CTS Practice Mixtures Problems

Dear Reader, below are three problems from mixtures. These questions are of similar nature and require you to remember a simple equation to solve these.
Question 1
Sharmila Begum, a tea merchant has two varieties of tea one costing Rs.62 per kg. and the other costing Rs.72 per kg. In what ratio she should mix these two varieties such that the mixture is worth Rs.64.50 per kg.
a) 3:1 b) 2:1 c)1:3 d)none of these.
Answer : a) 3 : 1
Solution :
Let us assume that the ratio of new mixture be a : b so that the new price becomes Rs. C. (in our case X = 64.5). Let us assume C1 be the cost per Kg of I variety of tea and C2 be the cost per Kg of the II variety. (In this problem, C1 = 62 and C2 = 72)
Then we have, a.C1 + b.C2 / (a + b) = C
Substituting C1 = 62, C2 = 72 and C = 64.5, we get
62a + 72b / a + b = 64.5 ---- eq 1
Now let us substitute the value of options one by one in this equation to check which one satisfies this equation.
Option I : a = 3, b = 1
Substitute a, b values in eq 1 we get,
62. 3 + 72.1 / 4 = 64.5
Or 258 = 258
Thus, option I satisfies the equation and hence it is the correct answer.


Question 2
Vinayaga Sait & son are the leading sugar dealers in Kolkata, a place known for milk sweets. Vinayaga Sait wanted to know how many kilogram of sugar costing Rs.18 per kg can be mixed with 27 kg of sugar costing Rs. 14 per kg such that there may be a gain of 10% on selling the mixture at Rs.18.48 per kg.?
a) 63 kg b)72 kg c)81 kg d) none of these.
Answer : a) 63 kg.
Solution :
Selling price of 1 kg of mixture = Rs. 18.48 at 10% profit
Cost price of 1kg of mixture = 100/110 x 18.48= Rs.16.80
Take a look at the solution for previous question. We are again going to use the familiar equation
a.C1 + b.C2 / (a + b) = C -> eq 1
Here, C = 16.8, C1 = 18, C2 = 14.
Also, a/b = Weight of I variety of Sugar/ Weight of II variety of sugar
a/b = x/27 , where x = weight of I variety of Sugar
Substituting values, eq 1 becomes,
18a + 14b = 16.8(a + b)
Divide by b on both sides
18a/b + 14 = 16.8(a/b + 1)
But a/b = x/27
Therefore, 18x/27 + 14 = 16.8(x/27 + 1)
Multiply by 27 on both sides,
18x + 378 = 16.8(x + 27)
18x – 16.8x = 453.6 – 378
1.2x = 75.6
x = 75.6/1.2 = 63 Kg
Question 3
Karimlal Sait is running a famous rice shop and he consults his Manager to find out that in what ratio must rice at Rs.9.30 per kg be mixed with rice at Rs.10.80 per kg so that the mixture be worth Rs.10 per kg?
a. 7 : 8 b. 8 : 7 c. 31 : 36 d. 36 : 31
Answer : b) 8 : 7
Solution :
This question is very similar to the first question.
Let us again use our familiar equation : a.C1 + b.C2 / (a + b) = C -> eq 1
Values as given in question values are C1 = 9.30, C2 = 10.80, C = 10.
Substituting above values in eq 1 we get,
9.3a + 10.8b = 10(a + b) -> eq 2
Consider Option I where a = 7 and b = 8.
Substituting these values in eq 2 we get
9.3 x 7 + 10.8 x 8 = 10 (7 + 8)
151.5 != 150.
Hence, option I does not satisfy equation no 2. Hence option I is not correct.
Consider Option II where a = 8 and b = 7
Substituting these values in eq 2 we get
9.3 x 8 + 10.8 x 7 = 10 (8 + 7)
150 = 150
Option II satisfies the equation no 2 perfectly and hence it is the right answer.