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Proportional Reasoning-2 - Dividing a Whole in a Given Ratio

Grade 8CBSE

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

🔑Concepts

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To divide a total quantity into a given ratio a:ba : b, the total quantity is treated as being made up of a+ba + b equal parts.

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The first share is calculated as aa+b\frac{a}{a+b} of the total, and the second share is ba+b\frac{b}{a+b} of the total.

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If a quantity is divided into three parts in the ratio a:b:ca : b : c, the sum of the terms is a+b+ca + b + c. Each part is then calculated as termsum of terms×Total Quantity\frac{\text{term}}{\text{sum of terms}} \times \text{Total Quantity}.

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The sum of all the individual parts must always equal the original whole quantity.

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Ratios do not have units, but the parts calculated from them will have the same units as the total quantity (e.g., kilograms, rupees, liters).

📐Formulae

Sum of terms=a+b\text{Sum of terms} = a + b

First Part=aa+b×Total Quantity\text{First Part} = \frac{a}{a + b} \times \text{Total Quantity}

Second Part=ba+b×Total Quantity\text{Second Part} = \frac{b}{a + b} \times \text{Total Quantity}

Value of nth part=Ratio of nth part∑Ratio terms×Total Quantity\text{Value of } n^{th} \text{ part} = \frac{\text{Ratio of } n^{th} \text{ part}}{\sum \text{Ratio terms}} \times \text{Total Quantity}

💡Examples

Problem 1:

Divide ₹ 25002500 between Rahul and Shweta in the ratio 3:23 : 2.

Solution:

  1. Sum of the ratio terms: 3+2=53 + 2 = 5.
  2. Rahul's share: 35×2500=3×500=₹1500\frac{3}{5} \times 2500 = 3 \times 500 = ₹ 1500.
  3. Shweta's share: 25×2500=2×500=₹1000\frac{2}{5} \times 2500 = 2 \times 500 = ₹ 1000. Check: 1500+10002500\begin{array}{r} 1500 \\ + 1000 \\ \hline 2500 \end{array}

Explanation:

We first find the total number of parts by adding the terms of the ratio (3+2=53+2=5). We then multiply the total amount by the fraction representing each person's share.

Problem 2:

The three angles of a triangle are in the ratio 1:2:31 : 2 : 3. Find the measure of each angle.

Solution:

  1. We know the sum of angles in a triangle is 180∘180^{\circ}.
  2. Sum of ratio terms: 1+2+3=61 + 2 + 3 = 6.
  3. First angle: 16×180∘=30∘\frac{1}{6} \times 180^{\circ} = 30^{\circ}.
  4. Second angle: 26×180∘=60∘\frac{2}{6} \times 180^{\circ} = 60^{\circ}.
  5. Third angle: 36×180∘=90∘\frac{3}{6} \times 180^{\circ} = 90^{\circ}.

Explanation:

Using the Angle Sum Property of a triangle, we identify the total as 180∘180^{\circ} and divide it into 66 parts according to the given ratio.

Problem 3:

A piece of wire 120 cm120 \text{ cm} long is cut into two pieces in the ratio 7:57 : 5. Find the difference in length between the two pieces.

Solution:

  1. Sum of terms: 7+5=127 + 5 = 12.
  2. Length of longer piece: 712×120=70 cm\frac{7}{12} \times 120 = 70 \text{ cm}.
  3. Length of shorter piece: 512×120=50 cm\frac{5}{12} \times 120 = 50 \text{ cm}.
  4. Difference: 70−5020\begin{array}{r} 70 \\ - 50 \\ \hline 20 \end{array} The difference is 20 cm20 \text{ cm}.

Explanation:

Calculate each part first by using the ratio and the total length. Then, subtract the smaller part from the larger part to find the difference.