Ratio and Proportion questions banks for class 7.

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Mathac.com is providing basic concepts, aptitude questions on Ratio & Proportion for class 7 with solution PDF. Our basic concepts, tips, and tricks are beneficial for solving aptitude-based questions for competitive examinations. Our experts provide extra questions on Ratio & Proportion with solutions in PDF format. The concepts in this chapter are very important for the next chapter (Direct and Indirect Inversition) and for competitive exams. In upcoming exams, most of the reasoning and case study questions will be based on the Ratio & Proportion chapter.

What do you mean by Ratio in Maths?

The ratio of two quantities of the same kind and in the same units is a fractions that shows how many times the one quantity is of the other. The ratio of two quantities a and b (b≠0) is a÷b or, \(\frac{a}{b}\) and is denoted by a : b. In this ratio a : b, the quantities (numbers) a and b are called the terms of the ratio. The first term, or antecedent, is a, and the second term, or consequent, is b.

Ratio in Maths

We know that a fraction doesn’t change if its numerator and denominator are multiplied or divided by the same nonzero number. Similarly, a ratio doesn’t change if its first and second terms are multiplied or divided by the same nonzero number.

Example:

  • 100 : 200 = 10 : 20 (Dividing the First term and Second term by 10)
  • 100 : 200 = 1000 : 2000 (Multiply The First term and Second term by 10)

Ratio in the Simplest Form(standard Form or lowest term).

A ratio a : b is said to be in simplest form if a and b have no common factor other than 1. If A ratio in simplest form is also called the ratio in the lowest terms.

Example: The ratio 10 : 20 is not in the simplest form because 10 is a common factor of its antecedent and consequent. The simplest form of this ratio is 1 : 2 (dividing the first and second terms by 10).

  • The ratio of two quantities determines how many times one quantity is contained by the other. However, the comparison becomes meaningless if the quantities being compared are not of the same kind or are not measured in the same units. For example, it is meaningless to compare 20 kilograms with 400 kilometers. Therefore, to find the ratio of two quantities, they must be expressed in the same units.
  • The ratio of two quantities of the same kind determines how many times one quantity is contained by the other. Therefore, the ratio of any two quantities of the same kind is an abstract quantity. The ratio has no units or it is independent of the units used in the quantities compared.
  • The order of the terms in a ratio is very important. For example, the ratio 3 : 2 is different from the ratio 2 : 3.

How do you find Equivalent Ratios and define the term Equivalent Ratio?

When a ratio obtained by multiplying or dividing the numerator and denominator of a given ratio by same number is called an equivalent ratio.

Example: Ratio 1 : 3.

\(\frac{1}{3} = \frac{1×2}{3×2} = \frac{2}{6}, \frac{1×3}{3×3} = \frac{3}{9}, \frac{1×4}{3×4} = \frac{4}{12}\) and so on.

Also, \(\frac{4}{12} = \frac{4÷2}{12÷2} = \frac{2}{6}\).

Thus, \(\frac{1}{3}, \frac{2}{6}, \frac{3}{9}, \frac{4}{9}, \frac{2}{6}\). are ratios equivalent to the ratio \(\frac{1}{3}\).

If a : b and c : d are two equivalent ratios, then \(\frac{a}{b} = \frac{c}{d}\).

How to Compare two Ratios?

In order to comparison of ratios, you should follow the given below steps:

Step 1: Obtain the given ratios.

Step 2: Express each one of them in form of a fraction in the simplest form.

Step 3: Find the L.C.M of the denominators of fractions obtained in above Steps 2.

Step 4: Obtain first fraction and its denominator. Divide the LCM obtained in Step 3 by the denominator to get a number x.

Step 5: Now, multiply the numerator and denominator of the fraction by x. Apply same procedure to get other fraction. Now, the denominator of all fractions will be same.

Step 6: Compare the numerators of the fractions obtained in Step 5. The fraction having larger numerator will be larger than other.

Question Compare the ratios: 5 : 12 and 3 : 8.

Solution: The given ratios as fractions,

5 : 12 = \(\frac{5}{12}\) and \(3 : 8 = \frac{3}{8}\)

Now, LCM of 12 and 8 is 24.

Making the denominator of each fraction equal to 24,

\(\frac{5}{12} = \frac{5×2}{12×2} = \frac{10}{24} and \frac{3}{8} = \frac{3×3}{8×3} = \frac{9}{24}\)

Clearly, From their numerator 10 > 9.

∴ \(\frac{10}{24} > \frac{9}{24} ⇒\frac{5}{12} > \frac{3}{8}\)

What do you mean by Proportion in Math?

An equality of two ratios is called a proportion.

Four numbers a, b, c, d are said to be in proportion, if the ratio of the first two is equal to the ratio of the last two, that is a : b = c : d.

If four numbers a, b, c, d are in proportion, then we write.

a : b :: c : d

Which is read as a‘ is to b as c is to d‘ or a‘ to b as c to d‘. Here a, b, c and d are the first, second, third and fourth terms of the proportion. The first and fourth terms of a proportion are called extreme terms or extremes. The second and third terms are called the middle terms or means.

structure of proportion

If four numbers are in proportion, then the product of the extreme terms is equal to the product of the middle terms.

∴ Product extreme terms = Product of middle terms.

In other words, a : b = c : d if and only if ad = bc.

Consider the two ratios 9 : 12 and 18 : 24.

9 : 12 = 1 : 3 and 18 : 24 = 1 : 3.

∴ 9 : 12 = 18 : 24 is a proportion.

Continued Proportion.

There numbers a, b, c, are said to be in continued proportion if a, b, b, c, are in proportion. Thus, if a, b, c are in continued proportion, then a, b, b, c are in proportion that is a : b :: b : c.

Structure of continued proportion

⇒Product of extreme terms = Product of means terms

⇒a × c = b × b

⇒ac = b\(^2\)

⇒b\(^2\) = ac

What do you mean by Mean Proportional?

If a, b, c, are in continued proportion, then b is called the mean proportional between a and c. if b is the mean proportional between a and c, then b\(^2\) = ac.

Q. Divide ₹1250 between Aman and Amit in the ratio 2 : 3.

Solution: We have,

The sum of terms of the ratios = (2 + 3) = 5

∴ Aman’s share = ₹\(\left( \frac{2}{5} \times 1250 \right)\) = ₹(2×2500) = ₹500

Amit’s share = ₹\(\left( \frac{3}{5} \times 1250 \right)\) = ₹(3×250) = ₹750.

Q. Divide ₹1500 among A, B, C in the ratio 3 : 5 : 2.

Solution: We have,

Sum of the terms of the ratio = (3 + 2 + 5) = 10

∴ A’s share = ₹\(\left( \frac{3}{10} \times 1500 \right)\) = ₹(3 × 150) = ₹450

B’s share = ₹\(\left( \frac{5}{10} \times 1500 \right)\) = ₹(5 × 150) = ₹750

C’s share = ₹\(\left( \frac{2}{10} \times 1500 \right)\) = ₹(2 × 150) = ₹300.

Q. A bag contains ₹187 in the form of ₹1, 50paise and 10paise coins in the ratio 3 : 4 : 5. Find the number of each type of coins.

Solution: Let the number of ₹1, 50paise and 10paise coins be 3x, 4x and 5x respectively. Then,

3x + 4x\(\times \frac{50}{100} + 5x \times \frac{10}{100}\) = 187

⇒ 3x + 2x + \(\frac{x}{2}\) = 187

⇒ \(\frac{11x}{2} = 187\)

⇒ x = 34

∴ 3x = 3×34 = 102, 4x = 4×34 = 136, 5×34 = 170

Hence, The number of ₹1, 50 paise and 10 paise coins are 102, 136 and 170 respectively.

Q. If x : y = 2 : 3, find the value of (3x + 2y) : (2x + 5y).

Solution: We have,

x : y = 2 : 3 ⇒ \(\frac{x}{y} = \frac{2}{3}\)

Now,

⇒(3x + 2y) : (2x + 5y)

⇒\(\frac{(3x + 2y)}{(2x + 5y)}\)

⇒\(\frac{\frac{(3x + 2y)}{y}}{ \frac{(2x + 5y)}{y}}\) [Dividing numerator and denominator by y]

⇒\(\frac{3(\frac{x}{y}) + 2}{ 2(\frac{x}{y}) + 5}\)

Where \(\frac{x}{y} = \frac{2}{3}\)

⇒\(\frac{3(\frac{2}{3}) + 2}{ 2(\frac{2}{3}) + 5}\)

⇒\(\frac{2 + 2}{\frac{4}{3} + 5}\)

⇒\(\frac{4}{\frac{19}{3}}\)

⇒\(\frac{12}{19}\) = 12 : 19

Q. If ( 4x + 5) : (3x + 11) = 13 : 17, find the value of x.

Solution: We have,

(4x + 5) : (3x + 11) = 13 : 17

⇒ \(\frac{4x + 5}{3x + 11} = \frac{13}{17}\)

⇒17(4x + 5) = 13(3x + 11)

⇒ 68x + 85 = 39x + 143

⇒ 68x – 39x = 143 – 85

⇒ 29x = 58

⇒ x = \(\frac{58}{29} = 2\)

Q. What must be added to each term of the ratio 2 : 5 so that it may become equal to 5 : 6?

Solution: Let the number to be added be x. Then,

(2 + x) : (5 + x) = 5 : 6

⇒ \(\frac{2 + x}{5 + x} = \frac{5}{6}\)

⇒6(2 + x) = 5(5 + x)

⇒12+ 6x = 25 + 5x

⇒ x = 13

Hence, the required number is 13.

Q. Are 36, 49, 6, 7 in proportion?

Solution: We know that,

⇒Product of extremes = Product of means

From question,

⇒36 × 7 = 49 × 6

⇒252 ≠ 294

Clearly, Product of extreme ≠ Product of means

Hence, 36, 49, 6, 7 are not in proportion.

Q. Are 4, 12, 36 in continued proportion?

Solution: We know that three numbers a, b, c are in continued proportion, if a, b, c are in proportion.

We have,

⇒Product of extreme = Product of means

⇒4 × 12 = 12 × 36

Here, Product of extreme = Product of means

⇒4, 12, 12, 36 are in proportion

⇒4, 12, 36 are in continued proportion.

Q. What should be added to the numbers, 6, 10, 14 and 22 so that they are in proportion?

Solution: Let the required number be x. then,

⇒ 6 + x, 10 + x, 14 + x, 22 + x, are in proportion.

⇒Product of extreme = Product of means

⇒(6 + x)(10 + x) = (10 + x)(14 + x)

⇒132 + 6x + 22x + x\(^2\) = 140 + 10x + 14x + x\(^2\)

⇒132 + 28x = 140 + 24x

⇒28x – 24x = 140 – 132

⇒4x = 8

⇒x = 2.

Hence, The required number is 2.

Q. The scale of a map is 1 : 3000000. What is the actual distance between the two towns, if they are 3cm apart on the map?

Solution: If the distance on the map is 1cm, then actual distance is 3000000 cm.

we have, 3000000 cm = \( \frac{3000000}{1000×100} km = 30 km \)

Thus, if the distance on the map is 1cm, the actual distance is 30 km.

Let the actual distance between the two towns be x km, we have the following information.

Distance On The MapActual Distance
1 cm30 km
3 cmx

∴ 1 : 3 = 30 : x

⇒1 × x = 3 × 30

⇒x = 90

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