Question : In a mixture of three varieties of tea, the ratio of their weights is 4 : 5 : 8. If 5 kg tea of the first variety, 10 kg tea of the second variety and some quantity of tea of the third variety are added to the mixture, the ratio of the weights of three varieties of tea becomes 5 : 7 : 9. In the final mixture, the quantity (in kg) of the third variety of tea was:
Option 1: 42
Option 2: 45
Option 3: 48
Option 4: 40
Correct Answer: 45
Solution :
Given:
The ratio of the three varieties of tea is 4 : 5 : 8.
Let the quantity of the first variety be $4x$ kg.
Quantity of the second variety of tea = $5x$ kg.
Quantity of the third variety of tea = $8x$ kg.
Let $y$ kg of the third variety of tea be mixed.
So, the resultant ratio = $(4x+5):(5x+10):(8x+y)= 5 : 7 : 9$
According to the question,
$\frac{4x+5}{5x+10}=\frac{5}{7}$
⇒ $28x+35=25x+50$
⇒ $3x=15$
$\therefore x=5$
Also, $\frac{5x+10}{8x+y}=\frac{7}{9}$
⇒ $\frac{5×5+10}{8×5+y}=\frac{7}{9}$
⇒ $\frac{35}{40+y}=\frac{7}{9}$
⇒ $40+y=9×5$
$\therefore y=45-40=5$
Required quantity of the third variety of tea
$=8x+y=8×5+5=45$ kg
Hence, the correct answer is 45.
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