C K1. STATISTICS (18)Periods Trigonometric ratios of an acute angle of a right-angled triangle....

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I. NUMBER SYSTEMS 11

II. ALGEBRA 23

III. GEOMETRY 17

IV. TRIGONOMETRY 22

V.

TOTAL 90

STATISTICS 17

(Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two

sides are divided in the same ratio.

(Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.

(Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the

triangles are similar.

(Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two

triangles are similar.

(Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are

proportional, the two triangles are similar.

(Motivate) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, the triangles on

each side of the perpendicular are similar to the whole triangle and to each other.

(Prove) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

(Prove) In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

(Prove) In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the angles opposite to

the first side is a right triangle.

Euclid’s division lemma, Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating

and motivating through examples, Proofs of irrationality of . Decimal representation of rational numbers in terms of

terminating/non-terminating recurring decimals.

Zeros of a polynomial. Relationship between zeros and coefficients of quadratic polynomials. Statement and simple problems

on division algorithm for polynomials with real coefficients.

Definitions, examples, counter examples of similar triangles.

Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency.

Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically — by

substitution, by elimination and by cross multiplication method. Simple situational problems. Simple problems on equations

reducible to linear equations.

First Term Marks: 90

UNITS MARKS

UNIT I : NUMBER SYSTEMS

1. POLYNOMIALS (7) Periods

2. PAIR OF LINEAR EQUATIONS INTWOVARIABLES (15) Periods

1. REAL NUMBERS (15) Periods

UNIT II : ALGEBRA

UNIT III : GEOMETRY

1. TRIANGLES (15) Periods

1.

2.

3.

4.

5.

6.

7.

8.

9.

C K

MATHEMATICS (CLASS–X)

SYLLABUS

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UNIT IV : TRIGONOMETRY

1. INTRODUCTIONTOTRIGONOMETRY (10) Periods

2. TRIGONOMETRIC IDENTITIES (15) Periods

UNIT V : STATISTICSAND PROBABILITY

1. STATISTICS (18) Periods

Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined); motivate the ratios

whichever are defined at 0° and 90°. Values (with proofs) of the trigonometric ratios of 30°, 45° 60°. Relationships between

the ratios.

and

Proof and applications of the identity sin A + cos A = 1. Only simple identities to be given. Trigonometric ratios of

complementary angles.

Mean, median and mode of grouped data (bimodal situation to be avoided). Cumulative frequency graph.

2 2

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All chapters as per

Syllabus and Textbook

NCERT

Concepts, Formulae,

Tips and Tricks

provides a comprehensive

summary of the concept

Every chapter divided

into Sub-topics

Super Refresher

NCERT Exemplar Problems

with complete solution to

supplement the entire

NCERT support material

Value Based Questions

to apply mathematical concepts

to real life situations with stress on

social values

Self Practice questions for

consolidation of each concept

Important Questions from

examination point of view

to ensure passing marks

NCERT

with detailed solution

Textbook Exercises

Highlights essential

information which

must be remembered

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C K

Mathematics

3

of 90 marks each

Model Question Papers

Self Assessment with answers

at the end of each chapter

HOTS (Higher Order Thinking Skills)

Questions with answers

Additional Questions with

answers at the end of each chapter

6



C K

TERM-I

1. Real Numbers

2. Polynomials

3. Pair of Linear Equations in two Variables

4. Triangles

5. Introduction to Trigonometry

6.

Model Question Papers for Practice

1–33

34–70

71–159

160–228

229–269

Statistics 270–340

(1-3) 341–351

CONTENTS

7

Prelims Pages.psC:\Users\amin\Desktop\Math -X Term-I\Prilim Image\MBD Super Refresher MATH-Gr-10-Term-I.cdrThursday, December 17, 2015 4:30:33 PMColor profile: Disabled


� Lemma: A lemma is a proven statement used for proving another statement.

� Algorithm: An algorithm is a series of well defined steps used for solving a type of problem.

� Theorem 1: Euclid’s Division Lemma: Let a and b are any two positive integers. Then there

exist two unique integers q and r such that:

a = bq + r, 0 ≤ r < b

Here, a is called the dividend, b is called the divisor, q is called the quotient and r is called the remainder.

Alternatively,

Dividend = Divisor × Quotient + Remainder

Euclid’s division algorithm is based on this lemma.

� Euclid’s Division Algorithm: It is a technique to compute HCF or GCD (greatest common divisor) of two

given positive integers say a and b (a > b) by successive use of Euclid’s division lemma given as:

Step I: Let a > b. Apply Euclid’s division lemma to a and b to find whole numbers q and r such that

a = bq + r, 0 ≤ r < b

Step II: If r = 0, then HCF (a, b) = b

But, if r ≠ 0, then we apply Euclid’s

division lemma to b and r i.e., we apply

again Step I.

Step III: We continue the division

process till the remainder becomes zero.

The divisor at the last stage would be

HCF of ‘a‘ and ‘b’. It can be summarisedin the form of flow chart.

Euclid’s Division Lemma

The Fundamental Theorem of Arithmetic

Revisiting Irrational Numbers

Revisiting Rational Numbers and Their Decimal

Expansions

Real Numbers

1

→

→

If

r = 0?

b is HCF of‘a’ and ‘b’

Continue withStep I taking b andr as base

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→

→→

→

Yes

No

b) a (q

bq

r

(NCERT Textbook Chapter No.1)

11

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2 MBD Super Refresher Mathematics-X

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� The Fundamental Theorem of Arithmetic (Unique Factorisation Theorem)

Before giving a formal statement of Fundamental Theorem of arithmetic. Let us recall what are prime andcomposite numbers.

Prime Number: A positive number p is prime, if p 1 and it has exactly two factors 1 and the number p itself. Inother words, a prime number is a natural number other than 1, which is divisible only by 1 and the number itself.

For example, 2, 3, 5, 7, 11, 13, 17, ... are prime numbers.

Composite number: A natural number is a composite number if it has more than two factors.

In other words, every positive integer other than 1, which is not prime is called a composite number.

For example, 4, 6, 8, 9, 10, 12, 14, 15, 16, ... are all composite numbers.

� Theorem 2: Fundamental theorem of Arthmetic: Every composite number can be factorised as a product of

primes, and this factorisation is unique, apart from the order in which prime factors occur.

� Determining the HCF and LCM of two numbers by prime factorisation method

HCF (a, b) = Product of the smallest powers of each common prime factors among the numbers.

LCM (a, b) = Product of greatest powers of each prime factors involved in the numbers.

Remember:

For any two numbers a and b,

HCF (a, b) × LCM (a, b) = a × b

Note: Above result is not valid for three or more numbers.

i.e., HCF (a, b, c) × LCM (a, b, c) a × b × c

� Theorem 3: If p is a prime number and p divides a2, then p also divides a where a is a positive integer.

If p is prime number then p is irrational.

� Revisiting Decimal Representation of Rational Numbers

In previous class, we have learnt that all rational numbers have either a terminating decimal representation ora non-terminating repeating (or recurring) decimal representation.

Under this section we will study the condition for a rational number p/q (q 0) to be terminating and conditionfor it to be non-terminating repeating (or recurring).

Let us consider few terminating decimal expansions:

(i) 0.25 (ii) 0.32

Now, we have

(i) 0.25 = 2 2 0

25 1 1 1

100 4 2 2 5= = =

×(ii) 0.32 = 2 0 2

32 8 8 8

100 25 5 2 5= = =

×

In each case, the terminating decimal expansion is first reduced to lowest form (i.e., of the form p/q (q 0), wherep and q are coprimes, and in its lowest form the prime factoristation of q is of the form 2m × 5n, where m and n arenon-negative integers.

We now, state the above result in the form of Theorem.

Theorem 4: Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the formp/q, where p and q are coprimes and prime factorisation of q is of the form 2m × 5m, where m and n are non-negativeintegers.

Theorem 5: (Converse of Theorem 4) : Let x = p/q be a rational number, such that the prime factorisation of q is ofthe form 2m × 5n, where m and n are non-negative integers, then x has decimal expansion which terminates.


Real Numbers 3

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Euclid’s Division Lemma

TEXTBOOK EXERCISE 1.1

Q. 1. Use Euclid’s division algorithm to find the HCFof:

(i) 135 and 225 (ii) 196 and 38220

(iii) 867 and 255

Sol. (i) By division AlgorithmStep 1: Since 225 > 135, we apply the division

lemma to 225 and 135, we get

225 = 135 × 1 + 90

Step 2: Since the remainder 90 ≠ 0, we apply the

division lemma to 135 and 90, we get

135 = 90 × 1 + 45

Step 3: Since the remainder 45 ≠ 0, we apply the

division lemma to 90 and 45, we get

90 = 45 × 2 + 0

Since the remainder has now become zero, sowe stop the procedure.

ä Divisor in the step 3 is 45

∴ HCF of 90 and 45 is 45

Hence, HCF of 135 and 225 is 45.

(ii) To find HCF of 196 and 38220

Since, 38220 > 196, we apply the division lemmato 196 and 38220, we get

38220 = 196 × 195 + 0


ä Divisor in the step 1 is 196

∴ HCF of 38220 and 196 is 196.


(iii) To find HCF of 867 and 255

Step 1: Since 867 > 255, we apply the divisionlemma to 867 and 255, we get

867 = 255 × 3 + 102

Step 2: Since remainder 102 ≠ 0, we apply thedivision lemma to 255 and 102, we get

255 = 102 × 2 + 51

Step 3: Since remainder 51 ≠ 0, we apply thedivision lemma to 51 and 102, by taking 102 asdividend, we get

102 = 51 × 2 + 0


ä Divisor in step 3 is 51

∴ HCF of 102 and 51 is 51.


Q. 2. Show that any positive odd integer is of the

form 6q + 1 or 6q + 3 or 6q + 5, where q is some

integer.

Sol. Let a be any positive odd integer, we apply thedivision algorithm with a and b = 6.Since 0 ≤ r < 6, the possible remainders are 0, 1,2, 3, 4 and 5.

i.e., a can be 6q or 6q + 1, or 6q + 2, or 6q + 3, or6q + 4, or 6q + 5 where q is quotient.

However, since a is odd

∴ a cannot be equal to 6q, 6q + 2, 6q + 4

ä All are divisible by 2

Therefore, any odd integer is of the form 6q + 1or 6q + 3 or 6q + 5.

Q. 3. An army contingent of 616 members is to

march behind an army band of 32 members in

a parade. The two groups are to march in the

same number of columns. What is the

maximum number of columns in which they

can march?

Sol. Total number of members in army = 616 and 32(A band of two groups)

Since two groups are to march in same numberof columns and we are to find out the maximum

number of columns.

∴ Maximum number of columns = HCF of 616and 32

Step 1: Since 616 > 32, we apply the divisionlemma to 616 and 32, to get

616 = 32 × 19 + 8

Step 2: Since the remainder 8 ≠ 0, we apply thedivision lemma to 32 and 8, to get

32 = 8 × 4 + 0


ä Divisor in the last step is 8

∴ HCF of 616 and 32 is 8

Hence, maximum number of columns in which

they can march is 8.

Q. 4. Use Euclid’s division lemma to show that thesquare of any positive integer is either of theform 3m or 3m + 1 for some integer m.

Sol. Let x be any positive integer and b = 3x = 3q + r

where q is quotient and r is remainder 0 ≤ r < 3

If r = 0 then x = 3q

If r = 1 then x = 3q + 1



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If r = 2 then x = 3q + 2

x is of the form 3q or 3q + 1 or 3q + 2

If x = 3q

Squaring both sides,

(x)2 = (3q)2

= 9q2 = 3 (3q2) = 3m

where m = 3q2

where m is also an integer

Hence, x2 = 3m ... (1)

If x = 3q + 1


x2 = (3q + 1)2

x2 = 9q2 + 1 + 2 × 3q × 1

x2 = 3 (3q2 + 2q) + 1

x2 = 3m + 1 ... (2)

where, m = 3q2 + 2q


If x = 3q + 2


x2 = (3q + 2)2

x2 = 9q2 + 6q + 4

x2 = 9q2 + 6q + 3 + 1

x2 = 3(3q2 + 2q + 1) + 1

x2 = 3m + 1 ... (3)

where, m = 3q2 + 2q + 1


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

x2 = 3m, 3m + 1

Hence, square of any positive integer is either of

the form 3m or 3m + 1 for some integer m.

Q. 5. Use Euclid’s division lemma to show that thecube of any positive integer is of the form 9m,9m + 1 or 9m + 8.

Sol. Let x be any positive integer then it is of the form3q, 3q + 1 or 3q + 2.If x = 3q

Cubing both sides,

x3 = (3q)3

x3 = 27q3 = 9 (3q3) = 9m

where, m = 3q3 and is an integer.

x3 = 9m ... (1)

If x = 3q + 1

Cubing both sides,

x3 = (3q + 1)3

x3 = 27q3 + 27q2 + 9q + 1

= 9 (3q3 + 3q2 + q) + 1 = 9m + 1

where, m = 3q3 + 3q2 + q and is an integer.

x3 = 9m + 1 ... (2)

Again, x3 = 9m + 1

If x = 3q + 2

Cubing both sides,

(x)3 = (3q + 2)2 = 27q3 + 54q2 + 36q + 8

x3 = 9 (3q3 + 6q3 + 4q) + 8

x3 = 9m + 8 ... (3)

where m = 3q3 + 6q2 + 4q is an integer

Again, x3 = 9m + 8

From (1), (2) and (3) we find that x3 can be of the

form 9m, 9m + 1, 9m + 8

Hence, cube of any positive integer can be of the

form 9m or 9m + 1 or 9m + 8.

SELF PRACTICE 1.1

1. Use Euclid’s division algorithm to find the HCFof

(a) 426 and 576 (b) 38540 and 48250

(c) 135 and 714 (d) 81445 and 687897

(e) 56 and 814

2. If the HCF of 45 and 210 is expressible in theform 210x + 45 × 5; find x.

3. Show that any positive integer is of the form 3qor 3q + 1 or 3q + 2 for some integer q.

4. Show that the square of an odd positive integeris of the form 8m + 1 for some integer m.

5. Find the greatest positive integer whichdivides 332 and 447 leaving remainder 2 ineach case.

6. For any positive integer n, prove that n3 – n isdivisible by 6.

7. There are 48 bottles of product X and 108bottles of product Y. These are to be packed inidentical boxes with same number of bottles,one particular product being in one box. Findthe minimum number of bottles that should bekept in our pack so that no bottle of product Aor B is left behind.

8. Show that the product of any two consecutivenumbers is divisible by 2.

Note: We write HCF of a and b briefly as HCF(a, b).

The Fundamental Theorem of Arithmetic


Q. 1. Express each number as a product of its primefactors:

(i) 140 (ii) 156 (iii) 3825

(iv) 5005 (v) 7429


Real Numbers 5

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Sol.

(i) Prime factorisation of 140 = (2)2 × (35)

= (2)2 × (5) × (7)

(ii) Prime factorisation of 156 = (2)2 × (39)

= (2)2 × (3) × (13)

(iii) Prime factorisation of 3825 = (3)2 × (425)

= (3)2 × (5) × (85)

= (3)2 × (5)2 × (17)

(iv) Prime factorisation of 5005 = (5) × (1001)

= (5) × (7) × (143)

= (5) × (7) × (11) × (13)

(v) Prime factorisation of 7429 = (17) × (437)

= (17) × (19) × (23).

Q. 2. Find the LCM and HCF of the following pairsof integers and verify that LCM × HCF =Product of the two numbers.

(i) 26 and 91 (ii) 510 and 92

(iii) 336 and 54

Sol. (i) Given numbers are 26 and 91

Prime factorisation of 26 and 91 are26 = (2) × (13)

and 91 = (7) × (13)

HCF (26, 91) = Product of least powers of

common factors

∴ HCF (26, 91) = 13

and LCM (26, 91) = Product of highest powers

of all the factors

= (2) × (7) × (13) = 182

Verification:

LCM (26, 91) × HCF (26, 91) = (13) × (182)

= (13) × (2) × (91) = (26) × (91)

= Product of given numbers

(ii) Given numbers are 510 and 92

Prime factorisation of 510 and 92 are

510 = (2) × (255) = (2) × (3) × (85)

= (2) × (3) × (5) × (17)

and 92 = (2) × (46) = (2)2 × (23)

HCF (510, 92) = Product of least powers of

common factors = 2

LCM (510, 92) = Product of highest powers of all

the factors

= (2)2 × (3) × (5) × (17) × (23) = 23460

Verification:

LCM (510, 92) × HCF (510, 92)

= (2) × (23460)

= (2) × (2)2 × (3) × (5) × (17) × (23)

= (2) × (3) × (5) × (17) × (2)2 × (23)

= 510 × 92

= Product of given numbers

(iii) Given numbers are 336 and 54

Prime factorisation of 336 and 54 are

336 = (2) × (168) = (2) × (2) × (84)

= (2) × (2) × (2) × (42)

= (2) × (2) × (2) × (2) × (21)

= (2)4 × (3) × (7)

and 54 = (2) × (27) = (2) × (3) × (9)

= (2) × (3) × (3) × (3) = (2) × (3)3

HCF (336, 54) = Product of least powers ofcommon factors

= (2) × (3) = 6

LCM (336, 54) = Product of highest powers of allthe factors

= (2)4 × (3)3 × (7) = 3024

Verification:

LCM (336, 54) × HCF (336, 54)

= 6 × 3024 = (2) × (3) × (2)4 × (3)3 × (7)

= (2)4 × (3) × (7) × (2) × (3)3 = 336 × 54

= Product of given numbers.

Q. 3. Find the LCM and HCF of the followingintegers by applying the prime factorisationmethod.

(i) 12, 15 and 21 (ii) 17, 23 and 29

(iii) 8, 9 and 25

Sol.

(i) Given numbers are 12, 15 and 21

Prime factorisation of 12, 15 and 21 are

12 = (2) × (6) = (2) × (2) × (3) = (2)2 × (3)

15 = (3) × (5)

21 = (3) × (7)

HCF (12, 15 and 21) = 3

LCM (12, 15 and 21) = (2)2 × (3) × (5) × (7) = 420

(ii) Given numbers are 17, 23 and 29.

These all are prime numbers and we know thatamong any two or more than two primes thereis just one common factor which is 1.

∴ HCF (17, 23, 29) = 1

LCM (17, 23 and 29) = 17 × 23 × 29 = 11339

(iii) Given numbers are 8, 9 and 25

Prime factorisation of 8, 9 and 25 are

8 = (2) × (4) = (2) × (2) × (2) = (2)3 × (1)

9 = (3) × (3) = (3)2

25 = (5) × (5) = (5)2

HCF (8, 9 and 25) = 1

LCM (8, 9 and 25) = (2)3 × (3)2 × (5)2 = 1800



CK

CK

Q. 4. Given that HCF (306, 657) = 9, find LCM (306,657).

Sol. Given numbers are 306 and 657.HCF (306, 657) = 9 (given)

ä HCF × LCM = Product of given number

∴ 9 × LCM (306, 657) = 306 × 657

or LCM (306, 657) = 306 657

9

×

= 34 × 657 = 22338

Q. 5. Check whether 6n can end with the digit 0 forany natural number n.

Sol. Let us suppose that 6n ends with the digit 0 forsome n ∈ N.

∴ 6n is divisible by 5.

But, prime factors of 6 are 2 and 3

∴ Prime factors of (6)n are (2 × 3)n

⇒ It is clear that in prime factorisation of 6n

there is no place for 5.

ä By Fundamental Theorem of Arithmetic,

Every composite number can be expressed as a

product of primes and this factorisation is

unique, apart from the order in which the prime

factors occur.

∴ Our supposition is wrong.

Hence, there exists no natural number n for which

6n ends with the digit zero.

Q. 6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4× 3 × 2 × 1 + 5 are composite numbers.

Sol. Consider, 7 × 11 × 13 + 13 = 13 [7 × 11 + 1]which is not a prime number because it has a

factor 13. So, it is a composite number.

Also, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

= 5 [7 × 6 × 4 × 3 × 2 × 1 + 1], which is not a prime

number because it has a factor 5. So it is a

composite number.

Q. 7. There is a circular path around a sports field.Sonia takes 18 minutes to drive one round ofthe field, while Ravi takes 12 minutes for thesame. Suppose they both start at the samepoint and at the same time, and go in the samedirection. After how many minutes will theymeet again at the starting point?

Sol. Time taken by Sonia to drive one round of thefield = 18 minutes

Time taken by Ravi to drive one round of same

field = 12 minutes

They meet again at the starting point

= LCM (18, 12)

Now, Prime factorisation of 18 and 12 are

18 = (2) × (9) = (2) × (3) × (3)

= (2) × (3)2

12 = (2) × (6) = (2) × (2) × (3)

= (2)2 × (3)

LCM (18, 12) = (2)2 × (3)2

= 4 × 9 = 36

Hence, after 36 minutes Sonia and Ravi will

meet again at the starting point.

SELF PRACTICE 1.2

1. Express each number as a product of its primefactors.

(a) 120 (b) 256

(c) 760 (d) 8008

2. The HCF of two numbers is 145 and their LCMis 2175. If one of the numbers is 725. Find theother number.

3. Check whether 4n, where n is a natural numberwill ends with the digit zero.

4. Find the LCM of 96 and 360 by usingfundamental theorem of arithmetic.

5. Can 12n end with digit zero for any naturalnumbers n?

6. Find HCF and LCM by prime factorisationmethod.

(a) 144, 180, 192 (b) 84, 90, 120

7. Find the LCM and HCF of each of the followingpairs of integers by prime factorisation methodand verify that HCF × LCM = Product of twonumbers.

(a) 28, 32 (b) 18, 35 (c) 85, 51

8. Find the least number that is divisible by all thenumbers between 2 and 10.

Revisiting Irrational Numbers


Q. 1. Prove that 5 is irrational. [CBSE 2009]

Sol. Let us suppose that 5 is rational so we canfind integers r and s where s ≠ 0

such that 5 =r

s

Suppose r and s have some common factor other

than 1, then divide r and s by the common factor

to get:

5 = a

b, where a and b are co-prime and b ≠ 0.

⇒ b 5 = a


MBD Super Refresher MathematicsClass-X Part-I CBSE /NCERT

Publisher : MBD GroupPublishers

ISBN : 9789385905469Author : V. K. Saxena, ShilpiAhuja

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