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Solution: Sequence of numerators ⇒1,3,5,7,9 Sequence of denominators ⇒2,4,8,16,32 :. Next fraction = 9/32 Then the option is (4) 9/32

Question: The area of a rectangle is x² + 12xy + 27y² and its length is (x + 9y). Find the breadth of the rectangle. (A) (X+2y)(B) (X+5y)(C) (X+3y)(D) (x-5y) Solution Area / Length= (x² + 12xy + 27y²) / (x + 9y)= x(x + 9y) + 3y(x + 9y) / (x + 9y)= (x

Question-Find the value of “p” from the polynomial x² + 3x + p, if one of the zeroes of the polynomial is 2.(A) 20(B) 30(C) -10(D) 10 Solution As 2 is the zero of the polynomial.We know that if α is a zero of the polynomial p(x), then p(α) = 0Substituting x = 2 in

Solution As 2 is the zero of the polynomial.We know that if α is a zero of the polynomial p(x), then p(α) = 0Substituting x = 2 in x2 + 3x + p,⇒ 22 + 3(2) + p = 0⇒ 4 + 6 + p = 0⇒ 10 + p = 0⇒ p = -10

(1) 13(2) 59(3) 60(4) 61 Solution

Solution Perimeter = 2 × (Length + Width)Perimeter = 2 × (12 units + 7 units)Perimeter = 2 × 19 unitsPerimeter = 38 units.

Solution Given AP: 21, 18, 15,…Here, a = 21,d = 18 – 21 = –3Let nth term of the given AP is -81.So, an = –81As we know,an = a + ( n – 1) dThus,– 81 = 21 + (n – 1)(– 3)– 81 = 24 – 3n– 105 = – 3nSo, n =

Solution Let the number be x.Hence, the two consecutive numbers are x and x+1.According to the given statement, the equation becomes⇒ x + x + 1 = 11⇒ 2x + 1 = 11⇒ 2x = 10⇒ x = 10/2 = 5If x = 5, then x + 1 = 5 + 1 = 6Hence, the

(1) 30cm2   (2) 60cm2 (3) 30√2cm2 (4) 60√2cm2 Solution: Semi-perimeter, s=9+10+11/2 = 15cm Area of triangle = √s(s-a)(s-b)(s-c) =√15(15-9)(15-10)(15-11) =√15*6*5*4 = 30√2cm2 Then the option is (3) 30√2cm2

(1) 0 (2) 1 (3) 2 (4) 5 Solution: (2b+1/b)2 = 4b2+1/b2+2*2b*1/b =2 + 4 = 6 ⇒ 2b+1/b =√6 .’. 8b3+1/b3  = (2b+1/b)3 -3*2b*1/b(2b+1/b) = (√6)3 -6(√6) = 6(√6) – 6(√6) =0 Then the option is (1)