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Proportional Reasoning-2 - Ratios with More than 2 Terms

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A ratio can compare more than two quantities, written as a:b:ca : b : c. This is known as a continued ratio.

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To simplify a ratio with more than two terms, divide all terms by their Highest Common Factor (HCF). For example, 15:20:2515 : 20 : 25 simplifies to 3:4:53 : 4 : 5 by dividing by 55.

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To combine two separate ratios like A:BA : B and B:CB : C into a single ratio A:B:CA : B : C, the value representing the common term (BB) must be made equal in both ratios using the Least Common Multiple (LCM).

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If a total quantity WW is divided in the ratio a:b:ca : b : c, the parts are calculated by dividing the total into (a+b+c)(a + b + c) equal parts and then multiplying by the respective ratio terms.

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The sum of the individual parts must always be equal to the total quantity: Part A+Part B+Part C=Total\text{Part } A + \text{Part } B + \text{Part } C = \text{Total}.

📐Formulae

Sum of ratio terms=a+b+c+…\text{Sum of ratio terms} = a + b + c + \dots

Value of one part=Total QuantitySum of ratio terms\text{Value of one part} = \frac{\text{Total Quantity}}{\text{Sum of ratio terms}}

Share of A=aa+b+c×Total Quantity\text{Share of } A = \frac{a}{a + b + c} \times \text{Total Quantity}

If A:B=x:y and B:C=y:z, then A:B:C=x:y:z\text{If } A:B = x:y \text{ and } B:C = y:z, \text{ then } A:B:C = x:y:z

If A:B=a:b and B:C=c:d, then A:B:C=(a×c):(b×c):(b×d)\text{If } A:B = a:b \text{ and } B:C = c:d, \text{ then } A:B:C = (a \times c) : (b \times c) : (b \times d)

💡Examples

Problem 1:

If P:Q=2:3P : Q = 2 : 3 and Q:R=4:5Q : R = 4 : 5, find the continued ratio P:Q:RP : Q : R.

Solution:

  1. Identify the common term, which is QQ.
  2. In the first ratio, Q=3Q = 3. In the second ratio, Q=4Q = 4.
  3. Find the LCM of 33 and 44, which is 1212.
  4. Adjust the first ratio: P:Q=(2×4):(3×4)=8:12P : Q = (2 \times 4) : (3 \times 4) = 8 : 12.
  5. Adjust the second ratio: Q:R=(4×3):(5×3)=12:15Q : R = (4 \times 3) : (5 \times 3) = 12 : 15.
  6. Since the value of QQ is now the same, P:Q:R=8:12:15P : Q : R = 8 : 12 : 15.

Explanation:

To link two ratios, we normalize the common variable by multiplying the ratios by factors that make the common variable's value equal to their LCM.

Problem 2:

Divide ₹3600₹ 3600 between A,B, and CA, B, \text{ and } C in the ratio 4:3:54 : 3 : 5.

Solution:

  1. Sum of ratio terms =4+3+5=12= 4 + 3 + 5 = 12.
  2. Total Amount =₹3600= ₹ 3600.
  3. AA's share =412×3600=4×300=₹1200= \frac{4}{12} \times 3600 = 4 \times 300 = ₹ 1200.
  4. BB's share =312×3600=3×300=₹900= \frac{3}{12} \times 3600 = 3 \times 300 = ₹ 900.
  5. CC's share =512×3600=5×300=₹1500= \frac{5}{12} \times 3600 = 5 \times 300 = ₹ 1500.

Explanation:

The total is divided into 12 equal parts. Each part is worth ₹300₹ 300. We then multiply this unit value by the ratio components for A,B, and CA, B, \text{ and } C.

Problem 3:

The sides of a triangle are in the ratio 2:3:42 : 3 : 4. If the perimeter is 54 cm54\text{ cm}, find the length of the longest side.

Solution:

  1. Let the sides be 2x2x, 3x3x, and 4x4x.
  2. Perimeter is the sum of all sides: 2x+3x+4x=542x + 3x + 4x = 54.
  3. Solve for xx: 9x=54x=549x=6\begin{array}{r} 9x = 54 \\ x = \frac{54}{9} \\ x = 6 \end{array}
  4. The longest side is 4x4x: 4×6=24 cm4 \times 6 = 24\text{ cm}.

Explanation:

By using a common multiplier xx, we can represent the terms of the ratio as actual lengths and solve the linear equation provided by the perimeter.