ICSE Class 8 Maths Arithmetical Problems Chapter 02 The Unitary Method Reference Content

The Unitary Method

The unitary method is a way of finding the value of the required number of things by first finding the value of one thing (or a unit) from the value of a given number of things.

Example

5 bottles of perfume cost Rs 1200. How much would 7 bottles cost?

Solution

5 bottles of perfume cost Rs 1200.

Therefore, 1 bottle of perfume costs Rs 1200 ÷ 5 = Rs \(\frac{1200}{5}\) = Rs 240.

7 bottles of perfume will cost Rs 240 × 7 = Rs 1680.

Direct Variation

In the example we have considered, the more the number of bottles, the more is the cost. Also, the less the number of bottles, the less is the cost. In other words, an increase in one quantity (the number) causes an increase in the other quantity (the cost) and a decrease in one quantity causes a decrease in the other quantity. We express this as the cost of the bottles varies directly as the number of bottles.

In general, two quantities x and y are said to vary directly or be in direct variation or in direct proportion if they change in such a way that the ratio of the two values of x is the same as the ratio of the corresponding two values of y. The quantity x (which causes the quantity y to change) is called the independent variable, while the quantity y is called the dependent variable.

Example

If 8 pencils cost Rs 24 then 6 such pencils will cost Rs 18.

Let the number of pencils = x and the cost = y.

So the ratio of the two values of x = \(\frac{8}{6}\) and the ratio of the corresponding values of y = \(\frac{24}{18}\).

Since \(\frac{8}{6} = \frac{24}{18}\), the cost of pencils varies directly with the number of pencils.

Inverse Variation

Two quantities x and y are said to vary indirectly or be in inverse variation or in inverse proportion if they change in such a way that the ratio of two values of x is the same as the inverse of the ratio of the corresponding two values of y. To put it more simply, when two quantities x and y are in inverse proportion, an increase in x (independent variable) causes a decrease in y (dependent variable) and while a decrease in x causes an increase in y.

Example

If 8 workers do a piece of work in 12 days then 16 workers will do the work in 6 days, provided they all work at the same rate. Here the number of workers (x) changes in the ratio 8 : 16, i.e., 1 : 2 while the number of days (y) changes in the ratio 12 : 6, i.e., 2 : 1. So, the two quantities are in inverse variation. More the number of workers, less is the number of days. Similarly, less the number of workers, more will be the number of days. So, when x (number of workers) increases, y (number of days) decreases and vice versa.

Multiplying Ratio

Problems related to both direct and inverse variation can be solved by finding what is known as the multiplying ratio.

Example

If 5 kg of sugar costs Rs 160, find the cost of 7 kg of sugar.

Solution

Since the cost increases when the weight increases, this is a case of direct variation.

Therefore, the ratio of the weights = the ratio of the corresponding costs.

\(\frac{5}{7} = \frac{160}{x}\), where Rs x is the cost of 7 kg of sugar.

x = \(\frac{7}{5}\) × Rs 160 = Rs 224.

Here, \(\frac{7}{5}\) is the multiplying ratio.

So, if two quantities change in the same ratio, the multiplying ratio is the inverse of this ratio.

Example

An army camp of 120 soldiers has enough ration to last 50 days. If 40 more soldiers join the camp after 10 days, how long will the remaining ration last?

Solution

At the end of 10 days the camp has enough ration to provide for 120 soldiers for 50 - 10 = 40 days. Since the number of soldiers increases, the ration will last for less days than planned. In other words, an increase in the number of soldiers will mean a decrease in the number of days. Hence, this is a case of inverse proportion.

Thus, the ratio of the number of soldiers = \(\frac{1}{\text{ration of number of days}}\)

\(\frac{120}{160} = \frac{1}{40}\), where x is the number of days for which the ration will last.

Or \(\frac{120}{160} = \frac{x}{40}\) or x = \(\frac{120}{160}\) × 40 = \(\frac{3}{4}\) × 40 = 30 days.

Here, \(\frac{3}{4}\) is the multiplying ratio.

In the case of inverse proportion, the multiplying ratio is the same as the ratio in which the first variable (or the independent variable) changes.

Teacher's Note

Understanding ratios and proportions helps students make sense of real-world scaling, like recipe adjustments or understanding discount percentages in shopping.

Solved Examples

Example 1

Anshuman earns Rs 2160 for a working week of 48 hours. If he did not work for 6 hours in a certain week, how much did he earn in that week?

Solution

This is a problem related to direct variation since an increase in the number of hours means an increase in earnings.

For 48 hours Anshuman's earnings = Rs 2160.

For 1 hour Anshuman's earnings = Rs \(\frac{2160}{48}\) [less hour less earnings - division].

For 42 hours Anshuman's earnings = Rs \(\frac{2160}{48}\) × 42 = Rs 1890 [more hour more earnings - multiplication].

Multiplying Ratio

The number of working hours decreases in the ratio 48 : 42, i.e., 8 : 7. Since this is a problem related to direct variation, the earnings decrease in the same ratio. Hence, the multiplying ratio is the inverse of this ratio (8 : 7) = \(\frac{7}{8}\).

Anshuman's earnings in that week = \(\frac{7}{8}\) × Rs 2160 = Rs 1890.

Example 2

If the wages of 9 labourers for 5 days are Rs 6300, find the wages of 15 labourers for 4 days.

Solution

The wages of 9 labourers for 5 days = Rs 6300.

Therefore, the wages of 9 labourers for 1 day = Rs \(\frac{6300}{5}\) = Rs 1260.

The wages of 1 labourer for 1 day = Rs \(\frac{1260}{9}\) = Rs 140.

The wages of 15 labourers for 1 day = Rs 140 × 15 = Rs 2100.

The wages of 15 labourers for 4 days = Rs 2100 × 4 = Rs 8400.

Multiplying Ratio

This is a case of direct proportion. First, the number of labourers increases in the ratio 9 : 15. So, the wages increase in the same ratio and the multiplying ratio = 15 : 9 = \(\frac{15}{9}\).

Second, the number of days decreases in the ratio 5 : 4, so the wages also decrease in the ratio 5 : 4.

Therefore, the multiplying ratio = \(\frac{15}{9}\) × \(\frac{4}{5}\) = \(\frac{4}{3}\).

The required wages = \(\frac{15}{9}\) × \(\frac{4}{5}\) × Rs 6300 = Rs 8400.

Example 3

A hostel has 150 students who consume 240 kg of chicken in 8 days. How many students will consume 180 kg of chicken in 9 days.

Solution

240 kg of chicken is consumed in 8 days by 150 students.

Therefore, 240 kg of chicken is consumed in 1 day by 8 × 150 students.

1 kg of chicken is consumed in 1 day by \(\frac{8 \times 150}{240}\) students.

180 kg of chicken will be consumed in 1 day by \(\frac{8 \times 150}{240}\) × 180 students.

180 kg of chicken will be consumed in 9 days by \(\frac{8 \times 150}{240}\) × \(\frac{180}{9}\) students that is, 100 students.

Teacher's Note

These examples show how unitary method breaks complex problems into simple, manageable steps - similar to how a cook adjusts ingredient quantities for different numbers of servings.

This is a preview of the first 3 pages. To get the complete book, click below.