Apply Euclid’s division lemma, to c and d [Given positive integers a and b, there exist unique integers q and r satisfying a = b*q + r, 0 ≤ r < b]. So, we find whole numbers, q and r such that c = d*q + r, 0 ≤ r < d.

If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.

Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

**Q.**Find the HCF of the integers 455 and 42

We start with the larger integer, that is, 455. 455 = 42 × 10 + 35

Consider the divisor 42 and the remainder 35, and apply the division lemma to get ; 42 = 35 × 1 + 7

Consider the divisor 35 and the remainder 7, and apply the division lemma to get 35 = 7 × 5 + 0

The remainder has become zero, and we cannot proceed any further. We claim that the HCF of 455 and 42 is the divisor at this stage, i.e., 7.

**Ans.**HCF = 7

**Q.**Find the HCF of 4052 and 12576.

Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get 12576 = 4052 × 3 + 420

Since the remainder 420 ≠ 0, we apply the division lemma to 4052 and 420, to get 4052 = 420 × 9 + 272

420 = 272 × 1 + 148

272 = 148 × 1 + 124

148 = 124 × 1 + 24

124 = 24 × 5 + 4

24 = 4 × 6 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.

**Ans.**HCF = 4

**Q.**A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to
stack them in such a way that each stack has the same number, and they take up the
least area of the tray. What is the maximum number of barfis that can be placed in
each stack for this purpose?

420 = 130 × 3 + 30

130 = 30 × 4 + 10

30 = 10 × 3 + 0

**Ans.**HCF of 420 and 130 is 10.

For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b. But product of three numbers is not equal to the product of their HCF and LCM

Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form p/q where p and q are coprime, and the prime factorization of q is of the form 2

^{n}5^{m}, where n, m are non-negative integers.LCM (p, q, r) = [ p*q*r* HCF(p,q,r) ] / [HCF(p,q) * HCF(q,r) * HCF(p,r)]

HCF (p, q, r) = [ p*q*r* LCM (p, q, r)] / [LCM(p,q) * LCM(q,r) * LCM(p,r)]

A real number k is said to be a zero of a polynomial p(x), if p(k) = 0.

In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x-axis at atmost n points. Therefore, a polynomial p(x) of degree n has at most n zeroes.

In general, if α and β are the zeroes of the quadratic polynomial p(x) = ax

^{2}+ bx +c = 0 then you know that (x – α) and ( x – β) are the factors of p(x). Therefore, (α + β) = -b/a and (α * β) = c/aIn general, if α , β and γ are the zeroes of the quadratic polynomial p(x) = ax

^{3}+ bx^{2}+cx+d = 0 then (α + β + γ) = -b/a and (α * β * γ) = -d/a

**Q.**Divide 3x^{3} + x^{2} + 2x + 5 by 1 + 2x + x^{2}

First arrange the terms of the dividend and the divisor in the decreasing order of their degrees. Recall that arranging the terms in this order is called writing the polynomials in standard form. In this example, the dividend is already in standard form, and the divisor, in standard form, is x

^{2}+ 2x+ 1To obtain the first term of the quotient, divide the highest degree term of the dividend (i.e., 3x

^{3}) by the highest degree term of the divisor (i.e., x^{2}). This is 3x. Then carry out the division process. What remains is – 5x^{2}– x + 5Now, to obtain the second term of the quotient, divide the highest degree termof the new dividend (i.e., –5x

^{2}) by the highest degree term of the divisor (i.e., x^{2}). This gives –5. Again carry out the division process with –5x^{2}– x + 5.What remains is 9x + 10. Now, the degree of 9x + 10 is less than the degree of the divisor x

^{2}+ 2x + 1. So, we cannot continue the division any further. So, the quotient is 3x – 5 and the remainder is 9x + 10.

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.

If the lines represented by the equation a

_{1}x + b_{1}y + c_{1}= 0 and a_{2}x + b_{2}y + c_{2}= 0are intersecting, then a

_{1}/ a_{2}≠ b_{1}/ b_{2}coincident, then a

_{1}/ a_{2}= b_{1}/ b_{2}= c_{1}/ c_{2}parallel, then a

_{1}/ a_{2}= b_{1}/ b_{2}≠ c_{1}/ c_{2}

**Q.**The sum of a two-digit number and the number obtained by reversing
the digits is 66. If the digits of the number differ by 2, find the number. How many such
numbers are there?

Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first number may be written as 1 0 x + y and its reverse as 10y + x According to the given condition.

(10x + y) + (10y + x) = 66

11(x + y) = 66 --- (1)

We are also given that the digits differ by 2, therefore, either x – y = 2 ---- (2) or y – x = 2 ---- (3)

f x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2. In this case, we get the number 42. If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4. In this case, we get the number 24. Thus, there are two such numbers 42 and 24.

**Ans.**42 and 24

For any pair of linear equations in two variables of the form: a

_{1}x+b_{1}y+c_{1}= 0 and a_{2}x+b_{2}y+c_{2}= 0To obtain the values of x and y as shown above:

x = ( b

_{1}c_{2}- b_{2}c_{1}) /( a_{1}b - a_{2}b_{1})y = ( c

_{1}a_{2}- c_{2}a_{1}) /( a_{1}b - a_{2}b_{1})

When ( a

_{1}/ a_{2}) ≠ (b_{1}/ b_{2}) ; we get a unique solution.When ( a

_{1}/ a_{2}) = (b_{1}/ b_{2}) = ( c_{1}/ c_{2}) there are infinitely many solutions.When ( a

_{1}/ a_{2}) = (b_{1}/ b_{2}) ≠ ( c_{1}/ c_{2}) there is no solution.

When an equation is given in the form below we can find x , y directly

a_{1} x + b_{1}y + c_{1} = 0

a_{2} x + b_{2} y + c_{2} = 0

**Q.** 5 / (x-1) + 1 / (y - 2) = 2

6 / (x-1) - 3 / (y - 2) = 1

we put 1 / (x-1) = p and 1 / (y - 2) = q

Now new equations can be written as : 5p + q = 2 and 6p – 3q = 1

By solving we get p = q = 1/3

so we have x=4; y=5

**Ans.**x=4; y=5

Thus, if b

^{2}- 4ac ≥ 0 then the roots of the quadratic equation ax^{2}+ bx + c = 0; are given byif b

^{2}- 4ac = 0 then the roots of the quadratic equation ax^{2}+ bx + c = 0; are given byx = (-b/2a)

b

^{2}- 4 a c is known as discriminant of the quadratic equation.