Ratio & Proportion, Alligation or Mixture

Question: Rafi, Nabil, and Hasan started a business together. Rafi invested one-fourth of the total capital, Nabil invested one-sixth, and the remaining capital was invested by Hasan. What is the ratio of their profits?

Solution:
Let the total capital be 12x
Then, Rafi's share = 12x × (1/4) = 3x
Nabil's share = 12x × (1/6) = 2x
Hasan's share = 12x - (3x + 2x) = 7x

So, required ratio = 3x : 2x : 7x = 3 : 2 : 7

Given k : l = 4 : 3 = 20 : 15
and, l : m = 5 : 3 = 15 : 9
∴ k : l : m = 20 : 15 : 9

Question: A vessel contains 140 litres of milk and water in the ratio of 4 : 3. If 20 litres of milk and 30 litres of water are added to the mixture, the difference between milk and water in the final mixture is Y. Find the value of 7Y?

Solution:
Given,
Total mixture = 140 litres
Ratio of milk and water = 4 : 3
Sum of the ratios = 4 + 3 = 7

Initially,
Quantity of milk = 140 × (4/7) = 80 litres
Quantity of water = 140 × (3/7) = 60 litres

After adding 20 litres of milk and 30 litres of water:
New quantity of milk = 80 + 20 = 100 litres
New quantity of water = 60 + 30 = 90 litres

According to the question, the difference between milk and water is Y:
∴ Y = |100 - 90| = 10 litres

∴ The value of 7Y = 7 × 10 = 70

Suppose B joined after x months
21000 × 12 = 36000 × (12 - x)
⇒ 36x = 180
⇒ x = 5

The ratio of original money numbers = 3:8
Common factor helps in finding actual values easily
So, take 'M' as a common factor.
∴ Original numbers will be 3M and 8M
Adding 5 to them, we get (3M + 5) and (8M+5)
∴ (3M + 5)/(8M + 5) = 2/5 ............ (Ratio of new numbers is 2:5)
∴ 15M + 25 = 16M + 10
∴ M = 15
Smaller money value is 3M = 3 x 15 = 45.

Required ratio:
= (65×8):(70×4)
= 520:280
= 13:7

Ration of A : B = 3 : 2
10 L taken out and replaced by B
So, A remain = 3x - 3×10/5 = 3x - 6 And
B remain = 2x - 2×10/5 +10
= 2x + 6
ATQ,
(3x - 6)/(2x + 6) = 2/3
Or, 9x - 18 = 4x + 12
Or, 5x = 30
or, x = 6
The total quantity of mixture = (3x + 2x) = 5x = 5 × 6 = 30 L [Answer.]

Question: How much should be added to each term of 4 : 7 so that it becomes 2 : 3?

Solution:
Given that,
Ratio of two numbers is 4 : 7
Let the number added to denominator and numerator be 'x' 

Now according to the question,
(4 + x) : (7 + x) = 2 : 3
⇒ (4 + x)/(7 + x) = 2/3
⇒ 12 + 3x = 14 + 2x
∴ x = 2 

∴ 2 will be added to make the term in the ratio of 2 : 3.

Question: Two containers contain milk and water in the ratios 5 : 2 and 9 : 5. What ratio should the mixtures be combined in to achieve a final ratio of 2 : 1 milk to water?

Solution:
Let,
P unit of the first mixture is added to Q unit of the second mixture.

So, in the P unit of the first mixture,
Amount of milk present = (5/7) × P = 5P/7
Amount of water present = (2/7) × P = 2P/7

In the Q unit of the second mixture,
Amount of milk present = (9/14) × Q = 9Q/14
Amount of water present = (5/14) × Q = 5Q/14

ATQ,
{(5P/7) + (9Q/14)}/{(2P/7) + (5Q/14)} = 2/1
⇒ {(10P + 9Q)/14}/{(4P + 5Q)/14} = 2
⇒ 10P + 9Q = 8P + 10Q
⇒ 2P = Q

∴ P : Q = 1 : 2

Here,                    A   :   B    :   C
Ratio of Profit → 2   :   3    :    7
Average gain= (2+3+7)/3 = 4 units
According to the question,
4 units= Tk. 8000
1 unit= Tk. 2000
3 units=3×2000= Tk. 6000
∴Share of B= Tk. 6000

Question: If 25% of (A + B) = 50% of (A - B), then find B : A - 

Solution: 
25% of (A + B) = 50% of (A - B)
⇒ (A + B) × 25/100 = (A - B) × 50/100
⇒ (A + B)/4 = (A - B)/2
⇒ 2 (A + B) = 4 (A - B)
⇒ 2A + 2B = 4A - 4B 
⇒ 4A - 2A = 2B + 4B
⇒ 2A = 6B
⇒ A/B = 6/2 = 3/1
∴ B/A = 1/3
= 1 : 3

Question: In a school there are 286 students where boys and girls are in the ratio 8:5. If 22 new girls are admitted, then new ratio of boys and girls:

Solution:
দেওয়া আছে,
একটি বিদ্যালয়ে মোট ছাত্র-ছাত্রীর সংখ্যা 286 জন
বিদ্যালয়ে বালক ও বালিকার সংখ্যার অনুপাত 8:5

মনে করি,
বিদ্যালয়ে বালকের সংখ্যা 8x জন
বিদ্যালয়ে বালিকার সংখ্যা 5x জন

প্রশ্নমতে,
8x + 5x = 286
⇒ 13x = 286
⇒ x = 286/13
∴ x = 22

∴ বিদ্যালয়ে বালকের সংখ্যা (8 × 22) = 176 জন
∴ বিদ্যালয়ে বালিকার সংখ্যা (5 × 22) = 110 জন

∴ যদি 22 জন বালিকা নতুন করে ভর্তি হয় তাহলে বালিকার সংখ্যা হবে = (110 + 22) = 132 জন

∴ বিদ্যালয়ে বালক ও বালিকার সংখ্যার নতুন অনুপাত = 176 : 132
= 4 : 3 [44 দ্বারা ভাগ করে]

Milk = 144 × 5/12 = 60L
Water = (144 - 60) = 84 L
ATQ,
(60 + x)/(144 + x) = 23/44
Or, 2640 + 44x = 3312 + 23x
Or, 21x = 672
Or, x = 672/21
or, x = 32L

Solution 2:
ATQ,
(60 + x)/84 = 23/21
Or, 60×21 + 21x = 23×84
Or, 21x = 672
or, x = 32 L

Question: The monthly incomes of two persons are in the ratio 5 : 4 and their monthly expenditures are in the ratio of 9 : 7. If each saves BDT 50 per month, what would be the total amount of their monthly expenditure?

Solution:
Let,
Their monthly income 5a and 4a
Their monthly expenses 9b and 7b

ATQ,
5a - 9b = 50 ..............(1)
4a - 7b = 50 ...............(2)

Multiply equation (1) and (2) by 4 and 5 respectively,
20a - 36b = 200 .............(3)
20a - 35b = 250 ..............(4)

(3) - (4) ⇒
20a - 36b- 20a + 35b = 200 - 250
or, - b = - 50
∴ b = 50

Hence, their total monthly expenditure = 9 × 50 + 7 × 50 = 450 + 350 = 800 tk

Question: 20 litres of a mixture contains milk and water 4 : 1. Then the amount of water to be added to the mixture so as to have milk and water in ratio 2 : 1 is-

Solution:
In 20 litres of mixture,
quantity of mik = 20 × (4/5) = 16 litres
quantity of water = 20 × (1/5) = 4 litres
Let,
The quantity of water be added m litres
ATQ,
16 : (4 + m) = 2 : 1
or, 16/(4 + m) = 2/1
or, 2m + 8 = 16
or, 2m = 16 - 8
or, 2m = 8
∴ m = 8/2 = 4

∴ 4 litres water to be added to the mixture.

Question: P and Q are two alloys of gold and copper prepared by mixing metals in the ratio 7 : 2 and 7 : 11 respectively. If equal amounts of both alloys are melted to form a third alloy R, then the ratio of gold and copper in R will be - 

Solution: 
Alloy P (Gold : Copper) = 7:2
Gold fraction in P = 7/9
Copper fraction in P = 2/9

Alloy Q (Gold : Copper) = 7:11
Gold fraction in Q = 7/18
Copper fraction in Q = 11/18

Let, 1 kg of each alloy is mixed; 
Gold in Alloy R = 7/9 + 7/18 = 21/18

Copper in Alloy R = 2/9 + 11/18 = 15/18 

∴ The ratio of Gold:Copper in R = (21/18) : (15/18)
= 21 : 15
= 7 : 5

salt = 16 × 1/4 = 4 Kg
water = 16 - 4
= 12 kg.
এখন water/salt = 4
∴ water = 4 × 4 = 16 Kg
∴ (16 - 12) = 4 Kg water add করতে হবে।

W : R : G = 2 : 3 : 5 [Initial]
W : R : G = 2 : 3 : 7 [Final]
Let, Initial Green = 5x Final Green = 7x
ATQ,
(7x - 5x) = 6
=> 2x = 6
=> x = 3
So, white marbles in the jar = (2 × 3) = 6

Question: Two brands of detergent are to be combined. Detergent A contains 30 percent bleach and 70 percent soap, while Detergent B contains 50 percent bleach and 50 percent soap. If the combined mixture is to be 40 percent bleach, what percent of the final mixture should be Detergent A?

Solution:
ধরি, মিশ্রণে Detergent A এর অংশ x% 
∴ Detergent B এর অংশ হবে (100 - x%)

এখানে,
ডিটারজেন্ট A থেকে প্রাপ্ত ব্লিচের পরিমাণ হলো (x% এর 30%)।
ডিটারজেন্ট B থেকে প্রাপ্ত ব্লিচের পরিমাণ হলো (100 - x)% এর 50%।
চূড়ান্ত মিশ্রণে ব্লিচের মোট পরিমাণ হলো (100% এর 40%)।

প্রশ্নমতে,
0.30x + 0.50 × (100 - x) = 0.40 × 100
⇒ 0.30x + 50 - 0.50x = 40
⇒ - 0.20x = - 10
⇒ x = - 10/(- 0.20)
∴ x = 50

অর্থাৎ, মিশ্রণের 50% হবে Detergent A

Question: A 42 liter mixture contains milk and water in the ratio 3 : 4. How many liters of milk must be added to the mixture so that the ratio of milk to water becomes 1 : 1?

Solution:
The ratio of milk to water is 3 : 4
Total portion = 3 + 4 = 7

Quantity of milk = 42 × (3/7) = 18 liters.
Quantity of water = 42 × (4/7) = 24 liters.

Let,
Quantity of milk to be added = x liters

According to the question,
(18 + x) : 24 = 1 : 1
⇒ (18 + x)/24 = 1/1
⇒ 18 + x = 24
⇒ x = 24 - 18
⇒ x = 6

∴ Quantity of milk to be added = 6 liters

Question: Silver is 17 times as heavy as water and copper is 7 times as heavy as water. In what ratio should these be mixed to get alloy 13 times as heavy as water?

Solution:
Given that,
Density of silver is 17 times as heavy as water
Density of copper is 7 times as heavy as water
Mixture should be 13 times as heavy as water

Let the weights of silver and copper be x and y respectively. Then we get, 
⇒ (17x + 7y)/(x + y) = 13
⇒ 17x + 7y = 13(x + y)
⇒ 17x + 7y = 13x + 13y 
⇒ 17x - 13x = 13y - 7y
⇒ 4x = 6y
⇒ x/y = 6/4 = 3/2
∴ x : y = 3 : 2

So Ratio of silver to copper = 3 : 2

ATQ,
x - y = 12 ...... (i)
x + y = 38 ........ (ii)
(i) + (ii), 2x = 50
Or, x = 25
So, y = 13
If 2 is added in both the numbers, then their ratio is:
x+2 / y+2
= 25+2 / 13+2
= 27/15
= 9/5

Question: 94 is divided into two parts such that the fifth part of the first and the eighth part of the second are in the ratio 3 : 4. Find the first part.

Solution:
Let the two parts be x and 94 - x.

According to the problem,
(x/5) : (94 - x)/8 = 3 : 4
⇒ (x/5)/{(94 - x)/8} = 3/4
⇒ 8x/5(94 - x) = 3/4
⇒ 32x = 15(94 - x)
⇒ 32x = 15 × 94 - 15x
⇒ 47x = 15 × 94
⇒ x = (15 × 94)/47
∴ x = 30

∴ First part is 30

In 1st mixture, water = 10/100 × 20 = 2 kg
So, Spirit = 20-2 = 18 kg
In 2nd mixture where the water is 25%,
75 kg of spirit is contained in 100 kg mixture
So, 18 kg spirit is contained in = (100×18)/75 = 24 kg
So, water to be added = 24-20 = 4 kg

ধরি,
বালক আছে x জন
বালিকা আছে = x এর 120% = 120x/100
= 1.2x জন।
প্রশ্নমতে, x + 1.2x = 66
⇒ 2.2x = 66
⇒ x = 66/2.2
⇒ x = 30
অতএব বালিকা আছে = 1.2x = 1.2 × 30 = 36
4 জন বালিকা ভর্তি হলে = 36 + 4 = 40 জন
∴ বালকঃ বালিকা = 30 : 40
= 3 : 4

Question: If x is 90% of y then what percent of x is y?

Solution: 
x = 90% of y
⇒ x = 90y/100
⇒ x = 9y/10
⇒ y/x = 10/9
= (10/9) × 100%
= 111.1%

A : B : C
= 10 × 7 : 12 × 5 : 15 × 3
= (2 × 7) : (12 × 1) : (3 × 3)
= 14 : 12 : 9
Amount that C should pay
= 175 × (9/35)
= 5 × 9
= 45.

Number of green ball = 24 × (3/8) = 9
Number of red ball= 24 × (5/8) = 15
since the required ratio is 1:1
so, additional green ball = (15 - 9) = 6

Question: In a mixture, the ratio of the milk and water is 6: 5. When 22 liter mixture is replaced by water, the ratio becomes 9 : 13. What is the quantity of water after replacement?

Solution:
Given that,
milk : water = 6 : 5
And 22 liter mixture are replaced by water

Now,
Let milk = 6x and water = 5x
In 22 liter mixture, milk removed = (6/11) × 22 = 12 liter
And water removed = (5/11) × 22 = 10 liter

According to question,
(6x - 12) : (5x - 10 + 22) = 9 : 13
⇒ 13(6x - 12) = 9(5x + 12) 
⇒ 78x - 156 = 45x + 108
⇒ 78x - 45x  = 108 + 156
⇒ 33x = 264
⇒ x = 8
∴ Initial water = 5x = 5 × 8 = 40 liters
Water removed in 22 L mixture = 10 liters
And water added back = 22 liters

∴ Water after replacement = Initial water - water removed + water added 
= 40 - 10 + 22
= 52 liters 

So the quantity of water after replacement is 52 liters.

In 1st mixture, water = 10/100 × 20 = 2 kg
So, Spirit = 20-2 = 18 kg
In 2nd mixture where the water is 25%,
75 kg of spirit is contained in 100 kg mixture
So, 18 kg spirit is contained in = (100×18)/75 = 24 kg
So, water to be added = 24-20 = 4 kg

Question: Two numbers are in the ratio 2 : 3. If 4 is subtracted from the first number, the ratio becomes 1 : 2. What are the numbers?

Solution:
Let the two numbers be: 2x and 3x

According to the question,
(2x - 4)/3x = 1/2
⇒ 2(2x - 4) = 3x
⇒ 4x - 8 = 3x
⇒ x = 8

∴ First number = 2 × 8 = 16
∴ Second number = 3 × 8 = 24

According to the question,
Acid : Water -
Vessel A - 4 : 3
Vessel B - 2 : 3
Now using alligation,

a² + b² + c² - ab - bc - ca = 0 .....(i)
Multiple equation (i) by 2 we get
⇒ 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0
⇒ (a² + b² - 2ab) + (b² + c² - 2bc) + (c² + a² - 2ca) = 0 [ (a + b)² = a² + b² + 2ab]
⇒ (a - b)² + ((b - c)² + (c - a)² = 0 [if x² + y² + z² = 0 then x = 0, y = 0, z = 0]
∴ a - b = 0
⇒ a = b
b - c = 0
⇒ b = c
c - a = 0
⇒ c = a
∴ a : b : c = 1 : 1 : 1

Question: A mixture contains two liquids 'A' and 'B' in the ratio 5 : 3. If 8 litres of the mixture is withdrawn and replaced with 8 litres of 'A', the ratio becomes 2 : 1. What was the initial quantity of 'B'?

Solution:
ধরি, প্রাথমিক মিশ্রণের পরিমাণ ছিল 8x লিটার।
যেখানে A এর পরিমাণ = 5x লিটার এবং B এর পরিমাণ = 3x লিটার।

8 লিটার মিশ্রণ তুলে নেওয়ার পর,
মিশ্রণে A এর পরিমাণ = 5x - (5/8) × 8 = 5x - 5 লিটার।
মিশ্রণে B এর পরিমাণ = 3x - (3/8) × 8 = 3x - 3 লিটার।

নতুন 8 লিটার 'A' যোগ করার পর,
A এর নতুন পরিমাণ = (5x - 5) + 8 = 5x + 3 লিটার।

প্রশ্নানুযায়ী, নতুন অনুপাত,
⇒ (5x + 3) / (3x - 3) = 2/1
⇒ 1(5x + 3) = 2(3x - 3)
⇒ 5x + 3 = 6x - 6
⇒ 3 + 6 = 6x - 5x
⇒ 9 = x

সুতরাং, প্রাথমিকভাবে B এর পরিমাণ ছিল = 3x = 3 × 9 = 27 লিটার।