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 βQβ = β15aβ So, βPβ = 15a Γ (2/3) = β10aβ And, βRβ = 15a Γ (4/5) = β12aβ Since, 6P + 2Q + 4R = 6 Γ 10a + 2 Γ 15a + 4 Γ 12a = 138a Out of the given options only 690 is a multiple of 138.
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