​In mixture A, there is 5 L of syrup in 6 L of water. In mixture B, there is 3.5 L of syrup in 5.5 L of water. In which of the two mixtures is the concentration of syrup greater?​

Correct option is D

Given:

Mixture A:

Volume of syrup: 5 L

Volume of water: 6 L

Mixture B:

Volume of syrup: 3.5 L

Volume of water: 5.5 L

Formula Used:

Concentration of syrup = $$\frac{\text{(Volume of syrup) }}{\text{(Total volume of mixture)}}$$​​

Solution:

Total Volume of each Mixture:

Mixture A: 5 L (syrup) + 6 L (water) = 11 L

Mixture B: 3.5 L (syrup) + 5.5 L (water) = 9 L

Concentration of Syrup in each Mixture:

Mixture A:

Concentration = $$\frac{5 L }{11 L}$$​​

Concentration ≈ 0.45 or 45%

Mixture B:

Concentration = $$\frac{3.5 L }{9 L}$$​​

Concentration ≈ 0.39 or 39%

Compare Concentrations:

Mixture A (45%) > Mixture B (39%)

Therefore, Mixture A has a greater concentration of syrup.

Option (d) is right answer.