ICSE Class 8 Maths Chapter 05 Unitary Method and its Applications

Chapter 5: Unitary Method And Its Applications

In the previous class, you learnt how to solve some real life simple problems using unitary method and also by using multiplying ratio method. We also solved problems on time and work. In this chapter, we shall refresh that knowledge and add a few tougher problems. We will also solve problems involving more than two different types of quantities.

Unitary Method

A method in which the value of a unit quantity is first obtained to find the value of any required quantity, is called unitary method.

In solving problems based on unitary method, we come across two types of variations:

(i) Direct variation (ii) Inverse variation

Direct Variation

Two quantities are said to vary directly if the increase (or decrease) in one quantity causes the increase (or decrease) in the other quantity.

For example:

(i) The cost of articles varies directly to the number of articles. More articles, more cost. Less articles, less cost.

(ii) The work done varies directly to the number of men at work. More men at work, more work done. Less men at work, less work done.

(iii) The work done varies directly to the working time. More work done in more time. Less work done in less time.

(iv) The distance covered by a moving object varies directly to its speed. More speed, more distance covered in same time. Less speed, less distance covered in same time.

In direct variation, the ratio of one kind of like terms is equal to the ratio of second kind of like terms.

Teacher's Note

When you buy multiple items at a store, the total cost increases proportionally with quantity - this is direct variation in everyday shopping.

Inverse Variation

Two quantities are said to vary inversely if the increase (or decrease) in one quantity causes the decrease (or increase) in the other quantity.

For example:

(i) The time taken to finish a work varies inversely to the number of men at work. More men at work, less time taken to finish the work. Less men at work, more time taken to finish the work.

(ii) The speed of a moving object varies inversely to the time taken to cover a certain distance. More speed, less time taken to cover the same distance. Less speed, more time taken to cover the same distance.

In inverse variation, the ratio of one kind of like terms is equal to the inverse ratio of second kind of like terms.

Sometimes, neither the idea of direct variation nor the idea of inverse variation applies.

For example:

(i) If the weight of a girl is 3 kg when she is 1 day old, we cannot say that her weight will be 15 kg when she is 5 days old.

(ii) If the height of a plant is 5 cm when it is 2 weeks old, we cannot say that its height will be 20 cm when it is 8 weeks old.

(iii) If your mother alone can cook an omelette in 10 minutes, we cannot say that your mother and father cooking together can do it in 5 minutes. Probably, it will still take 10 minutes to cook the omelette.

Thus, when using unitary method, we have to use common sense to see whether the direct variation applies or the inverse variation applies or there is no such variation.

A unitary method involving two different types of quantities is called a single unitary method and a unitary method involving more than two different types of quantities is called a compound unitary method.

Example 1

If 7 kg sugar costs \(\text{₹}115.50\), what is the cost of 12 kg sugar?

Solution.

Cost of 7 kg sugar is \(\text{₹}115.50\)

\(\therefore\) cost of 1 kg sugar = \(\text{₹}\frac{115.50}{7} = \text{₹}16.50\)

\(\therefore\) cost of 12 kg sugar = \(\text{₹}(16.50 \times 12) = \text{₹}198\).

Using multiplying ratio method

As the quantity of sugar increases in the ratio 7 : 12, the cost of sugar also increases in the ratio 7 : 12.

Multiplying \(\text{₹}115.50\) by the ratio \(\frac{12}{7}\),

the cost of 12 kg sugar = \(\frac{12}{7}\) of \(\text{₹}115.50 = \text{₹}\left(\frac{12}{7} \times 115.50\right)\)

= \(\text{₹}(12 \times 16.50) = \text{₹}198\).

Teacher's Note

When buying groceries, doubling the quantity doubles the cost - this direct variation helps you quickly estimate expenses while shopping.

Example 2

If one score eggs cost \(\text{₹}35\), how many eggs can be bought for \(\text{₹}63\)?

Solution.

Since for \(\text{₹}35\), the number of eggs bought is one score i.e. 20.

\(\therefore\) for \(\text{₹}1\), the number of eggs bought = \(\frac{20}{35}\)

\(\therefore\) for \(\text{₹}63\), the number of eggs bought = \(\frac{20}{35} \times 63 = 36\).

Multiplying ratio method

As the money increases in the ratio 35 : 63 i.e. 5 : 9, the number of eggs also increases in the ratio 5 : 9

Multiplying one score i.e. 20 by \(\frac{9}{5}\), the number of eggs bought = \(\frac{9}{5} \times 20 = 36\).

Teacher's Note

This shows how you can calculate discounts or quantity changes when prices vary - useful when comparing product values at the market.

Example 3

Sheetal has enough money to buy 5 kg mangoes at the rate of \(\text{₹}18\) per kg. How much quantity of mangoes she can buy in the same money if the price is increased to \(\text{₹}20\) per kg?

Solution.

The price of 5 kg mangoes at the rate of \(\text{₹}18\) per kg = \(\text{₹}(5 \times 18) = \text{₹}90\). Thus, Sheetal has \(\text{₹}90\).

Now for \(\text{₹}20\), the quantity of mangoes available = 1 kg

\(\therefore\) for \(\text{₹}1\), the quantity of mangoes available = \(\frac{1}{20}\) kg

\(\therefore\) for \(\text{₹}90\), the quantity of mangoes available = \(\left(90 \times \frac{1}{20}\right)\) kg = \(\frac{9}{2}\) kg = 4.5 kg.

Multiplying ratio method

As the price of mangoes increases in the ratio 18 : 20 i.e. 9 : 10, the quantity of mangoes decreases in the ratio 10 : 9

Multiplying 5 kg by the ratio \(\frac{9}{10}\), the quantity of mangoes that can be bought in the same money = \(\frac{9}{10}\) of 5 kg = \(\left(\frac{9}{10} \times 5\right)\) kg = \(\frac{9}{2}\) kg = 4.5 kg.

Teacher's Note

When prices increase, you get less quantity for the same money - understanding this inverse relationship helps in budget planning for household items.

Example 4

20 labourers can dig a pond in 12 days. How many days will it take 16 labourers to dig the same pond?

Solution.

20 labourers can dig a pond in 12 days

\(\therefore\) 1 labourer will dig the pond in (20 \times 12) days

\(\therefore\) 16 labourers will dig the pond in \(\frac{20 \times 12}{16}\) days

= 15 days.

Multiplying ratio method

As the number of labourers decrease in the ratio 20 : 16 i.e. 5 : 4, the number of days to dig the pond will increase in the ratio 4 : 5.

Multiplying 12 days by the ratio \(\frac{5}{4}\), the number of days required to dig the pond

= \(\frac{5}{4}\) of 12 days = \(\left(\frac{5}{4} \times 12\right)\) days = 15 days.

Teacher's Note

More workers on a project can complete it faster - this inverse relationship is why construction companies adjust team sizes based on project deadlines.

Example 5

A hostel had rations for 150 students for 60 days. After 12 days, 30 more students join the hostel. How long will the remaining ration last?

Solution.

After 12 days, the ration is sufficient for 150 students for (60 - 12) days i.e. 48 days.

After 12 days, 30 more students join the hostel. So number of students in the hostel = 150 + 30 = 180

Since for 150 students, the ration is sufficient for 48 days

\(\therefore\) for 1 student, the ration is sufficient for (150 \times 48) days

\(\therefore\) for 180 students, the ration is sufficient for \(\frac{150 \times 48}{180}\) days = 40 days.

Multiplying ratio method

After 12 days, the ration is sufficient for 150 students for 48 days. As 30 more students join the hostel, number of students in the hostel = 150 + 30 = 180

As the number of students increases in the ratio 150 : 180 i.e. 5 : 6, the number of days for which the ration lasts decreases in the ratio 6 : 5.

Multiplying 48 days by \(\frac{5}{6}\), the number of days for which the remaining ration lasts = \(\frac{5}{6}\) of 48 days = \(\left(\frac{5}{6} \times 48\right)\) days = 40 days.

Teacher's Note

Sharing limited resources among more people reduces how long those resources last - a principle applicable to food stocks in schools and institutions.

Example 6

If 3 men or 4 women can earn \(\text{₹}480\) in a day, find how much will 7 men and 11 women earn in a day?

Solution.

Since in a day, 3 men can earn \(\text{₹}480\)

\(\therefore\) in a day, 1 man will earn \(\text{₹}\frac{480}{3} = \text{₹}160\)

\(\therefore\) in a day, 7 men will earn \(\text{₹}(160 \times 7) = \text{₹}1120\).

Since in a day, 4 women can earn \(\text{₹}480\)

\(\therefore\) in a day, 1 woman will earn \(\text{₹}\frac{480}{4} = \text{₹}120\)

\(\therefore\) in a day, 11 women will earn \(\text{₹}(120 \times 11) = \text{₹}1320\)

\(\therefore\) Total earning of 7 men and 11 women in a day = \(\text{₹}(1120 + 1320) = \text{₹}2440\).

Teacher's Note

Different workers may have different productivity rates - understanding individual earning capacity helps calculate fair wages when mixing different worker types.

Example 7

If the wages of 15 labourers for 6 days are \(\text{₹}7200\), find the wages of 23 labourers for 5 days.

Solution.

Since wages of 15 labourers for 6 days are \(\text{₹}7200\).

\(\therefore\) wages of 1 labourer for 6 days = \(\text{₹}\frac{7200}{15} = \text{₹}480\)

\(\therefore\) wages of 1 labourer for 1 day = \(\text{₹}\frac{480}{6} = \text{₹}80\)

\(\therefore\) wages of 23 labourers for 1 day = \(\text{₹}(80 \times 23) = \text{₹}1840\)

\(\therefore\) wages of 23 labourers for 5 days = \(\text{₹}(1840 \times 5) = \text{₹}9200\).

Teacher's Note

Calculating labour costs requires breaking down rates per person per day - essential for project managers to estimate budgets accurately.

Example 8

If 7 typists typing 6 hours a day (at equal speeds) take 12 days to type the manuscript of a book, then how many days will 3 typists working 8 hours a day take to do the same job?

Solution.

Since 7 typists working 6 hours a day take 12 days to do the job.

\(\therefore\) 1 typist working 6 hours a day takes (12 \times 7) days = 84 days

\(\therefore\) 1 typist working 1 hour a day takes (84 \times 6) days = 504 days

\(\therefore\) 3 typists working 1 hour a day take \(\frac{504}{3}\) days = 168 days

\(\therefore\) 3 typists working 8 hours a day take \(\frac{168}{8}\) days = 21 days.

Note. (i) Less number of typists, more number of days required to complete the job. So, we have inverse variation. As the number of typists decreases in the ratio 7 : 3, the number of days will increase in the ratio 3 : 7.

(ii) More number of working hours, less number of days required to complete the job. So, we have inverse variation. As the number of working hours increases in the ratio 6 : 8 i.e. 3 : 4, the number of days required to complete the job will decrease in the ratio 4 : 3

Hence, the number of days required to complete the job = \(\frac{7}{3} \times \frac{3}{4} \times 12 = 21\).

Teacher's Note

Large projects with multiple workers and varying shifts require careful calculation of work completion times - understanding compound variations helps project planning.

Example 9

A contractor undertook to build a road in 180 days. He employed 150 men for the construction of road. After 60 days, he found that only one-fourth of the road could be built. How many additional men should be employed to complete the work in time?

Solution.

Portion of the road built = \(\frac{1}{4}\),

\(\therefore\) the portion of the road left = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\).

The number of days left for the completion of work = 180 - 60 = 120.

According to the given condition,

\(\frac{1}{4}\) of the road can be built in 60 days by 150 men,

\(\therefore\) \(\frac{1}{4}\) of the road can be built in one day by (150 \times 60) men = 9000 men

\(\therefore\) complete road can be built in one day by (9000 \times 4) men = 36000 men

\(\therefore\) complete road can be built in 120 days by \(\frac{36000}{120}\) men = 300 men

\(\therefore\) \(\frac{3}{4}\) of the road can be built in 120 days by \(\left(300 \times \frac{3}{4}\right)\) men = 225 men.

\(\therefore\) Number of additional men to be employed to complete the work in time = 225 - 150 = 75.

Teacher's Note

Construction projects often fall behind schedule, requiring additional workforce - this example shows how to calculate minimum staffing needs to meet deadlines.

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