Practice Allegations mixture based - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) A sugar solution of 3 litre contain 60% sugar. One liter of water is added to this solution. Then the percentage of sugar in the new solution is:
(a)
(b)
(c)
(d)
Quantity of sugar in the solution = ${3 × 60}/100 = 1.8$units
On adding 1 litre of water,
Required percent = $1.8/4 × 100 = 45%$
Q-2) 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)
(b)
(c)
(d)
In 12 litres salt solution,
Salt = ${7 × 12}/100 = 0.84$ units
Water = ${93 × 12}/100 = 11.16$ units
After evaporation,
Percentage of salt = $0.84/8 × 100 = 10.5$%
Q-3) 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)
(b)
(c)
(d)
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%
Q-4) An ore contains 25% of an alloy that has 90% iron. Other than this, in the remaining 75% of the ore, there is no iron. To obtain 60 kg of pure iron, the quantity of the ore needed (in kgs) is approximately :
(a)
(b)
(c)
(d)
In 4 kg of ore, iron = 0.9 kg.
Quantity of ore for 60 kg of iron
= ${60 × 4}/0.9$ = 266.67 kg
Q-5) A litre of pure alcohol is added to 6 litres of 30% alcohol solution. The percentage of water in the solution is
(a)
(b)
(c)
(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%
Q-6) In what ratio must 25% of alcohol be mixed with 50% of alcohol to get a mixture of 40% strength alcohol ?
(a)
(b)
(c)
(d)
Required ratio = $1/10 : 3/20 = 2 : 3$
Q-7) How much water must be added to 100 ml of 80 per cent solution of boric acid to reduce it to a 50 per cent solution ?
(a)
(b)
(c)
(d)
Let x ml of water be added.
${20 + x}/{100 + x} × 100$ = 50
40 + 2x = 100 + x ⇒ x = 60 ml
Q-8) In 50 gm alloy of gold and silver, the gold is 80% by weight. How much gold should be mixed to this alloy so that the weight of gold would become 95% ?
(a)
(b)
(c)
(d)
Initial quantity of gold
= ${50 × 80}/100$ = 40 gm
Let 'x ' gm be mixed.
$(40 + x) = (50 + x) × 95/100$
$40 + x =(50 + x) × 19/20$
800 + 20x = 950 + 19x
x = 150 gm
Q-9) The ratio in which two sugar solutions of the concentrations 15% and 40% are to be mixed to get a solution of concentration 30% is
(a)
(b)
(c)
(d)
Required ratio = 10 : 15 = 2 : 3
Q-10) 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)
(b)
(c)
(d)
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