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) 10 gm
(b) 200 gm
(c) 150 gm
(d) 50 gm
The correct answers to the above question in:
Answer: (c)
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
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Read more allegations mixture based Based Quantitative Aptitude Questions and Answers
Question : 1
40 litres of a mixture of milk and water contains 10% of water, the water to be added, to make the water content 20% in the new mixture is :
a) 5 litres
b) 6 litres
c) 6.5 litres
d) 5.5 litres
Answer »Answer: (a)
Water content in 40 litres of mixture = $40 × 10/100$ = 4 litres
Milk content = 40 – 4 = 36 litres
Let x litres of water is mixed.
Then, ${4 + x}/{40 + x} = 20/100$
${4 + x}/{40 + x} = 1/5$
20 + 5x = 40 + x
4x = 20 ⇒ x = 5 litres
Question : 2
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) 9 : 8
b) 2 : 3
c) 3 : 2
d) 8 : 9
Answer »Answer: (b)
Required ratio = 10 : 15 = 2 : 3
Question : 3
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 = $3/{15 + 3} × 100$
= $50/3 = 16{2}/3%$
Question : 4
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) 60 ml
b) 30 ml
c) 40 ml
d) 50 ml
Answer »Answer: (a)
Let x ml of water be added.
${20 + x}/{100 + x} × 100$ = 50
40 + 2x = 100 + x ⇒ x = 60 ml
Question : 5
300 grams of sugar solution has 40% of sugar in it. How much sugar should be added to make it 50% in the solution?
a) 80 gram
b) 40 gram
c) 10 gram
d) 60 gram
Answer »Answer: (d)
In 300 gm of solution,
Sugar = ${300 × 40}/100 = 120gm.$
Let x gm of sugar be mixed.
According to the question,
${120 + x}/{300 + x} =1/2$
240 + 2x = 300 + x
2x – x = 300 – 240 ⇒ x = 60 gm.
Question : 6
75 gm of sugar solution has 30% sugar in it. Then the quantity of sugar that should be added to the solution to make the quantity of the sugar 70% in the solution, is
a) 130 gm
b) 125 gm
c) 100 gm
d) 120 gm
Answer »Answer: (c)
Sugar in original solution
= ${75 × 30}/100$ = 22.5gm
Let x gm of sugar be mixed.
${22.5 + x}/{75 + x} × 100 = 70$
2250 + 100x = 75 × 70 + 70x
2250 + 100x = 5250 + 70x
30x = 5250 – 2250 = 3000
x = $3000/30$ = 100 gm
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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