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Ratio and Proportion - Unitary Method

Grade 6CBSE

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

🔑Concepts

A Ratio is a comparison of two quantities of the same kind by division, written as a:ba:b or ab\frac{a}{b}. For example, if you have 33 red pens and 99 blue pens, the ratio of red to blue pens is 3:93:9. Visually, this simplifies to 1:31:3, meaning for every 11 red pen, there are 33 blue pens.

Quantities in a ratio must be expressed in the same units before comparison. If you compare 5050 cm to 22 m, you must convert 22 m to 200200 cm first, making the ratio 50:20050:200 or 1:41:4. Imagine a balance scale where both sides must use the same weight standard to be accurate.

A ratio is in its simplest form if the HCF (Highest Common Factor) of its terms is 11. For example, the ratio 10:1510:15 is simplified to 2:32:3 by dividing both parts by 55. This is like looking at a large grid of 10×1510 \times 15 squares and seeing it as a simpler 2×32 \times 3 pattern.

Proportion exists when two ratios are equal, denoted by a:b=c:da:b = c:d or a:b::c:da:b :: c:d. This can be visualized using two similar shapes of different sizes; if the ratio of length to width is the same for both, their dimensions are in proportion.

In a proportion a:b::c:da:b :: c:d, the terms aa and dd are called 'Extremes' (the outer values) and bb and cc are 'Means' (the inner values). A fundamental property is that the product of extremes always equals the product of means: a×d=b×ca \times d = b \times c. Imagine a 'cross-multiplication' XX connecting the terms.

The Unitary Method is a technique to solve problems by first finding the value of a single unit. Think of it as a two-step process: 'Divide' to find the unit value, then 'Multiply' to find the total value for the required number of units.

To find the value of one unit (the unit rate), divide the total value by the total number of units. For instance, if 55 boxes weigh 2525 kg, then 11 box weighs 25÷5=525 \div 5 = 5 kg. Visualize breaking a large block into smaller, equal-sized pieces to see the value of just one piece.

To find the total value of multiple units, multiply the unit value by the number of units required. If you know 11 apple costs 10₹ 10, then 77 apples will cost 10×7=7010 \times 7 = ₹ 70. This is like building a tower of identical blocks, where the total height is the height of one block multiplied by the number of blocks used.

📐Formulae

Ratio of a to b=ab\text{Ratio of } a \text{ to } b = \frac{a}{b}

If a:b::c:d, then ab=cd\text{If } a:b :: c:d, \text{ then } \frac{a}{b} = \frac{c}{d}

Product of Extremes=Product of Means    a×d=b×c\text{Product of Extremes} = \text{Product of Means} \implies a \times d = b \times c

Value of one unit=Total ValueTotal Quantity\text{Value of one unit} = \frac{\text{Total Value}}{\text{Total Quantity}}

Value of n units=Value of one unit×n\text{Value of } n \text{ units} = \text{Value of one unit} \times n

💡Examples

Problem 1:

If the cost of 66 cans of juice is 210₹ 210, what will be the cost of 44 cans of juice?

Solution:

  1. Cost of 66 cans = 210₹ 210
  2. Using the unitary method, find the cost of 11 can: Cost of 1 can=2106=35\text{Cost of } 1 \text{ can} = ₹ \frac{210}{6} = ₹ 35
  3. Now, find the cost of 44 cans: Cost of 4 cans=35×4=140\text{Cost of } 4 \text{ cans} = ₹ 35 \times 4 = ₹ 140

Explanation:

First, we use division to find the 'unit price' (the cost of a single can). Once we have the unit price, we multiply it by the desired quantity (44) to find the total cost.

Problem 2:

A car travels 150150 km in 33 hours. How far will it travel in 55 hours if the speed remains constant?

Solution:

  1. Distance covered in 33 hours = 150150 km
  2. Distance covered in 11 hour = 1503=50\frac{150}{3} = 50 km
  3. Distance covered in 55 hours = 50×5=25050 \times 5 = 250 km

Explanation:

We first determine the speed per hour (distance in 11 hour) by dividing the total distance by total time. Then, we multiply this hourly distance by the target time (55 hours) to find the total distance.