T h e f i rs t j ug : v in e g a r o i l = 1 3 ⇒ t h e v o l u m e o f t h e o i l = 4 3 o f t h e j ug . ⇒ t h e v o l u m e o f o i l = 4 3 l i t re T h e seco n d j ug : v in e g a r o i l = 1 4 ⇒ t h e v o l u m e o f t h e o i l = 5 4 o f t h e j ug . ⇒ t h e v o l u m e o f o i l = 5 4 l i t re
T h e l a r g er j ug : 4 3 l i t re + 5 4 l i t re = 0.75 + 0.8 [ l i t re ] = 1.55 [ l i t re ] A n s . t h e v o l u m e o f t h e o i l i s 1.55 l i t res .
I n t h e f i rs t j ug i s 4 3 l o l i v e o i l . I n t h e seco n d j ug i s 5 4 l o l i v e o i l 4 3 l + 5 4 l 4 3 → t h e d e n o mina t or i s 4 l i s t t h e m u lt i pl es o f 4 : 0 ; 4 ; 8 ; 12 ; 16 ; 20 ; 24 ; 28 ; ... 5 4 → t h e d e n o mina t or i s 5 l i s t t h e m u lt i pl es o f 5 : 0 ; 5 ; 10 ; 15 ; 20 ; 25 ; 30 ; ... L e a s t C amm o n De n o mina t or o f 4 3 an d 5 4 i s 20
4 3 = 4 ⋅ 5 3 ⋅ 5 = 20 15 5 4 = 5 ⋅ 4 4 ⋅ 4 = 20 16 − − − − − − − − − − − − − − − − − 4 3 l + 5 4 l = 20 15 l + 20 16 l = 20 15 + 16 l = 20 31 l = 1 20 11 l = 1.55 l ← so l u t i o n
The total volume of oil in the larger jug is 1.55 liters, calculated by adding 0.75 liters from the first jug and 0.8 liters from the second jug.
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