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) 5 : 2
(b) 2 : 3
(c) 4 : 3
(d) 2 : 5
The correct answers to the above question in:
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
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Read more alligation and mixtures problems Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
There is 81 litres pure milk in a container. One-third of milk is replaced by water in the container. Again one-third of mixture is extracted and equal amount of water is added. What is the ratio of milk to water in the new mixture?
a) 2 : 1
b) 4 : 5
c) 1 : 1
d) 1 : 2
Answer »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 = $36/45 = 4/5 = 4 : 5$
Question : 3
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 : 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
Question : 5
A jar contained a mixture of two liquids A and B in the ratio 4 : 1. When 10 litres of the mixture was taken out and 10 litres of liquid B was poured into the jar, this ratio became 2 : 3. The quantity of liquid A contained in the jar initially was
a) 16 litres
b) 40 litres
c) 8 litres
d) 4 litres
Answer »Answer: (b)
Let the initial quantity of liquids A and B in the jar be 4x and x litres respectively.
After taking out 10 litres of the mixture,
Liquid A
= $4x - 4/5 × 10 = (4x - 8)$ litres
Liquid B
= $4x - 1/5 ×10 = (4x - 2)$ litres
After pouring 10 litres of liquid B,
${4x - 8}/{4x - 2 + 10} = 2/3$
12x - 24 = 8x + 16
4x = 40
$x = 40/4$ = 10
Quantity of liquid A = 4x
= 4 × 10 = 40 litres
Question : 6
In 40 litres mixture of milk and water the ratio of milk to water is 7 : 1. In order to make the ratio of milk and water 3 : 1, the quantity of water (in litres) that should be added to the mixture will be
a) 6$2/3$
b) 6$3/4$
c) 6$1/2$
d) 6
Answer »Answer: (a)
In 40 litres mixture,
Quantity of milk
= $7/8$ × 40 = 35 litres
Quantity of water = 5 litres
Let x litres of water be mixed
$35/{5 + x} =3/1$
3x + 15 = 35
3x = 20
$x = 20/3 = 6{2}/3$ litres
GET ratio & proportion PRACTICE TEST EXERCISES
type 1 basic concepts of ratio & proportion
type 2 age based ratio & proportion problems
type 3 addition subtraction product on ratio & proportion
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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