Class 10 Maths Full Syllabus Practice Paper

Class 10 - Mathematics 2 3 4 3 2 [1]

1. If a = (2 × 3 × 5 ) and b = (2 × 3 × 5) then HCF (a, b) = ? a) 540 b) 360 c) 180 d) 90 –

2. 2 - √ 3 is [1] a) an integer b) a whole number c) an irrational number d) a rational number 3 a [1]

3. If the prime factorisation of 2520 is 2 × 3 × b × 7, then the value of a + 2b is: a) 9 b) 10 c) 7 d) 12 x y z [1]

4. If 3825 = 3 × 5 × 17 , then the value of x + y - 2z is: a) 1 b) 3 c) 0 d) 2

5. The prime factorisation of 1728 is [1] 5 4 6 2 a) b) 2 × 3 2 × 3 c) 5 3 d) 6 3 2 × 3 2 × 3 2 2

6. If p and p are two odd prime numbers such that p 1 > p 2 , then p − p is [1] 1 2 1 2 a) an odd prime number b) a prime number c) an odd number d) an even number

7. Prime factorisation of 424 is: [1] a) 3 b) 4 2 × 53 2 × 53 c) 2 × 53 × 2 d) 2 × 53 × 4

8. If HCF (26,169) = 13, then LCM (26,169) = [1] a) 338 b) 52 c) 13 d) 26

9. If the HCF of 72 and 234 is 18, then the LCM (72, 234) is: [1] a) 936 b) 836 c) 324 d) 234 1

10. Four different electronic devices make a beep after every 30 minutes, 1 hour, 1 hour and 1 hour 45 minutes [1] 2 respectively. All the devices beeped together at 12 noon. They will again beep together at ________. a) 3 a.m. b) 12 midnight c) 9 a.m. d) 6 a.m. 11 . The LCM of two numbers is 1200. Which of the following cannot be their HCF? [1] a) 500 b) 200 c) 400 d) 600 12 . 7 × 11 × 13 + 13 is a/an: [1] a) odd number but not composite b) composite number c) prime number d) square number 13 . The LCM of smallest 2-digit number and smallest composite number is [1] a) 20 b) 4 c) 12 d) 40 14 . Which of the followings is an irrational number? [1] – 2 2 – 1 a) ( √ 2 − 1 ) b) ( 2 √ 3 − ) √ 3 – – ( √ 2 +5 √ 2 ) c) √ 2 − (2 + √ 2 ) d) √ 2 15 . The prime factorisation of the number 5488 is [1] a) 4 4 b) 4 3 2 × 7 2 × 7 3 4 3 3 c) d) 2 × 7 2 × 7 16 . A quadratic polynomial having zeroes -6 and 0 is: [1] a) 2 b) 2 6(x - x) x(x + 6) c) 2 d) 6x (x + 6) 6x - 1 2 [1] 17 . If -2 and 3 are the zeros of the quadratic polynomial x + (a + 1)x + b then a) a = -2, b = 6 b) a = -2, b = -6 c) a = 2, b = -6 d) a = 2, b = 6 – 2 [1] 18 . The sum of zeroes of the polynomial √ 2 x - 17 are given as:

17 √ 2 a) 0 b) 2 17 √ 2 c) 1 d) − 2 2 [1] 19 . If one root of the polynomial f(x) = 3x + 11x + p is reciprocal of the other, then the value of p is a) -3 b) 0 1 c) 3 d) 3 2 1 1 [1] 20 . If α and β are the zeros of the polynomial f(x) = x + px + q, then a polynomial having and is its zero is α β 2 2 a) b) qx + px + 1 x − px + q c) d) 2 2 x + qx + p px + qx + 1 2 [1] 21 . If α, β are the zeros of polynomial f(x) = x − p (x + 1) − c, then (α + 1) (β + 1) = a) c − 1 b) c c) 1 − c d) 1 + c 22 . The number of polynomials having zeros 1 and -2 is [1] a) more than 3 b) 2 c) 3 d) 1 2 [1] 23 . The zeroes of the polynomial 3x + 11x - 4 are:

−1 1 a) , -4 b) , -4 3 3 −1 1 c) , 4 d) , 4 3 3 24 . A quadratic polynomial with sum and product of its zeros as 8 and -9 respectively is [1] a) 2 b) 2 x + 8x - 9 x - 8x - 9 2 2 c) d) x - 8x + 9 x + 8x + 9 – 1 25 . A quadratic polynomial whose product and sum of zeroes are and √ 2 respectively is [1] 3 – – a) 2 b) 2 3x - x + 3 √ 2 x 3x + x - 3 √ 2 x – – c) 2 d) 2 3x + 3 √ 2 x + 1 3x - 3 √ 2 x + 1 2 [1] 26 . If one root of the polynomial p(y) = 5y + 13y + m is reciprocal of other, then the value of m is a) 5 b) 6 1 c) d) 0 5 α β 2 [1] 27 . If α and β are the zeroes of the polynomial ax + bx + c, then the value of + is α β 2 2 b c a) b) ac ab 2 2 b −2 ac a c) d) ac bc 2 2 2 [1] 28 . If α and β are the zeroes of the quadratic polynomial p(x) = x - ax - b, then the value of α + β is:

a) 2 b) 2 b + 2a a - 2b c) 2 d) 2 b - 2a a + 2b 2 [1] 29 . The sum and product of the zeroes of the polynomial x - 6x + 8 are respectively −3 a) 6 and 8 b) and – 1 2 −3 3 c) and 1 d) and 1 2 2 30 . The polynomial having zeroes -3 and 4 is: [1] a) 2 b) 2 x - x - 12 x - 2x + 1 c) 2 d) 2 x + 2x - 1 x + 2x + 1 2 31 . If one zero of the quadratic polynomial x + 3 x + k is 2, then the value of ‘k’ is [1] a) – 10 b) 10 c) 5 d) – 5 3 −1 32 . A quadratic polynomial whose zeros are and , is [1] 5 2 a) 2 b) 2 10x - x -3 10x - x + 3 c) 2 d) 2 10x + x + 3 10x + x - 3 2 2 [1] 33 . If one zero of the polynomial f(x) = (k + 4)x + 13x + 4k is reciprocal of the other, then k = a) 1 b) −2 c) 2 d) −1 2 [1] 34 . The zeroes of the polynomial p(x) = x + 3x + 2 are given as.

a) -2, 1 b) 2, -1 c) 1, 2 d) -2, -1 2 [1] 35 . The sum and product of zeroes of the polynomial p(x) = 3x - 5x + 2 are 2 −5 −2 a) 1, b) , 3 3 3 −5 5 2 2 c) , d) , 3 3 3 3 36 . At the end of the year 2002, Sam was half as old as his grandfather. The sum of the years in which they were [1] born is 3854. Age of Sam at the end of year 2003 is ________. a) 36 years b) 51 years c) 50 years d) 35 years 37 . Graphically, the pair of linear equations 3x - y + 8 = 0 and 3x - y = 24 represents two lines which are: [1] a) intersecting exactly at two points b) coincident c) parallel d) intersecting exactly at one point y 2 y 2 x 1 x [1] 38 . If − + = 0 and + = 3 then 3 2 6 2 3 a) x = -2, y = 3 b) x = - 2, y = -3 c) x = 2, y = 3 d) x = 2, y = -3 39 . The value of k, if (6, k) lies on the line represented by x - 3y + 6 = 0, is [1] a) 4 b) -12 c) 12 d) -4 40 . The value of k for which the pair of equations kx = y + 2 and 6x = 2y + 3 has infinitely many solutions, [1] a) is k = -3 b) does not exist c) is k = 3 d) is k = 4 41 . The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction [1] 1 becomes , then the fraction is 2 8 5 a) b) 7 7 6 4 c) d) 7 7 2 2 [1] 42 . The solution of the pair of equations x + y = a + b and ax - by = a - b is:

a) x = -a, y = b b) x = b, y = a c) x = a, y = b d) x = a, y = -b 43 . The area of the triangle formed by the lines 2x + 3y = 12 with the co – ordinate axis is [1] a) 12 sq. units b) 20 sq. units c) 10 sq. unit d) 16 sq. units 44 . Which of the following graphs represent the lines 2x + 4y = 8 and 3x - 4y = 12? [1] a) b) c) d) 1 45 . In a given fraction, if 1 is subtracted from the numerator and 2 is added to the denominator, it becomes . If 7 is [1] 2 1 subtracted from the numerator and 2 is subtracted from the denominator, it becomes . The fraction is 3 15 13 a) b) 26 24 16 16 c) d) 21 27 46 . If (-3, 2) is a solution of the linear equation 5x + 3 ky = 3, then the value of k is ________. [1] a) 6 b) 3 c) 5 d) 2 47 . The area of the triangle formed by the lines x = 3, y = 4 and x = y is [1] a) 3sq. unit b) 1/2 sq. unit c) 2sq. unit d) 1 sq. unit 48 . If a pair of linear equation is consistent, then the lines will be [1] a) parallel b) intersecting or coincident c) always coincident d) always intersecting 49 . Graphically, the pair of equations -6x - 2y = 21 and 2x - 3y + 7 = 0 represents two lines which are: [1] a) intersecting exactly at two points b) intersecting exactly at one point c) coincident d) parallel 50 . The pair of linear equations x + 2y + 5 = 0 and -3x - 6y + 1 = 0 has: [1] a) a unique solution b) exactly two solutions c) infinitely many solutions d) no solution 51 . If the lines represented by equations 3x + 2my = 2 and 2x + 5y + 1 = 0 are parallel, then the value of m is: [1] 3 15 a) b) 2 4 5 2 c) − d) 4 5 52 . If a pair of linear equations in two variables is consistent, then the lines represented by two equations are [1] a) always coincident b) intersecting or coincident c) always intersecting d) parallel 53 . If the pair of equations 3x - y + 8 = 0 and 6x - ry + 16 = 0 represent coincident lines, then the value of r is: [1] 1 1 a) − b) 2 2 c) -2 d) 2 54 . The value of k for which the pair of linear equations 5x + 2y - 7 = 0 and 2x + ky + 1 = 0 don't have a solution, is: [1] 5 a) 5 b) 4 4 5 c) d) 5 2 55 . If the system 6x – 2y = 3, kx – y = 2 has a unique solution, then [1] a) k = 3 b) k ≠ 3 c) k ≠ 4 d) k = 4 56 . Which of the given is a quadratic equation? [1] 1 2 2 2 a) x + = x b) 2 x − 5 x = ( x − 1 ) x 2 − − 1 c) x − 3 x + 2 = 0 d) x + = 5 √ 2 x 2 [1] 57 . If p and q are the roots of the equation x + px + q = 0, then a) p = - 2, q = 0 b) b = 0, 9 = 1 c) p = 1, q = - 2 d) p = - 2, q = l 2 [1] 58 . The roots of the quadratic equation x - 4 = 0 is/are:

a) 2 only b) -2, 2 c) -4, 4 d) 4 only 2 [1] 59 . If α and β are the roots of ax + bx + c = 0, then the wrong statement is c b a) αβ = b) α + β = a a 2 2 2 b −2 ac 1 1 − b c) d) + = α + β = 2 α β c a 2 [1] 60 . If x + 5kx + 16 = 0, has equal roots, then the value of k is 25 64 a) ± b) ± 64 25 8 5 c) ± d) ± 5 8 2 [1] 61 . If the roots of 5x -kx + 1 = 0 are real and distinct then – – – a) −2 √ 5 < k < 2 √ 5 b) k 2 √ 5 or k 2 √ 5 only 2 [1] 62 . If the equation x + 2(k + 2)x + 9k = 0 has equal roots then k = ?

a) -1 or 4 b) 1 or - 4 c) 1 or 4 d) -1 or - 4 63 . The perimeter of a right triangle is 70cm and its hypotenuse is 29cm. The area of the triangle is [1] a) 200 sq.cm b) 180 sq.cm c) 210 sq.cm d) 250 sq.cm 2 [1] 64 . If one root of the equation x + ax + 3 = 0 is 1, then its other root is a) 2 b) 3 c) -3 d) -2 65 . Which of the following is not a quadratic equation? [1] 2 2 2 2 a) 2( x − 1 ) = 4 x − 2 x + 1 b) x = x + 3 + 4 x – – 2 2 2 2 2 c) ( √ 2 x + √ 3 ) + x = 3 x − 5 x d) 2 x − x = x + 5 2 [1] 66 . The perimeter of a rectangle is 82 m and its area is 400 m . The breadth of the rectangle is a) 20 m b) 9 m c) 16 m d) 25 m 2 [1] 67 . The value(s) of k for which the quadratic equation 3x - kx + 3 = 0 has equal roots, is (are) a) -6 b) ±6 c) 6 d) 9 2 [1] 68 . The values of k for which the quadratic equation 2x – kx + k = 0 has equal roots is a) 0, 8 b) 8 only c) 0 only d) 4 69 . Which of the following equations has 2 as a root? [1] 2 2 a) b) x + 3x – 12 = 0 x – 4x + 5 = 0 c) 2 d) 2 2x – 7x + 6 = 0 3x – 6x – 2 = 0 2 [1] 70 . 3x + 2x - 1 = 0 have a) No Real roots b) Real roots c) real and equal root d) Real and Distinct roots 2 [1] 71 . If p = -7 and q = 12 and x + px + q = 0, Then the value of x is a) -3 and 4 b) -3 and -4 c) 3 and 4 d) 3 and -4 2 2 2 [1] 72 . (x + 1) – x = 0 has a) two real roots b) no real roots c) one real root. d) four real roots 2 [1] 73 . The ratio of the sum and product of the roots of the quadratic equation 5x - 6x + 21 = 0 is:

a) 5 : 21 b) 21 : 5 c) 7 : 2 d) 2 : 7 1 5 2 [1] 74 . If is a root of the equation x + kx – = 0, then the value of k is 2 4 1 a) b) 2 4 1 c) -2 d) 2 2 [1] 75 . The roots of the quadratic equation ax + bx + c = 0 are real and distinct, if: a) 2 b) 2 b - 4ac = 0 b - 4ac > 0 2 2 c) d) b - 4ac ≥ 0 b - 4ac < 0

Solution

Class 10 - Mathematics 1 .
(c) 180 Explanation: 2 3 4 3 2 It is given that: a = (2 × 3 × 5 ) and b = (2 × 3 × 5) 2 2 ∴ HCF (a, b) = Product of smallest power of each common prime factor in the numbers = 2 × 3 × 5 = 180 2 .
(c) an irrational number Explanation: – Let 2 - √ 3 be rational number – p 2 - √ 3 = where p and q are composite numbers q – p √ 3 = + 2 q – ( p +2 q ) √ 3 = q ( p +2 q ) since p, q are integers, so is rational q – ∴ √ 3 is an irrational number it shows our supposition was wrong – hence 2- √ 3 is an irrational number.

3 .
(d) 12 Explanation: 3 2 2520 = 2 × 3 × 5 × 7 on comparing a = 2, b = 5 So, a + 2b = 2 + 2 × 5 = 12 4 .
(d) 2 Explanation: 2 2 3825 = 3 × 5 × 17 On comparing x = 2, y = 2, z = 1 x + y - 2z = 2 + 2 - 2 × 1 = 4 - 2 = 2 5 . 6 3
(d) 2 × 3 Explanation: 6 3 2 × 3 6 .
(d) an even number Explanation: Let p and p be 5 two odd primes. 1 2 Then, 2 2 p − p = ( p 1 − p 2 )( p 1 + p 2 ) 1 2 We know that sum and difference of two odd numbers is even ∴ ( p − p ) and ( p + p ) are even numbers. 1 2 1 2 Also, we know that product of even numbers is an even number, therefore 2 2 p − p = ( p 1 − p 2 )( p 1 + p 2 ) , is an even number.

1 2 3 7 .
(a) 2 × 53 Explanation: 2 424 2 212 2 106 53 53 1 3 424 = 2 × 53 8 .
(a) 338 Explanation: HCF (26, 169) = 13 We have to find the value for LCM (26, 169) We know that the product of numbers is equal to the product of their HCF and LCM. Therefore, 13(LCM) = 26(169) 26(169) LCM = 13 LCM = 338 9 .
(a) 936 Explanation: (72×234) LCM (72, 234) = = 936 18 Therefore, the LCM of (72, 234) is 936. 10 .
(c) 9 a.m. Explanation: 2 2 L.C.M. (30, 60, 90, 105) = 2 × 3 × 5 × 7 = 1260 mins = 21 hours 11 .
(a) 500 Explanation:

It is given that the LCM of two numbers is 1200 . We know that the HCF of two numbers is always the factor of LCM. 500 is not the factor of 1200. So this cannot be the HCF. 12 .
(b) composite number Explanation: We have 7 × 11 × 13 + 13 = 13 (77 + 1) = 13 × 78. Since the given number has 2 more factors other than 1 and itself, therefore it is a composite number. 13 .
(a) 20 Explanation: As we know, the smallest two-digit number is 10 and the smallest composite number is 4. By prime factorisation, we get; 4 = 2 × 2 10 = 2 × 5 Now, LCM of 4, 10 = 2 × 2 × 5 = 20 Therefore, the LCM of the smalles two-digit number and the smallest composite number is 20.

– 2 14 .
(a) ( √ 2 − 1 ) Explanation: – 2 ( √ 2 − 1 ) 15 . 4 3
(b) 2 × 7 Explanation: 4 3 2 × 7 16 .
(d) 6x (x + 6) Explanation: 6x (x + 6) 17 .
(b) a = -2, b = -6 Explanation: α + β = 3 + (−2) = 1 and αβ = 3 × (−2) = −6 ∴ -(a + 1) = 1 ⇒ a + 1 = -1 ⇒ a = -2 Also, b = -6 18 .
(a) 0 Explanation: – 2 Given; P(x) = √ 2 x - 17 coeff of x Sum of zeroes = coeff of x 2 = 0 19 .
(c) 3 Explanation: Let one root be q. 1 ∴ Other root = q p p 1 ⇒ q × = ⇒ 1 = ⇒ p = 3 q 3 3 2 20 .
(a) qx + px + 1 Explanation: 2 Let α and β be the zeros of the polynomial f ( x ) = x + px + q .Then, − Coefficient of x p α + β = = − = − p Coefficient of x 2 1 Constant term q And αβ = = = q 2 1 Coefficient of x 1 1 Let S and R denote respectively the sum and product of the zeros of a polynomial whose zeros are and , then α β α + β − p 1 1 S = + = = α q β αβ 1 1 1 1 R = × = = α β αβ q Hence, the required polynomial g ( x ) whose sum and product of zeros are S and R is given by 2 x − Sx + R = 0 P 1 2 x + x + = 0 q q 2 q x + Px +1 = 0 q 2 ⇒ q x + px + 1 2 So g ( x ) = q x + px + 1 21 .

(c) 1 − c Explanation: 2 Since α and β are the zeros of quadratic polynomial f ( x ) = x − p ( x + 1) − c 2 = x − px − p − c − Coefficient of x α + β = 2 Coefficient of x − p = − ( ) = p 1 Constant term α × β = 2 Coefficient of x − p − c = = − p − c 1 We have ( α + 1)( β + 1) = αβ + β + α + 1 = αβ + ( α + β ) + 1 = − p − c + ( p ) + 1 = − c + 1 = 1 - c The value of ( α + 1)( β + 1) is 1 - c. 22 .
(a) more than 3 Explanation: Since, 1 and -2 Sum of Roots = 1+(-2) = -1 Product of roots = (1) (-2) = -2 2 Therefore, the polynomial (p(x)) is: [p(x) =K[x - (sum of roots)x + product of roots] 2 p(x) = K [ x - (-1)x+ (-2)] Therefore, There are infinitely many polynomials that can have (1) and (-2) as their zeros. We can multiply or divide the polynomial by any nonzero constant(k), and the zeros will remain the same. So, the required number of polynomials is infinite !

23 . 1
(b) , -4 3 Explanation: 2 Let f(x) = 3x + 11x - 4 2 f(x) = 3x + 12x - x - 4 f(x) = 3x(x + 4) - 1 (x + 4) f(x) = (x + 4)(3x - 1) Put both the factors equal to zero. x + 4 = 0, x = -4 1 3x - 1 = 0, x = 3 1 2 The zeroes of the polynomial 3x + 11x - 4 are and - 4. 3 24 . 2
(b) x - 8x - 9 Explanation: Given, α + β = 8 αβ = -9 2 p(x) = k(x - ( α + β )x + αβ ) 2 = k(x - (8)x + (-9)) 2 = k(x - 8x - 9) for k = 1, 2 p(x) = x - 8x - 9 25 . – 2
(d) 3x - 3 √ 2 x + 1 Explanation: √ 2 −(− √ 2 ) −(−3 √ 2 ) Given: α + β = = = 1 1 3 c 1 – And αβ = = On comparing, we get, a = 3, b = −3 √ 2 , c = 1 a 3 2 Putting these values in the general form of a quadratic polynomial ax + bx + c, 2 – we have 3x - 3 √ 2 + 1 26 .
(a) 5 Explanation:

2 p(y) = 5y + 13y + m. Given one root of p(x) is reciprocal of other 1 i.e. If α = a then β = a − b sum of roots ( α + β ) = a 1 13 a + = − a 5 c Product of roots ( α ⋅ β ) = a 1 m a ⋅ = . a 5 m 1 = . 5 m = 5 27 . 2 b −2 ac
(c) ac Explanation: Since 2 α 2 + β = αβ 2 ( α + β ) −2 αβ = αβ 2 − b c ( ) −2× a a = c a 2 b 2 c − a a 2 = c a 2 b −2 ac a = × 2 a c 2 b −2 ac = ac 28 . 2
(d) a + 2b Explanation: 2 Given, P(x) = x - ax - b α + β = a, αβ = -b 2 2 2 ( α + β ) = α + β + 2 αβ 2 2 2 a = α + β - 2b 2 2 2 α + β = a + 2b 29 .
(a) 6 and 8 Explanation:

− b 6 Sum of the zeroes of the polynomial = = = 6 a 1 c 8 And Product of the zeroes of the polynomial = = = 8 a 1 2 30 .
(a) x - x - 12 Explanation: A quadratic polynomial is always in the form of 2 x - (sum of zeros)x + (product of Zeros) hence the required polynomial is 2 x - (1)x + (-12) 2 = x - x - 12 31 .
(a) – 10 Explanation: 2 Given Polynomial is p ( x ) = x + 3 x + k According to question, p ( x ) = 0 (Put x = 2) p(2) = 0 2 ⇒ (2) + 3 × 2 + k = 0 ⇒ 4 + 6 + k = 0 ⇒ k = −10 2 32 .
(a) 10x - x -3 Explanation:

3 1 1 3 −1 −3 α + β = ( − ) = , αβ = × ( ) = 5 2 10 5 2 10 2 1 3 2 Required polynomial is x − x − , i.e., 10x - x - 3 10 10 33 .
(c) 2 Explanation: 2 2 We are given f ( x ) = ( k + 4 ) x + 13 x + 4 k then − Coefficient of x α + β = Coefficient of x 2 −13 = 2 k +4 Constant term α × β = 2 Coefficient of x 4 k = 2 k +4 One root of the polynomial is reciprocal of the other. Then, we have α × β = 1 4 k ⇒ = 1 2 k +4 2 ⇒ ( k − 2 ) = 0 2 ⇒ k − 4 k + 4 = 0 ⇒ k = 2 34 .
(d) -2, -1 Explanation: 2 P(x) = x + 3x + 2= 0 2 x + 2x + x + 2 = 0 x(x + 2) + 1(x + 2) = 0 (x + 1)(x + 2) = 0 x = -1, -2 hence, -1 & -2 are the zero of P(x) 35 .

5 2
(d) , 3 3 Explanation: Let α , β be the zero of Polynomial P(x) P(x) = 3x2 - 5x + 2 −(−5) − b 5 α + β = = = a 3 3 c 2 α β = = a 3 36 .
(b) 51 years Explanation: Let the year in which Sam was born be x and the year in which Sam's grandfather was born be y. 2002− y Then, according to question, 2002 - x = 2 ⇒ 2x - y = 2002 ...(i) and x + y = 3854 ....(ii) Solving (i) and (ii), we get ⇒ x = 1952 Thus in 2003, Sam's age would be 2003 - 1952 = 51 yrs 37 .
(c) parallel Explanation: parallel 38 .
(c) x = 2, y = 3 Explanation:

We have, y 2 x 1 − = − …(i) 3 2 6 x 2 y + = 3 …(ii) 2 3 Now, multiplying (i) and (ii) by 6 we get: 4x - 3y = - 1 …(iii) 3x + 4y = 18 …(iv) Now, multiplying (iii) by 4 and (iv) by 3 and adding them we get: 16x + 9x = - 4 + 54 50 x = = 2 25 Putting the value of x in (iv) we get: 3 × 2 + 4y = 18 18−6 y = 4 y = 3 39 .
(a) 4 Explanation: x - 3y + 6 = 0 6 - 3k + 6 = 0 ⇒ k = 4 40 .
(b) does not exist Explanation: does not exist 41 . 5
(b) 7 Explanation: x Let the fraction be y Where x is numerator and y be denominator.

ATQ. x + y = 12 ...(i) again New denominator is y + 3 x 1 ATQ. = y +3 2 ⇒ 2x = y + 3 using 2x - y = 3 ...(ii) By, Elimination method Add eq (i) & (ii) we get 15 x = 3 x = 5 put the value of x in eq. (i) we get 5 + y = 12 y = 12 - 5 = 7. y = 7 Hence Numerator = 5 denominator = 7. 5 fraction is . 7 42 .
(c) x = a, y = b Explanation: The given equations are x + y = a + b ...(i) 2 2 ax - by = a - b ...(ii) From (i) y = a + b - x Substituting y = a + b - x in (ii), we get 2 2 ax - b(a + b - x) = a - b 2 2 2 ⇒ ax - ab - b + bx = a - b 2 a + ab ⇒ x = = a a + b Now, substitute x = a in (i) to get a + y = a + b ⇒ y = b Hence, x = a and y = b.

43 .
(a) 12 sq. units Explanation: The triangle formed by the lines 2 x + 3 y = 12 with co-ordinate axes is shaded. The area of the shaded region, i.e., 2 x + 3 y = 12 1 Triangle OAB = × OA × AB 2 1 = × 6 × 4 = 12 sq. units 2 x 0 3 6 y 4 2 0 44 .
(d) Explanation: 15 45 .
(a) 26 Explanation: x Let the fraction be y According to the question, ( x −1) 1 = ( y +2) 2 2x - 2 = y + 2 y = 2x - 4 …(i) And, ( x −7) 1 = ( y −2) 2 3x - 21 = y - 2 3x = y + 19 …(ii) Using (i) in (ii) 3x = 2x - 4 + 19 x = 15 Using value of x in (i), we get y = 2 (15) - 4 y = 30 - 4 y = 26 15 Therefore, required fraction = 26 46 .

(b) 3 Explanation: Since, (-3,2) is the solution of 5x + 3/cy = 3. So (-3, 2) satisfies it. ∴ 5 x (-3) + 3 18 ⇒ −15 + 6 k = 3 ⇒ k = = 3 6 47 .
(b) 1/2 sq. unit Explanation: Given x = 3, y = 4 and x = y We have plotting points as (3,4), (3,3), (4,4) when x = y 1 1 1 1 Therefore, area of △ ABC = (Base × Height) = (AB × AC) = (1 × 1) = 2 2 2 2 1 Area of triangle ABC is square units. 2 48 .
(b) intersecting or coincident Explanation: If a consistent system has an infinite number of solutions, it is dependent. When you graph; the equations, both equations represent the same line. So for consistent line it has to be parallel or even they intersect at one point. If a system has no

solution, it is said to be inconsistent. The graphs of the lines do not intersect, so the graphs are parallel and there is no solution.

49 .
(b) intersecting exactly at one point Explanation: intersecting exactly at one point 50 .
(d) no solution Explanation: Here, a 1 = 1, b 1 = 2, c 1 = 5 a = -3, b = -6, c = 1 2 2 2 a 1 1 1 So, = = -( ) a 2 −3 3 b 1 2 1 = = -( ) −6 3 b 2 c 1 5 = c 2 1 a 1 b 1 c 1 = ≠ a 2 b c 2 2 Therefore, the pair of equations has no solution. 51 . 15
(b) 4 Explanation: Condition for the lines to be parallel is a b c 1 1 1 = ≠ a 2 b c 2 2 Here the equations are 3x + 2my = 2 and 2x + 5y + 1 = 0 So, a = 3, b = 2m, c = -2 and a = 2, b = 5, c = 1 1 1 1 2 2 2 a 1 3 b 1 2m c 1 −2 ∴ = , = and = = -2 a 2 2 5 c 2 1 b 2 2m 3 ∴ = 5 2 15 ∴ m = 4 52 .

(b) intersecting or coincident Explanation: If a pair of linear equations in two variables is consistent, then its solution exists. ∴ The lines represented by the equations are either intersecting or coincident. 53 .
(d) 2 Explanation: a b c 1 1 1 = = a 2 b 2 c 2 3 −1 8 ⇒ = = 6 − k 16 Taking, 3 −1 = 6 −k 1 1 ⇒ = 2 k ⇒ k = 2 −1 8 = −k 16 1 1 ⇒ = k 2 ⇒ k = 2 So, the answer is k = 2 54 . 4
(c) 5 Explanation: For no solution a 1 b 1 c 1 = ≠ a 2 b c 2 2 5 2 = 2 k 4 k = 5 55 .
(b) k ≠ 3 Explanation: a b 1 1 If the system has a unique solution, then ≠ a 2 b 2 Here a = 6, a = k , b = −2 1 2 1 and b 2 = −1 6 −2 ∴ ≠ ⇒ 3 k ≠ 6 ⇒ k ≠ 3 k −1 2k ≠ 6 k ≠ 3 56 .

2 2
(b) 2 x − 5 x = ( x − 1 ) Explanation: 2 2 2 2 2 2 x − 5 x = ( x − 1 ) using ( a − b ) = a + b − 2 ab 2 2 2 x − 5 x = x − 2 x + 1 2 2 2 x − 5 x − x + 2 x − 1 = 0 2 x − 3 x − 1 = 0 a = 1, b = -3 and c = -1 2 This is of the form ax + bx + c = 0 i.e. of degree 2(a ≠ 0, a, b, c are real numbers) Hence this is a quadratic equation. 57 .
(c) p = 1, q = - 2 Explanation: Given sum of roots, S = p + q = – p and product pq = q ⇒ q(p – 1) = 0 i.e. q = 0 or p = 1 Now If q = 0 then p = 0, this implies p = q If p = 1, then p + q = – p q = – 2p q = – 2(1) q = – 2 58 .

(b) -2, 2 Explanation: 2 x - 4 = 0 2 x = 4 x = ± 2 roots are +2, -2 59 . b
(b) α + β = a Explanation: 2 If α and β are the roots of ax + bx + c = 0, − b then α + β = a 60 . 8
(c) ± 5 Explanation: Here, a = 1, b = 5k, c = 16 2 If x + 5kx + 16 = 0 has equal roots, 2 then, b - 4ac = 0 2 ⇒ (5k) - 4 × 1 × 16 = 0 2 ⇒ 25k - 64 = 0 2 ⇒ 25k = 64 2 64 ⇒ k = 25 8 ⇒ k = ± 5 61 . – –
(c) either k > 2 √ 5 or k 0 ⇒ k − 20 > 0 – – This gives; k 2 √ 5 62 .

(c) 1 or 4 Explanation: Since the roots are equal, we have D = 0. 2 2 ∴ 4(k + 2) - 36k = 0 ⇒ {k + 2) - 9k = 0 2 2 k - 5k + 4 = 0 ⇒ k - 4k - k + 4 = 0 ⇒ k ( k − 4) − ( k − 4) = 0 ⇒ ( k − 4)( k − 1) = 0 ⇒ k = 4 or k = 1 . 63 .
(c) 210 sq.cm Explanation: Let base of the right triangle be x cm. Given: Perpendicular = x + 29 = 70 ⇒ Perpendicular = (41 − x ) cm Now, using Pythagoras theorem, 2 2 2 (29) = x + (41 − x ) 2 2 ⇒ 841 = 1681 + x − 82 x + x 2 ⇒ 2 x − 82 x + 840 = 0 2 ⇒ x − 41 x + 420 = 0 2 ⇒ x − 20 x − 21 x + 420 = 0 ⇒ x ( x − 20) − 21 ( x − 20) = 0 ⇒ ( x − 20) ( x − 21) = 0 ⇒ x − 20 = 0 and x − 21 = 0 ⇒ x = 20 and x = 21 Therefore, the two sides other than hypotenuse are of 20 cm and 21 cm.

1 1 ∴ Area of right triangle = × Base × Perpendicular = × 20 × 21 = 210 sq. cm 2 2 64 .
(b) 3 Explanation: 2 The given equation is x + ax + 3 = 0 One root = 1 c 3 and product of roots = = = 3 a 1 3 Second root = = 3 1 65 . – – 2 2 2
(c) ( √ 2 x + √ 3 ) + x = 3 x − 5 x Explanation: – – 2 2 2 In equation ( √ 2 x + √ 3 ) + x = 3 x − 5 x – 2 2 2 ⇒ 2 x + 3 + 2 √ 6 x + x = 3 x − 5 x – 2 2 ⇒ 3 x − 3 x + 5 x + 2 √ 6 x + 3 = 0 – ⇒ ( 5 + 2 √ 6 ) x + 3 = 0 It is not the quadratic equation because its degree is not 2.

66 .
(c) 16 m Explanation: 2(l + b) = 82 ⇒ l + b = 41 ⇒ l = (41 - b). And, lb = 400 ⇒ (41 - b)b = 400 2 2 ⇒ b - 41b + 400 = 0 ⇒ b - 25b - 16b + 400 = 0 ⇒ b(b - 25) - 16(b - 25) = 0 ⇒ (b - 25)(b - 16) = 0 ∴ b = 25 or b = 16. But for b = 25 we have l = (41 - 25) = 16 < b. ∴ breadth = 16 m. 67 .
(b) ±6 Explanation: For equal roots D = 0 2 b - 4ac = 0 2 (-k) - 4(3)(3) = 0 2 k - 36 = 0 2 k = 36 k = ±6 68 .
(a) 0, 8 Explanation: 2 2 If a quadratic equation ax + bx + c = 0, a ≠ 0 has two equal roots, then its discriminant value will be equal to zero i.e., D = b - 4ac = 0 2 Given, 2x – kx + k = 0 For equal roots, 2 D = b - 4ac = 0 2 ⇒ (-k) - 4(2)(k) = 0 2 ⇒ k - 8k = 0 ⇒ k (k - 8) = 0 ∴ k = 0,8 69 .

2
(c) 2x – 7x + 6 = 0 Explanation: 2 Given, 2x - 7x + 6 = 0 If 2 satisfies the above equation then 2 is a root. 2 Now, 2(2) - 7(2) + 6 = 0 ∴ 2 is a root of this equation 70 .
(d) Real and Distinct roots Explanation: 2 D = b - 4ac 2 D = 2 - 4 × 3 × (-1) D = 4 + 12 D = 16 D > 0. Hence Real and distinct roots. 71 .
(c) 3 and 4 Explanation: Putting the values of p and q in given equation, we get 2 x + (-7)x + 12 = 0 2 ⇒ x - 7x + 12 = 0 2 ⇒ x - 4x - 3x + 12 = 0 ⇒ x(x - 4) - 3(x - 4) = 0 ⇒ (x - 3)(x - 4) = 0 ⇒ x - 3 =0 and x - 4 = 0 ⇒ x = 3 and x = 4 72 .

(b) no real roots Explanation: Let, x² = y, then our given equation become 2 ( y + 1 ) − y = 0 2 ⇒ y + y + 1 = 0 2 2 D = b − 4 ac = 1 − 4 × 1 × 1 = 1 − 4 = −3 < 0 Hence no real root. 73 .
(d) 2 : 7 Explanation: − b Sum of roots = a −(−6) 6 = = 5 5 c product of roots = a 21 = 5 ATQ 6 Sum of roots 5 6 = = 21 21 prod. of roots 5 2 = 7 74 .
(b) 2 Explanation: 1 5 1 2 If is a root of the equation x + kx - = 0 then, substituting the value of in place of x should give us the value of k. 2 4 2 5 1 2 Given, x + kx - = 0 where, x = 4 2 1 2 1 5 ⇒ ( ) + k ( ) − = 0 2 2 4 k 5 1 ⇒ = − 2 4 4 ∴ k = 2 75 .

2
(b) b - 4ac > 0 Explanation: 2 2 A quadratic equation ax + bx + c = 0 has real and distinct roots, if b - 4ac > 0.