Ratio & Proportion, Alligation or Mixture

Let the third number be x.
First Number (120/100) × x = 6x/5
Second Number (150/100) × x = 3x/2
Ratio = 6x/5 : 3x/2
=> 4:5

Copper in 4 kg = 4/5 kg and
Zinc in 4 kg = 4 x (4/5) kg

Copper in 5 kg = 5/6 kg and
Zinc in 5 kg = 5 x (5/6) kg

Therefore, Copper in mixture = (4/5) + (5/6) = 49/30 kg
and Zinc in the mixture = (16/5) + (25/6) = 221/30 kg

Therefore the required ratio = 49/30 : 221/30
= 49 : 221.

Question: A mixture of 200 liters of milk and water contains 15% water. How much more water should be added so that water becomes 20% of the new mixture?

Solution:
Amount of water in the 200-liter mixture = 15% of 200
= (15/100) × 200
= 30 liters

Let,
P liters of water are added.
So, new amount of water = (30 + P)
and new total mixture = (200 + P)

ATQ,
(30 + P) = 20% of (200 + P)
⇒ 30 + P = (20/100) × (200 + P)
⇒ 30 + P = (1/5) × (200 + P)
⇒ 150 + 5P = 200 + P
⇒ 5P - P = 200 - 150
⇒ 4P = 50
∴ P = 12.5

∴ 12.5 liters more water should be added.

Question: Given that a shop opens at 10 a.m. and closes at 6:45 p.m., with a 20-minute tea break, what is the proportion of the break to the total working hours?

Solution:
Total working time = 6:45 - 10:00
= 8 hours 45 minutes
= (8 × 60) + 45
= 525 minutes

The ratio of the break to the total working period
= 20/525
= 4/105

Question: The present ratio of students to teachers at a certain school is 30 to 1. If the number of students were to increase by 50 and the number of teachers were to increase by 5, the ratio of students to teachers would then be 25 to 1. What is the present number of students?

Solution: 
Let the number of student and teacher be represented by x and y, respectively.
 x : y = 30 : 1
⇒ x/y = 30/1
⇒ x = 30y....................(1)

If the number of students increases by 50 and the number of teachers increases by 5
(50 + x) : (y + 5) = 25 : 1
⇒ (50 + x) / (y + 5) = 25/1
⇒ 50 + x = 25y + 125 
⇒ 50 + 30y = 25y + 125 
⇒ 30y - 25y = 125 - 50
⇒ 5y = 75
⇒ y = 75/5
⇒ y = 15

∴ Number of teachers = 15
And the number of students = 30 × 15 = 450

Given one star(S) equals four circles(C) and three circles equal four diamonds(D).
Then, S = 4C and 3C = 4D
Then, C = (4/3)D
∴ S = 4×(4/3)D
⇒ S = (16/3)D
Hence, S : D = 16 : 3

Question: The ratio of income and expenditure of a person is 5 : 2, If he saves Tk. 3000 per month, what is his monthly income?

Solution: 
Let, income = 5x
and expenditure = 2x 
Then,
Savings = 5x - 2x = 3000
⇒ 3x = 3000
∴ x = 1000

∴ monthly income = 5x = 5 × 1000 = Tk. 5000

Let number are = 2x, 3x, 4x
given,
LCM of (2×3×2)x = 12x
12x = 240
x = 20
∴ numbers are 2×20 = 40
3×20 = 60
4×20 = 80
∴ Smaller is 40

Let the salaries of A, B, C be x, 2x and 3x respectively.
Then, 2x + 3x = 6000
=> x = 1200.
A's salary = TK. 1200, B's salary = TK. 2400, and Cs salary TK. 3600.
Excess of C's salary over A's = [(2400/1200) x 100]
= 200%.

Present ratio of A & B
A : B
3 × (5/4) : 4 × (3/2)
= 15/4 : 12/4
= 15 : 24
= 5 : 8

A’s present salary = ( 4160 of 5/13) = 1600

.04 of x = .06 of y
⇒ x/y = .06/.4
⇒ x/y = 3/20
⇒ x : y = 3 : 20

Let man's rate upstream be x km/hr
Then, his rate downstream = 2x km/hr
∴ (speed in still water) : (Speed of stream)
(2x + x)/2 : (2x - x)/2
3x/2 : x/2
3 : 1

Question: In what ratio must a mixture of 30% alcohol strength be mixed with that of 50% alcohol strength so as to get a mixture of 45% alcohol strength?

Solution:
মনে করি, প্রথম প্রকারের x একক এবং দ্বিতীয় প্রকারের y একক মিশ্রিত করতে হবে।

প্রশ্নমতে,
প্রথম প্রকারের অ্যালকোহলের পরিমাণ = 30% of x = 0.30x
দ্বিতীয় প্রকারের অ্যালকোহলের পরিমাণ = 50% of y = 0.50y

মিশ্রণের মোট পরিমাণ = (x + y)
মিশ্রণে অ্যালকোহলের মোট পরিমাণ = 45% of (x + y) = 0.45(x + y)

শর্তমতে,
0.30x + 0.50y = 0.45(x + y)
⇒ 30x + 50y = 45(x + y)  [উভয় পক্ষকে 100 দ্বারা গুণ করে]
⇒ 30x + 50y = 45x + 45y
⇒ 50y - 45y = 45x - 30x
⇒ 5y = 15x
⇒ x/y = 5/15
⇒ x/y = 1/3
∴ x : y = 1 : 3

∴ The ratio is 1 : 3

According to the question,
Cost Price : Marked Price
(100 - Discount) : (100 + Profit)
100 - 10 : 100 + 12
90 : 112
45 : 56

Question: The expense of 9 pens and 5 pencils is the same as the expense of 7 pens and 8 pencils. What is the ratio between the price of one pen and one pencil?

Solution:
Let,
The price of one pen X Tk. 
The price of one pencil Y Tk. 

ATQ,
9X + 5Y = 7X + 8Y
⇒ 9X - 7X = 8Y - 5Y
⇒ 2X = 3Y
⇒ X/Y = 3/2
∴ X : Y = 3 : 2

Question: A and B are in the ratio of 6 : 5 and B and C are in the ratio of 4 : 3. What is the ratio of A : C?

 
Solution:
Given the ratio of,
A : B = 6 : 5 = (6 × 4) : (5 × 4) = 24 : 20

And,
B : C = 4 : 3 = (4 × 5) : (3 × 5) = 20 : 15

∴ A : C = 24 : 15 =  8 : 5

Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.
Then, the sum of their values = Tk. (25x/100) + (10 × 2x/100) + (5 × 3x/100) = Tk. 60x
Or, 60x/100 = 30
Or, x = (30 x 100)/60 = 50.
Hence, the number of 5 p coins = (3 x 50) = 150.

Question: A proficient worker is twice as efficient as an apprentice. After 10 days of joint work, they earn Tk. 45,000 together. What is the apprentice’s wage per day?

Solution: 
Given that, 
A seasoned workers efficiency is twice as much as an apprentice's.
They work together for 10 days
And earn Tk. 45000 together

Here, Efficiency of Seasoned Worker : Apprentice = 2 : 1
 Now, earning in 1 day = 45000/10 = Tk. 4500
So, Daily wage of Apprentice = 4500 × (1/3) = Tk. 1500

Thus, the daily wage of the apprentice is Tk. 1500.

Question: If A and B are in the ratio 3 : 4, and B and C are in the ratio 12 : 13. Then A and C will be in the ratio:

Solution:
দেওয়া আছে,
A : B = 3 : 4
⇒ A/B = 3/4

এবং
B : C = 12 : 13
⇒ B/C = 12/13

∴ (A/B) × (B/C) =  (3/4) × (12/13)
⇒ A/C = 9/13
∴ A : C = 9 : 13

Then A and C will be in the ratio is 9 : 13.

Question: Rafi weighs 72 kg. If he reduces his weight in the ratio 6 : 5, find his new weight​ in kg.

Solution:
ধরি, রাফির পূর্বের ওজন = 6x
রাফির পরের ওজন = 5x

প্রশ্নমতে,
6x = 72
⇒ x = 72 / 6 = 12

∴ ওজন কমে যাওয়ার পর হবে = 5x = 5 × 12 = 60 kg

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

Solution:
Let the quantity of liquid in the vessel = 6 units.
Out of this liquid, x units are replaced by water.

Amount of water in the new mixture = {2 - (2x/6) + x} unit
= 2 - (x/3) + x
= (6 - x + 3x)/3
= (2x + 6)/3 unit

Amount of juice in the new mixture = 4 - (4x/6) unit
= 4 - (2x/3) 
= (12 - 2x)/3 unit

ATQ,
(2x + 6)/3 = (12 - 2x)/3
⇒ 2x + 6 = 12 - 2x
⇒ 4x = 6
⇒ x = 3/2 
 
∴ The fraction of the mixture that is replaced is = (3/2) × (1/6) 
= 1/4

10% of X = 20% of Y
or, (10/100) × X = (20/100) × Y
or, X/10 = Y/5
or, 5X = 10Y
or, X = 2Y
or, X/Y = 2/1
So, X : Y = 2 : 1

Question: The ratio between the perimeter and the length of a rectangle is 5 : 2. If the area of the rectangle is 484 sq. cm, what is the length of the rectangle?

Solution: 
Let Length = l
& Breadth = b
Perimeter of a rectangle = 2(l + b)

Now,
2(l + b)/l = 5/2
⇒ 4(l + b) = 5l
⇒ l = 4b

Area, l × b = 484
⇒ 4b × b = 484
⇒ b² = 121
⇒ b = 11

∴ l = 4 × 11 = 44
The length of the rectangle is 44 cm. 

Current alcohol quantity = {(45/100) × 80}
= 36 Litres.

Let A be alcohol added.
So, 36 + A = (75/100) × (80 + A)
⇒ 36 + A = (3/4) × (80 + A)
⇒ 144 + 4A = 240 + 3A
∴ A = 96 Litres =

96 Litres is the additional quantity of alcohol to be added.

Question: The ratio of P : Q is 3 : 4 and the ratio of Q : R is 5 : 6. If P is equal to 9, what is the value of R?
Solution:
Given that,
P : Q = 3 : 4
Q : R = 5 : 6
And P = 9

Now,
P : Q = 3 : 4
or, P/Q = 3/4
or, 9/Q = 3/4
or, 3Q = 36
or, Q = 36/3
∴ Q = 12

And,
Q : R = 5 : 6
or, Q/R = 5/6
or, 12/R = 5/6
or, 5R = 72
or, R = 72/5
∴ R = 14.4

so, the value of R is 14.4

Question: A container has 64 liters of milk. From this container, 16 liters of milk was taken out and replaced with water. This process was repeated twice. What is the final ratio of milk and water in the container? 

Solution:
Initial quantity of milk = 64 liters
Removed quantity = 16 liters
Total quantity = 64 liters
Number of times the process is repeated = 2

Remaining milk = Initial quantity × {1 - (Removed Quantity/Total Quantity)}ⁿ
= 64 × {1 - (16/64)}²
= 64 × (3/4)²
= 36 liters

So, Final quantity of milk = 36 liters
Final quantity of water = (64 - 36) = 28 liters

Therefore, Ratio of milk to water = 36/28 = 9 : 7

Question: In a 90-liter mixture of milk and water, the ratio of milk to water is 2 : 1. How many liters of water must be added to make the ratio become 1 : 2?

Solution:
Total mixture = 90 litres
Given ratio (milk : water) = 2 : 1

Milk = 90 × (2/3) = 60 litres
Water = 90 − 60 = 30 litres

To make the ratio 1 : 2, x liters of water need to be added.
milk : water = 60 : (30 + x)

So,
60 / (30 + x) = 1 / 2

Cross-multiplying,
2 × 60 = 30 + x
120 = 30 + x
x = 120 − 30
∴ x = 90

Quantity of water to be added = 90 litres.

We know that
The ratio of Investment x Time = Ratio of Profit
∴ (A's investment × Time) : (B's investment × Time) = Profit of A : Profit of B

∴ Tk. 8000 × 12 months : Tk. 16000 × 9 months : Tk. 40000 × ? months = 6 : 9 : 5
∴ 96000 : 144000 : 40000 × ? = 6 : 9 : 5
By direct observation we can say, if common factor is K, then
6K = 96000;
∴ K = 16000;
and 5K = 40000 × ?
? = 5k/40000
= (5 × 16000)/40000
= 2 months.

প্রশ্ন: Haris and Sunny share some sweets in the ratio of 7 : 5. Haris has 12 more sweets than Sunny. How many sweets were there altogether?

সমাধান: 
Let,
Haris has 7x sweets 
Sunny has 5x sweets 
∴ Total sweets 7x + 5x = 12x

ATQ,
7x - 5x = 12
⇒ 2x = 12
∴ x = 6

∴ There were 12 × 6 = 72 sweets altogether.

Water in the mixture = 40×(10/100) = 4 litres
And, Milk in the mixture = 40 − 4 = 36 litres
Let x litres of water is mixed
⇒ (4 + x)/(40 + x) = 20/100
⇒ x = 5 litres

A : B = 100 : 75
B : C = 100 : 96

∴ A : C = A/B × B/C
= (100/75) × (100/96)
= 100/72
100 : 72

∴ A beats C by
100 - 72 = 28 m.

A : B : C = 7 : 8 : 11.
Hire charges paid by B = Tk. (520 × 8/26)
= Tk. 160.

Question: How much coffee, costing Tk. 100 per kg, should be mixed with 20 kg of cocoa priced at Tk. 300 per kg to get a blend worth Tk. 200 per kg?

Solution:
Ratio in which cocoa and coffee should be mixed
= 300 - 200 : 200 - 100
= 100 : 100
= 1 : 1

Let x be the quantity of coffee at 100/kg.

∴ 1 : 1 = x : 20
⇒ x = 20