In what ratio three kinds of rice costing 1.45rs, 1.54rs and 1.70rs mu...
Answer – c) 11:20:44 Explanation : By the rule of allegation, 145 154
…………165…………….
11 20
154 170
………165…………….
5 11
Final ratio = 11:20:44
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In what ratio three kinds of rice costing 1.45rs, 1.54rs and 1.70rs mu...
Question:
In what ratio three kinds of rice costing 1.45rs, 1.54rs and 1.70rs must be mixed so that the mixture can be sold at 1.65rs per kg.
Options:
a) 11:21:44
b) 20:11:44
c) 11:20:44
d) 44:20:11
e) None of these
Answer:
To solve this question, we will use the concept of weighted average.
Weighted average = (quantity of first item * price of first item + quantity of second item * price of second item + quantity of third item * price of third item) / (total quantity)
Let the quantities of first, second and third kind of rice be x, y and z respectively.
So, the weighted average of the mixture can be written as:
1.65 = (1.45x + 1.54y + 1.70z) / (x+y+z)
Simplifying this equation, we get:
1.65(x+y+z) = 1.45x + 1.54y + 1.70z
Multiplying by 100, we get:
165x + 165y + 165z = 145x + 154y + 170z
Simplifying further, we get:
20x = 9y + 5z
Now, we need to find the ratio of x:y:z. To do that, we will assume that x = 20k (where k is a constant). Substituting this value in the above equation, we get:
20k = 9y + 5z
Dividing by 5, we get:
4k = (9/5)y + z
We know that k, y and z are integers. To find the smallest possible values for k, y and z, we will assume k = 5 (as 20k needs to be divisible by 5). Substituting this value in the above equation, we get:
4(5) = (9/5)y + z
20 = 9y + 5z/5
20 = 9y + z
Now, we need to find two more equations in terms of y and z. To do that, we will assume that y = 11m (where m is a constant). Substituting this value in the above equation, we get:
20 = 9(11m) + z
z = 20 - 99m
Similarly, we will assume that z = 44n (where n is a constant). Substituting this value in the above equation, we get:
y = (20 - 44n)/11
Now, we have the values of x, y and z in terms of k, m and n. To find the ratio of x:y:z, we will assume k = 1, m = 1 and n = 2 (as these values will give us the smallest possible values for x, y and z).
Substituting these values, we get:
x = 20k = 20(1) = 20
y = 11m = 11(1) = 11
z = 44n = 44(2) = 88
Therefore, the ratio of x:y:
In what ratio three kinds of rice costing 1.45rs, 1.54rs and 1.70rs mu...
The correct option should be none of these. Because the correct answer is 5:5:31.
You can verify this answer by taking 5 units of 1.45rs rice, 5 units of 1.54rs rice and 31 units of 1.70rs rice and calculating the total cost. This cost will be (5 x 1.45) + (5 x 1.54) + (31 x 1.70) = 67.65rs.
Now the average cost = Total cost/(5+5+31) = 67.65/41 = 1.65rs.
And this is the same average cost that is required in the given question. Hence our ratio of 5:5:31 is correct.
I am not providing the solution here. You can think of it if you know the allegation mixture rule.