New IFSCA Phase 1 Results Out - Enroll Here for Phase 2

    Register

    Question

    Quantity I: Ratio of present ages of A and B is 6:7 respectively while the ratio of present ages of C and D is 8:9. Ratio of the age of B after two years to age of C seven years ago is 6:5. Sum of the present ages of A, B, C and D is 120, then find average of the ages of A and D.

    Quantity II: If the average of ‘q’ numbers is 20. If half of the numbers are decreased by 20 and half of the others are increased by 16, then what is the new average?

    A Quantity I > Quantity II Correct Answer Incorrect Answer
    B Quantity I ≥ Quantity II Correct Answer Incorrect Answer
    C Quantity I < Quantity II Correct Answer Incorrect Answer
    D Quantity I ≤ Quantity II Correct Answer Incorrect Answer
    E Quantity I = Quantity II or relation can't be established Correct Answer Incorrect Answer

    Solution

    Let the age of A = 6x years Therefore, age of B = 7x years Age of C=8 years Therefore, age of D = 36years (7x+2)/(8y–7) = 6/5… Equation(1) Sum of ages =120 13x+17y=120….. Equation(2) Solving Equation(1) & Equation(2) x=4 y=4 A=24years, D=36years Average of A and D = (24+36)/2=30years (quantity 1) Quantity 2 : Sum of q numbers = 20q So new sum = 20q + (0.5q*16) – (0.5q*20) = 18q New average = 18q/q = 18 Ans. A. Q1>Q2

    Practice Next