Subject: Compulsory Maths

A proportion is a name we give to a statement that two ratios are equal.In proportion, the first and fourth terms are called extremes and the second and third terms are called means.

When two ratios are equal, then it is called as a proportion. It is an equation that can be solved. It is a case where two fractions are equal. It can be written in two ways:

- Two equal fractions, \(\frac{a}{b}\) = \(\frac{c}{d}\)
- Using a colon, a : b = c : d

There are four quantities in proportions where the first and fourth terms are known as extremes and the second and third terms are called means. It shows or tells two fraction or ratios are equal.

Terms are the four different quantities in the proportion. In proportion a: b: :c: d, a, b, c, and d are its first term, second term, third term& fourth terms. The fourth term is called the fourth proportional to the numbers a, b, and c.

In a proportion, the first and fourth terms are called extremes and the second and third terms are called means.

**Examples**

- Find the fourth proportional of 6, 8, 9.

Solution:

Let the fourth proportional to 6, 8, 9

Then, \(\frac{6}{8}\) = \(\frac{9}{x}\)

or, 6x = 72

or, x = \(\frac{72}{6}\) = 12 - Find the value of x in 16 : 8 = x : 5

Solution:

16 : 8 = x : 5

or, \(\frac{16}{8}\) = \(\frac{x}{5}\)

or, 8x =16 \(\times\) 5

or, x = \(\frac{80}{8}\)

\(\therefore\) x = 10

- When two ratios are equal, a name is given to a statement which is called a proportion.
- Proportion is an equation where two ratios are equal to each other.
- In a proportion, the first and fourth terms are called extremes and the second and third terms are called means.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Find the third proportional to 2 and 22.

Solution:

Let the required third proportion be x.

2 : 22 = 22 : x

or, \(\frac{2}{22}\) = \(\frac{22}{x}\)

or, 2x = 22\(\times\) 22

or, x = \(\frac{22 \times 22}{2}\)

\(\therefore\) x = 242

Find the mean proportion between 2 and 32.

Solution:

Let the required mean proportion be x. Then,

2 : x = x : 32

or, \(\frac{2}{x}\) = \(\frac{x}{32}\)

or, x^{2} = 64

or, x = \(\sqrt{64}\)

\(\therefore\) x = 8

Find the fourth proportional to 5, 7 and 8.

Solution:

Let x be the proportional to 5, 7 and 8

Then, \(\frac{5}{7}\) = \(\frac{8}{x}\)

or, 5x =7 \(\times\) 8

or, x = \(\frac{56}{5}\)

\(\therefore\) x = 11.2

Find the value of x in 16 : 8 = x : 5

Solution:

16 : 8 = x : 5

or, \(\frac{16}{8}\) = \(\frac{x}{5}\)

or, 8x =16 \(\times\) 5

or, x = \(\frac{80}{8}\)

\(\therefore\) x = 10

If 7 pencils cost Rs 150, what is the cost of 12 pencils?

Solution:

Let the Cost of 12 pencils be Rs x.

Since the cost are proportional to the number of pencils, We have

\(\frac{7}{150}\) = \(\frac{12}{x}\)

Or, 7x = 12 \(\times\)150

Or, x = \(\frac{12\times150}{7}\)

= 257.14

Hence, the cost of 12 pencils = Rs 257.14

If 15 pens cost Rs 105, how much will 20 pens cost?

Solution:

Let the cost of 20 pens be x.

Since the costs are proportional to the number of pens, we have

\(\frac{15}{105}\) = \(\frac{20}{x}\)

or, 15 x = 20 \(\times\) 105

or, x = \(\frac{20 \times 105}{15}\)

or, x = \(\frac{2100}{15}\)

\(\therefore\) x = 140

Hence, the cost of 20 pens is Rs 140.

Given numbers 7, 9, z, 18. Find z.

Solution:

The ratio of first two numbers = \(\frac{7}{9}\)

The ratio of second two numbers = \(\frac{z}{18}\)

Now,

\(\frac{7}{9}\) = \(\frac{z}{18}\)

or, z = \(\frac{7\times18}{9}\)

or, z = 7\(\times\)2

\(\therefore\) z = 14

Find the value of x.

x:5 = 10:25

Solution:

Given, x:5 = 10:25

or, \(\frac{x}{5}\) = \(\frac{10}{25}\)

or, 25x = 5 \(\times\) 10

or, x = \(\frac{50}{25}\)

\(\therefore\) x = 2

Find the value of x.

3:7 = 21:x

Solution:

Given, 3:7 = 21:x

or, \(\frac{3}{7}\) = \(\frac{21}{x}\)

or, x = \(\frac{21\times7}{3}\)

\(\therefore\) x = 49

Given numbers 3, a, 9, 21. Find a.

Solution:

The first two number of ratio = \(\frac{3}{a}\)

Second two number of ratio = \(\frac{9}{21}\)

given numbers are in proportion, so

\(\frac{3}{a}\) = \(\frac{9}{21}\)

or, a = \(\frac {3\times21}{9}\)

\(\therefore\) a = 7

Find the value of x.

25 : 15 = x : 3

Solution:

Given,

25 : 15 = x : 3

or, \(\frac{25}{15}\) = \(\frac{x}{3}\)

or, x = \(\frac{25\times3}{15}\)

or, x = \(\frac{75}{15}\)

\(\therefore\) x = 5

Check, whether the following numbers form a proportion or not.

7, 8, 14, 20.

Solution:

The ratio of first and second numbers = \(\frac{7}{8}\) = 7:8

The ratio of third and fourth numbers = \(\frac{14}{20}\) = 7:10

Hence, the first ratio 7:8 and the second ratio 14:20 = 7:10 are not not in the proportional form.

Find the cost of 23 meters of cloth, if 5 meters of cloth costs Rs 625.

solution:

Let, the cost of 23 meters of cloth be x.

Since, the costs are proportional to the meters of cloth.

\(\frac{5}{625}\) = \(\frac{23}{x}\)

or, 5\(\times\)x = 23\(\times\)625

or, x = \(\frac{23\times625}{5}\)

\(\therefore\) x = 2875

Hence, the cost of 23 meters of cloth is Rs 2875.

The ratio of the students using pens and pencils in Dharan Higher Secondary School is 10:11. If 110 students are using pencils, find the number of students using pens?

Solution:

Ratio of students usingpens and pencils = 10:11

Number of students using pencils = 110

Number of students using pens =?

Let, the number of students using pen be x.

According to the question,

110:x = 10:11

or, \(\frac{110}{x}\) = \(\frac{10}{11}\)

or, x = \(\frac{110\times11}{10}\)

or, x = 11\(\times\)11

\(\therefore\) x = 121

Hence, 121 number of students are using pens.

The ratio of milk and sugar in a candy is 5:3. If there is 750 gm milk, how much is the sugar?

Solution:

Let, the quantity of sugar be x.

Quantity of milk= 750 gm

The ratio of milk and sugar = 5:3

\(\frac{750}{x}\) = \(\frac{5}{3}\)

or, x = \(\frac{750\times3}{5}\)

\(\therefore\) x = 450

Hence, There is 450gm sugar in the candy.

The Ratio of the Students passed in first division, Second division and Third division is 2:4:5, among 153 students. If 21 students failed the exam, how many passed in First division, Second division and Third division

Solution:

Number of Students participated in the exam = 153

Number of Failed Students = 21

Total passed students = 153 - 21 = 132

The Ratio of 132 students passed in first, second and third division = 2:4:5

Now, Number of students passed in First division = 2x

Number of students passed in Second division = 4x

Number of students passed in Third division = 5x

2x+ 4x+ 5x = 132

Or, 11x = 132

Or, x= \(\frac{132}{11}\) = 12

Hence,

Number of students passed in First division = 2x = 2\(\times\)12 = 24

Number of students passed in Second division = 4x = 4\(\times\)12 = 48.

Number of students passed in Third division = 5x = 5\(\times\)12 = 60

If the gravitational ratio of Earth and Moon is 1:6. If the weight of an object in Earth is 90N, find it's weight in the Moon ?

Solution:

The Gravitational Ratio of Earth and Moon = 1:6

Weight of an object in the Earth = 90N

Weight of that object in the Moon =?

Let, the weight in Moon be xN.

Now,

\(\frac{xN}{90N}\) = \(\frac{1}{6}\)

or, x = 90\(\times\)\(\frac{1}{6}\)

\(\therefore\) x = 15N

Hence, The weight in Moon is 15N.

How much should be added to both the numbers, to make the ratio 3:5 into 5:6 ?

Solution:

Let, x be the number to make the ratio 3:5 into 5:6.

Now,

\(\frac{3 + x}{5 + x}\) = \(\frac{5}{6}\)

or, 6( 3 + x) = 5( 5 +x)

or, 18 + 6x = 25 + 5x

or, 6x -5x = 25 - 18

\(\therefore\) x = 7

Hence, 7 should be added to both the numbers to make the Ratio3:5 into 5:6.

In the marriage ceremony, a cook can cook 20 kg meat in 30 minutes. How much time is required for them to cook 15 kg meat?

Solution:

Let, the time required to cook 15 kg meat be x minutes.

Now,

\(\frac{30}{x }\) = \(\frac{20}{15}\)

or, x = \(\frac{30\times15}{20}\)

or, x = \(\frac{450}{20}\)

\(\therefore\) x = 22.5 minutes

Hence, 22 minutes 5 sec is required to cook 15 kg meat.

Simran scored in the ratio of 10:12 in Mathematics and Science. If she scored 80 in Science, find the marks she got in Mathematics?

Solution:

Let, x be the marks she got in Mathematics.

The Ratio of Marks obtained in Mathematics and Science = \(\frac{10}{12}\)

Marks in Science = 80

Now,

\(\frac{x}{80}\) = \(\frac{10}{12}\)

or, x = \(\frac{10\times80}{12}\)

or, x = \(\frac{10\times20}{3}\)

or, x = \(\frac{200}{3}\)

\(\therefore\) x = 66\(\frac{2}{3}\)

She got 66\(\frac{2}{3}\) in Mathematics.

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