SSC MTS & Havaldar 2016 · Question 135 of 162
⇒ Let 3A = 4B = 12C = k ⇒ 3A = k, 4B = k, 12C = k ⇒ A = k/3, B = k/4, C = k/12 ⇒ A : B : C = k/3 : k/4 : k/12 ⇒ LCM (3, 4, 12) = 12 ⇒ A : B : C = (k/3) x 12 : (k/4) x 12 : (k/12) x 12 ∴A:B:C=4:3:1
Source: SSC MTS 2016 re-exam (held 10-Oct-2017 Shift 1) paper with key+explanations, via prepp.in · reliable-secondary
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