model 10 percentage with allegations & mixture Section-Wise Topic Notes With Detailed Explanation And Example Questions
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The following question based on percentage topic of quantitative aptitude
(a) 68 ml
(b) 60 ml
(c) 100 ml
(d) 128 ml
The correct answers to the above question in:
Answer: (c)
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
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Read more allegations mixture based Based Quantitative Aptitude Questions and Answers
Question : 1
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%
Answer »Answer: (b)
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 : 2
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%
Answer »Answer: (c)
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 : 3
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
Answer »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
Question : 4
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$1/7$%
Answer »Answer: (a)
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 : 5
In an alloy there is 12% of copper. To get 69 kg of copper, how much alloy will be required ?
a) 1736$2/3$kg
b) 424 kg
c) 575 kg
d) 828 kg
Answer »Answer: (c)
12 kg copper is contained in 100 kg of alloy
69 kg copper is contained in
$100/12 × 69$ = 575 kg of alloy
Question : 6
15 litres of a mixture contains alcohol and water in the ratio 1 : 4. If 3 litres of Water is mixed in it, the percentage of alcohol in the new mixture will be
a) 18$1/2$%
b) 15%
c) 16$2/3$%
d) 17%
Answer »Answer: (c)
Alcohol =15 × $1/5$ = 3 litres
Water =15 × $4/5$ = 12 litres
Required percentage = $3/{15 + 3} × 100$
= $50/3 = 16{2}/3%$
GET percentage PRACTICE TEST EXERCISES
model 1 simple percentage questions
model 2 net increase or decrease in %
model 3 reducing & exceeding prices
model 4 x & y comparison
model 5 income & expenditure
model 6 consumption & remaining
model 7 marks scored in examinations
Model 8 population based
model 9 voters & election
model 10 percentage with allegations & mixture
model 11 percentage with ratios
percentage Shortcuts and Techniques with Examples
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model 1 simple percentage questions
Defination & Shortcuts … -
model 2 net increase or decrease in %
Defination & Shortcuts … -
model 3 reducing & exceeding prices
Defination & Shortcuts … -
model 4 x & y comparison
Defination & Shortcuts … -
model 5 income & expenditure
Defination & Shortcuts … -
model 6 consumption & remaining
Defination & Shortcuts … -
model 7 marks scored in examinations
Defination & Shortcuts … -
Model 8 population based
Defination & Shortcuts … -
model 9 voters & election
Defination & Shortcuts … -
model 10 percentage with allegations & mixture
Defination & Shortcuts … -
model 11 percentage with ratios
Defination & Shortcuts …
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