New Enroll in RBI Grade B New Batch - Starts on May 21

    Question

    Read the information given below and answer the following questions  I. (a - 6)2 = (9a - P - a2) II. (P - b) 2 + (b + Q) 2 = 3(9b - Q) + 10 One root of equation 'I' is 2.5. Smaller root of equation II is (1/2) the larger root of equation I.

    What will be the product of smaller roots of both

    equations. 
    A 5 Correct Answer Incorrect Answer
    B 20 Correct Answer Incorrect Answer
    C 15 Correct Answer Incorrect Answer
    D 10 Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    Solution

    For I: (a - 6)2 = (9a - P - a2) Or, a2 + 36 - 12a = 9a - P - a2 Or, 2a2- 21a + (36 + P) = 0 Since, 2.5 is a root of the given equation, 2 × (2.5)2- (21 × 2.5) + (36 + P) = 0 2 × 6.25 - 52.5 + 36 + P = 0 Or, 12.5 - 52.5 + 36 + P = 0 Or, - 40 + 36 + P = 0 So, 'P' = 4 So, the given equation becomes, 2a2- 21a + 36 + 4 = 0 Or, 2a2- 21a + 40 = 0 So, product of the roots = (40/2) = 20 So, other root of equation I = 20 ÷ 2.5 = 8 So, smaller root of equation II = 8 ÷ 2 = 4 Now, for II: (P - b)2 + (b + Q)2 = 3(9b - B) + 10 Or, (4 - b)2 + (b + Q)2 = 27b - 3Q + 10 Or, 16 + b2- 8b + b2 + Q2 + 2Qb = 27b - 3Q + 10 Or, 2b2- 8b + 2Qb - 27b + 6 + 3Q + Q2 = 0 Or, 2b2- b(8 - 2Q + 27) + 6 + 3Q + Q2 = 0 Since, 4 is a root of the given equation, 2 × (4)2 - 4 × (35 - 2Q) + 6 + 3Q + Q2 = 0 Or, (2 × 16) - 140 + 8Q + 6 + 3Q + Q2 = 0 Or, 32 - 140 + 11Q + 6 + Q2 = 0 Or, Q2 + 11Q - 102 = 0 Or, Q2 + 17Q - 6Q - 102 = 0 Or, Q(Q + 17) - 6(Q + 17) = 0 Or, (Q + 17) (Q - 6) = 0 So, 'Q' = 6, or 'Q' = -17 Required product = 2.5 × 4 = 10

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