model 10 percentage with allegations & mixture Practice Questions Answers Test with Solutions & More Shortcuts
percentage PRACTICE TEST [11 - 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
Question : 16 [SSC CGL Prelim 2007]
1 litre of water is added to 5 litres of alcohol-water solution containing 40% alcohol strength. The strength of alcohol in the new solution will be
a) 33$1/3$%
b) 30%
c) 33%
d) 33$2/3$%
Answer »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.
Question : 17
In what ratio must a mixture of 30% alcohol strength be mixed with that of 50% alcohol strength so as to get a mixture of 45% alcohol strength ?
a) 3 : 1
b) 1 : 2
c) 1 : 3
d) 2 : 1
Answer »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 ⇒ $x/y = 1 : 3$
Question : 18
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) 300
b) 250.57
c) 266.67
d) 275.23
Answer »Answer: (c)
In 4 kg of ore, iron = 0.9 kg.
Quantity of ore for 60 kg of iron
= ${60 × 4}/0.9$ = 266.67 kg
Question : 19
A litre of pure alcohol is added to 6 litres of 30% alcohol solution. The percentage of water in the solution is
a) 40%
b) 50%
c) 65%
d) 60%
Answer »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%
Question : 20
A sample of 50 litres of glycerine is found to be adulterated to the extent of 20%. How much pure glycerine should be added to it so as to bring down the percentage of impurity to 5% ?
a) 149 litres
b) 155 litres
c) 150 litres
d) 150.4 litres
Answer »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.
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model 10 percentage with allegations & mixture Shortcuts »
Click to Read...model 10 percentage with allegations & mixture Online Quiz
Click to Start..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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