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 : 1 [SSC SO 2007]
200 litres of a mixture contains 15% water and the rest is milk. The amount of milk that must be added so that the resulting mixture contains 87.5% milk is
a) 45 litres
b) 30 litres
c) 35 litres
d) 40 litres
Answer »Answer: (d)
In 200 litres of mixture,
Quantity of milk = 85 100 ´ 200 = 170 litres
Quantity of water = 30 litres
Let the quantity of additional milk added be x litres.
According to the question,
${170 + x}/{200 + x} × 100 = 87.5$
(170 +x) × 100 = 17500 + 87.5x
100x – 87.5x = 17500 – 17000
12.5x = 500
x = $500/12.5$ = 40 litres
Question : 2 [SSC CGL Tier-I 2014]
In what ratio must 25% of alcohol be mixed with 50% of alcohol to get a mixture of 40% strength alcohol ?
a) 3 : 2
b) 1 : 2
c) 2 : 1
d) 2 : 3
Answer »Answer: (d)
Required ratio = $1/10 : 3/20 = 2 : 3$
Question : 3 [SSCCHSL 2011]
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
Question : 4 [SSC CGL Tier-II 2015]
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 : 5
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
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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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