In a party Sameer mixed two type of liquid in a glass, type A liquid contains `35%` of Rum and type B liquid contains `40%` of Rum. Sameer takes 6 ml from type A liquid and 4 ml from type B liquid, then find percentage of Rum in the glass?

Options

A. A)`27%`
B. B)`37%`
C. C)`35%`
D. D)`32%`

Answer

To find the percentage of Rum in the glass after Sameer mixes the two types of liquids, we can follow these steps:

### Step 1: Determine the amount of Rum in each type of liquid.
- **Type A Liquid:** Contains 35% Rum.
- Volume taken = 6 ml
- Amount of Rum = 35% of 6 ml = (35/100) * 6 = 2.1 ml

- **Type B Liquid:** Contains 40% Rum.
- Volume taken = 4 ml
- Amount of Rum = 40% of 4 ml = (40/100) * 4 = 1.6 ml

### Step 2: Calculate the total amount of Rum in the mixture.
- Total Rum = Amount of Rum from Type A + Amount of Rum from Type B
- Total Rum = 2.1 ml + 1.6 ml = 3.7 ml

### Step 3: Calculate the total volume of the mixture.
- Total volume = Volume from Type A + Volume from Type B
- Total volume = 6 ml + 4 ml = 10 ml

### Step 4: Calculate the percentage of Rum in the glass.
- Percentage of Rum = (Total Rum / Total Volume) * 100
- Percentage of Rum = (3.7 ml / 10 ml) * 100 = 37%

### Final Answer:
The percentage of Rum in the glass is **37%**.

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