type 8 alligation & mixtures based ratio & proportion problems Section-Wise Topic Notes With Detailed Explanation And Example Questions
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The following question based on ratio & proportion topic of quantitative aptitude
(a) 2 : 1
(b) 4 : 5
(c) 1 : 1
(d) 1 : 2
The correct answers to the above question in:
Answer: (b)
Quantity of milk in the last
= $81(1 - 27/81)^2 = 81(1 - 1/3)^2$
= $81 × 2/3 × 2/3$ = 36
Quantity of water in the last
= 81 - 36 = 45
Ratio = $36/45 = 4/5 = 4 : 5$
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Read more alligation and mixtures problems Based Quantitative Aptitude Questions and Answers
Question : 1
In a 729 litres mixture of milk and water, the ratio of milk to water is 7 : 2. To get a new mixture containing milk and water in the ratio 7 : 3, the amount of water to be added is
a) 56 litres
b) 50 litres
c) 71 litres
d) 81 litres
Answer »Answer: (d)
Quantity of milk in 729 litres of mixture
= $7/9 × 729$ = 567 litres
Quantity of water
= (729 - 567) litres = 162 litres.
Let x litres of water is mixed to get the required ratio of 7 : 3
$567/{162 + x} =7/3$
7x + 1134 = 1701
7x = 1701 - 1134 = 567
$x = 567/7$ = 81 litres
Question : 2
Two vessels A and B contain milk and water mixed in the ratio 4 : 3 and 2 : 3. The ratio in which these mixtures be mixed to form a new mixture containing half milk and half water is
a) 5 : 6
b) 4 : 3
c) 6 : 5
d) 7 : 5
Answer »Answer: (d)
Required ratio
= $1/10 : 1/14$ = 7 : 5
Question : 3
Vessels A and B contain mixtures of milk and water in the ratios 4 : 5 and 5 : 1 respectively. In what ratio should quantities of mixture be taken from A and B to form a mixture in which milk to water is in the ratio 5 : 4?
a) 5 : 2
b) 2 : 3
c) 4 : 3
d) 2 : 5
Answer »Answer: (a)
First of all we write the fraction of milk present in three mixtures.
In A; $4/9$ and In B; $5/6$
In combination of A and B; $5/9$
From alligation rule,
= $5/18 : 1/9$
$5/18 : 2/18$ ⇒ 5 : 2
So, ratio of
A : B = 5 : 2
Question : 4
A and B are two alloys of gold and copper prepared by mixing metals in ratios 7 : 2 and 7 : 11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be ;
a) 9 : 5
b) 5 : 7
c) 5 : 9
d) 7 : 5
Answer »Answer: (d)
Gold = $7/9 : 7/18 = 14/18 : 7/18$
Gold in new mixture
= 14 + 7 = 21
and copper = 18 × 2 - 21
= 36 - 21 = 15
Required ratio
= 21 : 15 = 7 : 5
GET ratio & proportion PRACTICE TEST EXERCISES
type 1 basic concepts of ratio & proportion
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type 4 income & expenditure based ratio & proportion problems
type 5 shares & partnership based ratio & proportion problems
type 6 fractions based ratio & proportion problems
type 7 finding sum difference product based ratio & proportion problems
type 8 alligation & mixtures based ratio & proportion problems
type 9 coins & rupees based ratio & proportion problems
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