 1) d

Substituting values of a and d from equation (i) and (ii), we get

a32 = 3 + (31)(5)

⇒ a32 = 3 + 155

⇒ a32 = 158

Hence the 20th term from the last term of the AP is 158.

OR

Given: 7 times the 7th term of an A.P is equal to 11 times its 11th term

To find: its 18th term

Explanation: The standard form of an AP term is

an = a + (n – 1) d

Now the 7th term is,

a7 = a + (7 – 1) d = a + 6d……..(i)

And the 11th term is,



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a11 = a + (11 – 1)d = a + 10d……..(ii)

And also the 18th term is,

a18 = a + (18 – 1)d = a + 17d……..(iii)

Now as per the given criteria,

7 times the 7th term of an A.P is equal to 11 times its 11th term

⇒ 7 × a7 = 11 × a11

Now substituting values form equation (i) and (ii), we get

⇒ 7 × (a + 6d) = 11 × (a + 10d)

⇒ 7a + 42d = 11a + 110d

⇒ 7a – 11a = 110d – 42d

⇒ – 4a = 68d

⇒ a = 68d

−4

⇒ a = – 17d…….(iv)

Now substituting equation (iv) in equation (iii), we get

a18 = a + 17d

⇒ a18 = (–17d) + 17d = 0

Hence the 18th term is 0.

9. Given: line segment A ( –2,5) and B(3, –5)

To find: the coordinates of the point P which divides the line segment AB into the ratio 2:3

Explanation: Given P divides the line segment AB into 2:3 ratio, let point P be denoted as P(x,y)

So, applying the section formula, we get

x = m1x2 + m2x1

m1 + m2, y =

m1y2 + m2y1

m1 + m2

In this case, m1 = 2, m2 = 3



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And x1 = – 2, x2 = 3, y1 = 5 and y2 = – 5

Substituting the above values in section formula, we get

x = m1x2 + m2x1

m1 + m2

⇒ x = 2 × 3 + 3 × (−2)

2 + 3

⇒ x = 6 − 6

5

⇒ x = 0 … . (i)

y = m1y2 + m2y1

m1 + m2

⇒ y = 2 × (−5) + 3 × 5

2 + 3

⇒ y = −10 + 15

5

⇒ y = 5

5 = 1 … . . (ii)

Hence the P (0, 1)

So, the coordinates of the point P which divides the line segment AB into the ratio 2:3 are 0 and 1.

10. Given: a card is drawn at random from a well shuffled deck of 52 cards

To find: the probability of getting neither a red card nor a queen

Explanation: In a well shuffled deck of 52 cards

There are 13 + 13 = 26 red cards

And 4 queens, but out of these 2 queens are red in color

Hence the numbers of cards which are red cards, or a queen are 26 + 2 = 28 cards, so out of these 28 cards a card is drawn.

So, the probability of getting either a red card or a queen is

P (getting either a red card or a queen)

= Number of red cards or queen

Total number of cards

= 28

52



19

= 7

13

So, probability of getting neither a red card nor a queen is

1 – P (getting either a red card or a queen)

1 −7

13

= 13 − 7

13

= 6

13

Hence the probability of getting neither a red card nor a queen is 6

13.

11. Given: two dice are thrown at the same time and the product of numbers appearing on them is noted

To find: the probability that the product is a prime number

Explanation: Total number of outcomes of one dice = 6

So total number of outcomes = 6 × 6 = 36

So, n(S) = 36

And all the outcomes of S are

S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Let E denote outcomes such that the product is prime (i.e., having only factors), then the favourable outcomes of event E are

E = {(1,2), (1,3), (1,5), (2,1), (3,1), (5,1)}

Hence number of outcomes of E are n(E) = 6

Hence the probability that the product is a prime number is

P (getting the product is a prime number)



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= n(E)

n(S)

Substituting corresponding values, we get

= 6

36

= 1

6

Hence the probability that the product is a prime number is 1

6.

12. given pair of linear equations are

(p – 3) x + 3y = p

⇒ (p – 3) x + 3y – p = 0

px + py = 12

⇒ px + py – 12 = 0

Comparing these with standard equation

a1x1 + b1x1 + c1 = 0 and a2x2 + b2x2 + c2 = 0

We get

a1 = (p – 3), b1 = 3, c1 = – p

a2 = p, b2 = p, c2 = – 12

Now it is given the linear pair of equations have infinitely many solutions, so the condition for it is

a1

a2 =

b1

b2 =

c1

c2


p − 3

p =

3

p =

−p

−12

Considering first equality, we get

p − 3

p =

3

p

⇒ p – 3 = 3

⇒ p = 6



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Considering the second equality, we get

3

p = −

p

−12

⇒ 3

p =

p

12

⇒ 3 × 12 = p2

⇒ p = √36

⇒ p = ±6

Hence p = + 6 satisfies both the conditions.

Hence for p = 6, the given pair of linear equations have infinitely many

solutions.

13. Using Euclid’s Division Algorithm, we find HCF of two positive numbers by repetitive division till we get 0 as the remainder.

In the given two numbers 726 and 275, 726 is greater, so we will divide 726 by 275, we get

We see that the remainder is 176≠0,

Now we divide 275 by 176, we get





22

We see that the remainder is 77≠0

Now we divide 99 by 77, we get,

We see that the remainder is 22≠0

Now we divide 77 by 22, we get,



We see that the remainder is 0.

Hence the HCF of the two numbers 726 and 275 is 11.

14. given polynomial is

5√5x2 + 30x + 8√5

Now factorizing by splitting the middle term.

Now we need to find two numbers whose sum is 30 and

Product is 5√5 × 8√5 = 5 × 8 × √5 × √5 = 40 × 5 = 200

So, the above given polynomial is

5√5x2 + 20x + 10x + 8√5

= 5√5x2 + 20x + (5 × 2) x + 8√5

= 5√5x2 + 20x + (√5 × √5 × 2) x + 8√5



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= 5x (√5x + 4) + 2√5(√5x + 4)

= (5x + 2√5) (√5x + 4)

So, the zeroes of the polynomial will be

5x + 2√5 = 0 and √5x + 4 = 0

⇒ 5x = – 2√5 and √5x = – 4

⇒ x = −2√5

5 and x =

−4

√5

⇒ x = −2√5

5 and x =

−4 × √5

√5 × √5

⇒ x = −2√5

5 and x =

−4√5

5

Hence the zeroes of the given polynomial are −2√5

5 and

−4√5

5.

15. Let the speed of first car be x km/hr

And the speed of the second car be y km/hr

Given A and B are 80km apart,

a car starts from A and another from B at the same time

Case 1: moving in same direction

The two cars will meet in 8 hours, so

time = 8hrs

Let the two cars meet at point C, so

So, distance travelled by first car = AC

And distance travelled by second car = BC



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So, the difference of distance travelled = AC – BC

⇒ AB = 80km is the relative distance

Relative speed will be speed of 1st car – speed of 2nd car

⇒ speed = (x – y) km/h

And we know

Distance = speed × time

Substituting the values, we get

80 = (x – y) × 8

⇒ x – y = 10……….(i)

Case 2: moving in opposite direction

The two cars will meet in 1 hour 20 minutes, so

time = 1 + 20

60hours

⇒ time = (60 + 20)

60hours

⇒ time = (80)

60hours

⇒ time = 2

3hours

Let the two cars meet at point C, so

So, distance travelled by first car = AC

And distance travelled by second car = BC

So the total distance travelled = AC + BC

⇒ AB = 80km is the relative distance

Relative speed will be speed of 1st car + speed of 2nd car

⇒ speed = (x + y) km/h

And we know

Distance = speed × time




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80 = (x + y) × 4

3

⇒ 80 × 3

4 = (x + y)

x + y = 60………(ii)

So, the pair of linear equations in two variables are

x + y = 60

x – y = 10

Now we will solve these pair of equations,

Adding the two equations we get

⇒ x = 35

Hence the speed of first car is 35 km/hr

Substituting the value of x in first equation, we get

x + y = 60

⇒ 35 + y = 60 ⇒ y = 60 – 35 ⇒ y = 25

Hence the speed of second car is 25 km/hr.

So, the speed of two cars is 35km/hr and 25km/hr.

16. The figure of the above condition is as shown below



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We know in a parallelogram, the diagonals bisect each other, so O is the midpoint of AC and BD (see the above figure)

Let the coordinates of O be x and y,

As O is midpoint of line BD, so the coordinates of O can be written as

x = x1 + x2

2 and y =

y1 + y2

2

Here x1 and y1 are the coordinates of point B(2,3)

And x2 and y2 are the coordinates of point D( – 4, – 3)


x = 2 + (−4)

2 and y =

3 + (−3)

2

x = −2

2 and y =

0

2

So, x = – 1, y = 0

So, the point O is ( – 1, 0)

Now we will consider the next diagonal AC,

Here also O ( – 1,0) is the midpoint

So, applying the midpoint formula

−1 = x1 + x2

2 and 0 =

y1 + y2

2

Here x1 and y1 are the coordinates of point A (1, – 2).

And x2 and y2 are the coordinates of point C(k,2).


−1 = 1 + k

2 and 0 =

−2 + 2

2 = 0

⇒ 1 + k = 2 × ( – 1)

⇒ k = – 2 – 1

⇒ k = – 3

So, the required value of k is – 3.

OR

Given: points (3k – 1, k – 2), (k,k – 7) and (k – 1, – k – 2) are collinear




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Explanation: collinear means all the points lie on same line.

Let the three points be A(3k – 1,k – 2), B(k,k – 7) and C(k – 1, – k – 2)

So, A, B, C will form a straight line and not a triangle

So, the area should be equal to 0

i.e., ar(ΔABC) = 0

now we know area of any triangle is given by

1

2[x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)]

This should be equal to zero,

From the given points,

x1 = 3k – 1, y1 = k – 2

x2 = k, y2 = k – 7

x3 = k – 1, y3 = – k – 2

Now substituting these values in area formula, we get

1

2[x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)] = 0

⇒ [(3k – 1) ((k – 7) – ( – k – 2)) + k(( – k – 2) – (k – 2) ) + (k – 1)((k – 2)

– (k – 7))] = 0

⇒ [(3k – 1) (k – 7 + k + 2) + k ( – k – 2 – k + 2) + (k – 1) (k – 2 – k + 7)]

= 0

⇒ [(3k – 1) (2k – 5) + k( – 2k) + (k – 1) (5)] = 0

⇒ (3k(2k – 5) – 1(2k – 5)) – 2k2 + 5k – 5 = 0

⇒ 6k2 – 15k – 2k + 5 – 2k2 + 5k – 5 = 0

⇒ 4k2 – 12k = 0

⇒ 4k(k – 3) = 0

⇒ 4k = 0 or k – 3 = 0

⇒ k = 0 or k = 3

Hence the required value of k is 0 and 3.

17. LHS = cot θ – tan θ

But we know cot θ = cos θ

sin θ, tan θ =

sin θ

cos θ ,



28

substituting these values in LHS, we get

= cos θ

sin θ−

sin θ

cos θ

= cos2 θ − sin2 θ

sin θ × cos θ

But we know cos2θ + sin2θ = 1

⇒ sin2θ = 1 – cos2θ, substituting this value in above equation, we get

= cos2 θ − (1 − cos2 θ)

sin θ × cos θ

= cos2 θ − 1 + cos2 θ

sin θ × cos θ

= 2 cos2 θ − 1

sin θ cos θ

= RHS

Hence proved

OR

To prove sin θ (1 + tan θ) + cos θ (1 + cot θ) = sec θ + cosec θ

Proof: LHS = sin θ (1 + tan θ) + cos θ (1 + cot θ)

But we know,

cot θ = cos θ

sin θ, tan θ =

sin θ

cos θ ,

substituting these values in LHS, we get

= sin θ (1 + sin θ

cos θ) + cos θ (1 +

cos θ

sin θ)

= sin θ (cos θ + sin θ

cos θ) + cos θ (

cos θ + sin θ

sin θ)

Now taking cos θ + sin θ common, we get

= (cos θ + sin θ) (sin θ

cos θ +

cos θ

sin θ)

= (cos θ + sin θ) (sin2 θ + cos2 θ

cos θ × sin θ)

But we know cos2θ + sin2θ = 1,

substituting this value in above equation, we get



29

= (cos θ + sin θ) (1

cos θ × sin θ)

= (cos θ + sin θ

cos θ × sin θ)

= (cos θ

cos θ × sin θ) + (

sin θ

cos θ × sin θ)

Cancelling the like terms, we get

⇒ = (1

sin θ) + (

1

cos θ)

But we know 1

sin θ = cosec θ ,

1

cos θ = sec θ ,

substituting these values in above equation, we get

= cosec θ + sec θ

= RHS

Hence proved

18. The figure for the above given condition is as shown below,

Now we will join OD, we get



30

So as per the given criteria,

AB = 26cm, BO = AO = 13cm, OD = 8cm……….(i)

Now consider smaller circle,

In this BD is tangent to the smaller circle given and OD is the radius of the smaller circle,

And we know in a circle tangent is perpendicular to the radius, i.e.,

OD⊥BD

⇒ ∠BDO = 90°….(ii)

Now consider bigger circle,

In this P is a point in the semicircle with radius AB,

And we know in a circle, angle in a semicircle is always a right angle, i.e.,

∠APB = 90°….(iii)

Now we will consider ΔABP and ΔOBD,

∠APB = ∠BDO = 90° (from equation(ii) and (iii))

∠ABP = ∠DBO (common angle)

Hence by AA similarity,

ΔABP~ΔOBD

And we know sides of similar triangles are proportional, hence in these two

triangles,

⇒ AP

OD =

AB

OB

Now substituting values from equation (i), we get

⇒ AP

8 =

26

13

⇒ AP = 2 × 8

⇒ AP = 16cm

Hence the length of AP is 16cm.

19. Given: ∠1 = ∠2

ΔNSQ≅ΔMTR

To prove: ΔPTS~ΔPRQ

Proof:



31

Now it is given ΔNSQ≅ΔMTR

So, by CPCT (corresponding parts of congruent triangles are equal), we have

∠NQS = ∠MRT

Or it can also be written as

∠PQR = ∠PRQ………….(i)

Now consider ΔPST,

We know in a triangle sum of all the angles is equal to 180°, hence

∠P + ∠PST + ∠STP = 180°

Or, ∠P + ∠1 + ∠2 = 180°

But given ∠1 = ∠2, substituting this value in above equation, we get

∠P + ∠1 + ∠1 = 180°

∠P + 2∠1 = 180°…………(ii)

Now consider ΔPQR,

We know in a triangle sum of all the angles is equal to 180°, hence

∠P + ∠PQR + ∠PRQ = 180°

Substituting value from equation (i)in above equation, we get

∠P + ∠PQR + ∠PRQ = 180°

∠P + 2∠PQR = 180° …………(iii)

Equating equation (ii) and (iii), we get

∠P + 2∠1 = ∠P + 2∠PQR

Now cancelling the like terms, we get

2∠1 = 2∠PQR

Or, ∠1 = ∠PQR…………(iv)

Or, ∠PST = ∠PQR

Now consider ΔPTS and ΔPRQ,

∠P = ∠P (common)

∠PST = ∠PQR (from equation (iv))

Hence the AA similarity,

ΔPTS ~ ΔPRQ

Hence proved



32

OR

Given: in the above figure ΔABC, if AD is the median

To show: AB2 + AC2 = 2(AD2 + BD2)

Explanation:

From above figure AE is perpendicular to BC,

Consider right – angled ΔABE,

Applying the Pythagoras theorem, we get

AB2 = AE2 + BE2………. (i)

Consider right – angled ΔAEC,


AC2 = AE2 + EC2………. (ii)

Consider right – angled ΔAED,


AD2 = AE2 + ED2………. (iii)

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

AB2 + AC2 = AE2 + AE2 + EC2 + BE2

AB2 + AC2 = 2AE2 + EC2 + BE2……….(iv)

But from figure

BE = BD – ED and EC = CD + ED

or EC = BD + ED (as AD is the median)

Substituting the above values in equation (iv), we get

AB2 + AC2 = 2AE2 + (BD + ED)2 + ( BD – ED)2

Applying the expansion of (a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 – 2ab, we get

AB2 + AC2 = 2AE2 + BD2 + ED2 + 2BD.ED + BD2 + ED2 – 2BD.ED

⇒ AB2 + AC2 = 2AE2 + 2ED2 + 2BD2

⇒ AB2 + AC2 = 2(AE2 + ED2 + BD2)

Now substituting value from equation (iii), we get

⇒ AB2 + AC2 = 2(AD2 + BD2)

Hence proved



33

20. The figure for the above question is

So, arc AB = 44cm, radius, r = BO = AO = 42cm………(i)

We know the length of the arc can be written as

AB = θ

360 × 2πr


44 = θ

360 × 2 ×

22

7 × (42)

⇒ 44 × 360 × 7

44 × 42 = θ

⇒ θ = 7 × 360

42

⇒ θ = 360

6

θ = 60°……(ii)

So, the angle of the minor segment is 60°.

The minor segment can be divided into two as shown in the figure.

So, we know the



34

area of segment ABX = area of sector AOBX – area of ΔAOB….(iii)

We know,

area of sector AOBX = θ

360 × πr2

Substituting the values from equation (i) and equation (ii), we get

area of sector AOBX = 60°

360 ×

22

7 × (42)2

area of sector AOBX = 1

6 ×

22

7 × 42 × 42

area of sector AOBX = 22 × 7 × 6

area of sector AOBX = 924cm2………..(iv)

Now consider ΔAOB, and draw a line AM perpendicular to AB, i.e.,

Here AO = OB,

∴ by RHS congruency ΔAMO≅ΔBMO

Now by CPCT (Corresponding parts of congruent triangles are equal), so AM = MB⇒ AB = 2AM………….(v)

Hence M is the mid – point of AB and

∠AOM = ∠BOM = 1

2 × 60° = 30°

Now consider right – angled ΔAMO, we know

cos θ = adjacent side

hypotenuse


cos 30° = OM

AO

= OM

42



35

But cos 30° = √3

2, substituting this value in above equation, we get

√3

2 =

OM

42

⇒ OM = 42 × √3

2

⇒ OM = 21√3cm…..(vi)

And also, in same triangle,

sin θ = opposite side

hypotenuse


sin 30° = AM

AO =

AM

42

But sin 30° = 1

2,

substituting this value in above equation, we get

1

2 =

AM

42

⇒ AM = 42 × 1

2

⇒ AM = 21cm….(vii)

Now the area of ΔABO

= 1

2 × base × height

= 1

2 × AB × OM

Substituting values from equation (v) and (vi), we get

Area of ΔABO = 1

2 × 2AM × 21√3

Substituting values from equation (vii), we get

Area of ΔABO = 1

2 × 2(21) × 21√3


Area of ΔABO = 441√3cm2……(viii)

So, from equation (iii), we have



36

area of segment ABX = area of sector AOBX –area of ΔAOB

Substituting values from equation (iv) and (viii) in above equation, we get

area of segment ABX = (924 – 441√3) cm2

Hence the area of the minor segment of the given circle is (924 – 441√3)cm2

21. Let the length of pipe filling the tank be x m.

Given the diameter of the pipe is 14cm

⇒ radius = 7cm

= 0.07m

Now the pipe is in shape of cylinder, so volume of the pipe, VP becomes,

VP = πr2h

Substituting the corresponding values, we get

VP = π(0.07)2(x)

VP = (0.0049) πx m3………..(i)

Now the water tank is rectangular, so it will be in form of cuboid.

So, given the length of the tank, l = 50 m long

and given the breadth of the tank, b = 44 m wide

We need to find the level of water in the tank will rise by 21cm, hence this will be the height, i.e., h = 21cm = 0.21m

So, the volume of the tank, VT, will be

VT = lbh


VT = 50 × 44 × (0.21)

⇒ VT = 462m3…………. (ii)

Now as the pipe is filling the water, so

Volume of the pipe should be equal to volume of the tank, i.e.,

VP = VT

Substituting values from equation(i) and (ii), we get

(0.0049) πx = 462

⇒ x = 462

0.0049 × π



37

⇒ x = 462

0.0049 × 227

⇒ x = 462 × 7

0.0049 × 22

⇒ x = 21

0.0007

⇒ x = 30000m = 30km

This is the length of the pipe.

And it is given water in the pipe is flowing at the rate of 15km/hr, means

15km travels in pipe in = 1hour

⇒ 1km travels in pipe in = 1

15hr

So, in 30km long pipe water travels in = 30 × 1

15hr

= 2 hr

So, in 2 hours the level of water in the tank will rise by 21cm.

OR

As the solid sphere is melted and recast into three small spherical balls, so

Volume of the big sphere = volume of small balls

⇒ Volume of big sphere = number of small balls × volume of small balls

……(i)

We know, volume of sphere = 4

3πr3

And let the radius big sphere, R = 3cm

And given diameter of small sphere = 0.6cm, so radius of each small sphere,

r = 0.6

2cm

= 0.3 cm

And let number of balls be n

Substituting this in equation (i), we get

⇒ 4

3πR3 = n ×

4

3πr3




38

R3 = n × r3

Substituting corresponding values, we get

(3)3 = n × (0.3)3

⇒ 27 = n × (0.027)

⇒ n = 27

0.027

⇒ n = 27 × 1000

0.027 × 1000

⇒ n = 27 × 1000

27

⇒ n = 1000

Hence 1000 number of balls can be made from the big sphere.

22. We know the mode can be calculated using the formula,

Mode = l + f1 − f0

2f1 − f0 − f2 × h … … (i)

Where l is lower limit of the modal class

h is class interval

f1 is frequency of the modal class

f0 is frequency of the class before modal class

f2 is frequency of the class after modal class

Here the modal class can be calculated as

Modal class = interval with highest frequency

From the given table the highest frequency is 12, so the corresponding interval is 200 – 250.

So the modal class = 200 – 250……(ii)

Now finding the corresponding values of various values for mode calculation

l = 200 (as lower limit of modal class is 200)

h = 250 – 200 = 50

f1 = 12 (frequency of modal class)

f0 = 5 (this is the frequency of class before modal class)

f2 = 2 (this is the frequency of class after modal class)



39

Substituting these values in equation (i), we get

Mode = l + f1 − f0

2f1 − f0 − f2 × h

Mode = 200 + 12 − 5

2(12) − 5 − 2 × 50

⇒ Mode = 200 + 7

24 − 7 × 50

⇒ Mode = 200 + 7

17 × 50

⇒ Mode = 200 + 350

17

⇒ Mode = 200 + 20.58

⇒ Mode = 220.58

Hence the modal daily expenditure on grocery is Rs. 220.58/ –

23. Let the speed of the train be x km/hr

The total journey distance, d = 300km….(i)

Case 1: with usual speed

usual speed of the train, s = x km/hr

We know Distance = speed × time

So usual time,

t = Distance

speed =

300

xhr … … (ii)

Case 2: with increased speed

new speed of the train, s1 = (x + 5) km/hr

We know Distance = speed × time

So increases time,

t1 = Distance

speed =

300

(x + 5)hr … … (iii)

Now it is given the train takes 2 hours less after increase in speed

So increased time = original time – 2 hours

i.e., t1 = t – 2



40

Substituting values from equation (ii) and (iii), we get

⇒ 300

x + 5 =

300

x− 2

⇒ 300

x + 5 =

300 − 2x

x

⇒ 300x = (x + 5) (300 – 2x)

⇒ 300x = x(300 – 2x) + 5(300 – 2x)


⇒ 300x = 300x – 2x2 + 1500 – 10x

⇒ 2x2 + 10x – 1500 = 0

Dividing throughout by 2, we get

⇒ x2 + 5x – 750 = 0

Now we will be factorizing this by splitting the middle term, we get

⇒ x2 + 30x – 25x – 750 = 0

⇒ x (x + 30) – 25(x + 30) = 0

⇒ (x + 30) (x – 25) = 0

⇒ (x + 30) = 0 or (x – 25) = 0

⇒ x = – 30 or x = 25

But x is the speed of the train and it cannot be negative, hence the usual

speed of the train is 25km/hr.

OR

1

a + b + x =

1

a +

1

b +

1

x

⇒ 1

a + b + x−

1

x =

1

a +

1

b

⇒ x − (a + b + x)

x(a + b + x) =

b + a

ab

⇒ −(a + b)

x(a + b + x) =

b + a

ab


⇒ −1

x(a + b + x) =

1

ab



41

⇒ – ab = x (a + b + x)

⇒ – ab = ax + bx + x2

⇒ x2 + (a + b) x + ab = 0

Comparing this with standard equation, i.e., Ax2 + Bx + C = 0, we get

A = 1, B = (a + b), C = ab

So, the value of x of a quadratic equation can be found by using the formula,

x = −B ± √B2 − 4ac

2A


x = −(a + b) ± √(a + b)2 − 4(1)(ab)

2(1)

But we know the expansion of (a + b)2 = a2 + b2 + 2ab, substituting this in above equation, we get

x = −(a + b) ± √a2 + b2 + 2ab − 4ab

2

x = −(a + b) ± √a2 + b2 − 2ab

2

But we know a2 + b2 – 2ab = (a – b)2, substituting this in above equation, we get

x = −(a + b) ± √(a − b)2

2

x = −(a + b) ± (a − b)

2

Now this has two possibilities,

x = −(a + b) + (a − b)

2 or x =

−(a + b) − (a − b)

2

x = −a − b + a − b

2 or x =

−a − b − a + b

2

x = −2b

2 or x =

−2a

2

⇒ x = – b or x = – a

These are the required values of x.



42

24. Given AP has 50 terms,

So general term can be written as,

an = a + (n – 1) d………..(i)

Now given the AP’s 3rd term is 12, substituting this in general term, we get

a3 = a + (3 – 1)d = 12

⇒ a + 2d = 12…………(ii)

It is also given the last term is 106, as we know the AP has 50 terms, so 50th term is 106, substituting this in general term, we get

a50 = a + (50 – 1)d = 106

⇒ a + 49d = 106…………(iii)

Now subtracting equation (ii) from equation (iii), we get

⇒ d = 2

Now substituting the value of d in equation (ii), we get

⇒ a + 2d = 12

⇒ a + 2(2) = 12

⇒ a + 4 = 12

⇒ a = 12 – 4

⇒ a = 8

So, the given AP has first term, a = 8 and the difference, d = 2, so the 29th term will have n = 29, substituting these values in equation (i), we get

⇒ a29 = a + (29 – 1)d

⇒ a29 = 8 + (28)2

⇒ a29 = 8 + 56

⇒ a29 = 64

Hence the 29th term of the given AP is 64.



43

25. Let the given right – angled triangle be ΔABC, right – angled at B.

We need to prove AC2 = AB2 + BC2

Now let’s draw a perpendicular line BD to line AC

i.e., BD⊥AC

Now we know,

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

Hence, ΔADB~ΔABC

Now applying the similar triangles condition, i.e., sides of similar triangles

are same in ratio, we get

AD

AB =

AB

AC

⇒ AD.AC = AB2………(i)

Now similarly consider the other set of similar triangles,

ΔBDC~ΔABC

Now applying the similar triangles condition, i.e., sides of similar triangles are same in ratio, we get

CD

BC =

BC

AC

⇒ CD.AC = BC2………(ii)

Now adding equation (i) and (ii), we get

AD.AC + CD.AC = BC2 + AB2

Now taking AC common on LHS, we get



44

⇒ AC (AD + CD) = BC2 + AB2

Now from figure we can see that AD + CD = AC, so above equation becomes

⇒ AC(AC) = BC2 + AB2

⇒ AC2 = BC2 + AB2

Hence in a right – angled triangle square of the hypotenuse is equal to sum of the squares of the other two sides.

Hence Proved.

26. Construction of ΔABC

Steps:

a. Draw a base line BC = 6 cm

b. From point B make an arc with radius 8cm. From point C make an arc with radius 9cm.The two arcs meets at a point; this point is point A. Join AB and AC.

c. Hence, given ΔABC is constructed.

Now we will construct a triangle similar to ΔABC whose sides are 3

5 of the

corresponding sides of ΔABC.

Steps of construction:



45

a. Draw a ray BX making an acute angle with AB on the side opposite to

vertex C. In 3

5 the greater number is 5, so we will make 5 points, namely B1,

B2, B3, B4 and B5, on ray BX at equal distance, i.e., BB1 = B1B2 = B2B3 = B3B4 = B4B5.

b. B3 is joined with C to form B3 C as 3 point is smaller. B5 C1 is drawn parallel to B3 C as 5 point is greater.

c. Now C1A1 is drawn parallel to CA.



46

Thus, A1BC1 is the required triangle.

27. The given scene can be represented as a figure shown below

Here AB is the tower with man on it

C is initial position of the car at 30° angle of depression,

D is final position of the car at 45° angle of depression

So, for the car to move from point C to D will take 12 minutes.

We know angle of elevation is equal to angle of depression, so

∠PAC = ∠ACB = 30° and ∠PAD = ∠ADB = 45°……….(i)

And also, the tower is vertical, hence



47

∠ABC = 90°……….(ii)

Now consider right – angled ΔADB, applying the trigonometric rule,

tan D = opposite side

adjacent side


tan 45° = AB

BD

But tan45° = 1, so above equation becomes,

1 = AB

BD

⇒ AB = BD………..(iii)

Now consider other right – angled ΔACB, applying the trigonometric rule,

tan C = opposite side

adjacent side

Now substituting the corresponding values we get

tan 30° = AB

BC

But tan30° = 1

√3, so above equation becomes,

1

√3 =

AB

BC

⇒ AB = BC

√3… … … … (iv)

Now comparing equation (iii) and (iv), we get

BD = BC

√3

⇒ BD = BD + DC

√3

⇒ √3 BD = BD + DC

⇒ DC = √3 BD – BD

⇒ DC = (√3 – 1) BD

⇒ BD = DC

√3 − 1… … (v)



48

But it is given time taken to cover DC = 12 minutes

So, the time taken to cover DC

√3−1 = 12 ×

1

√3−1 minutes

Now we will solve

12 × 1

√3 − 1

Multiply and divide by √3 + 1, we get

⇒ 12 × 1

√3 − 1 ×

√3 + 1

√3 + 1

⇒ 12(√3 + 1)

(√3 − 1)(√3 + 1)

⇒ 12(√3 + 1)

((√3)2

− 12)

⇒ 12(√3 + 1)

(3 − 1)

⇒ 12(√3 + 1)

2

⇒ 6(√3 + 1) minutes ⇒ 6(1.73 + 1)

= 6(2.73)

= 16.38 minutes

Hence the car will take 16.38 minutes to reach the observation tower from this point.

OR

The given scene can be represented as a figure shown below



49

Where BE is the water surface

A is 60m above the water surface ⇒ AB = 60m

C is the cloud in the sky

Given the angle of elevation of a cloud from a point 60m above the surface of the water of a lake is 30°, so

∠CAD = 30°

and also given the angle of depression of its shadow from the same point in water of lake is 60°, so

∠DAF = 60°

So, from figure CE will be the height of cloud above the water surface, and

EF will be depth of shadow below the water surface.

And AD is a line parallel to water surface BE, hence AD is perpendicular to CF.

So ΔCAD and ΔDAF are right – angled triangle.

Now consider right – angled ΔCAD, applying the trigonometric rule,

tan A = opposite side

adjacent side

Now substituting the corresponding values we get

tan 30° = CD

AD

But tan30° = 1

√3, so above equation becomes,

1

√3 =

CD

AD

⇒ AD = √3 CD….(i)

Now consider the other right – angled ΔDAF, applying the trigonometric rule,

tan A = opposite side

adjacent side


tan 60° = DF

AD

But tan60° = √3, so above equation becomes,



50

√3 = DF

AD

⇒ AD = DF

√3… . (ii)

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

√3CD = DF

√3

⇒ 3CD = DF

But from figure DF = DE + EF, so the above equation becomes,

⇒ 3CD = DE + EF

And we know height of cloud from surface of the water = depth of clouds shadow in the water, so CE = EF, substituting this in above equation we get

⇒ 3CD = DE + CE

But from figure CE = CD + DE, substituting this in above equation, we get

⇒ 3CD = DE + CD + DE

⇒ 3CD – CD = 2DE

⇒ 2CD = 2DE

⇒ CD = DE

Now as AD is parallel to BE, so DE = AB = 60m, substituting this in above

equation, we get

⇒ CD = DE = 60m

Hence From figure the height of cloud above the surface of the water is

CE = CD + DE = 60 + 60 = 120m

So, the height of the cloud from the surface of water is 120m.

28. Given total frequency of the given data is 100.

i.e., f = 100………..(i)

i.e., sum of frequency = 100

From table adding all the frequencies, we get

2 + 5 + x + 12 + 17 + 20 + Y + 9 + 7 + 4 = 100

⇒ x + Y + 76 = 100

⇒ x + Y = 100 – 76



51

⇒ x + Y = 24………..(ii)

Now we will rewrite the table with cumulative frequencies to find the median

So, from above table it is clear that the median class is 500 – 600

And we know, the median is calculated using the formula

Median = l +

n2

− cf

f × h

Where l = lower limit of median class = 500

h = class interval = 600 – 500 = 100

cf = cumulative frequency of the class before the median class = 36 + x

f = frequency of the median class = 20

n = ∑fi = 100

And given median = 525

Substituting these values in the above equation of median we get

525 = 500 +

1002

− (36 + x)

20 × 100

⇒ 525 − 500 = (100

2− (36 + x)) × 5

⇒ 25 = (50 × 5) – (36 × 5) – 5x

⇒ 5x = 250 – 180 – 25



52

⇒ 5x = 45

⇒ x = 9

Substituting the value of x in equation (i), we get

⇒ x + Y = 24

⇒ 9 + Y = 24

⇒ Y = 24 – 9

⇒ Y = 15

Hence the values of x and Y are 9 and 15 respectively.

OR

First, we will rewrite the table with cumulative frequencies.

Now we will draw less than table by writing cumulative frequencies under number of student column



53

Now we will draw less than type ogive for the above data,

We first draw the coordinate axes, with lower limits of the profit along the horizontal axis, and the cumulative frequency along the vertical axes.

Then, using less than table data, we plot the points (10,5), (20, 8), (30, 12), (40,15), (50, 18), (60, 22), (70, 29), (80, 38), (90, 45), (100, 53). We join these points with a smooth curve to get the ‘less than’ ogive, as shown below.

Now from the graph we see that total frequency = 53

⇒ N = ∑fi = 53

⇒ N

2 =

53

2 = 26.5

So, we will draw a line y = 26.5 parallel to x – axis, the intersection of this

line with the less than ogive curve gives the x coordinate.

And this x – coordinate is the median of the given data.



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Hence the required median mark of the given data is 66.4.

29. First, we will draw the bucket as per the given data,

Given the height of the bucket, h = 24cm

Radius of the upper end, r1 = 15cm

Radius of the lower end, r2 = 5cm

So, the slant height, l of the bucket can be calculated using the below formula



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l = √h2 + (r1 − r2)2


l = √(24)2 + (15 − 5)2

l = √576 + (10)2

l = √576 + 100

l = √676

l = 26cm

Now we know the curved surface area of a frustum is

= π(r1 + r2)l


= π (15 + 5) (26)

= π (20) (26)

= π (520)

= 3.14 × 520

= 1632.8≈ 1633cm2

Hence the curved surface area of the bucket is 1633cm2.

30. Given sec θ + tan θ = p

But we know sec θ = 1

cos θ, tan θ =

sin θ

cos θ,

substituting these values in the given equation, we get

⇒ 1

cos θ +

sin θ

cos θ = p

⇒ 1 + sin θ

cos θ = p

Now we will square on both sides, we get

⇒ (1 + sin θ

cos θ)

2

= p2

⇒ (1 + sin θ)2

cos2 θ = p2

Now we know sin2θ + cos2θ = 1⇒ cos2θ = 1 – sin2θ, substituting this in

above equation, we get



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⇒ (1 + sin θ)2

(1 − sin2 θ) = p2

Now expanding the numerator by applying the formula (a + b)2 = a2 + b2 + 2ab, we get

⇒ 12 + sin2 θ + 2 sin θ

(1 − sin2 θ) = p2

⇒ 1 + sin2 θ + 2 sin θ

(1 − sin2 θ) = p2

Now let x = sin θ, so the above equation becomes

⇒ 1 + x2 + 2x

(1 − x2) = p2

⇒ 1 + x2 + 2x = p2(1 – x2)

⇒ 1 + x2 + 2x = p2 – p2x2

⇒ 1 + x2 + 2x – p2 + p2x2 = 0

⇒ (1 + p2)x2 + 2x + (1 – p2) = 0

Comparing this with standard equation, i.e., Ax2 + Bx + C = 0, we get

A = (1 + p2), B = 2, C = (1 – p2)

So the value of x of a quadratic equation can be found by using the

formula,

x = −B ± √B2 − 4ac

2A


x = −2 ± √22 − 4(1 + p2)(1 − p2)

2(1 + p2)

x = −2 ± √4 − 4(12 − (p2)2)

2(1 + p2)

x = −2 ± √4 − 4(1 − p4)

2(1 + p2)

x = −2 ± √4 − 4 + 4p4

2(1 + p2)



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x = −2 ± √4p4

2(1 + p2)

x = −2 ± 4p2

2(1 + p2)

So the two possibilities are,

x = −2 + 2p2

2(1 + p2) or x =

−2 − 2p2

2(1 + p2)

x = 2(p2 − 1)

2(1 + p2) or x =

−2(p2 + 1)

2(1 + p2)

x = (p2 − 1)

(1 + p2) or x = −1

But we had assumed x = sin θ, substituting back we get

sin θ = (p2 − 1)

(1 + p2) or sin θ = −1

And we know

cosec θ = 1

sin θ

Substituting the value of sin θ, we get

cosec θ = 1

(p2 − 1)(1 + p2)

or cosec θ = 1

−1

cosec θ = (1 + p2)

(p2 − 1) or cosec θ = −1

These are the required value of cosec θ.

×××



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class x time allowed: 3 hrs max marks: 80 · class x mathematics sample question paper 2018-19 time...

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