IAS 2016 — CSAT Maths

Correct Option

To determine the relative sweetness of the solutions, the concentration of sugar in each vessel must be calculated. Concentration is expressed as the mass of sugar per unit volume of water.

• Vessel A: Sugar concentration = $\frac{\text{30g}}{\text{180ml}} = \frac{1}{6}\text{g/ml}$180ml30g​=61​g/ml
• Vessel B: Sugar concentration = $\frac{\text{40g}}{\text{280ml}} = \frac{1}{7}\text{g/ml}$280ml40g​=71​g/ml
• Vessel C: Sugar concentration = $\frac{\text{20g}}{\text{100ml}} = \frac{1}{5}\text{g/ml}$100ml20g​=51​g/ml

Comparing these concentrations:

• $\frac{1}{5}$51​ (Vessel C) = 0.2 g/ml
• $\frac{1}{6}$61​ (Vessel A) $\approx$≈ 0.1667 g/ml
• $\frac{1}{7}$71​ (Vessel B) $\approx$≈ 0.1428 g/ml

Thus, the order of sweetness from most to least is C > A > B. The solution in vessel B has a sugar concentration of $\frac{1}{7}\text{g/ml}$71​g/ml, which is lower than that in vessel C ($\frac{1}{5}\text{g/ml}$51​g/ml). Therefore, the solution in vessel B is less sweet than that in C.

Incorrect Options

• Option 1 (sweeter than that in C): This is incorrect because the sugar concentration in vessel B ($\frac{1}{7}\text{g/ml}$71​g/ml) is lower than that in vessel C ($\frac{1}{5}\text{g/ml}$51​g/ml). A lower concentration indicates less sweetness.

• Option 2 (sweeter than that in A): This is incorrect as the sugar concentration in vessel B ($\frac{1}{7}\text{g/ml}$71​g/ml) is lower than that in vessel A ($\frac{1}{6}\text{g/ml}$61​g/ml). Consequently, solution B is not sweeter than solution A.

• Option 3 (as sweet as that in C): This is incorrect because the sugar concentrations are distinct. Vessel B has a concentration of $\frac{1}{7}\text{g/ml}$71​g/ml while vessel C has $\frac{1}{5}\text{g/ml}$51​g/ml. These values are not equal, indicating differing levels of sweetness.