# Model 10 Percentage With Allegations & Mixture Practice Questions Answers Test With Solutions & More Shortcuts

#### PERCENTAGE PRACTICE TEST [11 - EXERCISES]

Question : 11 [SSC CHSL 2013]

One type of liquid contains 20% water and the second type of liquid contains 35% of water. A glass is filled with 10 parts of first liquid and 4 parts of second liquid. The water in the new mixture in the glass is

a) 24\$2/7\$%

b) 37%

c) 46%

d) 12\$1/7\$%

In 10 litres of first type of liquid,

Water = \$1/5×10\$ = 2 litres

In 4 litres of second type of liquid,

Water = \$4 × 35/100 = 7/5\$ litres

Total amount of water = 2 + \$7/5 = 17/5\$ litres

Required percentage = \${17/5}/14 × 100\$

= \$170/7 = 24{2}/7%\$

Question : 12 [SSC CGL Prelim 2003]

How much pure alcohol has to be added to 400 ml of a solution containing 15% of alcohol to change the concentration of alcohol in the mixture to 32% ?

a) 68 ml

b) 60 ml

c) 100 ml

d) 128 ml

Alcohol = \$(15/100 × 400)\$ml = 60 ml.

Water = 340 ml.

Let x ml of alcohol be added.

Then, \${60 + x}/{400 + x} × 100 = 32\$

or \${60 + x}/{400 + x} = 32/100 = 8/25\$

or 1500 + 25x = 3200 + 8x

or 17x = 1700

or x = 100 ml

Question : 13 [SSC CPO S.I.2009]

If 4 litres of water is evaporated on boiling from 12 litres of salt solution containing 7 percentage salt, the percentage of salt in the remaining solution is

a) 13%

b) 10.5%

c) 11.5%

d) 12%

In 12 litres salt solution,

Salt = \${7 × 12}/100 = 0.84\$ units

Water = \${93 × 12}/100 = 11.16\$ units

After evaporation,

Percentage of salt = \$0.84/8 × 100 = 10.5\$%

Question : 14

8 litres of water is added to 32 litres of a solution containing 20% of alcohol in water. What is the approximate concentration of alcohol in the solution now ?

a) 24%

b) 16%

c) 8%

d) 12%

In 32 litres of solution,

Alcohol = \${32 × 20}/100\$ = 6.4 litres

Water = 32 – 6.4 = 25.6 litres

On adding 8 litres of water,

Required percent = \$6.4/40 × 100\$ = 16%

Question : 15 [SSC CHSL 2011]

A vessel has 60 litres of solution of acid and water having 80% acid. How much water be added to make it a solution in which acid forms 60% ?

a) None of these

b) 48 litres

c) 20 litres

d) 36 litres

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

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