model 10 percentage with allegations & mixture Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (d)

In 30% alcohol solution,

Alcohol = $30/100 × 6$ =1.8litres

Water = 4.2 litres

On mixing 1 litre of pure alcohol,

Percentage of water =$4.2/7 × 100$ = 60%

Answer: (c)

Glycerine in mixture = 40 litres

Water = 10 litres

Let x litres of pure glycerine is mixed with the mixture.

${40 + x}/{50 + x} = 95/100 = 19/20$

800 + 20x = 950 + 19x

x = 950 – 800 = 150 litres.

Answer: (c)

Quantity of sugar in the solution = ${3 × 60}/100 = 1.8$units

On adding 1 litre of water,

Required percent = $1.8/4 × 100 = 45%$

Answer: (c)

Let x litres of first mixture is mixed with y litres of the second mixture.

According to the question,

${x × 30 /100 + y × 50/100}/{x × 70/100 + y × 50/100} = 45/55$

${0.3x + 0.5y}/{0.7x + 0.5y} = 9/11$

6.3x + 4.5y = 3.3x + 5.5y

6.3x – 3.3x = 5.5y – 4.5y

3x = y ⇒ $x/y = 1 : 3$

Answer: (a)

Alcohol in original solution = $40/100 × 5$ = 2 litres

Water in original solution = 3 litres

On adding 1 litre water, water becomes 4 litres.

Now, 6 litres of solution contains 2 litres of alcohol.

100 litres of solution contains = $2/6 × 100$

= $100/3 = 33{1}/3%$ alcohol.

Answer: (c)

In 60 litres of solution,

Water = ${60×20}/100$ = 12 litres

On adding x litres of water,

${12 + x}/{60 + x}$× 100 = 40

60 + 5x = 120 + 2x

3x = 60 ⇒ x = 20 litres