Ratio & Proportion, Alligation or Mixture

Question: In a class, the number of girls is 20% more than that of the boys. The strength of the class is 66. If 4 more girls are admitted to the class, the ratio of the number of boys to that of the girls is-

Solution:
Let the number of boys be 5k.
Then the number of girls = 20% more than the boys = 5k × 1.2 = 6k

Total students = 66
5k + 6k = 66
⇒ 11k = 66
⇒ k = 66/11 
∴ k = 6

∴ Number of boys = 5 × 6 = 30
∴ Number of girls = 6 × 6 = 36

Now, when 4 more girls are admitted than New number of girls = 36 + 4 = 40

∴ Ratio of boys to girls = 30 : 40 = 3 : 4 (dividing both by 10)

So the ratio of the number of boys to girls is 3 : 4

Let, these two numbers be 3x and 4x then their LCM = 12x
Now, according to question,
12x =  60
Or, x = 5
Thus, the numbers are (3x = 3 × 5) = 15 and (4x = 4 × 5) = 20

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively
Quantity of A in mixture left = {7x - (7/12) × 9} = 7x - 21/4
Quantity of B in mixture left = {5x - (5/12) × 9} = 5x - 15/4

According to the question,
(7x - 21/4)/{(5x - 15/4) + 9} = 7/9
⇒ (28x - 21)/(20x + 21) = 7/9
⇒ 252x - 189 = 140x + 147
⇒ 252x - 140x = 147 + 189
⇒ 112x = 336
⇒ x = 3.

So, they can contain 21 litres of A.

Question: How many kilogram of sugar costing Tk. 9 per kg must be mixed with 27 kg of sugar costing Tk. 7 per kg so that there may be a gain of 10% by selling the mixture at Tk. 9.24 per kg?

Solution:
Let,
Quantity of sugar costing Tk. 9 per kg = x
S.P. of 1 kg of mixture = Tk. 9.24,
Gain 10%.
C.P. of 1 kg of mixture = Tk. (100/110) × 9.24 = Tk. 8.40

By the rule of allilation, we have:

⇒ 27 : x = (9 - 8.40) : (8.40 - 7) = 0.60 : 1.40 = 6 : 14 = 3 : 7
⇒ 27/x = 3/7
⇒ x/27 = 7/3
⇒ x = (7 × 27)/3
∴ x = 63

Question: In 2 kg mixture of copper and aluminum, 30% is copper. How much aluminum powder should be added to the mixture so that the quantity of copper becomes 20%?

Solution:
According to the question,
Mixture of copper and aluminum = 2 Kg = (2 × 1000) = 2000 gm
30% of this mixture is copper,
= (30/100) × 2000 gm
= 600 gm copper
∴ In 2 kg mixture of copper and aluminum, aluminum is = (2000 - 600) = 1400 gm

Let x be the mass of aluminum added. The new total mass is (2000 + x).
 We want the 600 gm of copper to be 20% of this new total.
⇒ 600 = 0.20(2000 + x)
⇒ 600 = 400 + 0.2x
⇒ 0.2x = 200
∴ x = 1000 gm

Question: How much water should be added to 40 liters of pure milk to gain an extra 20% profit when selling the mixture at the price of pure milk?

Solution:
Assume,
Price of pure milk per liter = 100 Taka
So, the price of 40 liters of pure milk = 40 × 100 = 4000 Taka

Let, the amount of water added = x liters
Total mixture = (40 + x) liters

Since the mixture is sold at the price of pure milk,
Selling price of (40 + x) liters = 100(40 + x) Taka

According to the question,
100(40 + x) = 4000 + 4000 of 20%
⇒ 4000 + 100x = 4000 + (4000 × 20/100)
⇒ 4000 + 100x = 4000 + 800
⇒ 100x = 800
⇒ x = 800/100
⇒ x = 8

∴ Required water: 8 liters.

Question: When 25% of the first number is added to the second number, the second number becomes 3/2 times the first number. What is the ratio of the first number to the second number?

Solution:
Let the first number = x
and the second number = y.

According to the question,
y + 25% of x = (3/2)x
⇒ y + (25/100)x = (3/2)x
⇒ y + (1/4)x = (3/2)x
⇒ y = (3/2)x - (1/4)x
⇒ y = (6 - 1)x/4
⇒ y = 5x/4
∴ y = 5x/4

Therefore, x : y = 4 : 5 

Let, B invested the money for t months
Then the ratio of investment = (12×11 : 11× t) = 12 : t
So, 12/t = 4/1
⇒ t = 3 months

Solution: 
Let, present ages of three persons 4x, 7x, 9x 

Eight years ago, the sum of their ages was 56 years
At present, the sum of their ages is = 56 + 8 + 8 + 8 
= 56 + 24 years 
= 80 years 

4x + 7x + 9x = 80 
⇒ 20x = 80
⇒ x = 4 

The present age of the eldest person is = 4 × 9 = 36 years 

Question: The ratio of milk to water in a mixture is 5:3. When adding 4 liters of water, the ratio becomes 5:5. What was the quantity of milk in the original mixture? 

Solution:
Let the initial quantity of 
Milk = 5x liters
Water = 3x liters

When 4 liters of water are added, the new quantity of water becomes = (3x+4) liters
The new ratio becomes 5:5, which simplifies to 1:1. This means the amount of milk and water are now equal. 
5x=3x+4
2x = 4 
∴ x = 2

So, the initial quantity of Milk = 5 × 2 = 10 liters

We know that, sinθ = AB/AC
⇒ sinθ = 5/13
⇒ θ = sin^-1(5/13)
∴ θ = 22.62°

Question: A jar is filled with liquid, 2 parts of which are water and 4 parts milk. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half milk?

Solution:
মনে করি,
পাত্রটিতে তরল পদার্থ আছে = 6 ইউনিট
এবং এই তরলের x ইউনিট পানি দ্বারা প্রতিস্থাপন করা হল।

∴ নতুন মিশ্রণে পানির পরিমাণ = (2 - (2x/6) + x) 
= 2 - (x/3) + x 
= (6 - x + 3x)/3
= (2x + 6)/3 ইউনিট 

এবং, নতুন মিশ্রণে দুধের পরিমাণ = 4 - (4x/6)
= 4 - (2x/3)
= (12 - 2x)/3 ইউনিট

প্রশ্নমতে,
(2x + 6)/3 = (12 - 2x)/3
⇒ 2x + 6 = 12 - 2x
⇒ 2x + 2x = 12 - 6
⇒ 4x = 6
⇒ x = 3/2 
 
∴ মিশ্রণের প্রতিস্থাপিত অংশের পরিমাণ = (3/2) × (1/6)
= 1/4 

Question: The ratio of orange and water in a mixture is 2 : 3, then the percentage of water in mixture is-

Solution: 
The ratio of orange and water in a mixture is 2 : 3
Total = 5

∴ percentage of water = (3/5) × 100%
= 60%

Question: If two times A is equal to three times of B and also equal to four times of C, then A : B : C is -

Solution:
2A = 3B
Or, B = 2A/3
and 2A = 4C
Or, C = A/2

Hence, A : B : C = A : 2A/3 : A/2
= 1 : 2/3 : 1/2
= 6 : 4 : 3

A:: - G : S = 40 : 60
B:: - G : S : C = 35 : 40 : 25
New, G : S = {(1×40) + (4×35)} : {(40×4) + (1×60)}
= 180 : 220
= 18 : 22
= 9 : 11 [Answer. ]

Nirmal:Kapil = 9000×12:(12000×6 + 6000×6)= 1:1
Kapil's share = Tk [4600 ×(1/2)) = Tk. 2300

Question: After a 25% discount, the price of a refrigerator is Tk. 18,000. What was the price before the discount?

Solution:
In 25% discount,
Discount price = 75 when original price = 100

∴ Discount price 1 = 100/75 of original price

∴ Discount price 18,000 = (18,000 × 100)/75
= 24,000

∴ Original price = Tk. 24,000

Question: If 2/3 of X = 75% of Y = 0.4 of Z, then find the ratio X : Y : Z.

Solution:
According to the question,
(2/3) × X = 75% of Y
⇒ 2X/3 = 75Y/100
⇒ 2X/3 = 3Y/4
⇒ X/Y = 9/8

Now,
75% of Y = 0.4 of Z
⇒(75Y/100) = (2Z/5)
⇒ (3Y/4) = (2Z/5)
⇒ 15Y = 8Z
⇒ Y/Z = 8/15

X : Y : Z = (9/8) : 1 : (15/8)
Multiply all terms by 8:
X = 9 × 1 = 9
Y = 1 × 8 = 8
Z = (15/8) × 8 = 15

Therefore,
X : Y : Z = 9 : 8 : 15

Let,
the full marks for each paper be x.
Let the marks obtained in the five papers be 6y, 7y, 8y, 9y and 10y respectively.
Then, (6y + 7y + 8y + 9y + 10y)/5x = 60/100
⇒ 40y/5x = 3/5
⇒ 40y = 3x
⇒ x = (40/3)y
50% of x = (50/100) × (40/3)y
= 20y/3
= 6(2/3)y.
Clearly, the student got more than 50% marks in each of the last 4 papers.

Number of liters of water in 125 liters of the mixture = 20% of 150 = 1/5 of 150 = 30 liters
Let us assume that other ‘P’ liters of water are added to the mixture to make water 25% of the new mixture.
So, the total amount of water becomes (30 + P) and
the total volume of the mixture becomes (150 + P)
Thus,
(30 + P) = 25% of (150 + P)
Solving, we get P = 10 liters

Question: Seats for Chemistry, Bangla and English in a school are in the ratio 5 : 8 : 10. There is a proposal to increase these seats by 20%, 30% and 40% respectively. What will be the ratio of increased seats?

Solution:

Originally, let the number of seats for Chemistry, Bangla and English be 5x, 8x and 10x respectively

Number of increased seats are (120% of 5x), (130% of 8x) and (140% of 10x)
= {(120/100) × 5x}, {(130/100) × 7x} and {(140/100) × 8x}
= 6x, 52x/5 and 14x

∴ The required ratio = 6x : 52x/5 : 14x
= 30 : 52 : 70
=15 : 26 : 35

Let x be the amount of water to be added.
New total amount of water = 4 + x
Total amount of chemical = 4
∴ New total = 4 + 4 + x = 8 + x
Final ratio required (for water) = 5/(5 + 3) = 5/8
Thus, (4 + x)/(8 + x) = 5/8
Or, 32 + 8x = 40 + 5x
Or, 3x = 8
Or, x = 8/3

Question: Rahim weighs 74.5 kg. If he reduces his weight in the ratio 5 : 3, find his reduced weight in kg.

Solution:
মনেকরি,
রহিমের পূর্বের ওজন = 5x
রহিমের বর্তমান ওজন = 3x

প্রশ্নমতে,
⇒ 5x = 74.5
⇒ x = 74.5/5
∴ x = 14.9

∴ ওজন কমে যাওয়ার পর হবে = 3 × 14.9
= 44.70 kg

Question: In a partnership, three friends invested Tk. 2700, Tk. 8100, and Tk. 7200 respectively. After one year, the third partner received Tk. 3600 as his share of the profit. How much total profit did the business make?

Solution:
Let the total profit be x

Here,
ratio of investment,
= first partner : second partner : third partner 
= 2700 : 8100 : 7200
= 3 : 9 : 8

Then, third partner's share = 8x/20 = 2x/5

According to the question,
2x/5 = 3600
⇒ x = (3600 × 5)/2 
⇒ x = 9000

So, The total profit is Tk. 9000

Question: X and Y started a partnership business investing some amount in the ratio of 4 : 7. Z joined after 8 months with an amount equal to that of X. In what proportion should the profit at the end of one year be distributed among X, Y, and Z?

Solution:
Let the initial investments of X and Y be 4b and 7b.

X : Y : Z = (4b × 12) : (7b × 12) : (4b × 4)
= 48 : 84 : 16
= 12 : 21 : 4

∴ Required proportion - 12 : 21 : 4

Question: A 60-liter mixture contains milk and water in the ratio 5:1. How many liters of water must be added to the mixture so that the new ratio of milk to water becomes 2:1?

Solution:
Milk in the mixture = 60 × (5/6) = 50 liters
Water in the mixture = 60 - 50 = 10 liters

Let x liters of water be added.
Then the new ratio should be 2 : 1.

So,
50/(10 + x) = 2/1
⇒ 50 = 2(10 + x)
⇒ 50 = 20 + 2x
⇒ 2x = 30
⇒ x = 15

∴ 15 liters of water should be added.

Now,
The total parts of the mixture = 4 + 3 = 7 parts.
Quantity of milk in the mixture = (4/7) × 630 = 360 liters of milk.
Quantity of water in the mixture = (3/7) × 630 = 270 liters of water.

When 140 liters of the mixture is taken out, the ratio of milk and water in the 140 liters will also be 4 : 3.
Milk taken out = (4/7) × 140 = 80 liters.

So, the remaining milk = 360 - 80 = 280 liters.

∴ The quantity of milk now is 280 liters.

ধরি, C এর ইনকাম 100 টাকা
তাহলে B এর ইনকাম (100 - 100 এর 20%) = 80 টাকা
এবং A এর ইনকাম (80 + 80 এর 10 %) = 88 টাকা।
∴ A : B : C = 88 : 80 : 100 = 22 : 20 : 25

Question: If the ratios A : B = 1 : 2, B : C = 4 : 3, hold true, and their total sum is 630, determine C.

Solution: 
Given,
A : B = 1 : 2 = 2 : 4
B : C = 4 : 3

∴ A : B : C = 2 : 4 : 3

∴ Value of C = 630 × 3/9
= 210

Number of rounds to be completed = 4 / 1/4 = 16 times
Since the ratio of speeds of A and B is 5 : 4
⇒ When A covers 5 rounds, B will complete 4 rounds and they will meet.
⇒ A will pass B after 5^th, 10^th & 15^th round.
∴ The winner passes the other thrice.

Let the numbers be 7x and 4x
ATQ,
(7x + 8) : (4x + 8) = 13 : 8
Or, 56x + 64 = 52x + 104
Or, 4x = 40

The smaller number is 40

Let the number A, B and C be 12a , 15a and 25a respectively
Then from the question 12a + 15a + 25a = 312
⇒ 52a = 312
⇒ a = 312/52
⇒ a = 6
Required ratio = {(15 × 6) - (12 × 6)}/{(25 × 6) - (15 × 6)}
= 3/10
= 3 : 10.

Question: The ratio between two numbers is 7 : 9. If 25% of the smaller number is added to 30% of the larger number, the result is 178. What is the greater number?

Solution:
Let the numbers be 7x and 9x.
Then, 
25% of the smaller = 0.25 × 7x = 1.75x
30% of the larger = 0.30 × 9x = 2.7x

According to the question,
1.75x + 2.7x = 178
⇒ 4.45x = 178
⇒ x = 178/4.45
∴ x = 40

∴ The greater number is (9 × 40) = 360