To solve the problem, let's define the prices and quantities of the different varieties of tea being mixed.
Given:
Let the price of the third variety of tea be x Rs./kg.
The total mixture is worth Rs. 153/kg. Since the ratio of the tea varieties is 1:1:2, we can assume the quantities of each type of tea in the mixture to be:
The total cost of the mixture can be expressed as the weighted average cost of the individual varieties based on their respective quantities.
Let's compute the cost per unit for the mixture:
The total quantity of the mixture is:
The average price per kg of the mixture is given as Rs. 153. Therefore, we set up the equation:
First, simplify the numerator:
Next, substitute this back into the equation and solve for x:
Multiply both sides by 4 to clear the denominator:
Now, isolate x:
Thus, the price of the third variety of tea is Rs. 175.50 per kg.
0.0169 / 0.0130 = 169 / 130
= 13 / 10
Given Exp. = 4 / 7 + {(2q - p) / (2q + p)}
Dividing numerator as well as denominator by q,
Exp = 4/7 + {2-p/q) / (2 + p/q)}
= 4/7 + {(2 - 4/5) / (2 + 4/5)}
= 4/7 + 6/14
= 4/7 + 3/7
=7/7
=1.
25% of 25% = (25/100) x (25/100) = 625/10000 = 0.625
Let y% of 20 = .05
Then, (y x 20)/100 = .05
? y = .25
2 | 24 - 36 - 40 -------------------- 2 | 12 - 18 - 20 -------------------- 2 | 6 - 9 - 10 ------------------- 3 | 3 - 9 - 5 ------------------- | 1 - 3 - 5 L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
81/3% = (25/3 x 1/100) = 1/12
.025 = (25/1000) x 100% = 2.5%
Let N% of 2/7 is 1/35
? (2/7 x N) / 100 = 1/35
? N = 1/35 x 7/2 x 100 = 10%
Let the money interest at 8% interest be ? P .
Then, the money interest at 10% interest = ?(4000 - P)
According to the question,
(P x 8 x 1)/100 + [(4000 - P) x 10 x 1]/100 = 352
? 8P + 40000 - 10P = 35200
? 40000 - 35200 = 2P
? P = 4800/2 = ? 2400
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