Question
The natural numbers 'P', 'Q', and
'R' satisfy the ratios P:Q = 2:3 and Q:R = 5:4. Which of the following could be a possible value of (6P + 2Q + 4R)?Solution
ATQ, Let ‘Q’ = ‘15a’ So, ‘P’ = 15a × (2/3) = ‘10a’ And, ‘R’ = 15a × (4/5) = ‘12a’ Since, 6P + 2Q + 4R = 6 × 10a + 2 × 15a + 4 × 12a = 138a Out of the given options only 690 is a multiple of 138.
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