New RBI Grade B 2025 Notification will come Soon! Enroll Here

    Question

    'F' distributed sweets among his four children and

    retained some for himself. The three eldest girls received sweets in the ratio of 3:11:7. The total number of sweets with 'F' and the youngest boy is three times the total sweets with the three eldest girls. The ratio of sweets with 'F' to the total sweets with all the children is 3:4. Determine the total number of sweets if the youngest boy has 81 sweets with him.
    A 125 Correct Answer Incorrect Answer
    B 265 Correct Answer Incorrect Answer
    C 150 Correct Answer Incorrect Answer
    D 252 Correct Answer Incorrect Answer
    E none of these Correct Answer Incorrect Answer

    Solution

    ATQ, Let the children be P, Q, R and S Sweets with P : Q : R= 3 : 7 : 11 Let the number of sweets be 3k, 7k and 11k Total sweets with three eldest girls child = 21k sweets with F and S  = 3 × 21k = 63k Total sweets  = (21k + 63k)  = 84k Sweets with F : (P + Q + R + S) = 3 : 4 Total 7 units of sweets  = 84 k 1 unit = 12k sweets with F = 3 × 12k = 36k sweets with S =  (63k – 36k) = 27k 27 k = 81  →  k=3 Total number of sweets = 84k = 84 × 3 = 252

    Practice Next

    Relevant for Exams: