37
CBSE CLASS X MATHEMATICSCHAPTER 1 REAL NUMBERS RECAP FROM
PREVIOUS CLASS/ES1.NUMBER LINE: Representation of fractions on a
number line2.Whole Numbers (W): 0, 1, 2, 3, ..3. Natural Numbers
(N): 1, 2, 3, ..4. Integers (Z): .. 3, -2, -1, 0, 1, 2, 3, .(Comes
from German word, Zahlen, which means To Count.5. Rational Numbers
(r): Can be written in the form of p/q, where, q 0, As, anything
divided by 0 is not defined. ( These Include N, W & Z). p &
q have no common factors other than 1, ie. They are
co-prime.Equivalent Rational Numbers or Fractions: Their Standard
form is same. Eg. 1/2 = 2/4 = 10/20.Finding a Rational Number
between Two Given Rational Numbers, a & b: (a + b)/2, is the
required number; Proceeding in the same way, we may find more
Rational Numbers between the given two Rational Numbers. There are
infinitely many Rational Numbers between the given two Rational
Numbers.The other way is to convert the given rational numbers into
like fractions ( ie. With the same denominators). 6. Irrational
Numbers (s) are the ones which cant be written in the form of p/q,
where p & q are integers, and, q0.Pythagoreans were the first
ones to discover irrational numbers.7. Real Numbers (R): The
Rational and Irrational Numbers together form the Real Numbers. A
Real Number is either a Rational or Irrational Number. Every Real
Number is represented by a unique point on a number line. And,
also, every point on a number line, represents a unique real
number. (Shown by German Scientists Cantor & Dedekind.)Thus,
the number line is also called Real Number Line.The decimal
expansion of a rational number is either terminating or non -
terminating recurring (which can be pure recurring or mixed
recurring - Explained Later) ; & vice - versa.The decimal
expansion of a irrational number is non - terminating non
recurring; & vice - versa.Let a > 0, be a real number, and,
n be a positive integer, then, = b, if bn= a. The Symbol is called
the Radical Sign.8. Laws of Indices: Let a and b be positive real
numbers, then,(a, n, m are natural numbers; a is called the base; m
& n are the exponents)Now, a0 = 1; so, 1/an = a0 / an = a0-n =
a-n.1/2 = 2 / 2; which is half of 2; and Thence 1/2 can be
represented on a Number Line.When the denominator of an expression
contains a term with a square root (or a number with a radical
sign), the process of converting it to an equivalent expression
whose denominator is a rational number is called Rationalising the
Denominator.
9. FEW RELEVANT POINTS RELATED TO THE CHAPTER:Note the Following
points about the Real Numbers:1.Sum or Difference of a rational and
an irrational number is irrational [eg. (3 + 7) or (7 - 3) is
irrational.]2.The Product or Quotient of a non - zero rational
number with an irrational number is irrational [eg. The Expressions
22 OR 32 are irrational.]3.If we add, subtract, multiply or divide
two irrationals, the result may be rational or irrational [eg. 2 -
2 is 0, which is rational; but, 7 - 2 is irrational].Square root of
all positive integers which are not perfect squares; cube roots of
all integers which are not perfect cubes, and so on, are all
irrational numbers.Pure Recurring Decimals are the ones, in which,
all the digits after the decimal point are repeated; eg. 0.32323232
= 0.32.Mixed Recurring Decimals are the ones, in which, at least
one digit after the decimal point is not repeated; eg.
18.33249494949 = 18.33249Prime Numbers are the numbers, other than
1, whose only factors are1 and the number itself; Eg. 3, 13,
17Composite Numbers are the numbers which have more than 2 common
factors; Eg. 12, 15 etc.Co - Prime Numbers are those two Natural
Numbers (Not necessary prime numbers), ehich have their Highest
Common Factor as One. Eg. (3, 10); (15, 33) etc.1 (One) is a Unique
Number; ie. It is neither a Prime or a Composite Number.-CBSE CLASS
X CHAPTER ONE: REAL NUMBERS1. A non-zero integer a is said to
divide an integer b, if there exists an integer c such that: b =
acEuclids Division Lemma: Let there be two positive integers a and
b. Then, there exist unique integers q and r such that a = bq + r,
0 r < b. Euclids Division Lemma is a restatement of the long
division process, and the integers q & r are called the
Quotient and Remeinder.((Find integers q and r for follg. pairs of
positive integers a and b: (i) 10, 3 (ii) 4, 19 (iii) 81, 3))The
Fundamental Theorem of Arithmetic: Every composite number can be
expressed as a product of primes (or the powers of primes) in a
unique way (ie. The factorization is unique); apart from the order
in which the prime factors occur. Composite Number = Product of
Primes.This Theorum is used: (i) to prove the irrationality of many
of the numbers, and,(ii) to find out when exactly the decimal
expansion of a rational number is terminating, and when it is
non-terminating, repeating.
An algorithm is a series of well defined steps which gives a
procedure for solving a type of problem.A lemma is a proven
statement used for proving another statement.2. Finding HCF &
LCM by prime Factorisation Method: We can find HCF and LCM of two
positive integers using the Fundamental Theorem of Arithmetic. This
method is also called the prime factorisation method.HCF = Product
of the smallest power of each common prime factor in the
numbers.LCM = Product of the greatest power of each prime factor,
involved in the numbers.For any two positive integers (a & b),
HCF (a, b) X LCM (a, b) = a X b
Let x = p/q be a rational Number (where p q are co-prime):If q =
2n 5m, then p/q is a Terminating Decimal Expansion. If q 2n 5m,
then p/q is a Non-Terminating Repeating Decimal Expansion.(n& m
are non-negative integers)3. Finding HCF of two Positive Integers
using Euclids Division Lemma:Let a & b be two positive
integers.Obtain two whole numbers, q1 & r1, such that a = bq1 +
r1; 0 r1 < b.If r1 = 0, b is the HCF of a & b.If r1 0, Apply
Euclids Division Lemma to b & r1, and Obtain two whole numbers,
q2 & r2, such that b = r1q2 + r2; Proceed as such, till the
remainder becomes zero. The divisor at this stage is the HCF of a
& b. 4. Theorum: Let p be a prime number. If p divides a2, then
p divides a, where a is a positive integer.Proof: Let P1, P2,
P3...Pn be factors of a. Thus, a = P1.P2.P3...Pn ; where, P1, P2,
P3...Pn are primes, not necessarily all distinct.Thus, a2 =
(P1.P2.P3...Pn ). (P1.P2.P3...Pn ); Thus, a2 = P12.P22.P32...Pn2P
divides a2 (i) => P is a Prime factor of a2, from The
Fundamental Theorem of Arithmetic.Prime factors of a2 are only
(P1.P2.P3...Pn ) (ii) (Uniqueness of factors from Fund. Th. Of
Arith.)From (i) & (ii), we have P is a prime factor of
(P1.P2.P3...Pn ) => P divides a.
HANDLING INDICES1.Introduction:Why Exponents were
Invented:10,000,000,000,000 = Ten trillion, or simply 1013;
7x7x7x7x7x7x7x7 = 78; 6,000,000 = 6 x 1,000,000 = 6 x 106 In am or
an , a is the base, and m & n are called the exponents;a0 = 1,
ie. Anything raised to the power 0 is One (1); Thus, 30=1, x0=1,
100=1.When no exponent is mentioned on to a number, The Exponent is
taken to be as 1, ie. 3 = 31; x = x1; 7 = 71A Negative Index
implies the Reciprocal of a number, ie. 3-2 = 1/32 = 1/9; x-4 =
1/x4; 2-3 = 1/23 = 1/8; an = 1/a-n; a-n = 1/an.A root is presented
as such: 2 = 21/2 [A second root of 2, or Square root of 2, or
simply, under-root of 2]; 33 = 31/3 [A Third root of 3, or Cube
root of 3]; nx = x1/n [ nth root of x] 2.Laws of Indices:am. an =
am+n [32. 33= 3x3x3x3x3 = 35 = 32+3];am / an = am-n [56 / 54 =
(5x5x5x5x5x5) / (5x5x5x5) = 5x5 = 52 = 56-4];(am)n = amn [(73)2 =
(7x7x7)2 = 7x7x7 x 7x7x7 = 76 = 73x2 ];(ab)m = am x bm [(2x5)3 =
2x5 x 2x5 x 2x5 = 2x2x2x5x5x5 = 23 x 53 ];(a/b)m = am / bm [ (3/5)2
= 3/5 x 3/5 = (3x3) / (5x5) = 32 / 52 ][ 1/an = a0 / an = a0-n =
a-n ]3. Applying Indices to Roots:ab = a b [(ab)1/2 = a1/2 . b1/2 =
a . b];a/b = a / b [(a/b)1/2 = a1/2 / b1/2 = a / b]; (a - b) x (a +
b) = a - b [ (a - b) x (a + b) = (a)2 - (b)2 = (a1/2)2 - (b1/2)2 =
a1/2 x 2 - b1/2 x 2 = a1 - b1 = a - b]
CBSE CLASS X MATHEMATICSCHAPTER 1 REAL NUMBERS - NCERT EXERCISES
SOLUTIONSEXERCISE 1.1 (P. 7)Q1. Use Euclids division algorithm to
find the HCF of :(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and
255.A.(i) 225 = 135 X 1 + 90; 135 = 90 X 1 + 45; 90 = 45 X 2 + 0;
HCF = 45 (ii)HCF = 196 (iii) 867 = 255 X 3 + 102; 255 = 102 X 2 +
51; 102 = 51 X 2 + 0;HCF (867, 255) = HCF (255, 102) = HCF (102,
51) = 51.Q2. Show that any positive odd integer is of the form 6q +
1, or 6q + 3, or 6q + 5, where q is some integer.A. By Euclids
division algorithm, a = bq + r (i), 0 r < b(The value of r will
always be less than the value of b; ie. Less than the value of
divisor)(Also. The degree of r is always less than the degree of
the divisor)Putting b = 6 in (i), we get a = 6q + r [ 0 r < 6;
ie. r = 0, 1, 2, 3, 4, 5]If r = 0, a = 6q, 6q is divisible by 6
=> 6q is even.(Divisibility Criteria: If a number is divided by
2 & 3, it is divisible by 6!) If r = 1, a = 6q + 1, 6q +1 is
not divisible by 2.If r = 2, a = 6q + 2, 6q + 2 is even and is
divisible by 2.If r = 3, a = 6q + 3, 6q + 3 is not divisible by
2.If r = 4, a = 6q + 4, 6q + 4 is even and is divisible by 2.If r =
5, a = 6q + 5, 6q + 5 is not divisible by 2.As 6q, 6q + 2, 6q + 4
are even; therefore, 6q + 1, 6q + 3, 6q + 5 are odd.Q3. 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?A. To find the maximum number of columns, we have to find the
HCF of 616 and 32.By Euclids division Lemma, HCF of 616 & 32 is
8.Hence, maximum number of columns is 8.Q4. Use Euclids division
lemma to show that the square of any positive integer is either of
the form 3m or 3m + 1 for some integer m.A. By Euclids division
algorithm, a = bq + r, 0 r < b (i);On putting, b = 3 in (i), we
get a = 3q + r, 0 r < 3, ie. R = 0, 1, 2.If r = 0, a = 3q=>
a2 = 9q2;(ii)If r = 1, a = 3q + 1=> a2 = 9q2 + 6q + 1;(iii)If r
= 2, a = 3q + 2=> a2 = 9q2 + 12q + 4;(iv)From (ii), 9q2 is a
square of the form 3m, where m = 3q2;From (iii), 9q2 + 6q + 1, ie.
3(3q2 + 2q) + 1 is a square of the form 3m + 1, where m = 3q2 +
2q;From (iv), 9q2 + 12q + 4, ie. 3(3q2 + 4q + 1) + 1 is a square of
the form 3m + 1, where m = 3q2 + 4q + 1.Thus, square of any
positive integer is either of the form 3m or 3m + 1.Q5. Use Euclids
division lemma to show that the cube of any positive integer is of
the form 9m, 9m + 1 or 9m + 8.A. Let x be any positive integer.
Then, it is of the form 3m, 3m + 1, 3m + 2. now, we have to prove
that the cube of each of these can be rewritten in the form 9q, 9q
+ 1, 9q + 8.Now, (3m)3 = 27m3 = 9(3m3) = 9q, where q = 3m3;(3m +
1)3 = 27m3 + 27m2 + 9m + 1 = 9(3m3 + 3m2 + m) + 1 = 9q + 1, where q
= 3m3 + 3m2 + m;(3m + 2)3 = 27m3 + 54m2 + 36m + 8 = 9(3m3 + 3m2 +
m) + 8 = 9q + 8, where q = 3m3 + 6m2 + 4m.EXERCISE 1.2 (P. 11) Q1.
Express each number as a product of its prime factors:(i) 140 (ii)
156 (iii) 3825 (iv) 5005 (v) 7429. (Use Division Method)A. (i) 140
= 22 X 5 X 7(ii) 156 = 22 X 3 X 13(iii) 3825 = 32 x 52 x 17(iv)
5005 = 5 X 7 X 11 X 13(v) 7429 = 17 X 19 X 23Q2. Find the LCM and
HCF of the following pairs of integers and verify that LCM HCF =
product of the two numbers.(i) 26 and 91 (ii) 510 and 92 (iii) 336
and 54A.Do this solution yourself.Q3. Find the LCM and HCF of the
following integers by applying the prime factorisation method. (i)
12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25A.(i) 12 = 2X2X3;
15 = 3X5; 21 = 3X7; HCF = 3; LCM = 2x2x3x5x7 = 420(ii) The given
three numbers don't have any common factor; They are primes.HCF =
1; LCM = 17X23X29 = 11339.(iii) Do this yourself!Q4. Given that HCF
(306, 657) = 9, find LCM (306, 657).A.For any two positive integers
a and b, HCF (a, b) LCM (a, b) = The product of the numbers (a
b)Q5. Check whether 6n can end with the digit 0 for any natural
number n.A.If the number 6n ends with the digit zero; then, it is
divisible by 5. Therefore, the prime factorisation of 6n contains
the prime number 5. This is not possible, as the only prime in the
factorisation of 6n is 2 & 3; and, the uniqueness of the
fundamental theorum of arithmetic guarantees that there are no
other prime in the factorisation of 6n .So, there is no value of n
in natural numbers for which 6n ends with the digit zero. Q6.
Explain why 7 11 13 + 13 and 7 6 5 4 3 2 1 + 5 are composite
numbers.A. (i) 7 11 13 + 13 = 1001 + 13 = 1014; and, 1014 =
2X3X13X13. Thus, 1014 is the product of prime factors. Hence, it is
a composite number.(ii) 7 6 5 4 3 2 1 + 5 = 5040 + 5 = 5045 = 5 X
1009. Thus, it is a product of prime factors 5 and 1009. Hence, it
is a composite number.Q7. There is a circular path around a sports
field. Sonia takes 18 minutes to drive one round of the field,
while Ravi takes 12 minutes for the same. Suppose they both start
at the same point and at the same time, and go in the same
direction. After how many minutes will they meet again at the
starting point?A. They will be again at the starting point after
common multiples of 18 & 12.18 = 2x3x3; 12 = 2x2x3; Thus, LCM
of 18 & 12 = 36. In 36 mts. Ravi arrives at the starting point
after making 3 rounds, as he takes 12 mts. to drive one round &
Sonia also arrives at the starting point after making 2 rounds, as
he takes 18 mts. to drive one round. Thus, they will meet again at
the starting point after 36 minutes.EXERCISE 1.3 (P. 14) Q1. Prove
that 5 is irrational.A. Let 5 is a rational number. Then, 5 can be
written in the form p/q, where p, q are integers and have no common
factor (other than 1), q 0. ie. 5 = p/q; Squaring both sides: 5 =
p2 / q2 => p2 = 5q2 (i)=> 5 is a factor of p2, & thus
divides p2.This implies that 5 divides p (ii). Now, p = 5m => p2
= 25m2. Putting this in equation (i), we get 25m2 = 5q2 => 5m2 =
q2 => 5 divides q2 => 5 divides q (iii).From (ii), 5 divides
p, and from (iii) 5 divides q. This means 5 is a common factor of p
and q. This contradicts the supposition that there is no common
factor of p and q. Hence, 5 is an irrational number. Q2. Prove that
3 + 25 is irrational.A. Let 3 + 25 be a rational number. Now, let 3
+ 25 = a/b, where a and b are co-prime & b 0.So, 25 = a/b - 3;
Or, 5 = a/(2b) - 3/2.Since, a and b are integers, therefore a/(2b)
- 3/2 is a rational number. And, thus 5 is a rational number. But 5
is an irrational number. Thus, our supposition is wrong.Hence, 3 +
25 is an irrational number. Q3. Prove that the following are
irrationals :(i) 1/2, (ii) 75,(iii) 6 + 2(From Previous Class:
Square root of all the numbers which are not perfect squares; and,
Cube root of all the numbers which are not perfect cubes are
Irrational Numbers.)A. (i) Let 1/2 be a rational number. Now, let
1/2 = a/b, where a and b are co-prime & b 0.Or, (1 X 2 )/(2 X2
) = a / b; Or, 2 / 2 = a / b; Or, 2 = 2a/b.((Always try to isolate
the Irrational Component))Since a and b are integers, 2a/b is
rational and so 2 is rational. But, 2 is irrational.Thus, 1/2 is
Irrational.(ii) Let 75 be a rational number. Now, let 75= a/b,
where a and b are co-prime & b 0.Or, 5= a / 7b; ((Always try to
isolate the Irrational Component))Since a and b are integers, a/7b
is rational and so 5 is rational. But, 5 is irrational.Thus, 75is
Irrational.(iii) Let 6 + 2 be a rational number. Now, let 6 + 2=
a/b, where a and b are co-prime & b 0.Or, a/b - 6 = 2; ((Always
try to isolate the Irrational Component))Since a and b are
integers, a/b - 6 is rational and so 2 is rational. But, 2 is
irrational.Thus, 6 + 2 is
Irrational.======================================================================EXERCISE
1.4 (P. 17) Q1. Without actually performing the long division,
state whether the following rational numbers will have a
terminating decimal expansion or a non-terminating repeating
decimal expansion:i.13 / 3125. 3125 = 20 x 55. Thus, Denominator is
of the form 2m X 5n. Thus, 13 / 3125 is a Terminating Decimal.ii.17
/ 8. 8 = 23 x 50. Thus, Denominator is of the form 2m X 5n. Thus, 8
is a Terminating Decimal.iii.64 / 455. 455 = 5 X 7 X 13. Thus,
Denominator is not of the form 2m X 5n. Thus, 64 / 455 is a Non -
Terminating Repeating Decimal.iv.15 / 1600. 1600 = 26 x 52. Thus,
Denominator is of the form 2m X 5n. Thus, 15 / 1600 is a
Terminating Decimal.v.29 / 343.Denominator is not of the form 2m X
5n. Thus, 64 / 455 is a Non - Terminating Repeating Decimal. (As
343 is a odd number, it is not divisible by 2, and hence its
denominators factors cant have powers of 2)vi.23 / 23 52.
Denominator is of the form 2m X 5n. Thus, 23 / 23 52 is a
Terminating Decimal.vii.129 / 225775. Denominator is not of the
form 2m X 5n. Thus, 129 / 225775 is a Non - Terminating Repeating
Decimal. viii. 6/15 = 2/5. 5 = 20 x 51. Thus, Denominator is of the
form 2m X 5n. Thus, 6 / 15 is a Terminating Decimal.ix.35 / 50 = 7
/ 10 = 7 / (21 x 51). Thus, Denominator is of the form 2m X 5n.
Thus, 35 / 50 is a Terminating Decimal. x.77 / 210. 210 = 7x3x2x5.
Thus, Denominator is not of the form 2m X 5n. Thus, 77 / 210 is a
Non - Terminating Repeating Decimal.Q2. Write down the decimal
expansions of those rational numbers in Question 1 above which have
terminating decimal expansions.A. (i) 13 / 3125 = 13 / (5x5x5x5x5)
= (13x2x2x2x2x2) / (5x2x5x2x5x2x5x2x5x2) = 416/100000 =
0.00416.(ii) 17 / 8 = 17 / (23 x 50) = (17x53) / (23 x 53) =
(17x125)/103 = 2125/1000 = 2.125.(iii) Non - Terminating
Repeating.(iv) 15 / 1600 = 15 / (26 x 52) = 15 / (24x22x52) =
(15x54)/(24x54x22x52) = (15x625)/106 = 9375/1000000 = 0.009375.(v)
Non - Terminating Repeating.(vi) 23 / 23 52 = (23x5)/2x5x22x52) =
115/103 = 0.115(vii) Non - Terminating Repeating.(viii) 6/15 = 2/5
= (2x2)/(5x2) = 4/10 = 0.4.(ix) 35/50 = 35/(5x10) = (35x2)/(2x5x10)
= 70/100 = 0.7(x) Non - Terminating Repeating.Q3. The following
real numbers have decimal expansions as given below. In each case,
decide whether they are rational or not. If they are rational, and
of the form p/q, what can you say about the prime factors of q?(i)
43.123456789 (ii) 0.120120012000120000. . . (iii) 43.123456789A.
(i) 43.123456789 is terminating, and hence is a Rational Number.
43.123456789 = 43123456789 = p/q. q = 1000000000 = 109 = 29x59.
1000000000 (ii) 0.120120012000120000. . . is Non - Terminating Non
- Repeating; and hence is Irrational. (iii) 43.123456789 is Non -
Terminating but Repeating; and hence is Rational.43.123456789 =
4312345646 = p/q. Hence, q = 999999999. 999999999
CBSE CLASS X MATHEMATICSCHAPTER 2 POLYNOMIALSRECAP FROM PREVIOUS
CLASS/ES1.A Function p(x) of the form p(x) = a0 + a1x + a2x2 + a3x3
+.+ an xn is called a polynomial, where (i) a0, a1, a2, ..an are
real numbers, and, (ii) n is a non - negative (ie. 0 or positive)
integer. ((Thus, (2/x + 3), (x2 + 2x + 4/x) are not Polynomials.a0,
a1, a2, ..an are called co - efficient of the polynomial. The
expression of this form is called Polynomial in one variable. We
can also have polynomials in more than one variable, eg. x2 + y2 +
xyz is a polynomial in three variables.2. Terms: The polynomial 3x2
+ 4x + 5 has three terms.3. Co - efficient: Each term of the
polynomial has a co-efficient. In 3x2 + 4x + 5, the co-efficients
of x2 , x and x0 are 3, 4, and 5.4. Constant Polynomial: -3, 0, 2,
5.. are examples of Constant Polynomials. The Constant Polynomial 0
is called a Zero Polynomial.A Polynomial written either in the
descending or ascending powers of x is called the Standard Form of
a Polynomial. A Polynomial can be Monomial, Binomial and so on.5.
Degree of a Polynomial: The highest exponent in various terms of
one variable is called its degree. Eg. A Polynomial can be Linear,
Quadratic, Cubic, Biquadratic, etc. If a0 = a1 = a2 = ..=an = 0,
ie. If all the constants are zero, we get a Zero Polynomial, which
is denoted by 0. The Degree of a Zero Polynomial is not defined. 6.
Zeroes of a Polynomial: The value of a polynomial p(x) at x = a, is
p(a), obtained on replacing x by a. In general, a Zero of a
Polynomial p(x) is a, such that p(a) = 0. Consider a polynomial
p(x) = x - 1; Now, p(1) = 0, ie. 1 is a Zero (or root) of the
Polynomial p(x), which gives x=1 Now, consider a constant
polynomial 5. It has no zero because replacing x by any number in
5x0, still gives 5. Thus, a non - zero constant polynomial has no
zero.We will be dealing with Polynomials in one variable
only.Zeroes of a Zero (0)Polynomial: By Convention, every Real
Number is a zero of the Zero Polynomial.Number of Zeroes of a
polynomial of degree n is n.7. Division of a Polynomial: Say, on
dividing polynomial p(x) by another polynomial g(x), we get
quotient q(x) & remainder r(x). Dividend = ( Divisor X
Quotient) + Remainder; ie. P(x) = g(x) . Q(x) + r(x). The degree of
the remainder r(x) is always less than the degree of the divisor
g(x), ie. Degree of r(x) < Degree of g(x).If the divisor is
linear (ie. One degree), then the degree of remainder will be 0,
ie. The remainder will be a constant.2x3 + x2 + x = x(2x2 + x + 1),
We say that x and (2x2 + x + 1) are Factors of (2x3 + x2 + x ), and
(2x3 + x2 + x ) is a multiple of x as well as (2x2 + x + 1) 8.
Remainder Theorum: Let p(x) be any polynomial of degree greater
than or equal to one, and, let a be any real number. If p(x) is
divided by a linear polynomial (x-a), then the remainder is
p(a).======================================================================Remember
The Following Points:A zero of a polynomial need not be zero.Every
Linear Polynomial has one and only one zero.A polynomial can have
more than one zero.Number of Zeroes is the degree of the
polynomial.[Division of a polynomial by other polynomial using long
division method has to be thoroughly practiced by the
student.]Proof of Remainder Theorum: Let p(x) be any polynomial
with degree 1. Lets say, when p(x) is divided by (x-a), the
quotient is q(x), and, the remainder is r(x); ie. P(x) = (x-a) .
q(x) + r(x).Since, the degree of (x-a) is one, and the degree of
r(x) is less than the degree of (x-a), the degree of r(x) = 0. This
means that r(x) is a constant, say r. Thus, for every value of x,
r(x) = r.Therefore, p(x) = (x-a) . q(x) + r. In particular, if x=a,
this equation gives us p(a) = (a-a) . q(a) + r = r; which proves
the theorem.9. Factor Theorum: If p(x) is a polynomial of degree n
1, and a is any real number, then:(i) (x-a) is a factor of p(x), if
p(a) = 0, and,(ii) p(a) = 0, if (x-a) is a factor of p(x). This
actually follows from the Remainder Theorum.10. Factorisation of
Polynomials:A) Splitting the Middle Term: (ie. Quadratic
Equations): To factorise ax2 + bx + c, we have to write b as the
sum of two numbers whose product is ac. Let p & q be those two
numbers, ie. p+q = b, and, p.q = c.[If on multiplying a & c, we
get + sign, well get the middle term by adding; Sign will be that
of the middle term.]
B) By Factor Theorum: Let f(x) be a given polynomial. 1.Proceed
as follows: (Factorising equations of Degree 3) Write separately,
the constant term of the polynomial. Write down its factors. If the
number is big, and has many factors, write down the smallest, first
few factors.Let a be the zero of polynomial; ie. (x-a) is factor of
polynomial f(x).Divide the given polynomial f(x) by (x-a), and get
the Quotient.Factorise the Quotient by splitting the middle
term.You would thus obtain the factors of the given polynomial.
READ THE FOLLOWING EXAMPLE:Let f(x) = x3 - 3x2 - 9x - 5. The
constant term is 5, so, its factors are 5, 1 Now, p(-1) = 0; Thus
(x+1) is a factor of p(x). Now, do the long division. = (x2 - 4x -
5) (Remainder is zero);Now, by splitting the middle term, (x2 - 4x
- 5) = (x+1) (x - 5).Thus, x3 - 3x2 - 9x - 5 = (x+1)(x+1) (x - 5) =
(x+1)2 (x-5)In Regards to this method, remember the following
important points:If the sum of coefficients of odd powers of any
polynomial p(x) is the same as that the sum of its coefficients of
even powers, then, (x+1) is always the factor of p(x).If the sum of
the coefficients of given expression p(x) is zero, then, (x-1) is
always the factor of p(x).C) Factorisation by Algebric Identities:
Read the following identies. Understand them, prove them if you
have to, and Remember them.(a + b)2= a2+ 2ab + b2 (a - b)2= a2- 2ab
+ b2 a2 b2 = (a + b) (a b) (x + a)(x + b) = x2+ (a + b)x + ab (a +
b + c)2= a2+ b2+ c2+ 2ab + 2ac + 2bc (a + b)3= a3+ b3 + 3ab(a+b) (a
- b)3= a3- b3 - 3ab(a-b) a3+ b3= (a + b) (a2 ab + b2) a3 b3= (a b)
(a2+ ab + b2) a3+ b3+ c3- 3abc = (a + b + c)(a2+ b2+ c2- ab - bc -
ca); Thus, a3+ b3+ c3= 3abc, if (a + b + c) = 0.11. Least Common
Multiple (LCM) of Polynomials = (Highest Power of Common Term) X
Remaining Terms. --
CBSE CLASS X CHAPTER TWO: POLYNOMIALS1.A Quadratic Polynomial
can have either two distinct zeroes, two equal zeroes or no zeroes;
ie. A polynomial of degree 2 has at the most two zeroes.Similarly,
a Cubic Polynomial can have at the most 3 zeroes. Consider the
following:If k is a zero of p(x) = ax + b; then p(k) = ak + b = 0,
ie. k = (-b) / a. Thus, the zero of the linear polynomial [ak + b]
is [(-b) / a] = (-)Constant Term / Coefficient of x.
Type of PolynomialGeneral Form No. of Zeroes Relationship
between ZeroesLinear ax + b, a0 1 k = = Constant term Co-efficient
of xQuadratic ax2 + bx + c; a0 2 Sum of Zeroes ( + )= (Let , be two
zeroes) Product of Zeroes (.) = .Cubic Polynomial ax3 + bx2 + cx +
d 3 Sum of Zeroes ( + +)(Let , , be three zeroes) = = -Co-efficient
of x2 Co-efficient of x3 Product of Zeroes (..) = =
Sum of the product of zeroes taken two at a time (For Cubic
Polynomial) = (. + . + ) = = 2. To form a Quadratic Polynomial with
its given zeroes:Let , be the zeroes of a Quadratic Polynomial. x =
, => x - = 0; x = , => x - = 0. Thus, (x - ).(x - ) is the
Quadratic Polynomial, ie. x2 - ( + )x + ; ie. x2 - (Sum of Zeroes)
x + Product of Zeroes.3. Division Algorithm for Polynomial: If p(x)
& g(x) are any two polynomials with g(x) 0, then we can find
polynomials q(x) & r(x) such that p(x) = g(x) X q(x) + r(x);
where r(x) = 0, or degree of r(x) is less than the degree of
g(x).Dividend = Divisor X Quotient + Remainder. 4. Geometric
Meaning of the zeroes of the Polynomial. a.Graph of y = ax + b is a
straight line which intersects the x-axis at ( , 0). This
x-Coordinate (of the point of intersection of the graph with
x-axis) is the zero of the polynomial y = ax + b .
b.Graph of the Quadratic Equation y = ax2 + bx + c, a 0, can
have two distinct zeroes (graph cutting the x-axis at two distinct
points), one zero (or two equal zeroes), or no zeroes (graph not
cutting the x-axis at all (in which case, it would not be possible
to factorise the quadratic polynomial).In fact, for any quadratic
polynomial ax2 + bx + c, a 0, the graph of the corresponding
equation y = ax2 + bx + c has one of the two shapes either open
upwards or open downwards depending on whether a > 0 or a <
0. (These curves are called parabolas.)c. A cubic polynomial of the
form y = a3 can have at the most 3 zeroes.In general, a polynomial
p(x) of degree n has at the most n zeroes. N.B.: Formation of a
cubic polynomial: Let , , be three zeroes of the Polynomial. Then,
the required cubic polynomial is (x - ) (x - )(x - ).
CBSE CLASS X MATHEMATICSCHAPTER 2 POLYNOMIALS - NCERT EXERCISES
SOLUTIONSEXERCISE 2.1 (P. 28)Q1. The graphs of y = p(x) are given
in the Fig. below, for some polynomials p(x). Find the number of
zeroes of p(x), in each case. A.Self Explanatory.-EXERCISE 2.2 (P.
33)Q1. Find the zeroes of the following quadratic polynomials and
verify the relationship between the zeroes and the
coefficients.Compare the quadratic equations given with the general
quadratic equation, ie. ax2 + bx + c. A.(i) x2 2x 8 = (x-4)(x+2).
The zeroes are x = 4, -2Sum of Zeroes = 4 - 2 = 2 = -(-2)/1 = -
Coefficient of x = (-b)/a Coefficient of x2Product of zeroes = -8 =
Constant Term Coefficient of x2 (ii) 4s2 4s + 1 (iii) 6x2 3 7x (iv)
4u2 + 8u (v) t2 15 (vi) 3x2 x 4Q2. Find a quadratic polynomial each
with the given numbers as the sum and product of its zeroes
respectively.A. Let the polynomial be ax2 + bx + c, and its zeroes
be & .(i) 1/4, -1; + = 1/4; . = -1The polynomial formed is x2 -
(Sum of Zeroes) x + Product of Zeroes = x2 - 1/4 x + (-1) = x2 -
x/4 - 1. The other possible polynomials would be k(x2 - x/4 - 1).If
k = 4, then the polynomial is 4x2 - x - 4. (ii) 2, 1/3; + = 2; . =
1/3The polynomial formed is x2 - (Sum of Zeroes) x + Product of
Zeroes = x2 - 2x + 1/3. The other possible polynomials would be
k(x2 - 2x + 1/3).If k = 3, then the polynomial is 3x2 - 32x +
1.(iii) 0, 5; + = 0; . = 5The polynomial formed is x2 - (Sum of
Zeroes) x + Product of Zeroes = x2 - 0.x + 5 = x2 + 5. (iv) 1, 1; +
= 1; . = 1The polynomial formed is x2 - (Sum of Zeroes) x + Product
of Zeroes = x2 - 1.x + 1 = x2 - x + 1.(v) -1/4, 1/4; + = -1/4; . =
1/4The polynomial formed is x2 - (Sum of Zeroes) x + Product of
Zeroes = x2 - (-1/4)x + 1/4 = x2 + x/4 + 1/4. The other possible
polynomials would be k(x2 + x/4 + 1/4).If k = 4, then the
polynomial is 4x2 + x + 1.(vi) 4, 1; + = 4; . = 1The polynomial
formed is x2 - (Sum of Zeroes) x + Product of Zeroes = x2 - 4x +
1.--
EXERCISE 2.3 (P. 36) Q1. Divide the polynomial p(x) by the
polynomial g(x) and find the quotient and remainder in each of the
following :(i) p(x) = x3 3x2 + 5x 3, g(x) = x2 2(x3 3x2 + 5x 3) =
(x2 2).(x-3) + (7x - 9). (ii) p(x) = x4 - 3x2 + 4x + 5, g(x) = x2 x
+ 1; (x4 + 0x3 - 3x2 + 4x + 5) = (x2 x + 1).(x2 + x - 3) + (8).
(iii) p(x) = x4 + 0.x2 - 5x + 6, g(x) = 2 - x2 = -x2 + 2(x4 + 0.x2
- 5x + 6) = (-x2 + 2).(-x2 - 2) + (-5x + 10). Q2. Check whether the
first polynomial is a factor of the second polynomial by dividing
the second polynomial by the first polynomial:Remainder is 0,
So,(t2 - 3) is a factor of (2t4 + 3t3 - 2t2 - 9t - 12). (ii) x2 +
3x + 1; 3x4 + 5x3 - 7x2 + 2x + 2Remainder is 0, So,(x2 + 3x + 1) is
a factor of (3x4 + 5x3 - 7x2 + 2x + 2).(iii) x3 - 3x + 1; x5 - 4x4+
+ x2 + 3x + 1As Remainder is not 0, So,(x3 - 3x + 1) is not a
factor of (x5 - 4x4+ + x2 + 3x + 1).Q3. Obtain all other zeroes of
3x4 + 6x3 2x2 10x 5, if two of its zeroes are (5/3), and -
(5/3)A.Since the two zeroes are , therefore, is a factor of
p(x).Now, applying division algorithm to the given polynomial and
3x2 - 5.
Now, x2 + 2x + 1 = (x + 1)2. Zeroes of (x + 1)2 are -1,
-1.Hence, all its zeroes are (5/3), (-)(5/3), -1,-1.Q4. On dividing
x3 3x2 + x + 2 by a polynomial g(x), the quotient and remainder
were (x 2) and (2x + 4), respectively. Find g(x).A. By Division
Algorithm, p(x) = q(x).g(x) + r(x)Thus, (x3 3x2 + x + 2) = (x
2).g(x) + (2x + 4);Or, (x 2).g(x) = (x3 3x2 + x + 2) - (2x + 4) =
(x3 3x2 + 3x - 2)Thus, g(x) = (x3 3x2 + 3x - 2) (x 2). This gives
g(x) as x2 - x + 1Q5. Give examples of polynomials p(x), g(x), q(x)
and r(x), which satisfy the division algorithm and (i) deg p(x) =
deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0A. Accordingto
the division algorithm, ifp(x) andg(x) are two polynomials withg(x)
0, then we can find polynomialsq(x) andr(x) such thatp(x) =g(x)
q(x) +r(x),wherer(x) = 0 or degree ofr(x) < degree ofg(x)Degree
of a polynomial is the highest power of thevariablein the
polynomial (i) deg p(x) = deg q(x) Degree of quotient will be equal
to degree ofdividendwhendivisoris constant ( i.e., when any
polynomial isdividedby a constant) Let p(x) = 12x2 + 8x + 24; q(x)
= 3x2 + 2x + 6; g(x) = 4; & r(x) =0.Degree ofp(x) andq(x) is
the same i.e., 2. Checking fordivision algorithm, p(x) =g(x) q(x)
+r(x);(12x2 + 8x + 24) = 4(3x2 + 2x + 6) + 0; Thus, the division
algorithm is satisfied.(ii) deg q(x) = deg r(x) (Here, we have to
remember that the degree of r(x) is always less than the degree of
g(x) )Let us assume the division ofx3+ xbyx2,Here, p(x) =x3+ xg(x)
=x2q(x) =xandr(x) =xClearly, thedegreeofq(x) andr(x) is the same
i.e., 1.Checking fordivision algorithm,p(x) =g(x) q(x) +r(x)x3+ x=
x2 x+xx3+ x = x3+ xThus, the division algorithm is satisfied.(iii)
deg r(x) = 0Degree of remainder will be 0 when remainder comes to a
constant.Let us assume the division ofx3+1byx2.Here, p(x) =x3+1g(x)
=x2q(x) =xandr(x) = 1Clearly, the degree ofr(x) is 0.Checking
fordivision algorithm,p(x) =g(x) q(x) +r(x)x3+1 = (x2) x+ 1x3+1=
x3+1Thus, the division algorithm is satisfied.
CBSE CLASS X MATHEMATICSCHAPTER 3 LINEAR EQUATIONS IN TWO
VARIABLESRECAP FROM PREVIOUS CLASS/ES1. An Equation is a statement
of equality of two algebraic expressions involving one or more
unknown quantities called the variables.The Equations involving
only one variable are called equations in one variable, eg. 3x -
6=0; y3 - 8 =0.An Equation involving only linear polynomial is
called a linear equation; eg. 3x+7=0.The value of the variable,
which when substituted for the variable (in the equation), makes
both sides of the given equation equal, is called a solution, or
root of the equation; eg. x=3 is root of the equation 3x+10 = 19.2.
Properties of Equation: We can add, subtract, to both sides of the
equation by the same number,We can perform multiplication or
division to both sides of the equation by the same non-zero number,
without changing the equality.3. Linear equation in two variables:
An Equation of the form ax + by + c = 0, where a, b, c are real
numbers, and a 0, b 0, is called a linear equation in two
variables, eg. x - y = 0. A linear equation in two variables has
infinitely many solutions . Solutions to a Linear Equation lie on a
straight line.Often, the condition, a & b are not both zero is
denoted by a2 + b2 0The reason that a degree one polynomial
equation ax + by + c = 0 is called a linear equation is that its
geometrical representation is a straight line.4. Drawing graph of
the equation ax + by + c = 0:Express the equation in the form y= -
ax + c ; b Give integral values to x, and find corresponding values
of y;Plot the points (a1,b1), (a2,b2), (a3,b3) so obtained on the
graph paper;Join the points to get a line which represents the
equation ax + by + c = 0.Each solution (x, y) of a linear equation
in two variables, ax + by + c = 0, corresponds to a point on the
line representing the equation, and vice versa.CBSE CLASS X CHAPTER
THREE: LINEAR EQUATIONS IN TWO VARIABLES1. The general form for a
pair of linear equations in two variables x and y is a1x + b1y + c1
= 0and a2x + b2y + c2 = 0, where,a1 , b1, c1, a2, b2, c2 are all
real numbers, and a12 + b12 0; a22 + b22 0For any two given lines
in a plane, only one of the following three possibilities can
happen:The two lines will intersect at one point,The two lines will
not intersect, i.e., they are parallel, &The two lines will be
coincident.(( When solving the linear equation in two variable,
when one of the variable is put to zero, the equation reduces to a
linear equation in one variable, which is solved easily.))A pair of
linear equations in two variables is said to form a system of
Simultaneous Linear Equations. A pair of values of x & y
satisfying each one of the equation in x & y is called a
solution of the system.2. Graphical Method of Solution of a Pair of
Linear Equations:A pair of linear equations which has no solution,
is called an inconsistent pair of linear equations. A pair of
linear equations in two variables, which has a solution, is called
a consistent pair of linear equations. A pair of linear equations
which are equivalent has infinitely many distinct common solutions.
Such a pair is called a dependent pair of linear equations in two
variables. Note that a dependent pair of linear equations is always
consistent.Lines representing a pair of linear equations in two
variables:The lines may intersect in a single point. In this case,
the pair of equations has a unique solution (consistent pair of
equations with unique solution),The lines may be coincident. In
this case, the equations have infinitely many solutions [dependent
(consistent) pair of equations with infinite solutions],The lines
may be parallel. In this case, the equations have no solution
(inconsistent pair of equations with no solution).
Summarizing the above facts:Again, consider the general form for
a pair of linear equations in two variables x and y a1x + b1y + c1
= 0and a2x + b2y + c2 = 0, where,a1 , b1, c1, a2, b2, c2 are all
real numbers, and a12 + b12 0; a22 + b22 0 (a & b are not both
zero)Intersecting lines (Consistent pair of equations with unique
solution): a1/a2 b1/b2Coincident lines (Consistent pair of
equations with infinite solutions): a1/a2 = b1/b2 = c1/c2 Parallel
Lines (Inconsistent pair of equations with no solution): a1/a2 =
b1/b2 c1/c2 3. Algebraic solution of a system of linear equations:
The graphical method is not convenient in cases when the point
representing the solution of the linear equations has non-integral
coordinates like ( 3, 27 ), (1.05, 3.8) etc.Three algebraic methods
available to solve a pair of linear Equations are:3.1 Substitution
Method: Consider the following steps to understand this method of
solving a given pair of equations: Step 1 : Find the value of one
variable, say y in terms of the other variable, i.e., x from either
equation, whichever is convenient.Step 2 : Substitute this value of
y in the other equation. This reduces it to an equation in one
variable, i.e., in terms of x, which can now be solved. Sometimes,
we may get statements with no variable. If the statement obtained
is true (eg. 23 = 23), we can conclude that the pair of linear
equations has infinitely many solutions. If the statement obtained
is false, then the pair of linear equations is inconsistent (&
has no solutions).Step 3 : Substitute the value of x obtained in
Step 2 in any of the original equation or the one obtained in Step
1 to obtain the value of the other variable.Remark : We have
substituted the value of one variable by expressing it in terms of
the other variable to solve the pair of linear equations. That is
why the method is known as the substitution method.3.2 Elimination
Method: This is the method of eliminating (i.e., removing) one
variable. The steps involved are:Step 1: First multiply both the
equations by some suitable non-zero constants to make the
coefficients of one variable (either x or y) numerically equal.Step
2: Then add or subtract one equation from the other so that one
variable gets eliminated. If we get an equation in one variable, go
to Step 3.If in Step 2, we obtain a true statement involving no
variable, then the original pair of equations has infinitely many
solutions.If in Step 2, we obtain a false statement involving no
variable, then the original pair of equations has no solution,
i.e., it is inconsistent.Step 3: Solve the equation in one variable
(x or y) so obtained to get its value.Step 4: Substitute this value
of x (or y) in either of the original equations to get the value of
the other variable.3.3 Cross - Multiplication Method: Consider the
two linear equations in their general form:a1x + b1y + c1 = 0 (i),
and, a2x + b2y + c2 = 0 (ii). Follow the following steps:Step 1:
Multiply equation (i) by b2, & Equation (ii) by b1. We
get:b2a1x + b2b1y + b2c1 = 0 (iii), and, b1a2x + b1b2y + b1c2 = 0
(iv). Step 2: Subtracting (iv) from (iii), we get: x=(b1c2 -
b2c1)/(a1b2 - a2b1); where, a1b2 - a2b1 0 (v)Step 3: Substituting
this value of x in (i) or (ii), we get,y=(c1a2 - c2a1)/(a1b2 -
a2b1) (vi); Now, Here two cases may arise:Case 1. a1b2 - a2b1 0.
Thus, a1/a2 b1/b2 => The pair of linear equations has a unique
solution.Case 2. a1b2 - a2b1 = 0. If we write a1/a2 = b1/b2 = k,
then, a1 = ka2; b1 = kb2.Putting the values of a1 & a2 in (i),
k(a2x + b2y) + c1 = 0 (vii)Now, Equations (vii) & (ii) can both
be satisfied only if c1 = kc2 => c1/c2 = k.If c1 = kc2, , any
solution of Equation (ii) will satisfy the Equation (i), and vice
versa. So, if, a1/a2 = b1/b2 = c1/c2 = k, then there are infinitely
many solutions to the pair of linear equations given by (i) &
(ii).If c1 kc2 , then any solution of Equation (i) will not satisfy
Equation (ii) and vice versa. Therefore the pair has no solution.
Thus, we have:When a1/a2 b1/b2; we get a unique solution;When a1/a2
= b1/b2 = c1/c2; we get infinitely many solutions;When a1/a2 =
b1/b2 c1/c2; we don't get any solutions (Inconsistent pair of
equations).==>The solutions given in equations (v) & (vi),
may be written as: x = y = 1 (viii) b1c2 - b2c1 c1a2 - c2a1 a1b2 -
a2b1Use the following diagram to memorise the above
result.:xy1b1c1a1b1b2c2 a2b2 Thus, we follow the following steps,
in solving a pair of linear equations by the cross-multiplication
method:Step 1 : Write the given equations in the form (i) and
(ii),Step 2 : Taking the help of the diagram above, write Equations
as given in (viii),Step 3 : Find x and y, provided a1b2 - a2b1
0
CBSE CLASS X MATHEMATICSCHAPTER 3 LINEAR EQUATIONS IN TWO
VARIABLES - NCERT EXERCISES SOLUTIONSEXERCISE 3.1 (P. 44)Q1. Aftab
tells his daughter, Seven years ago, I was seven times as old as
you were then. Also, three years from now, I shall be three times
as old as you will be. (Isnt this interesting?) Represent this
situation algebraically and graphically.A. Let thepresentage of
Aftab bex, and, present age of his daughter =ySeven years ago: Age
of Aftab =x 7, and, Age of his daughter =y 7Three years hence, Age
of Aftab =x+ 3 Age of his daughter =y+ 3Q2. The coach of a cricket
team buys 3 bats and 6 balls for Rs 3900. Later, she buys another
bat and 2 more balls of the same kind for Rs 1300. Represent this
situation algebraically and geometrically.A. Let the cost of a bat
be Rsx, and of ball be RsyAccording to the question, Preparing
solution table for the above two equations:y = (3900 - 3x)/6; and,
y = (1300 - x)/2 Q3. The cost of 2 kg of apples and 1kg of grapes
on a day was found to be Rs 160. After a month, the cost of 4 kg of
apples and 2 kg of grapes is Rs 300. Represent the situation
algebraically and geometrically.A. Let the cost of 1 kg of apples
be Rsx, & cost of 1 kg ofgrapes= Rsy. Acc. to the question,
Preparing solution table for the above two equations: y = 160 - 2x;
and, y = (300 - 4x)/2, we get the above graph.EXERCISE 3.2 (P.
49)Q1. Form the pair of linear equations in the following problems,
and find their solutions graphically.(i) 10 students of Class X
took part in a Mathematics quiz. If the number of girls is 4 more
than the number of boys, find the number of boys and girls who took
part in the quiz.A. Let the number of girls bexand the number of
boys bey.Accordingto the question, x+y= 10 (ie. y = 10 - x); and
xy= 4 (ie. y = x - 4).Drawing the solution table for the two
equations, we get the required graph:(ii) 5 pencils and 7 pens
together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs
46. Find the cost of one pencil and that of one pen.A. Let the cost
of 1 pencil be Rsxand the cost of 1 pen be Rsy.According to the
question, 5x+ 7y= 50 (ie. y = (50 - 5x)/7); and, 7x+ 5y= 46 (ie. y
= (46 - 7x)/5)Drawing the solution table for the two equations, we
get the required graph:Q2. On comparing the ratios a1/a2, b1/b2
& c1/c2 , find out whether the lines representing the following
pairs of linear equations intersect at a point, are parallel or
coincident:(i) 5x 4y + 8 = 0; 7x + 6y 9 = 0Comparing the equations
with general form, we have Since, a1/a2 b1/b2; Hence, the lines
representing the given pair of equations have a unique solution and
the pair of linesintersectsat exactly one point. (ii) 9x + 3y + 12
= 0; 18x + 6y + 24 = 0Comparing the equations with general form, we
have Since, a1/a2 = b1/b2 = c1/c2; Hence, the lines representing
the given pair of equations are coincident and there are
infinitepossiblesolutionsfor the given pair of equations. (iii) 6x
3y + 10 = 0 2x y + 9 = 0Comparing the equations with general form,
we have Since, a1/a2 = b1/b2 c1/c2; Hence, the given pair of
Equations are inconsistent (Parallel Lines) with no solution.
Q3. On comparing the ratios a1/a2, b1/b2 & c1/c2 , find out
whether the following pair of linear equations are consistent, or
inconsistent.(i) 3x+ 2y= 5; 2x 3y= 7 Comparing the equations with
general form, we have Since, a1/a2 b1/b2; Hence, the lines
representing the given pair of equations have a unique solution and
the pair of linesintersectsat exactly one point. (ii) 2x 3y = 8; 4x
6y = 9Comparing the equations with general form, we have Since,
a1/a2 = b1/b2 c1/c2 ; Hence, the given pair of Equations are
inconsistent (Parallel Lines) with no solution.(iii) 3/2 x + 5/3 y
= 7; ; 9x 10y = 14Comparing the equations with general form, we
have Since, a1/a2 b1/b2; Hence, the lines representing the given
pair of equations have a unique solution and the pair of
linesintersectsat exactly one point. Hence, the pair of linear
equations is consistent. (iv) 5x 3y = 11; 10x + 6y = 22Comparing
the equations with general form, we have Since, a1/a2 = b1/b2 =
c1/c2 ; Hence, the lines representing the given pair of equations
are Coincident lines (ie. Consistent pair of equations with
infinite solutions).(v) 4/3 x + 2y = 8; 2x + 3y = 12Comparing the
equations with general form, we have Since, a1/a2 = b1/b2 = c1/c2 ;
Hence, the lines representing the given pair of equations are
Coincident lines (ie. Consistent pair of equations with infinite
solutions).Q4. Which of the following pairs of linear equations are
consistent/inconsistent? If consistent, obtain the solution
graphically:A. (i)x+y= 5; 2x+ 2y= 10 Comparing the equations with
general form, Since, a1/a2 = b1/b2 = c1/c2 ; Hence, the lines
representing the given pair of equations are Coincident lines (ie.
Consistent pair of equations with infinite solutions).x+y= 5 =>
y = 5 - x; 2x+ 2y= 10 => y = (10 - 2x)/2Drawing the solution
table for the two equations, we get the required graph:It can be
seen that the two lines are overlapping each other. (ii) x y = 8,
3x 3y = 16Comparing the equations with general form, we haveSince,
a1/a2 = b1/b2 c1/c2 ; Hence, the given pair of Equations are
inconsistent (Parallel Lines) with no solution.(iii) 2x + y 6 = 0,
4x 2y 4 = 0Comparing the equations with general form, we haveSince,
a1/a2 b1/b2; Hence, the lines representing the given pair of
equations have a unique solution and the pair of linesintersectsat
exactly one point (ie. They are consistent). 2x + y 6 = 0, ie. y =
- 2x + 6; 4x 2y 4 = 0, ie. y = (4x - 4)/2 Drawing the solution
table for the two equations, we get the required graph:(iv) 2x 2y 2
= 0, 4x 4y 5 = 0Since, a1/a2 = b1/b2 c1/c2 ; Hence, the given pair
of Equations are inconsistent (Parallel Lines) with no solution.(To
demonstrate by drawing a graph, that the two lines are indeed
parallel.)Q5. Half the perimeter of a rectangular garden, whose
length is 4 m more than its width, is 36 m. Find the dimensions of
the garden.A. Let the length of the garden bex, & the width be
y.According to the question, x= y + 4 (1)y+x= 36 (2) Drawing the
solution table for the two equations, we will get the required
graph.The point where the two lines will intersect, will give us
the required dimensions of the garden.Length = x = 20m; Width = y =
16m.Q6. Given the linear equation 2x + 3y 8 = 0, write another
linear equation in two variables such that the geometrical
representation of the pair so formed is: (i) intersecting lines
(ii) parallel lines (iii) coincident linesA. (i) For intersecting
lines: 6x + 12y - 8 = 0. (ii) For parallel lines: 4x+ 6y 8 = 0.
(iii) For coincident lines: 6x+ 9y 24 = 0.
Q7. Draw the graphs of the equations x y + 1 = 0 and 3x + 2y 12
= 0. Determine the coordinates of the vertices of the triangle
formed by these lines and the x-axis, and shade the triangular
region.A. x y + 1 = 0, ie. y = x + 1; and, 3x + 2y 12 = 0, ie. y =
-3x + 12.Drawing the solution table for the two equations, we will
get the required graph:From the figure, the three lines are
intersecting each other at point (2, 3) andx-axis at (1, 0) and (4,
0). Therefore, the vertices of the triangle are (2, 3), (1, 0), and
(4, 0).--
EXERCISE 3.3 (P. 53)Q1. Solve the following pair of linear
equations by the substitution method. (i) x + y = 14 (1); x - y = 4
(2).From (1), we obtain y = 14 - x(3); Putting this value in eq.
(2), x - (14 - x) = 4; ie. x = 9. Putting the value of x in eq.
(3), we get y = 5.(ii) s t = 3 (1); s/3 + t/2 = 6 (2).From (1), we
obtain s = t + 3(3); Putting this value in eq. (2), (t + 3)/3 + t/2
= 6; ie. t = 6. Putting the value of t in eq. (3), we get s =
9.(iii) 3x y = 3 (1); 9x 3y = 9 (2).From (1), we obtain y= 3x 3 (3
); Putting this value in eq. (2), 9x - 3(3x 3) = 9; ie. 9 = 9. This
is always true. Hence, the given pair of equations has infinite
possible solutions One of its possible solutions isx= 0,y= -3.(iv)
0.2x + 0.3y = 1.3 (1); 0.4x + 0.5y = 2.3 (2).From (1), we obtain y
= (1.3 - 0.2x)/0.3(3); Putting this value in eq. (2), 0.4x +
0.5(1.3 - 0.2x)/0.3 = 2.3; ie. x = 2. Putting the value of x in eq.
(3), we get y = 3.(v) 2 x + 3 y = 0 (1); 3 x - 8 y = 0 (2).From
(1), we obtain y = (-2 x)/3 (3); Putting this value in eq. (2), 3 x
- 8 ((-2 x)/3) = 0; ie. x = 0. Putting the value of x in eq. (3),
we get y = 0.(vi) 3x/2 - 5y/3 = -2 (1); x/3 + y/2 = 13/6 (2).DO IT
ON YOUR OWN!!x = 2, y = 3.Q2. Solve 2x + 3y = 11 and 2x 4y = 24 and
hence find the value of m for which y = mx + 3.A. 2x + 3y = 11 (1);
2x - 4y = -24 (2).Solving for x & y, using substitution method,
we get x = -2; y = 5.y = mx + 3; ie. 5 = m (-2) + 3; or, 2 = -2m;
or, m = -1.Q3. Form the pair of linear equations for the following
problems and find their solution by substitution method.(i) The
difference between two numbers is 26 and one number is three times
the other. Find them.A. Let the first number bexand the other
number beysuch thaty>x.According to giveninfo., y = 3x (1); y -
x = 26 (2)Substitutingthe value ofyfrom equation (1) into equation
(2), 3x - x = 26 => x = 13 (3) Putting this value in eq. (1), y
= 3 x 13 = 39.(ii) The larger of two supplementary angles exceeds
the smaller by 18 degrees. Find them.A. Let the larger angle
bexandsmallerangle bey. As per question. x - y = 180 (1), and as
the two are supplementary angles, x + y = 1800 (2)From (1) y = x -
180 (3). Put this value in eq. (2), x + x - 180 = 1800; Thus, x =
990.Putting the value of x in eq. (3), we get y = 990 - 180 =
810.(iii) The coach of a cricket team buys 7 bats and 6 balls for
Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the
cost of each bat and each ball.A. Let the cost of a bat and a ball
bexandy.According to the given info. 7x + 6y = 3800 (1); and, 3x +
5y = 1750 (2)From (1), y = (3800 - 7x)/6 (3); Putting this value in
eq. (2) => 3x + 5 [(3800 - 7x)/6] = 1750, which gives x = 500
(4).Putting this value in eq. (3), we get y = 50.(iv) The taxi
charges in a city consist of a fixed charge together with the
charge for the distance covered. For a distance of 10 km, the
charge paid is Rs 105 and for a journey of 15 km, the charge paid
is Rs 155. What are the fixed charges and the charge per km? How
much does a person have to pay for travelling a distance of 25
km?A. Let the fixedchargebe Rsxand per km charge be Rsy.As per the
question, x + 10y = 105 - (i); and, x + 15y = 155 - (ii)From (i),
we get, x = 105 - 10y - (iii). Putting this in eq. (ii), we get,
105 - 10y + 15y = 155,which gives, y = 10 - (iv) Putting this in
eq. (iii), we get, x = 5.Hence, fixed charge = Rs 5, and, Per km
charge = Rs 10.Charge for 25 km = x + 25y = Rs 255(v) A fraction
becomes 9/11, if 2 is added to both the numerator and the
denominator. If, 3 is added to both the numerator and the
denominator it becomes 5/6. Find the fraction. A. Let the fraction
be x/y. As per the given information,(x+2)/(y+2) = 9/11 => 11x -
9y = (-)4 - (i);(x+3)/(y+3) = 5/6 => 6x - 5y = (-)3 - (ii)From
(i), x = (-4 + 9y)/11 - (iii). Substituting this in equation (ii),
we get, y = 9 - (iv)Substituting y = 9 in equation (iii), we get, x
= 7. Thus the fraction is 7/9(vi) Five years hence, the age of
Jacob will be three times that of his son. Five years ago, Jacobs
age was seven times that of his son. What are their present ages?A.
Let the age of Jacob bexand the age of his son bey.According to the
given information, (x + 5) = 3(y + 5) => x - 3y = 10 - (i) (x -
5) = 7(y - 5) => x - 7y = -30 - (ii)From, (i), x = 3y + 10 -
(iii). Substituting this in equation (ii), we get, y = 10 -
(iv)Substituting y = 10 in equation (iii), we get, x = 40. Hence,
the present age of Jacob is 40 years, and the present age of his
son is 10 years.--EXERCISE 3.4 (P. 56) Q1. Solve the following pair
of linear equations by the elimination method and the substitution
method :(i) x + y = 5 and 2x 3y = 4By Elimination Method:x + y = 5
(i); 2x 3y = 4 (ii)Multiplying (i) by 2, we get, 2x + 2y = 10
(iii)Subtracting (ii) from (iii), we get, y = 6/5 (iv). Putting
this in eq. (i), we get, x = 19/5By Substitution Method:x + y = 5
(i); 2x 3y = 4 (ii)x + y = 5 (i), gives, x = 5 - y (iii). Putting
this in (ii), we get, y = 6/5. Putting y = 6/5 in (iii), we get, x
= 19/5(ii) 3x + 4y = 10 and 2x 2y = 2; (iii) 3x 5y 4 = 0 and 9x =
2y + 7; and, (iv) x/2 + 2y/3 = -1 and x - y/3 = 3These three pair
of equations can be solved following the method described above.Q2.
Form the pair of linear equations in the following problems, and
find their solutions (if they exist) by the elimination method :(i)
If we add 1 to the numerator and subtract 1 from the denominator, a
fraction reduces to 1. It becomes 1/2 if we only add 1 to the
denominator. What is the fraction?A. Let the fraction be x/y.
According to given information,(x + 1) / (y - 1) = 1=> x - y =
-2 (i)x / (y + 1) = 1/2 => 2x - y = 1 (ii)Subtracting (i) from
(ii), we get x = 3 (iii). Putting this in (i), we get y = 5. Thus,
the fraction is 3/5(ii) Five years ago, Nuri was thrice as old as
Sonu. Ten years later, Nuri will be twice as old as Sonu. How old
are Nuri and Sonu?A.Do it yourself! Age of Nuri = 50 years; and,
Age of Sonu = 20 years.(iii) The sum of the digits of a two-digit
number is 9. Also, nine times this number is twice the number
obtained by reversing the order of the digits. Find the number.A.
Let the unit digit and tens digits of the number
bexandyrespectively. Then, number (written as yx) = 10y+x Number
afterreversingthe digits = 10x+yAccording to the given
information,x+y= 9 (i); 9(10y+x) = 2(10x+y) => 88y 11x= 0 =>
x+ 8y=0 (ii)Adding equation (i) and (ii), we get 9y= 9, or, y= 1
(iii)Substituting the value in equation (1), we get, x= 8Hence, the
number is 10y+x= 10 1 + 8 = 18.(iv) Meena went to a bank to
withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs
100 notes only. Meena got 25 notes in all. Find how many notes of
Rs 50 and Rs 100 she received.A. Let the number of Rs 50 and Rs 100
notes bexandy.According to the given information, x + y = 25 (i);
50x + 50y = 2000 (ii)Multiplying equation (i) by 50, we get, 50x +
50y = 1250 (iii)Subtracting equation (iii) from equation (ii), we
get, 50y = 750 => y = 15.Substituting y = 15, in equation (i),
we get, x= 10Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs
100. (v) A lending library has a fixed charge for the first three
days and an additional charge for each day thereafter. Saritha paid
Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the
book she kept for five days. Find the fixed charge and the charge
for each extra day.A. Let the fixedchargefor first three days and
each day charge thereafter be Rsxand Rsy.According to the given
information, x + 4y = 27 (i); x + 2y = 21 (ii)Subtracting equation
(ii) from equation (i), we get, 2y = 6 => y = 3
(iii)Substituting in equation (i), we get, x + 12 = 27 => x =
15Hence, fixed charge = Rs 15, and, Charge per day = Rs 3
--EXERCISE 3.5 (P. 62)Q1. Which of the following pairs of linear
equations has unique solution, no solution, or infinitely many
solutions. In case there is a unique solution, find it by using
cross multiplication method.(i) x 3y 3 = 0; 3x 9y 2 = 0a1 / a2 =
1/3; b1 / b2 = 1/3; c1 / c2 = 3/2. Therefore, a1 / a2 = b1 / b2 c1
/ c2 Therefore, the given sets of lines are parallel to each other.
Therefore, they will not intersect each other and thus, there will
not be any solution for these equations.(ii) 2x + y = 5; 3x + 2y =
8a1 / a2 = 2/3; b1 / b2 = 1/2; c1 / c2 = 5/8. Therefore, a1 / a2 b1
/ b2 . Therefore, they will intersect each other at a unique point
and thus, there will be a unique solution for these equations.By
cross-multiplication method, x = y = 1____ b1c2 - b2c1 c1a2 - c2a1
a1b2 - a2b1 This gives, x/2 = y/1 = 1. Therefore, x = 2, and, y =
1(iii) 3x 5y = 20; 6x 10y = 40a1 / a2 = 1/2; b1 / b2 = 1/2; c1 / c2
= 1/2. Therefore, a1 / a2 = b1 / b2 = c1 / c2Therefore, the given
sets of lines are coincident to each other and thus, there
areinfinitesolutions possible for these equations. (iv) x 3y 7 = 0;
3x 3y 15 = 0a1 / a2 = 1/3; b1 / b2 = 1; c1 / c2 = 7/15. Therefore,
a1 / a2 b1 / b2 Therefore, they will intersect each other at a
unique point and thus, there will be a unique solution for these
equations. By cross-multiplication, x = y = 1____ b1c2 - b2c1 c1a2
- c2a1 a1b2 - a2b1 Thus, x = y = 1____ 45 - 21 -21 - (-15) -3 -
(-9)Or, x / 24 = y / (-6) = 1/6; This gives, x = 4 & y = -1Q2.
(i) For which values of a and b does the following pair of linear
equations have an infinite number of solutions?2x + 3y = 7; (a b) x
+ (a + b) y = 3a + b 2A. a1 / a2 = 2/(a - b); b1 / b2 = 3/(a + b);
c1 / c2 = 7/(3a +b - 2). For infinitely many solutions, a1 / a2 =
b1 / b2 = c1 / c22/(a - b) = 7/(3a +b - 2), which gives, a - 9b =
-4 (i)2/(a - b) = 3/(a + b), which gives, a - 5b = 0 (ii)Solving
equations (i), and (ii), we get, a = 5, and, b = 1; for which the
given equations will have infinitely many solutions.Q2. (ii) For
which values of k will the following pair of linear equations have
no solutions?3x + y = 1; (2k 1) x + (k 1) y = 2k + 1A. a1 / a2 =
3/(2k - 1); b1 / b2 = 1/(k - 1); c1 / c2 = 1/(2k + 1). For no
solution, a1 / a2 = b1 / b2 c1 / c2 ; which gives k = 2.Hence, for
k = 2, the given equation will have no solution.Q3. Solve the
following pair of linear equations by the substitution and
cross-multiplication methods :8x + 5y = 9; 3x + 2y = 4A. Do It
Yourself!Q4. Form the pair of linear equations in the following
problems and find their solutions (if they exist) by any algebraic
method :i) A part of monthly hostel charges is fixed and the
remaining depends on the number of days one has taken food in the
mess. When a student A takes food for 20 days she has to pay Rs
1000 as hostel charges whereas a student B, who takes food for 26
days, pays Rs 1180 as hostel charges. Find the fixed charges and
the cost of food per day.A. Letxbe the fixed charge of the food
andybe the charge for food per day.According to the
giveninformation, x + 20y = 1000 (i); x + 26y = 1180
(ii)Subtracting equation (i) from equation (ii), we get, 6y = 180
=> y = 30Substituting this value in equation (i), we get, x =
400Hence, fixed charge = Rs 400; and charge per day = Rs 30 Q4.(ii)
A fraction becomes 1/3 when 1 is subtracted from the numerator and
it becomes 1/4 when 8 is added to its denominator. Find the
fraction.A. Let thefractionbex/y.According to the given info., (x -
1) / y = 1/3 => 3x - y = 3 (i); x / (y + 8) = 1/4 => 4x - y =
8 (ii)Subtracting equation (i) from equation (ii), we get, x = 5.
Puttingthis (i), we get, y = 12Hence, the fraction is5/12.Q4.(iii)
Yash scored 40 marks in a test, getting 3 marks for each right
answer and losing 1 mark for each wrong answer. Had 4 marks been
awarded for each correct answer and 2 marks been deducted for each
incorrect answer, then Yash would have scored 50 marks. How many
questions were there in the test?A. Let the number of right answers
and wrong answers bexandy.According to the given information, 3x -
y = 40 (i); 4x - 2y = 50 => 2x - y = 25 (ii)Subtracting equation
(ii) from equation (i), we get, x = 15 (iii). Puttingthis in (ii),
y = 5Therefore, number of right answers = 15; and, number of wrong
answers = 5.Total number ofquestions= 20Q4.(iv) Places A and B are
100 km apart on a highway. One car starts from A and another from B
at the same time. If the cars travel in the same direction at
different speeds, they meet in 5 hours. If they travel towards each
other, they meet in 1 hour. What are the speeds of the two cars?A.
Let the speed of 1stcar and 2ndcar beukm/h andvkm/h. Relative speed
of both cars while they are travelling in same direction = (u - v)
km/h. (u > v)Relative speed of both cars while they are
travelling in opposite directions i.e., travelling towards each
other = (u + v) km/hAccording to the given information, 5(u - v) =
100 => u - v = 20 (i); and, 1(u + v) = 100 (ii)Adding both the
equations, we get, u = 60 km/h (iii)Substituting this value in
equation (2), we get, v = 40 km/hHence, speed of one car = 60 km/h
and speed of other car = 40 km/h.Q4.(v) The area of a rectangle
gets reduced by 9 square units, if its length is reduced by 5 units
and breadth is increased by 3 units. If we increase the length by 3
units and the breadth by 2 units, the area increases by 67 square
units. Find the dimensions of the rectangle.A. (v) Let length and
breadth of rectangle bexunit andyunit. Area =xy. Acc. to the
question,(x - 5)(y + 3) = xy - 9 => 3x - 5y - 6 = 0 (i); (x +
3)(y + 2) = xy + 67 => 2x + 3y - 61 = 0 (ii)By
cross-multiplication method, we get, x = y = 1____ ; b1c2 - b2c1
c1a2 - c2a1 a1b2 - a2b1 Or, x = y = 1___; 305 - (-18) -12 - (-183)
9 - (-10) Or, x / 323 = y / 171 = 1 / 19 => x = 17, and, y =
9Hence, the length and breadth of the rectangle are 17 units and 9
unitsrespectively.
EXERCISE 3.6 (P. 67) Q1. Solve the following pairs of equations
by reducing them to a pair of linear equations:Q1. (i) 1 + 1 = 2; 1
+ 1 = 13/6; 2x 3y 3x 2y Let 1/x = p, & 1/y = q; The equations
thus become,p/2 + q/3 = 2 => 3p + 2q - 12 = 0 (i)p/3 + q/2 =
13/6 => 2p + 3q - 13 = 0 (ii)Using cross multiplication method,
we get, p = 2, and, q = 3; which gives, x = 1/2, and, y = 1/3.Q1.
(ii) 2/(x) + 3/(y) = 2; /(x) - 9/(y) = -1Putting 1/(x) = p, and,
1/(y) = q. The equations thus become,2p + 3q = 2 (i); 4p - 9q = -1
(ii)Solving, we get, p (=1/(x)) = 1/2; q (=1/(y)) = 1/3Solving for
x, and y, we get, x = 4; y = 9Q1. (iii) 4/x + 3y = 14; 3/x - 4y =
23Putting 1/x = p, the equations become,4p + 3y = 14 (i); 3p - 4y =
23 (ii)Solving, we get, p (=1/x) = 5, => x = 1/5; y = -2Q1. (iv)
5/(x - 1) + 1/(y - 2) = 2; 6/(x - 1) - 3/(y - 2) = 1Putting 1/(x -
1) = p, and, 1/(y - 2) = q. The equations thus become,5p + q = 2
(i); 6p - 3q = 1 (ii)Solving, we get, p (=1/(x - 1) ) = 1/3; which
gives, x = 4Also, we get, q (=1/(y - 2)) = 1/3; which gives, y =
5Therefore, x = 4, and, y = 5Q1. (v) (7x - 2y)/xy = 5 => 7/y -
2/x = 5 (i); (8x + 7y)/xy = 15 => 8/y + 7/x = 15 (ii)Putting 1/x
= p, and, 1/y = q. The equations thus become,-2p + 7q = 5 (iii); 7p
+ 8q = 15 (iv)Solving, we get, p (=1/x) = 1; which gives, x = 1;
Also, we get, q (=1/y) = 1; which gives, y = 1Therefore, x = 1,
and, y = 1Q1. (vi) 6x + 3y = 6xy => 6/y + 3/x = 6 (i); 2x + 4y =
5xy => 2/y + 4/x = 5 (ii)Putting 1/x = p, and, 1/y = q, the
equations thus become, 3p + 6q - 6 = 0 (iii); 4p + 2q - 5 = 0
(iv)Solving, using cross multiplication method, we get, p (=1/x) =
0 => x = 1; q (=1/y) = 1/2 => y = 2Q1. (vii) 10/(x + y) +
2/(x - y) = 4 (i); 15/(x + y) - 5/(x - y) = -2 (ii)Putting 1/(x+y)
= p, and, 1/(x-y) = q, the equations become, 10p + 2q = 4(iii); 15p
- 5q = -2 (iv)Solving, using cross multiplication method, we get, p
(=1/(x+y)) = 1/5 => x + y = 5 (v); and, q (= 1/(x - y) = 1 =>
x - y = 1 (vi)Solving equations (v) and (vi), we get, x = 3, and y
= 2Q1. (viii) 1/(3x + y) + 1/(3x - y) = 3/4 (i); 1/[2(3x + y)] -
1/[2(3x - y) = -1/8 (ii)Putting 1/(3x+y) = p, and, 1/(3x-y) = q,
the equations become, p + q = 3/4(iii); p/2 - q/2 = -1/8 => p -
q = -1/4 (iv)Solving, we get, p [= 1/(3x+y) ] = 1/4 => 3x + y =
4 (v), and, q (= 1/(3x - y) = 1/2 => 3x - y = 2 (vi)Solving
equations (v) and (vi), we get, x = 1, and y = 1Q2. Formulate the
following problems as a pair of equations, and hence find their
solutions:(i) Ritu can row downstream 20 km in 2 hours, and
upstream 4 km in 2 hours. Find her speed of rowing in still water
and the speed of the current.A. Let the speed of Ritu in still
water be x km/h, and the speed of stream beykm/h. Speed of Ritu
whilerowing: Upstream =(x - y)km/h; Downstream =(x +
y)km/hAccording toquestion: 2(x + y) = 20 => x + y = 10 (i); 2(x
- y) = 4 => x - y = 2 (ii)solving the equations, we get, x = 6,
and, y = 4.Hence,Ritusspeed in still water is 6 km/h and the speed
of thecurrentis 4 km/h. (ii) 2 women and 5 men can together finish
an embroidery work in 4 days, while 3 women and 6 men can finish it
in 3 days. Find the time taken by 1 woman alone to finish the work,
and also that taken by 1 man alone.A. Let the number of days taken
by a woman and a man, to complete the work bexandy.Therefore, work
done by a woman in 1 day = 1/x, and,Work done by a man in 1 day
=1/yAccording to the question: 2/x + 5/y = 1/4 (i); and, 3/x + 6/y
= 1/3 (ii)Putting1/x = p, and 1/y = q in these equations, we get,
2p + 5q = 1/4 => 8p + 20q = 1 (iii), and, 3p + 6q = 1/3 => 9p
+ 18q = 1 (iv)Solving equations (iii) & (iv), by
cross-multiplication, we get, p (= 1/x) = 1/18 => x = 18; and, q
(= 1/y) = 1/36 => y = 36 Hence, number of days taken by a woman
to finish the work = 18, and, Number of days taken by a man to
finish the work= 36(iii) Roohi travels 300 km to her home partly by
train and partly by bus. She takes 4 hours if she travels 60 km by
train and the remaining by bus. If she travels 100 km by train and
the remaining by bus, she takes 10 minutes longer. Find the speed
of the train and the bus separately.A. Let the speed of train and
bus beukm/h andvkm/h. According to the giveninformation,60/u +
240/v = 4 (i); [4Hrs10Mts = 25/6Hrs]; 100/u + 200/v = 25/6
(ii).Putting 1/u = p, and1/v = q, in these equatns, we get,60p +
240q = 4 (iii), 600p + 1200q = 25 (iv)Solving equations (iii) and
(iv), we get, p = 1/60; q = 1/80Hence, Speed of train = u = 60km/h,
and, Speed of bus = 80km/h.
CBSE CLASS X MATHEMATICSCHAPTER 4 QUADRATIC EQUATIONS1.
Introduction: Quadratic polynomial is of the form ax2 + bx + c, a
0. When we equate this polynomial to zero, we get a quadratic
equation.2. Quadratic Equations: A quadratic equation in the
variable x is an equation of the form ax2 + bx + c = 0, where a, b,
c are real numbers, a 0. For example, 2x2 + x 300 = 0 is a
quadratic equation.In fact, any equation of the form p(x) = 0,
where p(x) is a polynomial of degree 2, is a quadratic equation.
But when we write the terms of p(x) in descending order of their
degrees, then we get the standard form of the equation. That is,
ax2 + bx + c = 0, a 0 is called the standard form of a quadratic
equation.Q. Check whether the following are quadratic equations:(i)
(x 2)2 + 1 = 2x 3;(ii) x(x + 1) + 8 = (x + 2) (x 2);(iii) x (2x +
3) = x2 + 1;(iv) (x + 2)3 = x3 43. Solution of a Quadratic Equation
by Factorisation:A real number is called a root of the quadratic
equation ax2 + bx + c = 0, a 0 if a2 + b + c = 0. We also say that
x = is a solution of the quadratic equation, or that satisfies the
quadratic equation. Note that the zeroes of the quadratic
polynomial ax2 + bx + c and the roots of the quadratic equation ax2
+ bx + c = 0 are the same. Any quadratic equation can have at-most
two roots.We can find the roots of a quadratic equation by
factorising it into two linear factors (by splitting the middle
term) and equating each factor to zero.Ex. Find the roots of the
quadratic equation 6x2 x 2 = 0.A. 6x2 x 2 = (3x 2)(2x + 1)The roots
of 6x2 x 2 = 0 are the values of x for which (3x 2)(2x + 1) =
0.Therefore, 3x 2 = 0 or 2x + 1 = 0, ie. x = 2/3; or, x = -1/2We
verify the roots, by checking that 2/3 & -1/2 satisfy 6x2 x 2 =
0 4. Solution of a Quadratic Equation by Completing the
SquareHerein, the term containing x is brought completely inside a
square, and thence, the roots are found easily by taking the square
roots. We can convert any quadratic equation to the form (x + a)2
b2 = 0 and then we can easily find its roots. The process is:x2 +
4x = (x + 4/2)2 - 4ie. Make the term containing x2, a perfect
square (by dividing or multiplying appropriate number); Half the
co-efficient of x, and bring it in the term containing perfect
square.Few Examples: (i) Solve the equation 3x2 - 5x + 2 =
0.Dividing both sides by 3; x2 - 5/3x + 2/3 = 0 => (x - 5/6)2 +
2/3 - 25/36 = 0Solving we get, (x - 5/6)2 = (1/6)2 => x = 1, or
x = 2/3; which are the roots of the given equation.(ii) Solve the
equation 2x2 - 5x + 3 = 0 => x2 - 5/2 x + 3/2 = 0 => (x -
5/4)2 + 3/2 - 25/16 = 0;Or, (x - 5/4)2 = 1/16 => x - 5/4 = 1/4
=> x = 3/2 or x = 1
(iii) Find the roots of 4x2 + 3x + 5 = 0 by the method of
completing the square.A. 4x2 + 3x + 5 = 0 => x2 + 3/4x + 5/4 = 0
=> (x + 3/8)2 - 9/64 + 5/4 = 0 => (x + 3/8)2 = (-71)/64 <
0.But, (x + 3/8)2 cant be negative for any real value of x (as ve x
ve = +ve). So, there is no real value of x satisfying the given
equation. Therefore, the given equation has no real
roots.Generalising the above method of completing the
squares:Consider the quadratic equation ax2 + bx + c = 0 (a 0).
Dividing throughout by a, we get x2 + b/a x + c/a = 0=> (x +
b/2a)2 - (b/2a)2 + c/a =0=> (x + b )2 - (b2 - 4ac) => 2a
4a2=> (x + b )2 = b2 - 4ac (i) 2a 4a2If b2 4ac 0, then by taking
the square roots in (i), we get (x + b/2a) = ((b2 - 4ac) / 2aThus,
x = -b ((b2 - 4ac) [which are the two roots of quadratic equation
ax2 + bx + c = 0 (a 0) 2a if (b2 - 4ac) 0. This is also known as
the Quadratic Formula.]Solve the following equations by quadratic
formula (find the roots if they exist):(i) x2 + 2x 143 = 0 [ x =
11, or x = -13 ](ii) x2 - 3x - 4 = 0 [ x = 4, or x = -1 ](iii) x2 +
4x + 5 = 0 [ b2 4ac < 0 ](iv) 2x2 22 x + 1 = 0 [ x = 1/2, or, x
= 1/2 ]5. Nature of Roots:The roots of the equation ax2 + bx + c =
0 are given by x = - b (b2 - 4ac). 2aIf b2 - 4ac > 0, we get two
distinct real roots:x = - b + (b2 - 4ac), and, x = - b - (b2 -
4ac). 2a 2aIf b2 - 4ac = 0, then x = -b/2a 0, ie. x = -b/2a or
b/2a.So, the roots of the equation ax2 + bx + c = 0 are both b/2a.
Thus, the quadratic equation ax2 + bx + c = 0 has two equal real
roots in this case.If b2 4ac < 0, then there is no real number
whose square is b2 4ac. Therefore, there are no real roots for the
given quadratic equation in this case.Since b2 4ac determines
whether the quadratic equation ax2 + bx + c = 0 has real roots or
not, b2 4ac is called the discriminant of this quadratic
equation.So, a quadratic equation ax2 + bx + c = 0 has(i) two
distinct real roots, if b2 4ac > 0,(ii) two equal real roots, if
b2 4ac = 0, (iii) no real roots, if b2 4ac < 0.Eg. Find the
Discriminant of the equation 3x2 - 2x + 1/3 = 0, and hence find the
nature of its roots. Find them if the are real.Ans. a = 3, b = -2,
c = 1/3. Therefore, Discriminant b2 4ac = 0Hence, the given
quadratic equation has two equal real roots.The roots are b/2a,
-b/2a; ie. 1/3, 1/3
CBSE CLASS X MATHEMATICSCHAPTER 4 QUADRATIC EQUATIONS- NCERT
EXERCISES SOLUTIONSEXERCISE 4.1 (P. 73):Q1. Check whether the
following are quadratic equations :(i) (x + 1)2 = 2(x 3) => x2 +
7 = 0.It is of the form ax2 + bx + c = 0; and hence is a quadratic
equation.(ii) x2 2x = (2) (3 x) => x2 4x + 6 = 0.It is of the
form ax2 + bx + c = 0; and hence is a quadratic equation.(iii) (x
2)(x + 1) = (x 1)(x + 3) => 3x - 1 = 0It is not of the form ax2
+ bx + c = 0; and hence is not a quadratic equation.(iv) (x 3)(2x
+1) = x(x + 5) => x2 10x - 3 = 0.It is of the form ax2 + bx + c
= 0; and hence is a quadratic equation.(v) (2x 1)(x 3) = (x + 5)(x
1) => x2 11x + 8 = 0.It is of the form ax2 + bx + c = 0; and
hence is a quadratic equation.(vi) x2 + 3x + 1 = (x 2)2 => 7x -
3 = 0.It is not of the form ax2 + bx + c = 0; and hence is not a
quadratic equation.(vii) (x + 2)3 = 2x (x2 1) => x3 - 6x2 - 14x
- 8 = 0.It is not of the form ax2 + bx + c = 0; and hence is not a
quadratic equation.(viii) x3 4x2 x + 1 = (x 2)3 => 2x2 - 13x + 9
= 0.It is of the form ax2 + bx + c = 0; and hence is a quadratic
equation.Q2. Represent the following situations in the form of
quadratic equations :(Assume what is to be found out in the given
question)(i) The area of a rectangular plot is 528 m2. The length
of the plot (in metres) is one more than twice its breadth. We need
to find the length and breadth of the plot.A. Let thebreadthof the
plot bexm => lengthof the plot is (2x+ 1) m.Area of a rectangle
=Length Breadth; 528 =x(2x+ 1) => 2x2 + x - 528 = 0(ii) The
product of two consecutive positive integers is 306. We need to
find the integers.A. Let the consecutive integers bexandx+ 1. It is
given that their product is 306. x(x + 1) = 306 => x2 + x - 306
= 0.(iii) Rohans mother is 26 years older than him. The product of
their ages (in years) 3 years from now will be 360. We would like
to find Rohans present age.A. Let Rohans age bex. Hence, his
mothers age =x+ 26.Three years hence: Rohans age =x+ 3; Mothers age
=x+ 26 + 3 =x+ 29It is given that the product of their ages after 3
years is 360. (x + 3)(x + 29) = 360 => x2 + 32x - 273 = 0.(iv) A
train travels a distance of 480 km at a uniform speed. If the speed
had been 8 km/h less, then it would have taken 3 hours more to
cover the same distance. We need to find the speed of the train.A.
Let the speed of train bexkm/h. Time taken to travel 480 km = 480/x
hrs.If speed = (x - 8) km/h; time taken to cover 480 kms is (480/x
+ 3) hrs.Now, distance = speed X time => 480 = (x - 8) X (480/x
+ 3)=> x2 - 8x + 1280 = 0.EXERCISE 4.2 (P. 76):Q1. Find the
roots of the following quadratic equations by factorisation: (ie.
By factorizing the middle term)(i) x2 3x 10 = 0; or, (x - 5)(x + 2)
= 0.Roots of thisequationare thevaluesfor which (x - 5)(x + 2) = 0
=> x - 5 = 0; or x + 2 = 0.Thus, i.e.,x= 5 orx= 2
(ii) 2x2 + x 6 = 0; or, (x + 2)(2x - 3) = 0. Roots of this
equation are the values for which (x + 2)(2x - 3) = 0; => (x +
2) = 0; or, (2x - 3) = 0i.e.,x= 2 orx= 3/2(iii) 2 x2 + 7x + 52 = 0;
or, (2x + 5)(x + 2) = 0.Roots of this equation are the values for
which (2x + 5)(x + 2) = 0; => (2x + 5)= 0; or, (x + 2) = 0;
i.e.,x= 5/2 orx= -2(iv) 2x2 - x + 1/8 = 0; or, 1/8(16x2 - 8x + 1) =
1/8 (4x - 1)2 = 0Roots of this equation are the values for which
(4x - 1)2 = 0; => (4x - 1)= 0; or, (4x - 1) = 0; i.e.,x= 1/4, or
x = 1/4(v) 100x2 20x + 1 = 0; or, (10x - 1)2 = 0.Roots of this
equation are the values for which (10x - 1)2 = 0; => (10x - 1)=
0; or, (10x - 1) = 0; i.e.,x= 1/10, or x = 1/10Q2. Solve the
problems given in Example 1.Example 1:(i) Required representation
of the problem: x2 - 45x + 324 = 0. (Solve by splitting the middle
term)(ii) Required representation of the problem: x2 - 55x + 750 =
0. (Solve by splitting the middle term)Q3. Find two numbers whose
sum is 27 and product is 182.A. Let the first number bex; then,
thesecondnumber is 27 x.As per question,x(27 x) = 182 => x2 -
27x + 182 = 0 => (x - 13)(x - 14) = 0.Thus, either (x - 13) = 0;
or, (x - 14) = 0; ie. x = 13, or x = 14.If x = 13, the other number
is 14, and vice versa. Thus, the numbers are 13 and 14.Q4. Find two
consecutive positive integers, sum of whose squares is 365.A. Let
theconsecutivepositiveintegersbexandx+ 1. Thus, x2 + (x + 1)2 = 365
=> (x + 14)(x - 13) = 0; Thus, x = -14; or, x = 13.Since the
integers are positive, x = 13. Thus, the two consecutive positive
integers are 13 and 14.Q5. The altitude of a right triangle is 7 cm
less than its base. If the hypotenuse is 13 cm, find the other two
sides.A. Let the base of the righttrianglebexcm. Then, its altitude
= (x 7) cm.By Pythagoras Theorum, x2 + (x - 7)2 = 132 => (x -
12)(x + 5) = 0Thus, x = 12, or, x = -5. As sides cant be negative,
x = 12.Thus, the base of the triangle is 12cm, and the altitude is
5cm.Q6. A cottage industry produces a certain number of pottery
articles in a day. It was observed on a particular day that the
cost of production of each article (in rupees) was 3 more than
twice the number of articles produced on that day. If the total
cost of production on that day was Rs 90, find the number of
articles produced and the cost of each article.A. Let the number of
articles produced bex.Therefore, cost of production of each article
= Rs (2x+ 3)It is given that the total production is Rs 90. Thus,
x(2x+ 3) = 90 => (2x + 15)(x - 6) = 0Thus, x = -15/2, or, x =
6.As the number of articles produced can only be a positive
integer, therefore,xcan only be 6.Hence, number of articles
produced = 6, and, Cost of each article = 2 6 + 3 = Rs 15
EXERCISE 4.3 (P. 87):Q1. Find the roots of the following
quadratic equations, if they exist, by the method of completing the
square:(i) 2x2 7x + 3 = 0 => x2 - 7/2x + 3/2 = 0 => (x -
7/4)2 - 49/16 + 3/2 = 0 => (x - 7/4)2 = 49/16 - 3/2 => (x -
7/4) = 5/4 => x = 3, or x = 1/2(ii) 2x2 + x 4 = 0 => x2 + x/2
- 2 = 0 => (x + 1/4)2 - 1/16 - 2 = 0=> (x + 1/4)2 = 1/16 + 2
=> (x + 1/4)2 = 33/16. Solving, x = (33 - 1)/4 or ( - 33 -
1)/4(iii) 4x2 + 43 x + 3 = 0; (x = -3/2; x = -3/2)(iv) 2x2 + x + 4
= 0; (No real roots, as the square of a number cannot be
negative)Q2. Find the roots of the quadratic equations given in Q.1
above by applying the quadratic formula.A. Compare the equation
with ax2 + bx + c = 0. Get the value of a, b, and c.Find the value
of (b2 - 4ac), and proceed. Part (iv) is being solved:(iv) 2x2 + x
+ 4 = 0. Using quadratic formula, x = - b (b2 - 4ac), we get, 2ax =
-1 (-31). But, the square of a number cant be negative. Thus, there
are no real roots for the 4 given equation.Q3. Find the roots of
the following equations:(i) x - 1/x = 3, x 0. Use quadratic
formula, x = (3 13) / 2Form an equation, factorise it by splitting
the middle term.We get x = 1 or x = 2Q4. The sum of the reciprocals
of Rehmans ages, (in years) 3 years ago and 5 years from now is
1/3. Find his present age.A. Let x be Rehmans present age. As per
question, 1/(x - 3) + 1/(x + 5) = 1/3.Forming a quadratic equation,
and factorizing it by splitting the middle term, we get,x = 7, or x
= -3. But, as age cant be negative, x = 7.Q5. In a class test, the
sum of Shefalis marks in Mathematics and English is 30. Had she got
2 marks more in Mathematics and 3 marks less in English, the
product of their marks would have been 210. Find her marks in the
two subjects.A. Let the marks in Maths bex. Then, the marks in
English will be 30 x. Acc. to the question, (x + 2)(30 - x - 3) =
210. Factorizing we get, (x - 12)(x - 13) = 0 => x = 12, or x =
13.If the marks in Maths are 12, then marks in English will be 30
12 = 18If the marks in Maths are 13, then marks in English will be
30 13 = 17 ==>Q6. The diagonal of a rectangular field is 60
metres more than the shorter side. If the longer side is 30 metres
more than the shorter side, find the sides of the field.A. Let the
shorter side of the rectangle bexm.Then, larger side of the
rectangle = (x+ 30) m. Diagonal is given to be 60 more than the
shorter side.
However, side cannot be negative. Therefore, the length of the
shorter side will be 90 m.Hence, length of the larger side will be
(90 + 30) m = 120 m.
Q7. The difference of squares of two numbers is 180. The square
of the smaller number is 8 times the larger number. Find the two
numbers.A. Let the larger and smaller number
bexandyrespectively.
According to the given question, However, the larger number
cannot be negative as 8 times of the larger number will be negative
and hence, the square of the smaller number will also be negative
which is not possible.Therefore, the larger number is 18. Thus, y2
= 144 => y = 12.Thus, the numbers are 18 & 12; or, 18 &
-12. Q8. A train travels 360 km at a uniform speed. If the speed
had been 5 km/h more, it would have taken 1 hour less for the same
journey. Find the speed of the train.A. Let the speed of the train
bexkm/hr. Time taken to cover 360 km = 360/x hrsAccording to the
given question,
However, speed cannot be negative. Therefore, the speed of train
is 40 km/h.Q9. Two water taps together can fill a tank in9 3/8
hours. The tap of larger diameter takes 10 hours less than the
smaller one to fill the tank separately. Find the time in which
each tap can separately fill the tank.A. Let the time taken by the
smaller pipe to fill the tank bexhr.Time taken by the larger pipe =
(x 10) hrPart of tank filled by smaller pipe in 1 hour = 1/xPart of
tank filled by larger pipe in 1 hour = 1 / (x - 10)It is given that
the tank can be filled in9 3/8 = 75/8hours by both the pipes
together. Therefore, Part of tank filled by both the pipes together
in 1 hour = 8/75Thus, 1/x + 1/(x - 10) = 8/75 => (x - 25)(8x -
30) = 0 => x = 25, or, x = 30/8Time taken by the smaller pipe
cannot be30/8 = 3.75 hours. As in this case, the time taken by the
larger pipe will be negative, which is logically not
possible.Therefore, time taken individually by the smaller pipe and
the larger pipe will be 25 and 25 10 =15 hours respectively. Q10.
An express train takes 1 hour less than a passenger train to travel
132 km between Mysore and Bangalore (without taking into
consideration the time they stop at intermediate stations). If the
average speed of the express train is 11km/h more than that of the
passenger train, find the average speed of the two trains.A. Let
the average speed of passenger train bexkm/h.Average speed of
express train = (x+ 11) km/hIt is given that the time taken by the
express train to cover 132 km is 1 hour less than the passenger
train to cover the same distance. Now, Speed cannot be
negative.Therefore, the speed of the passenger train will be 33
km/h and thus, the speed of the express train will be 33 + 11 = 44
km/h.
Q11. Sum of the areas of two squares is 468 m2. If the
difference of their perimeters is 24 m, find the sides of the two
squares.A. Let the sides of the two squares bexm andym. Therefore,
their perimeter will be 4x and 4yrespectively and their areas will
bex2andy2respectively.It is given that, 4x 4y= 24 => xy= 6 =>
x=y+ 6
Thus, y = -18; y = 12 However, side of a square cannot be
negative.Hence, the sides of the squares are 12 m and (12 + 6) m =
18 m
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EXERCISE 4.4 (P. 91):Q1. Find the nature of the roots of the
following quadratic equations. If the real roots exist, find
them:A. For a quadratic equationax2+bx+c= 0,discriminantisb2
4ac.(A) Ifb2 4ac> 0 twodistinctreal roots(B) Ifb2 4ac= 0 two
equal real roots(C) Ifb2 4ac< 0 no real roots (i) 2x23x+ 5 =
0Comparing this equation withax2+bx+c= 0, we obtain, a= 2,b= 3,c=
5Discriminant =b2 4ac= ( 3)2 4 (2) (5) = 9 40 = 31Asb2 4ac< 0.
Therefore, no real root ispossiblefor the given equation. (ii) 3x2
- 43 x + 4 = 0Comparing this equation withax2+bx+c= 0, we obtain,
a= 3,b= - 43,c= 4Discriminant =b2 4ac= (- 43)2 4 (3) (4) = 48 48 =
0 Asb2 4ac= 0. Therefore, real roots exist for the given equation
and they are equal to each other.The roots are: -b/2a, and b/2a. [
x = -b/2a = -(- 43)/(2x3) = 2/3]; that is 2/3 & 2/3(iii) 2x2 -
6x + 3 = 0Comparing this equation withax2+bx+c= 0, we obtain, a=
2,b= 6,c= 3Discriminant =b2 4ac= ( 6)2 4 (2) (3) = 36 24 = 12Asb2
4ac> 0, therefore, distinct real roots exist for this equation
asfollows:Therefore, the roots are, (3 + 3)/2, or (3 - 3)/2Q2. Find
the values of k for each of the following quadratic equations, so
that they have two equal roots.A. If an equationax2+bx+c= 0 has two
equal roots, its discriminant (b2 4ac) will be 0.(i) 2x2+kx+ 3 =
0Comparing this equation withax2+bx+ c = 0, we obtain, a= 2,b=k,c=
3Discriminant =b2 4ac=(k)2 4(2) (3) = k2 24. For equal roots,
Discriminant = 0 => k2 24 = 0=> k2= 24 => k = 24 = 26(ii)
kx (x 2) + 6 = 0 => kx2 - 2kx + 6 + 0Comparing this equation
withax2+bx+c= 0, we obtain, a=k,b= 2k,c= 6Discriminant =b2 4ac= (
2k)2 4 (k) (6) = = 4k2 24kFor equal roots, b2 4ac= 0 => 4k2 24k=
0 => 4k(k 6) = 0=> Either 4k= 0 ork= 6 = 0 => k= 0 ork=
6However, ifk= 0, then the equation will not have the terms x2 and
x.Therefore, if this equation has two equal roots,kshould be 6
only. Q3. Is it possible to design a rectangular mango grove whose
length is twice its breadth, and the area is 800m2? If so, find its
length and breadth.A. Let the breadth of mango grove bex. Thus,
length of mango grove = 2x.Area of mango grove = (2x)(x) =
2x2.Given, 2x2 = 800 => x2 - 400 = 0.Comparing this
withax2+bx+c= 0, we obtain, a= 1,b= 0,c= 400Discriminant =b2 4ac=
(0)2 4 (1) ( 400) = 1600Here,b2 4ac> 0 => The equation will
have real roots. And hence, the desired rectangular mango grove can
be designed.x = 20. However, length cannot be negative.Therefore,
breadth of mango grove = 20 mLength of mango grove = 2 20 = 40 mQ4.
Is the following situation possible? If so, determine their present
ages. The sum of the ages of two friends is 20 years. Four years
ago, the product of their ages in years was 48.A. Let the age of
onefriendbexyears => Age of the other friend = (20 x) years.Four
years ago, age of 1stfriend = (x 4) yrs; &, age of 2ndfriend =
(20 x 4)= = (16 x) yrsGiven that, (x 4) (16 x) = 48 => 16x- x2 -
64 + 4x= 48 => x2+ 20x 112 = 0=> x2 20x+ 112 = 0. Comparing
this equation withax2+bx+c= 0, we obtain,a= 1,b= 20,c=
112Discriminant =b2 4ac= ( 20)2 4 (1) (112) = 400 448 = 48Asb2
4ac< 0, no real root is possible for this equation and hence,
this situation is not possible.Q5. Is it possible to design a
rectangular park of perimeter 80 m and area 400 m2? If so, find its
length and breadth.A. Let the length and breadth of the park
bexandy.Perimeter = 2(x+y) = 80 => x+y= 40 => y = 40 - xArea
=xy=x(40 x) => Area = 40xx2 As per the question, 40xx2= 400
=> x2 - 40x + 400 = 0Comparing this equation withax2+bx+c= 0, we
obtain, a= 1,b= 40,c= 400Discriminate =b2 4ac= ( 40)24 (1) (400) =
1600 1600 = 0Asb2 4ac= 0, therefore, this equation has equal real
roots. And hence, this situation is possible.Root of this equation
are: x = -b/2a = - (-40)/2(1) = 20.Therefore, length of park,x= 20
m, and, breadth of park,y= 40 x= 40 20 = 20CBSE CLASS X
MATHEMATICSCHAPTER 5 ARITHMETIC PROGRESSIONS1. Introduction: Some
of the patterns observed in everyday life are such that, the
succeeding terms are obtained by adding a fixed number, or by
multiplying with a fixed number, or we find that they are squares
of consecutive numbers, and so on.Here-in, we shall learn about a
pattern in which succeeding terms are obtained by adding a fixed
number to the preceding terms. And, also, to find their nth terms
and the sum of n consecutive terms, and use this knowledge in
solving some daily life problems.2. Arithmetic Progressions:
Consider the following lists of numbers :1, 2, 3, 4, . . .100, 70,
40, 10, . . .Each of the numbers in the list is called a term. In
each of the above two lists, we see that successive terms are
obtained by adding a fixed number to the preceding terms. Such list
of numbers is said to form an Arithmetic Progression (AP). So, an
arithmetic progression is a list of numbers in which each term is
obtained by adding a fixed number to the preceding term except the
first term.This fixed number is called the common difference of the
AP, which can be positive, negative or zero.Let us denote the first
term of an AP by a1, second term by a2, . . ., nth term by an and
the common difference by d. Then the AP becomes a1, a2, a3, . . .,
an.So, a2 a1 = a3 a2 = . . . = an an - 1 = d.a, a + d, a + 2d, a +
3d, . . . represents an arithmetic progression where a is the first
term and d the common difference. This is called the general form
of an AP.A finite AP has finite number of terms (The last term is
known). An infinite Arithmetic Progression does not have a last
term.3. nth term of an A.P.: Let a