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Grade XMathematics

Exam Preparation Booklet

Chapter WiseImportant Questions-Solutions

Real Numbers_Questions

Q. No.1

On dividing a number X by 126, the remainder is 47. The remainder when the same number is divided by 18 is

A. 7B. 9C. 11D. 13

Q. No.2

What is the number of zeroes at the end of , if ?

A. 0B. 1C. 2D. Can't be determined

Q. No.3

If x is the value of the sum of exponents of the prime factors of 584, then the value of x2 is

A. 9B. 16C. 25D. 36

Q. No.4

If the decimal expansion of the rational number terminates after n decimal places, then the least prime factor of n is

A. 1B. 2C. 3D. 5

Q. No.5

The square of the smallest number by which should be multiplied so as to get a positive rational number is

A. an irrational numberB. an odd number

a

100a = 24 × 35 × 53 × 72

12345375

√294

C. an even numberD. a multiple of 4

Q. No.6

The smallest number which leaves the same remainder 5 when divided by 21, 39 and 42 separately, is _____.

Q. No.7

The last digit of 92x + 1 − 5x, where x is a natural number, is _____.

Q. No.8

If the LCM(70, 125, 175) = 17 , then the value of is _____.

Q. No.9

If n is the H.C.F of 18 and 21, then is _____ divisible by 5, 7, and 19.

Q. No.10

If a and b are two consecutive positive integers such that , then the difference between their squares is always equal to _____.

(x + 1) − 1 x

2

32n − 22n

a < b

Real Numbers_SolutionsSolution 1

We have,Let the number X when divided by 126 gives x as quotient and 47 as the remainderThus, X = 126x + 47 =

Thus, when X is divided by 18, it gives 11 as the remainder.Hence, the correct answer is C.

Solution 2

We have,

Since there is a factor of 10 in and the exponent of 10 is 1, so there will be one zero at the end of Hence, the correct answer is option B.

Solution 3

We have,The prime factorization of 584 = Thus, x = (3 + 1) = 4So, x2 = 16Hence, the correct answer is option B.

Solution 4

We have,

So, n = 2Thus, the least prime factor of n is 2.Hence, the correct answer is option B.

Solution 5

We have,

Now, in order to get a positive rational number, the smallest number which must be multiplied to is

Thus, the square of the smallest number by which should be multiplied so as to get a positive rational number is .

Hence, the correct answer is option C.

Solution 6

We have,The required number = LCM(21, 39, 42) + 5Now, LCM(21, 39, 42) = LCM( ) = = 546So, the required number is (546 + 5) = 551Thus, the smallest number which leaves the same remainder 5 when divided by 21, 39 and 42 separately, is .

Solution 7

We have,For every natural number x, 2x + 1 is always oddTherefore, 92x + 1 will always end with 9 and 5x will always end with 5So, the last digit of 92x + 1 − 5x will always be 4, for any value of xThus, the last digit of 92x + 1 − 5x, where x is a natural number, is .

Solution 8

We have,LCM(70, 125, 175) = 1750 = Comparing it with 17 , we get

(18 × 7x) + (18 × 2 + 11)⇒ X = 18 × {(7x + 2)} + 11

a = 24 × 35 × 53 × 72

⇒ a = 2 × 23 × 53 × 35 × 72

⇒ a = 2 × 103 × 35 × 72

⇒ = = 2 × 35 × 72 × 10a

1002×35×72×1000

100

a

100

a

100

23 × 73

= = = = = 32. 9212345375

12345125×3

12345×8125×3×8

987603×1000

98.763

√294 = √49 × 6 = 7√6√294 √6

√294 (√6)2

= 6

3 × 7, 3 × 13, 2 × 3 × 7 2 × 3 × 7 × 13

551

4

17 × 103 − 1 = 17 (102 + 1) − 1(x + 1) − 1

1 102 1

Thus, if the LCM(70, 125, 175) = 17 , then the value of is .

Solution 9

H.C.F of 18 and 21 can be found by the prime factorization of 18 and 21

Therefore, H.C.F of 18 and 21 = 3Now,

Thus, if n is the H.C.F of 18 and 21, then is always divisible by 5, 7, and 19.

Solution 10

Since a and b are two consecutive positive integers,Therefore, one of them is even and the other one is odd.Using Euclid's algorithm, we know that every positive even integer is of form 2p, and every positive odd integer is of form 2p + 1,where p is some positive integer.

Thus, .

Thus, if a and b are two consecutive positive integers such that , then the difference between their squares is always equal to .

x + 1 = 102 + 1x = 102

⇒ = 51x

2

(x + 1) − 1x

251

18 = 2 × 3 × 321 = 3 × 7

32n − 22n = 36 − 26 = (33)2

− (23)2

= (27)2 − (8)2

= (27 − 8) (27 + 8)

= 19 × 35 = 19 × 5 × 7

32n − 22n

∴ a = 2p, b = 2p + 1

b2 − a2 = (2p + 1)2

− (2p)2

= 4p + 1⇒ 4p + 1 = (2p) + (2p + 1) = a + b

a < b

a + b

Polynomials_Questions

Q. No.1

Let is a quadratic polynomial. If is one of the zeroes of the given polynomial then the value of k is

A.

B.

C.

D.

Q. No.2

The graph of the polynomial , where a > 0 and c < 0 is

A.

B.

kx2 − (k − 1)x + 3 −2

−5

2

5

2

−5

3

5

3

P(x) = ax2 + bx + c

C.

D.

Q. No.3

If is one of the zeroes of the polynomial then the value of k is

A.

B.

C.

D.

Q. No.4

Let is a cubic polynomial. If one of the zeroes of the given polynomial is zero then the sum of the other two is

A.

B.

C.

D.

−1 kx2 − kx + 3

−3

2

−2

3

2

3

3

2

px3 + bx2 + cx + d

cp

−b

p

bp

dp

Q. No.5

If the zeroes of the polynomial are positive then p and q are

A. both positiveB. both negativeC. p is negative and q is positiveD. both are equal

Q. No.6

Let p and q are the zeroes of the polynomial then the value of is _____.

Q. No.7

If a polynomial of degree 2n is divided by a polynomial of degree n then the degree of the quotient is _____.

Q. No.8

If a polynomial of odd degree n touches the x-axis at two different points where n > 3, then the number of pairs of equal zeroes is_____.

Q. No.9

If in a quadratic polynomial p(x) there is no linear term and the constant term is negative, then the zeroes of p(x) have the _____ sign.

Q. No.10

If a cubic polynomial intersects the x-axis at three distinct points and another quadratic polynomial intersects the x-axis at two distinctpoints then the sum of the numbers of real zeroes of both the polynomial is _____.

x2 + px + q

2x2 + 3x − 6p+q

pq

Polynomials_SolutionsSolution 1

The given polynomial is f(x) = .Since is one of the zeroes of the given polynomial, so .

Hence, the correct answer is option A.

Solution 2

If a > 0 it means the graph of the polynomial should be upward parabolic.Again the graph cuts the y-axis at P(0, y)By putting x = 0 in , we get y = cSince c < 0 , so that y < 0.Hence, the correct answer is option D.

Solution 3

If is one of the zeroes of the polynomial then it satisfies the given polynomial.


Solution 4

Let are the three zeroes of the given polynomial.

We know that the sum of the zeroes of a cubic polynomial

Thus,

Hence, the correct answer is option B.

Solution 5

Let are the zeroes of the given polynomial.

If are positive then p should be negative and q should be positive.Hence, the correct answer is option C.

Solution 6

We know that, and

Thus,

Let p and q are the zeroes of the polynomial then the value of is .

Solution 7

Let the degree of polynomial p(x) = 2ndegree of polynomial g(x) = nAccording to the division algorithm,

kx2 − (k − 1)x + 3−2 f (−2) = 0

k(−2)2

+ (k − 1) (−2) + 3 = 0

⇒ 4k − 2k + 5 = 0

⇒ 2k + 5 = 0

⇒ k =−5

2

y = ax2 + bx + c

−1 kx2 − kx + 3

k(−1)2 − k(−1) + 3 = 0

⇒ 2k + 3 = 0

⇒ k =−3

2

α, β, γ

= =− coefficient of x2

coefficient of x3

−b

p

α + β + γ =

⇒ α + β = (∵ γ = 0)

−b

p

−b

p

α and β∴ α + β = −p and

αβ = q

α and β

p + q = =− coeffcient of x

coeffcient of x2

−3

2

pq = =constant term

coeffcient of x2

−6

2

= × =p+q

pq

−3

2

2

−6

1

2

2x2 + 3x − 6p+q

pq1

2

Degree of quotient = degree of dividend degree of divisor Degree of quotient = degree of p(x) degree of g(x)

= 2n n = nIf a polynomial of degree 2n is divided by a polynomial of degree n then the degree of the quotient is n.

Solution 8

When a curve touches the x-axis at one point, it means it has two common zeros at that point.Since the given polynomial touches the x-axis at two different points thus it has two pairs of equal zeroes.Hence, if a polynomial of odd degree n touches the x-axis at two different points where n > 3, then the number of pairs of equal zeroes istwo.

Solution 9

Let is a polynomial. If the linear term is zero and the constant term is negative, then

Thus, are the zeroes of the given polynomial which have the opposite sign.

Hence, if in a quadratic polynomial p(x) there is no linear term and the constant term is negative, then the zeroes of p(x) have the oppositesign.

Solution 10

If a cubic polynomial intersects the x-axis at three distinct points, then it means it has three real zeroes.Similarly for quadratic polynomial number of real zeroes is two.Thus, the sum of the numbers of real zeroes of both the polynomial is 2 + 3 = 5Hence, If a cubic polynomial intersects the x-axis at three distinct points and another quadratic polynomial intersects the x-axis at twodistinct points then the sum of the numbers of real zeroes of both the polynomial is five.

−⇒ −

−

ax2 + bx + cp(x) = ax2 − c

±√c

a

Pair of Linear Equations in Two Variables_Questions

Q. No.1

If 2x + 3y = 4 and 8x + by = 6 has no solution, then the value of b2 is

A. 100B. 121C. 144D. 169

Q. No.2

What is the value of ?

A. 1B. 2

C.

D.

Q. No.3

Maria's age will be 3 times the age of her granddaughter Janeth after 6 years. If 4 years ago the age of Maria was 4 times the age ofJaneth, then the difference of their present ages is

A. 45B. 50C. 55D. 60

Q. No.4

What is the value of , if the lines given by the equations and are parallel?

A.B. 0C. 1D. 2

Q. No.5

If the difference between a two-digit number and the sum of its tens and units digit is 72 and the difference between the digits of thenumber is 6, then the required number is

A. 72

(x2 − y2), if 4x+y = 4x−y = √64

9

4

3

( − 1)3n

10−4x + 3ny = 7 2x − 5y = 11

−1

B. 82C. 84D. 93

Q. No.6

Aman has a total of 45 coins with him in the form of Rs 5 and Rs 2. If the total sum of the money he has is Rs 195, then the value of

is _____.

Q. No.7

The system of linear equations where has _____ solution/s.

Q. No.8

The equation 4x + y = 0 passes through the point of intersection of equations _____.

Q. No.9

If the non-parallel lines and do not bind any area with the x or y-axis together, then the value of p is_____.

Q. No.10

If there are two distinct solutions of and , then the value of is _____.

( )xy

10

x + ky + 1 = 0 and ax + ly + b = 0 l = ak = bk

x − 2y = 9 and 2x + y =

3x + 5y = 7 3x + py = 7

+ = a + bx

a

y

b+ = 2x

a2

y

b2a − b

Pair of Linear Equations in Two Variables_SolutionsSolution 1

We have,

For no solution:

Now,

Thus b2 = 144Hence, the correct answer is option C.

Solution 2

We have,


Solution 3

We have,Let Maria's age present be x years and Janeth's present age be y yearsNow, according to the question

From (1) and (2), we havex = 84 and y = 24Thus, Hence, the correct answer is option D.

Solution 4

We have,

A pair of linear equations will represent parallel lines, i.e has no solution, if


Solution 5

We have,Let the tens and units digits be and respectively.Thus, the number can be expressed as .Now, according to the question

= ≠a1

a2

b1

b2

c1

c2

a1 = 2, b1 = 3, c1 = −4, a2 = 8, b2 = b, c2 = −6

∴ = ≠

⇒ b = = 12

2

8

3

b

−4

−6

8×32

4x+y = 4x−y = √64

⇒ 4x+y = 4x−y = 4√4

⇒ 4x+y = 4x−y = 4

⇒ x + y = x − y =

⇒ (x2 − y2) = (x + y) (x − y) = ( )( ) =

3

2

3

2

32

32

9

4

x + 6 = 3 (y + 6) ⋯ (1)

x − 4 = 4 (y − 4) ⋯ (2)

x − y = 84 − 24 = 60

a1 = −4, b1 = 3n, c1 = −7, a2 = 2, b2 = −5, c2 = −11

= ≠a1

a2

b1

b2

c1

c2

⇒ = ≠

⇒ 6n = 20

⇒ n = =

⇒ ( − 1) = ( × − 1) = 0

−42

3n

−57

11

20

6

10

3

3n

10

3

10

10

3

x y10x + y

(10x + y) − (x + y) = 72

⇒ 9x = 72 ∴ x = 8

Also

Thus, the required number is Hence, the correct answer is option B.

Solution 6

We have,Let Aman has x coins of Rs 5 and y coins of Rs 2Now, according to the question

From (1) and (2), we get

Thus, Aman has a total of 45 coins with him in the form of Rs 5 and Rs 2. If the sum total of the money he has is Rs 195, then the value of

is .

Solution 7

We have,

Now,

From (1), (2), and (3)

Thus, the system of linear equations where has infinitely many solutions.

Solution 8

The given two equations are

Now, using the substitution method, we get Putting the value (1, −4) in 2x + y, we get 2x + y = −2Thus, the equation 4x + y = 0 passes through the point of intersection of equations .

Solution 9

We have,Since the given lines are not parallel, so there are only two conditions left, i.e.(i) The lines intersect each other. (ii) The lines are coincident. Now, it is given that the lines do not bind any area with the x or y-axis, therefore the lines are coincident. For coincident lines, we have

Thus, if the non-parallel lines and do not bind any area with the x or y-axis together, then the value of p is .

Solution 10

We have,The given equations can be written as and

x − y = 6

⇒ y = x − 6

⇒ y = 8 − 6 = 2

10x + y = 10 × 8 + 2 = 82.

x + y = 45 . . … (1)

and 5x + 2y = 195 . . … (2)

x = 35y = 10

⇒ ( ) = ( ) = 35xy

1035×10

10

( )xy

1035

a1 = 1, b1 = k, c1 = 1, a2 = a, b2 = l, c2 = b

= = [ ∵ ak = bk ⇒ a = b] ⋯(1)

= = = = [∵ l = ak] ⋯(2)

= = ⋯(3)

a1

a2

1a

1b

b1

b2

k

l

k

ak

1a

1b

c1

c2

1b

1a

= =a1

a2

b1

b2

c1

c2

x + ky − 1 = 0 and ax + ly − b = 0 l = ak = bk

4x + y = 0 and x − 2y = 9

x = 1, y = −4

x − 2y = 9 and 2x + y = −2

= =

⇒ = =

⇒ = 1

⇒ p = 5

a1

a2

b1

b2

c1

c2

3

3

5p

−7

−7

5p

3x + 5y = 7 3x + py = 7 5

bx + ay − ab(a + b) = 0 b2x + a2y − 2a2b2 = 0

Now, a pair of linear equation has only one of the following three possibilities(i) No solution.(ii) Only one solution.(iii) Infinitely many solutions.Since there are two distinct solutions to the given pair of linear equations, this can be possible only when the given pair of linear equationshas infinitely many solutions.Therefore,

Thus, if two distinct points are the solutions of and , then the value of is .

= =

⇒ = =

⇒ a = b

b

b2

a

a2

−ab(a+b)

−2a2b2

1b

1a

a+b

2ab

⇒ a − b = 0+ = a + b

x

a

y

b+ = 2

x

a2

y

b2 a − b 0

Quadratic Equations_Questions

Q. No.1

Let and are the factors of the quadratic equation , then a + b is a

A. composite numberB. even numberC. prime numberD. multiple of 3

Q. No.2

Let is a quadratic equation. If a and c have the opposite sign then the roots of the given equation are

A. always realB. always imaginaryC. equalD. none of the above

Q. No.3

If the equation has no distinct roots then

A. a = 0B. a = 8C. both A and CD. none of the above

Q. No.4

If one root of the equation is twice the other root, then

A.B.C.D.

Q. No.5

What will be the value of ?

A. 2B. 3C. 5

(x − a) (x − b) x2 − 5x + 6 = 0

ax2 + bx + c = 0

ax2 + ax + 2 = 0

x2 − bx + c = 0

b = c

2b2 = c

b2 = 9c

2b2 = 9c

u = √2 + √2 + √2+. . .

D. 7

Q. No.6

If the sum of the roots is equal to the product of the roots of the quadratic equation , then the number of values ofa is _____.

Q. No.7

If the sum of the roots of an equation is half the product of the roots of the equation , then thevalue of a is _____.

Q. No.8

If has no real roots, then _____.

Q. No.9

If the equation has equal roots and the equation has imaginary roots then the value of k is_____.

Q. No.10

If one of the roots of the quadratic equation is , then the value of k is _____.

x2 − 2ax + a

2 = 0

x2 + 2x = a(x − 1) x

2 − a = x + 4

(a + b)x2 + 2abx + a − b = 0 a2 <

9x2 + 6kx + 4 = 0 x2 + 2x − k = 0

2x2 + kx − 5 = 0 −1

Quadratic Equations_SolutionsSolution 1

Thus, a = 3 and b = 2 a + b = 5, which is a prime number.


Solution 2

The roots of the given equation are real if If a and c have the opposite sign then their product is always negative.Here b2 is always positive thus, the overall value of is always positive so the roots of the given equation are always real.Hence, the correct answer is option A.

Solution 3

If has no distinct roots, then


Solution 4

Let are the roots of the given equation.

Now,

Hence, the correct answer is option D.

Solution 5

Given,

Since, so, it cannot have a negative value. So, Hence, the correct answer is option A.

Solution 6

x2 − 5x + 6 = 0

⇒ x2 − 3x − 2x + 6 = 0

⇒ x (x − 3) − 2 (x − 3) = 0

⇒ (x − 3) (x − 2) = 0

∴

b2 − 4ac > 0

b2 − 4ac

ax2 + ax + 2 = 0a2 − 8a = 0

⇒ a(a − 8) = 0

⇒ a = 0, 8

α and β∴ α = 2β

α + β = b

⇒ β =

and αβ = c

⇒ α =

∴ =

⇒ 2b2 = 9c

b

3

3c

b

3c

b

2b

3

u = √2 + √2 + √2+. . .

⇒ u = √2 + u

⇒ u2 = 2 + u

⇒ u2 − u − 2 = 0

⇒ u2 − 2u + u − 2 = 0

⇒ u (u − 2) + (u − 2) = 0

⇒ (u + 1) (u − 2) = 0

⇒ u = −1, 2

u = √2 + √2 + √2+. . . u = 2

The sum of the roots of the given quadratic equation = 2aThe product of the roots of the given quadratic equation = a2

According to the given condition,

Hence, if the sum of the roots is equal to the product of the roots of the quadratic equation , then the number ofvalues of a is 2.

Solution 7

Sum of the roots of the equation = Now,

Product of the roots = Now,

Hence, if the sum of the roots of an equation is half the product of the roots of the equation , thenthe value of a is 0.

Solution 8

If has no real roots, then

Hence, If has no real roots, then .

Solution 9

For the equation

Now, for the equation

Thus, the value of k which satisfy both the equation is .Hence, if the equation has equal roots and the equation has imaginary roots then the valueof k is .

Solution 10

If one of the roots of the quadratic equation is , then

Hence, if one of the roots of the quadratic equation is , then the value of k is .

a2 = 2a

⇒ a2 − 2a = 0

⇒ a(a − 2) = 0

⇒ a = 0, 2

x2 − 2ax + a2 = 0

x2 + 2x = a(x − 1)

⇒ x2 + x(2 − a) + a = 0

∴ a − 2

x2 − a = x + 4

⇒ x2 − x − 4 − a = 0

∴ −4 − a

a − 2 =

⇒ 3a − 4 = −4⇒ a = 0

−4−a

2

x2 + 2x = a(x − 1) x2 − a = x + 4

(a + b)x2 + 2abx + a − b = 0D < 0

⇒ 4a2b2 − 4 (a + b) (a − b) < 0

⇒ a2b2 − a2 + b2 < 0

⇒ a2(b2 − 1) + b2 < 0

⇒ a2 <b2

1−b2

(a + b)x2 + 2abx + a − b = 0 a2 <b2

1−b2

9x2 + 6kx + 4 = 0D = 0

⇒ 36k2 − 4 × 36 = 0

⇒ k2 = 4⇒ k = ±2

x2 + 2x − k = 0D < 0⇒ 4 + 4k < 0

⇒ k < −1

−29x2 + 6kx + 4 = 0 x2 + 2x − k = 0

−2

2x2 + kx − 5 = 0 −12(−1)2 + k (−1) − 5 = 0

⇒ 2 − k − 5 = 0

⇒ k = −3

2x2 + kx − 5 = 0 −1 −3

Arithmetic Progressions_Questions

Q. No.1

How many multiples of 7 lie between 20 and 360?

A. 48B. 49C. 50D. 51

Q. No.2

If S is the sum of the first 20 odd natural numbers, then the value of is

A. 9B. 16C. 25D. 36

Q. No.3

If 7, x, y, z, 63 are in AP, then the value of is

A.

B.

C.

D.

Q. No.4

The sum of n terms of the AP is

A.

B.

C.

D.

( )2

S

100

( )x+y

z

788

79

88

9

√3, √12, √27, √48, . . .

√3n(n+√3)

2

√3n(√3n+1)

2

√3n(n+2√3)

2√3n(n+1)

2

Q. No.5

If the sum of first n terms of an AP is 3n2 − n, and its common difference is 4, then its first term is

A.B.C.D.

Q. No.6

The difference of 8th term from the end of the AP 3, 6, 9, 12,..., 393 and the total number of terms in the given AP is _____.

Q. No.7

The _____ term of the AP given as 57, 54, 51, 48,..., −57 is zero.

Q. No.8

If the first and last terms of an AP are 8 and 128, and the sum of all the terms is 272, then the total number of terms in the AP are_____.

Q. No.9

The 11th term of the given sequence 1, 8, 3, 11, 5, 14, ... is _____.

Q. No.10

There are total _____ numbers divisible by both 4 and 8 in between 31 and 409.

n − 1n

n + 1n + 2

Arithmetic Progressions_QuestionsSolution 1

We have,The multiples of 7 lying between 20 and 360 are 21, 28, 35, 42,..., 357 Now, these numbers form an AP with the first term(a) = 21, the common difference(d) = 7, and the last term(an) = 357Using an = a + (n − 1)d, we get

Thus, there are 49 multiples of 7 between 20 and 360.Hence, the correct answer is option B.

Solution 2

We have,The first 20 odd natural numbers are1, 3, 5, 7, 9,... up to 20 termsNow, the above sequence forms an AP with the first term(a) = 1, the common difference(d) = 2, and the total number of terms(n) = 20Using sum of an AP =


Solution 3

We have,a = 7a5 = 63

Thus, x = 7 + 14 = 21, y = 21 +14 = 35, z = 35 + 14 = 49

So,


Solution 4

We have,The given AP can be rewritten as

Now,


Solution 5

We have,Let the first term of the AP be a

357 = 21 + (n − 1)7

⇒ n − 1 = = 48

⇒ n = 49

336

7

(2a + (n − 1)d)n

2

S = (2 × 1 + (20 − 1)2)

⇒ S = 10 (2 + (19 × 2))

⇒ S = 400

∴ ( )2

= ( )2

= (4)2 = 16

202

S

100400100

⇒ a + 4d = 63

⇒ 4d = 63 − 7 = 56

⇒ d = = 14564

( ) = ( ) = =x+y

z

21+35

49

56

49

8

7

√3, 2√3, 3√3, 4√3, . . .a = √3, d = 2√3 − √3 = √3

Sn = (2a + (n − 1)d)

⇒ Sn = (2√3 + (n − 1)√3)

⇒ Sn = (√3 + n√3)

⇒ Sn =

n

2

n

2

n

2

√3n(n+1)

2

Now, Sum of first n terms of an AP = Sn =


Solution 6

We have,nth term from the end of an AP = , where l is the last term, d is the common differenceNow,8th term from the end = l8 =

So, the required difference = l8 − N = Thus, the difference of 8th term from the end of the AP 3, 6, 9, 12,..., 393 and the total number of terms in the given AP is .

Solution 7

We have,Let the nth term of the given AP be zeroNow, a = 57, d = (54 − 57) = −3According to the question0 = 57 + (n −1)(−3)

Thus, the term of the AP given as 57, 54, 51, 48,..., −57 is zero.

Solution 8

We have,Sum of first n terms of an AP = Sn = Now, a = 8 and l = 128

Thus, if the first and last terms of an AP are 8 and 128, and the sum of all the terms is 272, then the total number of terms in the AP are .

Solution 9

We have,Splitting the given sequence, we get two APsThe first AP is 1, 3, 5, ...The second AP is 8, 11, 14, ...So, the 11th term of the given sequence will be the 6th term of the first APNow, for the first AP

6th term = a + 5d= 1 + 5(2)=11Thus, the 11th term of the sequence 1, 8, 3, 11, 5, 14, ... is .

Solution 10

We have,The numbers divisible by both 4 and 8 in between 31 and 409 are32, 40, 48, 56,..., 408

(2a + (n − 1)d)n

2

⇒ 3n2 − n = (2a + (n − 1)4)

⇒ 6n2 − 2n = 2an + 4n2 − 4n

⇒ 2n2 + 2n − 2an = 0

⇒ n (n + 1 − a) = 0

∵ n ≠ 0

⇒ n + 1 − a = 0∴ a = n + 1

n

2

l − (n − 1)d

393 − (8 − 1)3 = 393 − 21 = 372

l = a + (N − 1)d

⇒ 393 = 3 + (N − 1)3

⇒ N = 131

372 − 131 = 241

241

⇒ −3n + 3 + 57 = 0⇒ 3n = 60⇒ n = 20

20 th

(a + l)n

2

⇒ 272 = (8 + 128)

⇒ 136n = 544⇒ n = 4

n

2

4

d = 5 − 3 = 3 − 1 = 2

11

Now, the above sequence forms an AP with the first term(a) = 32 and common difference(d) = 8Also, an = a + (n − 1)d

Thus, there are total numbers divisible by both 4 and 8 in between 31 and 409

⇒ 408 = 32 + (n − 1)8

⇒ n − 1 =

⇒ n = 47 + 1 = 48

3768

48

Triangles_Questions

Q. No.1

In an isosceles triangle ABC if AB2 = 2AC2 and AC is equal to BC then is

A.B.C.D.

Q. No.2

From an external point P, a tangent of length 12 cm is drawn to a circle of radius 5 cm. The distance between the centre of the circleand point P is

A. 10 cmB. 11 cmC. 12 cmD. 13 cm

Q. No.3

In the figure given below, if the perimeter of = 25 cm then the perimeter of is

A. 25 cmB. 50 cmC. 75 cmD. 100 cm

Q. No.4

In a triangle ABC, DE is parallel to BC. If AD = 2 cm, BD = 3 cm, AE = 4 cm and BC = 7 cm, then perimeter of triangle ABC is

△ ∠A

90°

60°

45°

30°

△ ABC △ PQR

A. 22 cmB. 23 cmC. 24 cmD. 25 cm

Q. No.5

If in ∆ABC and ∆DEF, , then ∆ABC ∼ ∆DFE when

A.B.C.D.

Q. No.6

Let and the areas of given triangles are 81 cm2 and 36 cm2 respectively. If the angle bisector of the is 6cm, then the corresponding angle bisector of the is _____.

Q. No.7

In a certain city, two multi-storey buildings of height 24 m and 30 m are _____ apart given that the distance between their roof is 10 m.

Q. No.8

In ∆ABC, a line DE parallel to BC cuts AB at D and AC at E. If BE bisects ∠DEC and ∠AED = 40 , then ∠EBC = _____.

Q. No.9

In an isosceles triangle ABC, AD BC. The ratio between BD and DC is equal to _____.

Q. No.10

=AC

DE

BC

EF

∠B = ∠E

∠C = ∠F

∠C = ∠E

∠F = ∠E

△ ABC ~ △ PQR △ ABC△ PQR

°

⊥

If PQR is an equilateral triangle and PA is the median on QR, then the perimeter of PQR is _____PA.△ △

Triangles_SolutionsSolution 1

Given:AB2 = 2AC2 andAC = BC

AB2 = AC2 + AC2

AB2 = AC2 + BC2

By converse of Pythagoras theorem, is a right-angled isosceles triangle where = 90 degreeSince AC = BC,

(Angles opposite to equal sides)Now,


Solution 2

In the above figure, PQO is a right-angled triangle.Thus, from Pythagoras theorem


Solution 3

From the given figure,

We know that the ratio of corresponding sides of similar triangles is equal to the ratio of their perimeters.Hence,


Solution 4

From basic proportionality theorem,


Solution 5

We know that if in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the two trianglesare similar.Then,

∴

⇒△ ACB ∠C

⇒ ∠A = ∠B

∠A + ∠B + ∠C = 180°

⇒ 2∠A = 90°⇒ ∠A = 45°

△

OP2 = PQ2 + OQ2

⇒ OP2 = 144 + 25 = 169⇒ OP = 13 cm

∠A = ∠P and ∠B = ∠Q∴△ ABC ~ △ PQR (AA criteria)

=

⇒ P (PQR) = = 50 cm

ABPQ

P(ABC)

P(PQR)

25×10

5

=

⇒ EC = × 4 = 6 cm

Thus, perimeter of triangle ABC = 2 + 3 + 4 + 6 + 7 = 22 cm

ADBD

AEEC

32

∠C = ∠E

Hence, ΔABC is similar to ΔDFE, we should have .Hence, the correct answer is option C.

Solution 6

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their angle bisectors.

Hence, Let and the areas of given triangles are 81 cm2 and 36 cm2 respectively. If the angle bisector of the is 6 cm, then the corresponding angle bisector of the is 4 cm.

Solution 7

Let CD be the building with a height of 24 m.AB is the building with a height of 30 m, the distance between their roof i.e. AC is 10 m.Let us assume a point E on the building AB which is 24 m from the base of AB.Thus,AE = AB − 24 = 30 − 24 = 6 mNow in right triangle AEC, Applying Pythagoras theoremAC2 = AE2 + EC2

102 = 62 + EC2 EC2 = 100 36EC = 8 m

Hence, In a certain city, two multi-storage buildings of height 24 m and 30 m are 8 m apart given that the distance between their roof is 10m.

Solution 8

Since DE||BCSo EBC = BED (Alternate angles)

EBC = BEC

∠C = ∠E

= ( )2

⇒ =

⇒ (angle bisector of △ PQR) = = 4 cm

ar(△ABC)

ar(△PQR)

angle bisector of△ABC

angle bisector of△PQR

81

36

36

(angle bisector of△PQR)2

6×6

9

△ ABC ~ △ PQR △ ABC△ PQR

⇒⇒ −⇒

∠ ∠⇒∠ ∠

Now, AED = ECB (Corresponding angles)Thus, EBC + BEC + ECB = 180

2 EBC =180 40 = 140EBC = 70

Hence, in ∆ABC, a line DE parallel to BC cuts AB at D and AC at E. If BE bisects ∠DEC and ∠AED = 40 , then ∠EBC = 70 .

Solution 9

From figure,

Hence, in an isosceles triangle ABC, AD BC. The ratio between BD and DC is equal to 1.

Solution 10

In an equilateral triangle, PQ = QR = PRAnd, QA = AR ( AP is the median)Also, in an equilateral triangle the median, angle bisector and altitude are the same straight line.Thus, in ,

Hence, if PQR is an equilateral triangle and PA is the median on QR, then the perimeter of PQR is PA.

∠ ∠∠ ∠ ∠ °

⇒ ∠ ° − ° °⇒∠ °

° °

∠ADB = ∠ ADC (Each 90°)

∠ABD = ∠ ACD (Angles opposite to the equal sides)

Thus, △ ADC ~ △ ADB (By AA creiteria)

We know that, ratio of corresponding sides of similiar triangles are equal

= = = 1AD

AD

BD

DC

AB

AC

⊥

∵

△ PAQ

PQ2 = PA2 + AQ2

PQ2 = PA2 + (AQ = )

PQ2 = PA2 + (QR = PQ)

3 PQ2 = 4 PA2

PQ =

Thus, perimeter = 3 × = 2√3 PA

QR2

4

QR

2

PQ2

4

2 PA

√3

2 PA

√3

△ △ 2√3

Coordinate Geometry_Questions

Q. No.1

The point P(3, a) divides the line segment joining the points X(7, 4) and Y(3, 4) such that XP : PY = 3 : 5, the value of a is

A. 3B. 4C.D.

Q. No.2

The points (3, 6), (2, 5) and (x, 3) are collinear if the value of x is

A.B. 0C. 1D. 2

Q. No.3

The coordinates of the center of a circle given that the coordinates of the end-points of its diameter are (2, 8) and (6, 4) are

A.B.C.D.

Q. No.4

y-axis divides the line segment joining the points ( 3, 6) and (6, 9) in the ratio

A.B.C.D.

Q. No.5

The distance between the points (3, 2) and (y, ) is units, then the value of y is

A. 2B. 8C. Both A and BD. None of these

−3−4

−1

(−4, 6)

(4, − 6)

(−4, − 6)

(4, 6)

− −

1 : 31 : 22 : 33 : 2

− −1 √26

−

Q. No.6

The distance of the point P from the origin is _____ units.

Q. No.7

In an equilateral triangle PQR, PS is the altitude from point P to the line segment QR. The coordinates of point Q are (5, 2) and point Rare . The coordinates of point S are _____.

Q. No.8

The length of the line segment joining the centers of two circles whose coordinates are X( 3, 3) and Y(5, 3), is _____ units.

Q. No.9

The vertices of a triangle are , then its area is _____ square units.

Q. No.10

The points (3, 6), ( 1, 2) and (6, 12) _____ lie on the same line.

(−3, − 4)

(7, −4)

− −

(−4, 4), (2, 6) and (−7, − 3)

− −

Coordinate Geometry_SolutionsSolution 1

We have,Using the section formula, the value of a is given as:

Hence, the correct answer is option B

Solution 2

We have,Since the given points are collinear


Solution 3

We have,The coordinates of the center of a circle is the midpoint of the end-points of its diameter

Thus, the coordinates of the center of the circle are


Solution 4

We have,Let y-axis divides the line segment joining the points ( 3, 6) and (6, 9) in the ratio Now, any point on the y-axis is given as (0, y)Using section formula, we get the coordinates of the required point as

Thus, y-axis divides the line segment joining the points ( 3, 6) and (6, 9) in the ratio 1 : 2Hence, the correct answer is option B.

Solution 5

We have,Using distance formula, the distance between the given points is given by


Solution 6

We have,Coordinates of origin are (0, 0)Now, using the distance formula, we get

The distance of point P from the origin =

Thus, the distance of the point P from the origin is units.

Solution 7

We have,PS is the altitude from the point P to the line segment QR.Since PQR is an equilateral triangle, S is the mid-point of QR

a = = = 4(3×4)+(5×4)

3+5328

⇒ 3 (5 − 3) + 2 (3 − 6) + x (6 − 5) = 0

⇒ 6 − 6 + x = 0⇒ x = 0

( , ) = (4, 6)2+6

2

8+4

2

− − λ : 1

( , ) = (0, y)

⇒ = 0

⇒ λ =

6λ−3

λ+1

9λ−6

λ+1

6λ−3

λ+1

12

− −

√(3 − y)2 + (−2 + 1)2 = √26

⇒ (3 − y)2 + (−2 + 1)2 = 26

⇒ (3 − y)2 = 25

⇒ 3 − y = ±5

∴ y = −2, 8

√(−3 − 0)2 + (−4 − 0)2 = √9 + 16 = √25 = 5 units

(−3, − 4) 5

△

Hence, the coordinates of S are

Thus, in an equilateral triangle PQR, PS is the altitude from point P to the line segment QR. The coordinates of point Q are (5, 2) and pointR are . The coordinates of point S are .

Solution 8

We have,Using distance formula, the length of the line segment joining the centers of the two circles is given as

XY = units.

Thus, the length of the line segment joining the centers of two circles whose coordinates are X( 3, 3) and Y(5, 3), is units.

Solution 9

We have,

Since area cannot be negativeThus, the required area is 18 square unitsHence, the vertices of a triangle are , then its area is square units.

Solution 10

We have,Area of the triangle formed by the given points is given as

Thus, the given points are collinear.Hence, the points (3, 6), ( 1, 2) and (6, 12) always lie on the same line.

( , ) = (6, − 1)5+72

2−42

(7, −4) (6, − 1)

√(−3 − 5)2 + (3 + 3)2 = √64 + 36 = √100 = 10

− − 10

Area of triangle = {−4 (6 + 3) + 2 (−3 − 4) − 7 (4 − 6)}

= {−36 − 14 + 14}

= × (−36) = −18

12

1

212

(−4, 4), (2, 6) and (−7, − 3) 18

{3 (−2 − 12) − 1 (12 − 6) + 6 (6 + 2)}

= (−42 − 6 + 48)

= 0

12

1

2

− −

Introduction to Trigonometry_Questions

Q. No.1

If , then the value of is

A. 0B. 1

C.

D. 2

Q. No.2

If , where 3x < 90 then the value of cot2x is equal to

A. 1B.

C.

D. not defined

Q. No.3

If , then the value of is

A. 1

B.

C.

D.

Q. No.4

If , then is equal to

A.B.C.D.

Q. No.5

In a right-angled triangle ABC, right angled at B, , then the value of cosC is

1 + cos2 x = sec2 x sin x + sin2 x

3

2

sin 3x − cos x = 0 °

√31

√3

cos x = 23

+ cosec xtan x−1

sin x

32

252

tan α =p

q

p2+q2

pq

cot α

tan α − cot α

tan α + cot α

tan α × cot α

sin A = 35

A.

B.

C.

D.

Q. No.6

Let A and B are the complementary angles, then the value of p in pcosAcosecB =1 is _____.

Q. No.7

If , then the value of p is equal to _____.

Q. No.8

If , then the value of k is equal to _____.

Q. No.9

If A + B = 90 , then the value of k in the equation is _____.

Q. No.10

The value of is _____.

35453

42

3

= p2(1−sin2 θ) sec2 θ

cot2 θ(sec2 θ−1)

= k (1 + sin θ)100 cos2 θ

1−sin θ

° sec2 A − 1 = k cot2 B

3 cosec2 θ −3

tan2 θ

Introduction to Trigonometry_SolutionsSolution 1

Hence, the correct answer is B.

Solution 2


Solution 3


Solution 4

Given:


Solution 5

In a right-angled triangle Thus, Hence, the correct answer is option A.

Solution 6

If A and B are the complementary angles, then A + B = 90Now, pcosAcosecB = 1pcosecB = secA

Hence, let A and B are the complementary angles, then the value of p in pcosAcosecB = 1 is 1.

Solution 7

1 + cos2 x = sec2 x

⇒ cos2 x = tan2 x [From sec2 x − 1 = tan2 x]

⇒ cos4 x = sin2 x

Taking square root both sides, we get

cos2 x = sin x

⇒ 1 − sin2 x = sin x

⇒ sin x + sin2 x = 1

sin 3x − cos x = 0⇒ sin 3x = cos x

⇒ sin 3x = sin (90 − x)

⇒ 3x = 90 − x

⇒ x =

∴ cot 2x = cot 45 = 1

45

2

+ cosec x

= + = = =

tan x−1

sin x

tan x−1

sin x

1

sin x

tan x

sin x

1cos x

3

2

tan α =

∴ cot α = (∵ tan α = )

tan α + cot α = + =

p

q

q

p1

cot α

p

q

q

p

p2+q2

pq

∠C = 90° − ∠A

cos C = cos (90° − A) = sin A = 3

5

°

p = (∵ A + B = 90°)

p = = 1

cosec(90°−B)

cosec B

cosec B

cosec B

( )

Hence, If , then the value of p is equal to 2.

Solution 8

Hence, If , then the value of k is equal to 100.

Solution 9

Hence, If A + B = 90 , then the value of k in the equation is 1.

Solution 10

We know that, Therefore,

Hence, the value of is 3.

= p

⇒ = = 2 = p

2(1−sin2 θ) sec2 θ


2 cos2 θsec2θ

cot2 θtan2θ

2×1

1

= p2(1−sin2 θ) sec2 θ


= k (1 + sin θ)

⇒ = k

⇒ k = 100

100 cos2 θ

1−sin θ

100(1−sin2 θ)

(1−sin θ)(1+sin θ)

= k (1 + sin θ)100 cos2 θ

1−sin θ

sec2 A − 1 = k cot2 B

⇒ tan2 A = k cot2 B

⇒ tan2(90° − B) = k cot2 B

⇒ cot2 B = k cot2 B⇒ k = 1

° sec2 A − 1 = k cot2 B

3 cosec2 θ − = 3 cosec2 θ − 3 cot2 θ3

tan2 θ

cosec2 θ − cot2 θ = 1

3 cosec2 θ − = 3 cosec2 θ − 3 cot2 θ = 3 (cosec2 θ − cot2 θ) = 33

tan2 θ

3 cosec2 θ −3

tan2 θ

Some Applications of Trigonometry_Questions

Q. No.1

The height of a boy standing vertically on the ground is equal to the length of his shadow on the ground, the angle of elevation of thesource of light is

A.B.C.D.

Q. No.2

Kim makes an angle of 30 with the ground when leaning against a pole. If Kim's feet is 3 m away from the pole, the height of Kim is

A. mB. 2 mC. 3 mD. 3 m

Q. No.3

The elevation of a lamp post changes from 45 to 60 , the difference between the lengths of shadows of a girl 3 m tall is

A.

B.

C.

D.

Q. No.4

25 m away from the foot of a pole, the angle of elevation of the top of the pole is , the height of the pole is

A. m

B. 25 m

C. m

D. m

Q. No.5

0°30°45°90°

°

√3

√3

√3

° °

3(√3 − 1) m

√3(√3 + 1) m

3(1 + √3) m

√3(√3 − 1) m

60°

50

√3

√325√3

225

√3

Rahul is looking at a window such that the angle of elevation is 30 . He moves 30 m closer towards the window on a straight path, theangle of elevation now becomes 45 . The height of the window from ground is

A. 15 m

B. 15 m

C. m

D. 30 m

Q. No.6

If the angle of elevation of the sun is 60 , then the ratio of the length of a vertical pole to its shadow is _____.

Q. No.7

If the distance of Rahul from the foot of a 120 m high lamp-post is 120 m, then the angle of depression of Rahul from the top ofthe lamp-post is _____.

Q. No.8

The angle of elevation of the top of a tree 45 m high from a point on the ground l m away from the base of the tree is 60 , the value of lis _____ m.

Q. No.9

A tower x meters high casts a shadow of length 5 m on the ground when the elevation of the source of light is 30 . The value of

_____.

Q. No.10

The angle formed by the line of sight with the horizontal when the object is below the _____ level is called the angle of _____.

°°

(√3 − 1)

(√3 + 1)

30(√3 − 1)

(√3 + 1)

°

√3

°

√3 °

isx

2

5

Some Applications of Trigonometry_SolutionsSolution 1

We have,

Let PQ be the height of boy and PR be its shadowNow,PQ = PRIn PQR, we have


Solution 2

We have,

Let XY be Kim and YZ be the poleNow, let XY = a m


Solution 3

Δ

tan θ = = 1

⇒ θ = 45°

PQ

PR

⇒ = cos 30°

⇒ =

⇒ a = 2√3 m

XZXY

3a

√3

2

We have,

Let AB be the girl and AC and AD be her shadows Now,

From (1) and (2), the required difference is


Solution 4

We have,

Let LM be the pole and its height be x meters, LN = 25 m such that Now,


Solution 5

∠ACB = 45° and ∠ADB = 60°

= cot 45°

⇒ = 1

⇒ AC = 3 m … (1)

AC

AB

AC

3

= cot 60°

⇒ =

⇒ AD = √3 m … (2)

AD

AB

AD

31

√3

AC − AD = 3 − √3 = √3(√3 − 1) m

∠LNM = 60°

= tan 60°

⇒ = √3

⇒ x = 25√3 m

LM

LN

x

25

We have,

Let C be Rahul's initial position, D be his position after moving 30 m, AD = x m, and AB = h m.Now, According to the question


Solution 6

We have,

Let AB be the length of the pole, such that AB = l, and AC be the length of its shadow, such that AC = vNow, Also,

Thus, if the angle of elevation of the sun is 60 , then the ratio of the length of a vertical pole to its shadow is .

Solution 7

= cot 45° = 1

⇒ x = h … (1)

Also,

= cot 30° = √3

⇒ x = h√3 − 30 ⋯ (2)

From (1) and (2), we get

⇒ h = 15(√3 + 1) m

x

h

x+30

h

∠ACB = 60°

tan (∠ACB) = =

⇒ = tan 60° = √3

AB

AC

l

v

lv

° √3 : 1

We have,

Let PQ be the tower, such that PQ = 120 m, R be Rahul's position, such that PR = 120 m, and Now,

Hence, If the distance of Rahul from the foot of a 120 m high lamp-post is 120 m, then the angle of depression of Rahul from the top ofthe lamp-post is .

Solution 8

We have,

Let XY be the tree and Z be the point l meters away from the base of the tree, such that XZ = l mNow,

Hence, the angle of elevation of the top of a tree 45 m high from a point on the ground l m away from the base of the tree is 60 , the valueof l is m.

Solution 9

We have,

Let LM be the tower of height x meters and LN = .Now,

√3 ∠PRQ = θ

cot θ = =

⇒ cot θ = √3

∴ θ = 30°

PR

PQ

120√3

120

√330°

tan 60° =

⇒ l = = 15√3 m

XY

XZ

45

√3

°

15√3

5√3 m

x

Thus, a tower x meters high casts a shadow of length 5 m on the ground when the elevation of the source of light is 30 . The value of

.

Solution 10

We have,The angle formed by the line of sight with the horizontal when the object is below the horizontal level is called the angle of depression.

tan 30° =

⇒ x = × 5√3 = 5 m

x

5√3

1

√3

⇒ = = 5x2

5255

√3 °

isx2

55

Circles_Questions

Q. No.1

If in a circle the angle between the radii and line joining the centre and an external point P is 60 degree, then the angle between thetangent from point P to the circle and the line joining the centre and an external point P is

A. 30 degreeB. 45 degreeC. 60 degreeD. 90 degree

Q. No.2

If the length of the tangent from point P to the circle at the point of contact is 5 cm and the radius of the circle is 5 cm, then the anglewhich the tangent makes with the line joining the point P and the centre of the circle is

A. 30 degreeB. 45 degreeC. 60 degreeD. 90 degree

Q. No.3

AP and AQ are tangents drawn from a point A to a circle with centre O and radius 3 cm. If OA = 5 cm, then the perimeter of APOQ is

A. 12 cmB. 13 cmC. 14 cmD. 15 cm

Q. No.4

In the given figure, two circles touching each other at point T and QR is a common tangent to the given circles. The tangent at T meets

QR at P. If PT = 6 cm, then the value of isPQ+PR+PT

3

A. 6 cmB. 9 cmC. 12 cmD. 18 cm

Q. No.5

If PQ is a tangent to a circle with centre O at the point P, then can not be a

A. isosceles triangleB. right angle triangleC. equilateral triangleD. scalene triangle

Q. No.6

In the given figure, PQ is the tangent on a circle with centre O, PQ || MO and then is _____.

Q. No.7

In the given figure, a circle is inscribed in a ΔABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. Ifthe lengths of sides AB, BC and AC are 6 cm, 10 cm and 8 cm respectively, then the lengths of AD is _____.

Q. No.8

If two circles of different radius are touching each other externally then the maximum number of common tangents is _____.

Q. No.9

△ OPQ

∠POM = 30° ∠ OPQ

Given two concentric circles, the chord of the larger circle which is the tangent to the smaller circle is of length 6 cm. If the radius of thelarger circle is 5 cm then the diameter of the smaller circle will be _____.

Q. No.10

In the given figure, PB is the diameter of the circle with centre O and tangent XY meets the circle at point P.If AP = AB then the measure of XPA is _____.

∠

Circles_SolutionsSolution 1

We know that tangent is perpendicular to the radius through the point of contact.

Thus,


Solution 2

Let QPO = .From the figure, in


Solution 3

Let us first put the given data in the form of a diagram.

We know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore,

Since tangents drawn from an external point will be equal in length,AP = AQ = 4 cmThus, the perimeter of APOQ = 4 + 4 + 3 + 3 =14 cmHence, the correct answer is option C.

Solution 4

PT = 6 cm. Now, PQ and PT are tangents drawn to the same circle from an external point P.∴ PQ = PT = 6 cm PR and PT are tangents drawn to the same circle from an external point P.∴ PR = PT = 6 cm

∠P + 60° + 90° = 180°

⇒ ∠P = 30°

∠ θ

△ PQO

tan θ = = = 1

⇒ θ = 45°

OQ

PQ

5

5

AP2 = OA2 − OP2 = 25 − 9 = 16

⇒ AP = 4 cm

Now,


Solution 5

We know that the tangent is perpendicular to the radius through the point of contact.Thus, It means can not be an equilateral triangle.Hence, the correct answer is option C.

Solution 6

We know that tangent is perpendicular to the radius through the point of contact. Thus, PQ || MO so, PQO + MOQ = 180 as they are co-interior anglesSo, MOQ = 90And In ,

Hence, In the given figure, PQ is the tangent on a circle with centre O and then is .

Solution 7

Let AD = AF = x (Lengths of tangents from an external point to a circle are equal)Now, DB = BE = 6 x and CF = CE = BC = BE + CE =

Hence, In the given figure, a circle is inscribed in a ΔABC, such that it touches the sides AB, BC and CA at points D, E and F respectively.If the lengths of sides AB, BC and AC are 6 cm, 10 cm and 8 cm respectively, then the lengths of AD is 2 cm.

Solution 8

We have,As clearly seen from the figure

Two circles of different radius touching each other externally can have a maximum of 3 common tangents.Thus, if two circles of different radius are touching each other externally then the maximum number of common tangents is 3.

Solution 9

= = 6 cmPQ+PR+PT

3

6+6+6

3

∠OPQ = 90°△ OPQ

∠OQP = 90°∠ °

∠ °∠POQ = 90° − ∠POM = 90° − 30° = 60°

△ OQP∠OPQ = 180° − 90° − 60° = 30°

∠POM = 30° ∠ OPQ 30°

− 8 − x

6 − x + 8 − x = 10⇒ 2x = 14 − 10 = 4⇒ x = 2 cm

Chord length of larger circle which is tangent to smaller circler = 6 cmCB = In COB,

Since, radius = 4 cm, so diameter = 8 cmThus, given two concentric circles, the chord of the larger circle which is the tangent to the smaller circle is of length 6 cm. If the radius ofthe larger circle is 5 cm then the diameter of the smaller circle will be 8 cm.

Solution 10

We know that diameter of a circle subtends 90 at any point of the circle.Also, AP = AB so, APB = ABP = 45 .Since, tangent is perpendicular to the radius at the point of contactThus, measure of XPA = 45 .

= = 3 cmAB

2

6

2

△

CO2 + CB2 = OB2

⇒ CO2 + (3)2 = (5)2

⇒ CO2 + 9 = 25

⇒ CO2 = 16⇒ CO = 4 cm

°∠ ∠ °

∠ °

Areas Related to Circles_Questions

Q. No.1

If r is the radius of a semicircular park and its perimeter is 72 cm, then the value of is

A. 7B. 14C. 21D. 28

Q. No.2

The area swept by the minute hand of a clock in 15 minutes, if the length of the minute hand is 14 cm is

A. 110 cm2

B. 132 cm2

C. 154 cm2

D. 176 cm2

Q. No.3

The ratio of perimeters of a circle and a square is , the ratio of their areas is

A. 2 : 3B. 3 : 4C. 1 : 1D. 2 : 1

Q. No.4

Two circles C1 and C2 have their radii as r1 and r2 and their sum is equal to radii r of a third circle C, then

A. area(C) area(C1) + area(C2)B. area(C) area(C1) + area(C2)C. area(C) area(C1) + area(C2)D. area(C) area(C1) area(C2)

Q. No.5

The perimeter of a rectangle made up using copper wire is 220 cm. If the same wire is used to form a circle, then the radius of thecircle formed is

A. 14 cmB. 21 cm

r2

7

√π : 2

=

>

<

= −

C. 28 cmD. 35 cm

Q. No.6

The area of a quadrant of a circle whose circumference is 44 cm is _____ cm2.

Q. No.7

The diameter of a wheel is 14 m, it covers a distance of _____ km in 1200 revolutions.

Q. No.8

The circumference of a sector of angle 90 of a circle with a radius of 14 cm is _____ cm.

Q. No.9

If the length of an arc in a circle with radius 14 cm, is 28 cm, then the angle subtended by it at the center is _____.

Q. No.10

The radius of a circle is increased by 25%, its circumference is increased by _____%.

°

Areas Related to Circles_SolutionsSolution 1

We have,The perimeter of the semicircular park = Now,


Solution 2

We have,Area swept by the minute hand in 60 minutes =

area swept by the minute hand in 15 minutes =

Thus, the required area = Hence, the correct answer is option C.

Solution 3

We have,Let r be the radius of the circle and l be the length of the side of the squareNow,

The ratio of their areas =


Solution 4

We have,

Sum of areas of circles C1 and C2 =

Area of circle C = Since, Thus, area(C) area(C1) + area(C2)Hence, the correct answer is option B.

Solution 5

We have,Let the radius of the circle formed be rThe perimeter of the circle formed = perimeter of the rectangle


Solution 6

We have,Let the radius of the circle be r cmNow,

πr + 2r

πr + 2r = 72

⇒ r (π + 2) = 72

⇒ r = = = 14

⇒ = = = 28

72

+222

7

72×736

r2

7

(14)2

7

196

7

πr2

∴ × 15πr2

60

= = 154 cm2( )(14)222

7

422×14×2

4

=

⇒ = √π

⇒ = ⋯ (1)

2πr4l

√π

2πr

l

r

l

1

√π

= π( )2

= = 1 [From (1)]πr2

l2

r

l

π

π

r = r1 + r2

πr12 + πr2

2

πr2 = π(r1 + r2)2 = πr12 + πr2

2 + 2πr1r2

r1 > 0 and r2 > 0>

2πr = 220

⇒ r =

⇒ r = = 35 cm

2202π

220×72×22

Thus, the area of quadrant =

Hence, the area of a quadrant of a circle whose circumference is 44 cm is cm2

Solution 7

We have,Distance covered by the wheel in one revolution = =

Thus, distance covered by the wheel in 1200 revolutions = Hence, the diameter of a wheel is 14 m, it covers a distance of km in 1200 revolutions.

Solution 8

We have,

Circumference of the sector =

Thus, the circumference of a sector of angle 90 of a circle with a radius of 14 cm is cm.

Solution 9

We have,Length of an arc that subtends an angle at the center = Now,

Thus, if the length of an arc in a circle with radius 14 cm, is 28 cm, then the angle subtended by it at the center is .

Solution 10

We have,Let r be the radius and C be the circumference of the original circle Now, after a 25% increase, the new radius is,

Thus, the new circumference =

Thus, the % increase =

Hence, the radius of a circle is increased by 25%, its circumference is increased by %.

2πr = 44

⇒ r = = 7 cm44×72×22

(πr2) = × × (7)2 = = = 38. 5 cm214

14

227

1544

772

38. 5

πd × 14 = 44 m22

7(1200 × 44) m = 52800 m = 52. 8 km

52. 8

+ 2r = + (2 × 7) = + 14 = 36 cm2πrθ

360

2× ×14×9022

7

360

88

4

° 36

θ2πrθ360

= 28

⇒ θ = = 114. 54°

2× ×14×θ22

7

360

28×360

88

114. 54°

r + (r) = r25

100

5

4

2πr = 2 × π × r = (2πr) = C5

4

5

4

5

4

× 100 = × 100 = 25%C−C

5

4

C

14

25

Surface Areas and Volumes_Questions

Q. No.1

The diameters of two circular ends of the bucket are 30 cm and 16 cm. If the capacity of the bucket is 1.227 lit. then the height of thebucket is

A. 9 cmB. 10 cmC. 11 cmD. 12 cm

Q. No.2

The ratio of surface areas of two spheres is 16 : 81. The ratio of their volumes is

A. 4 : 9B. 9 : 4C. 729 : 64D. 64 : 729

Q. No.3

The ratio of lateral surface area to the total surface area of a solid cone with a base diameter 10 m and height 12 cm is

A. 11 : 25B. 12 : 25C. 13 : 25D. 24 : 25

Q. No.4

If the ratio of total surface area to the curved surface area of a solid cylinder is 25 : 16 then the ratio between the radius and the heightof cylinder is

A. 11 : 16B. 10 : 16C. 11 : 15D. 9 : 16

Q. No.5

A metallic solid cone is melted to form a solid cylinder. If the height of the cone was 16 cm and the ratio of the radius of the cone to thecylinder is 3 : 2, then the height of the cylinder is

A. 10 cmB. 12 cm

π

C. 18 cmD. 24 cm

Q. No.6

If 27 identical marbles of radius r are melted to make a bigger marble of radius R, then R : r is equal to _____.

Q. No.7

Let two identical solid-hemispheres are joined together along with their bases, if the curved surface area of one solid hemisphere is 24cm2 then the curved surface area of this new solid is _____.

Q. No.8

In a right circular cone, the cross-section made by a plane perpendicular to the base and passing through the axis of the cone is a_____.

Q. No.9

A cone is enclosed in a right circular cylinder of radius r and height h(h < 2r) then the maximum height of the cone is _____.

Q. No.10

The number of solid cylinders, each of diameter 2 cm and height 10 cm that can be made by melting a solid sphere of diameter 60 cmare _____.

Surface Areas and Volumes_SolutionsSolution 1

The bucket is in the form of a frustum.The diameters are respectively, cmRadii of the circular ends =


Solution 2

The ratio of the surface area of the spheres will be

Thus, the ratio of the radii = 4 : 9Ratio of volume of the spheres is

So, the ratio of their volume is 64 : 729Hence the correct answer is option D.

Solution 3

and The ratio of lateral surface to the total surface area of the cone


Solution 4

Let the radius of the cylinder is r and the height of cylinder is h.Thus,


Solution 5

Let the height of the cone be h.The volume of cylinder = Volume of the cone


d1 = 30 cm and d2 = 16 cmr1 = 15 cm and r2 = 8 cm

Volume, V = πh (r21 + r1r2 + r2

2)

V = π × h (152 + 15 × 8 + 82)

1. 227 × π × 103 = × π × h × (409)

h = 9 cm

13

13

13

= ( )2

= = ( )2

4πr21

4πr22

r1

r2

16

8149

⇒ V1 : V2 = πr31 : πr3

2

⇒ ( )3

= ( )3

=

43

43

r1

r2

49

64729

r = = 5 cm102

l = √h2 + r2 = √144 + 25 = 13 cm

= = =a1

a2

π×r×l

π×r×(l+h)

1313+12

1325

=

⇒ + 1 =

⇒ = − 1 =

2πr(r+h)

2πrh2516

r2

rh

25

16r

h

25

16

9

16

⇒ πr12h = πr2

2 × 16

⇒ h = × ( )2

⇒ h = × = 12 cm

13

16

3

r2

r1

16

3

9

4

Solution 6

The radius of the bigger marble = R cmThe radius of smaller marbles = r cm

Hence, if 27 identical marbles of radius r are melted to make a bigger marble of radius R, then R : r is equal to 3 : 1.

Solution 7

Since the two hemispheres are joined end to end, it becomes a complete sphere. The curved surface area of the new solid = total surface area of the sphere.

Hence, let two identical solid-hemispheres are joined together along with their bases, if the curved surface area of one solid hemisphere is24 cm2 then the curved surface area of this new solid is 48 cm2.

Solution 8

From Figure, it is clear that the cross-section obtained is a triangle.Hence, In a right circular cone, the cross-section made by a plane perpendicular to the base and passing through the axis of the cone isa triangle.

Solution 9

If a cone is enclosed in a right circular cylinder of radius r and height h(h < 2r) then the base of the cone should lie with the base of thecylinder thus the maximum height of the cone can be the height of the cylinder.Hence, a cone is enclosed in a right circular cylinder of radius r and height h(h < 2r) then the maximum height of the cone is h.

Solution 10

Let the number of solid cylinders be n.Now, Volume of a solid sphere = Volume of n cylinder

Hence, the number of solid cylinders, each of diameter 2 cm and height 10 cm that can be made by melting a solid sphere of diameter 60cm are 3600.

= = = 1

⇒ R = 3r

⇒ =

Volume of bigger marble

Volume of small marbles

πR34

3

27× πr343

R3

(3r)3

R

r

3

1

Curved surface area of the new solid = 24 + 24 = 48 cm2

⇒ × × ( )3

= n × × ( )2

× 10

⇒ × 27 × 1000 = n × 1 × 10

⇒ n = 3600

43

22

7

602

22

7

22

43

Statistics_Questions

Q. No.1

The mean of n observations is m. If each even term is increased by 2 and each odd term is decreased by 2 then the new mean is

A. m

B.

C.

D. Both A and B

Q. No.2

What would be the mode of the data whose particulars are given as follows and the variables have usual meaning:

A. 6B. 8C. 10D. 12

Q. No.3

The midpoints of the class intervals of a grouped data are and the corresponding frequencies are . If the mean is , then the

value of is

A. 0B. 1C. 2D. 4

Q. No.4

If there are total 120 observations, then the cumulative frequency of median class can be

A. 60B. 65C. 120D. 125

Q. No.5

The median of squares of first 6 natural numbers is

A. 12

m − 2n

m + 2n

u = 12, f1 − f0 = 16, f1 − f2 = 10, h = 6

xi's fi's x(∑ fi(xi− x ))2

2

B. 12.5C. 13D. 13.5

Q. No.6

If the mean of 30 integers is 0, then the maximum number of positive integers in the dataset is _____.

Q. No.7

For a _____ frequency distribution, mean = median = mode.

Q. No.8

For the data given as: 4, 2, 7, 6, 4, 6, 1, 2, 4. If x × (mean) = y × (median) = z × (mode), then the value of is _____.

Q. No.9

If for a grouped data, mode = median, then the value of {mode + mean 2(median)} is _____.

Q. No.10

If there are total 25 observations and the mean of the first 12 observations is 10, last 12 observation is 12 and the median is 36, thenthe value of the mean is _____.

xy + yz + zx

−

Statistics_SolutionsSolution 1

Let a1, a2, ..., an are n observations, then

m = Now if the numbers of observations are even, thennew mean = = mNow if the numbers of observations are odd, then

new mean = = Hence, the correct answer is option D.

Solution 2

Given The lower limit = l =

Mode = =


Solution 3

If are the midpoints of the class and are the corresponding frequencies, then

Now,


Solution 4

We have, The cumulative frequency of median class is greater than (and nearest to) .Now, N = 120

Hence, the correct answer is option B

Solution 5

We have,Squares of first 6 natural numbers can be written as1, 4, 9, 16, 25, 36Now, since the number of observations is even Thus, median = mean of 3rd and 4th observation = Hence, the correct answer is option B.

Solution 6

We have,Mean of 30 integers = 0Thus, the sum of these 30 integers = 0Now, at a maximum 29 out of these 30 integers can be positive and if their sum is x, the 30th integer is ( ).Thus, if the mean of 30 integers is 0, then the maximum number of positive integers in the dataset is .

Solution 7

We have,mean = median = mode, only if the frequency distribution is symmetricalThus, for a symmetrical frequency distribution, mean = median = mode.

a1+a2+...+an

n

a1+a2+...+an+n−n

n

a1+a2+...+an+n−(n+2)

nm − 2

n

u = 12, f1 − f0 = 16, f1 − f2 = 10, h = 6u − h = 12 − 6 = 6

∴ l + ( )× hf1−f0

2f1−f0−f2l + [ ]× h = 6 + [ ]× 6 = 6 + ( )× 6 = 6 + 4 = 10

f1−f0

(f1−f0)+(f1−f2)

16

16+10

12

18

xi's fi's

= x

⇒ ∑fixi = x ∑ fi … (1)

∑ fixi

∑ fi

∑ fi(xi − x ) = ∑ fixi − x ∑ fi

⇒ ∑ fi(xi − x ) = x ∑ fi − x ∑ fi = 0 [Using (1)]

⇒ = = 0(∑ fi(xi− x ))2

2

(0)2

2

N

2

⇒ = 60N

2

= = 12. 509+16

2

25

2

−x29

Solution 8

We have,The given data can be rearranged as 1, 2, 2, 4, 4, 4, 6, 6, 7.Now,Mean =

Median =

Mode = 4Thus, mean = median = mode

Hence, for the data given as: 4, 2, 7, 6, 4, 6, 1, 2, 4. If x(mean) = y(median) = z(mode), then the value of is .

Solution 9

We have,For a grouped data, the Empirical Relationship is given by3 Median = 2 Mean + ModeWe can easily verify the result that if two of them are equal, then the third measure will also be equal to both of them.Now, let mode = median = aThus, mean = aTherefore, {mode + mean 2(median)} = a + a 2a = 0.Hence, if for a grouped data, mode = median, then the value of {mode + mean 2(median)} is .

Solution 10

We have,Sum of the first 12 observations = Sum of the last 12 observations = Since the number of observations is odd, so median = termNow, n = 25

median = term = 36

Thus, mean = Hence, if there are total 25 observations and the mean of the first 12 observations is 10, last 12 observation is 12 and the median is 36,then the value of the mean is .

= = 41+2+2+4+4+4+6+6+7

9

36

9

( ) th observation = 5 th observation = 49+1

2

⇒ x = y = z = 1

∴ xy + yz + zx = 3

xy + yz + zx 3

− −− 0

12 × 10 = 12012 × 12 = 144

thn+1

2

⇒ th = 13 th25+1

2

= 12120+144+36

25

12

Probability_Questions

Q. No.1

A bag contains 2 blue marbles, 4 red marbles, and 5 black marbles. A marble is drawn from the bag at random. If the probability ofgetting a blue marble is p and the probability of getting a black marble is q then p + q is equal to

A.

B.

C.

D.

Q. No.2

Two dice are thrown together. The probability of getting the different numbers on both dice is

A.

B.

C.

D. 1

Q. No.3

In a family of 3 children, the probability of having at most one girl is

A.

B.

C.

D.

Q. No.4

A bag contains cards numbered from 1 to 20. A card is drawn at random from the bag. The probability that the number on this card ismultiple of 3 is

A.

B.

C.

D.

2

11

3

11

5

11

7

11

1

6

1

3

5

6

1

2

5

8

3

4

7

8

1

10

3

20

3

10

1

4

Q. No.5

All ace cards are removed from the pack of 52 cards. A card is drawn at random from the leftover cards. Find the probability ofselecting a red king.

A.

B.

C.

D.

Q. No.6

From a well-shuffled pack of cards, a card is drawn at random. The probability of getting a number card with number less than 6 is_____.

Q. No.7

Rahul is trying to hit a bull's eye on a dartboard. If he misses 9 shots out of 10 then the probability of hitting the target is _____.

Q. No.8

Rahul visits Chandigarh 3 times by bus and 7 times by train out of 10 visits. The probability that Rahul will take a bus on his next visit is____.

Q. No.9

A letter of the English alphabet is chosen at random. The difference between the probability of the chosen letter being a consonantand the probability of the chosen letter being a vowel is _____

Q. No.10

A number is selected at random from the numbers 4, 4, 4, 6, 6, 6, 10, 10, 10, 10. The probability that the selected number is greaterthan their average will be _____.

1

13

2

13

5

24

1

24

Probability_SolutionsSolution 1

Total number of outcomes = 2 + 4 + 5 = 11Favourable number of outcomes for blue marbles = 2Favourable number of outcomes for black marbles = 5Thus, p = and q =

Therefore, Hence, the correct answer is option D.

Solution 2

When two dice are thrown together, all possible outcomes are

(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)

∴ Total number of outcomes = 36

The non-favorable outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

So, the number of non-favorable outcomes is 6.

∴ P(getting the different numbers on both dice) =


Solution 3

The possible outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG.


The favorable outcomes are BBB, BBG, BGB, GBB.

So, the favorable number of outcomes are 4.

∴ P(at most one girl) =


Solution 4

The total number of cards in the bag is 20.


The numbers from 1 to 20 which are multiples of 3 are 3, 6, 9, 12, 15, 18.

So, the favourable number of outcomes is 6.

∴ P(number on the drawn card is multiples of 3) =


Solution 5

Total cards in the pack = 52Cards left after removing all the ace cards = 52 4 = 48

2

11

5

11

p + q = + =2

11

5

11

7

11

1 − = 1 − =non−favourable number of outcomes

Total number of outcomes

6

36

5

6

= =Favourable number of outcomes


4

8

1

2

= =Favourable number of outcomes


6

20

3

10

−

Number of red kings in the pack = 2Probability of selecting a red king from the leftover cards = Hence, the correct answer is option D.

Solution 6

The total number of the cards are 52.Number card with a number less than 6 is = 20

We know that probability =

Thus, the probability of getting a number card with a number less than 6 is = Hence, from a well-shuffled pack of cards, a card is drawn at random. The probability of getting a number card with number less than 6 is

.

Solution 7

The favorable number of shots = 10 9 = 1Total number of shots = 10Thus, the required probability = Hence, Rahul is trying to hit a bull's eye on a dartboard. If he misses 9 shots out of 10 then the probability of hitting the target is 0.1.

Solution 8

Required probability = = 0.3Hence, Rahul visits Chandigarh 3 times by bus and 7 times by train out of 10 visits. The probability that Rahul will take a bus on his nextvisit is 0.3

Solution 9

Total number of outcomes = 26

Out of 26 letters, 21 are consonants and 5 are vowels.∴ P(chosen letter is a vowel) =

Thus, P(chosen letter is a consonant) =

Required difference =

Hence, a letter of the English alphabet is chosen at random. The probability that the chosen letter is a vowel is .

Solution 10

The given numbers are 4, 4, 4, 6, 6, 6, 10, 10, 10, 10.Their average = 7The number greater than 7 will be 10 among the given numbers.Thus, the required probability =

=2

48

1

24

4 × 5

=20

52

5

13

5

13

−

= 0. 11

10

=Rahul traveled by bus

Total number of visits

3

10

=Favourable number of outcomes


5

26

=Favourable number of outcomes


21

26

− = =21

26

5

26

16

26

8

13

8

13

=4

10

2

5

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