Question
Download Solution PDFA mixture of 100 L contains kerosene and turpentine oil in the ratio 3 : 2. What is the minimum quantity of kerosene in litres (whole number) that should be mixed in the mixture so that the resulting mixture has 20% of kerosene ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
The initial mixture contains kerosene and turpentine oil in the ratio 3:2.
Total mixture = 100 L
Kerosene = (3/5) × 100 = 60 L
Turpentine oil = (2/5) × 100 = 40 L
Final kerosene percentage = 20%
Formula used:
Final kerosene in the mixture = Total mixture × (Final kerosene percentage / 100)
Additional kerosene to be added = Final kerosene - Initial kerosene
Calculation:
Let the additional kerosene to be added = x L.
Total mixture after adding kerosene = 100 + x L
Kerosene in the new mixture = 60 + x L
For 20% kerosene in the final mixture:
⇒ (60 + x) / (100 + x) = 20 / 100
⇒ (60 + x) = (20 / 100) × (100 + x)
⇒ 60 + x = 20 × (100 + x) / 100
⇒ 60 + x = (2000 + 20x) / 100
⇒ 100(60 + x) = 2000 + 20x
⇒ 6000 + 100x = 2000 + 20x
⇒ 6000 - 2000 = 20x - 100x
⇒ 4000 = -80x
⇒ x = -4000 / 80
⇒ x = -50
∴ It is not possible to achieve 20% kerosene in the final mixture. The correct answer is option (4).
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