Question

Raju, a cake seller, mixes flour, cream, and sugar in a ratio of 5:3:2 and sells it at the cost price. Their prices are in the ratio 4:10:5. What ratio should the three be mixed in to get a 10% profit selling at the earlier proportions' price?
i) The proportion of sugar should remain the same.
ii) The price of sugar is Rs. 30 per kg

Hint:

In data sufficiency problems involving ratios and percentages, assigning a variable (like 'k' here) to the ratio can clarify whether absolute values are needed. If the variable cancels out from the final equation, then knowing its actual value is not necessary.

Answer 1

Step 1: Understanding the Concept:
This is a data sufficiency question involving mixtures, ratios, and profit. We need to determine if the given statements provide enough information to find a new mixture ratio. The key is to work with ratios and proportional values; the actual prices might not be necessary.
Step 2: Analyze the Main Question:
Let the prices of flour, cream, and sugar be \(P_F, P_C, P_S\). Given price ratio: \(P_F:P_C:P_S = 4:10:5\). Let the prices be \(4k, 10k, 5k\) per unit weight.
Initial mixture ratio (Mix 1): Flour:Cream:Sugar = 5:3:2. Let's find the cost price (CP) of 10 units (5+3+2) of Mix 1. CP of Mix 1 = \(5 \times P_F + 3 \times P_C + 2 \times P_S = 5(4k) + 3(10k) + 2(5k) = 20k + 30k + 10k = 60k\). CP per unit of Mix 1 = \(\frac{60k}{10} = 6k\).
Selling Price (SP) of Mix 1 = CP of Mix 1 = \(6k\). This is the "earlier proportions' price".
New Goal (Mix 2): Let the new ratio be \(f:c:s\). The SP of Mix 2 is the same as the SP of Mix 1, so \(SP_{new} = 6k\). We want a 10% profit. This means \(SP_{new} = 1.10 \times CP_{new}\). \[ 6k = 1.1 \times CP_{new} \implies CP_{new} = \frac{6k}{1.1} = \frac{60k}{11} \] The cost price of the new mixture is given by: \[ CP_{new} = \frac{f \times P_F + c \times P_C + s \times P_S}{f+c+s} = \frac{f(4k) + c(10k) + s(5k)}{f+c+s} \] Equating the two expressions for \(CP_{new}\) and cancelling \(k\): \[ \frac{60}{11} = \frac{4f + 10c + 5s}{f+c+s} \] This is one equation with three unknowns (\(f, c, s\)). We need more information to find the ratio \(f:c:s\).
Step 3: Analyze the Statements:
Statement (1): The proportion of sugar should remain the same.
The proportion of sugar in Mix 1 was \(\frac{2}{5+3+2} = \frac{2}{10} = \frac{1}{5}\). So, for Mix 2, we have \(\frac{s}{f+c+s} = \frac{1}{5}\). This gives us a second equation: \(5s = f+c+s \implies 4s = f+c\). Now we have a system of two equations:

\(\frac{60}{11} = \frac{4f + 10c + 5s}{f+c+s}\)
\(f+c = 4s\)
Substitute \(f+c+s = 5s\) from the proportion info into equation 1: \[ \frac{60}{11} = \frac{4f + 10c + 5s}{5s} \] \[ \frac{300s}{11} = 4f + 10c + 5s \implies 245s = 44f + 110c \] Now substitute \(s = \frac{f+c}{4}\) into this new equation: \[ 245\left(\frac{f+c}{4}\right) = 44f + 110c \] \[ 245f + 245c = 176f + 440c \] \[ 69f = 195c \implies \frac{f}{c} = \frac{195}{69} = \frac{65}{23} \] We have found the ratio \(f:c\). Since we also know \(s\) in terms of \(f\) and \(c\), we can find the full ratio \(f:c:s\). For example, if \(f=65, c=23\), then \(s = (65+23)/4 = 22\). The ratio is 65:23:22. Thus, Statement (1) ALONE is sufficient.
Statement (2): The price of sugar is Rs. 30 per kg.
This information allows us to find the value of \(k\). \(P_S = 5k = 30 \implies k = 6\). This means we know the actual prices: \(P_F=24\), \(P_C=60\), \(P_S=30\). The main equation becomes: \[ \frac{60(6)}{11} = \frac{4f(6) + 10c(6) + 5s(6)}{f+c+s} \] The factor of 6 (\(k\)) cancels from both sides, leaving us with the same single equation we had in the beginning: \(\frac{60}{11} = \frac{4f + 10c + 5s}{f+c+s}\). This is still one equation with three unknowns. Knowing the actual prices doesn't help determine the ratio. Thus, Statement (2) ALONE is not sufficient.
Step 4: Final Answer:
Statement (1) is sufficient, but statement (2) is not.