Question
The difference between the present ages of a mother and
her son is 32 years. Six years ago, the mother was 4 years younger than twice the present age of her daughter. If the sum of the present ages of the son and the daughter is 34 years, find the present age of the father, who is 5 years older than the mother.Solution
ATQ, Let the present age of the daughter be 'z' years. Present age of the mother = '2z - 4' years Present age of the son = 'z - 32' years z + (z - 32) = 34 2z = 66 So, z = 33 years Therefore, present age of the father = Mother's age + 5 = (2z - 4) + 5 = (2 × 33) - 4 + 5 = 66 - 4 + 5 = 67 years.
Statement: E> T ≥ H = N> I ≤ C
Conclusion:
i) E> i
ii) T
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Statement: ...
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