Class 7 Exam  >  Class 7 Notes  >  RD Sharma Solutions for Class 7 Mathematics  >  RD Sharma Solutions (Part - 2) - Ex - 9.1, Ratio And Proportion, Class 7, Math

RD Sharma Solutions (Part - 2) - Ex - 9.1, Ratio And Proportion, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics PDF Download

QUESTION 12:

The boys and the girls in a school are in the ratio 7 : 4. If total strength of the school be 550, find the number of boys and girls.

ANSWER 12:

We have the ratio boys : girls = 7 : 4.
So, let there be 7x boys and 4x girls. It is given that there are a total of 550 students in the school.
 Therefore, 7x + 4x = 550
                 11x = 550
                    x = 550/11 = 50
Hence, the number of boys = 7x = 7×× 50 = 350, and the number of girls = 4x = 4 ×× 50 = 200.


QUESTION 13:

The ratio of monthly income to the savings of a family is 7 : 2. If the savings be of Rs 500, find the income and expenditure.

ANSWER 13:

We have the ratio of income : savings = 7 : 2.
 So, let the income be 7x and the savings be 2x. It is given that the savings are Rs 500.
   Therefore, 2x = 500
                    x = Rs 500/2 = Rs 250
            Thus, the income = 7x = 7 × 250 = Rs 1750.
     Now, expenditure = Income −- savings = Rs 1750 − Rs 500 = Rs 1250.
 Thus, the income = Rs 1750, and the expenditure = Rs 1250.


QUESTION 14:

The sides of a triangle are in the ratio 1 : 2 : 3. If the perimeter is 36 cm, find its sides.

ANSWER 14:

We have the ratio of the sides of the triangle = 1 : 2 : 3.
 Now, let the sides of the triangle be x, 2x and 3x, respectively.
 Therefore, the perimeter = x + 2x + 3x = 36
                                    ⇒ 6x = 36
                                    ⇒ x = 36/6 = 6
 Thus, the sides of the triangle = x = 6 cm; 2x = 2××6 = 12 cm; 3x = 3 ××6 = 18 cm.
    So, the sides of the triangle = 6 cm, 12 cm and 18 cm.


Question 15:

A sum of Rs 5500 is to be divided between Raman and Aman in the ratio 2 : 3. How much will each get?

Answer 15:

We have
Sum of the terms of the ratio = 2 + 3 = 5, and the total sum = Rs 5500
Therefore, Raman's share = (2/5×5500) = Rs 2200
Aman's share = (3/5×5500) = Rs 3300


Question 16:

The ratio of zinc and copper in an alloy is 7 : 9. If the weight of the copper in the alloy is 11.7 kg, find the weight of the zinc in the alloy.

Answer 16:

We have
Weight of zinc : weight of copper = 7 : 9
So, let the weight of zinc in the alloy be '7x' kg and the weight of copper in the alloy be '9x' kg.
But the weight of copper in the alloy is given to be 11.7 kg.
Therefore, 9x = 11.7
                  x = 11.7/9 = 1.3
 Hence, the weight of zinc in the alloy = 7x = 7××1.3 = 9.1 kg.


Question 17:

In the ratio 7 : 8, if the consequent is 40, what is the antecedent?

Answer 17:

In a ratio a : b, 'a' is known as the antecedent and 'b' is known as the consequent.
In the given ratio, let the antecedent be 7x and the consequent be 8x, respectively,
But consequent = 8x = 40
                              x = 40/8 = 5
Therefore, the antecedent = 7x = 7××5 = 35.


Question 18:

Divide Rs 351 into two parts such that one may be to the other as 2 : 7.

Answer 18:

We have
            Sum of the ratio of the terms = 2 +7 = 9
           Therefore, first part = Rs. (2/9×351) =  Rs. 78
         Similarly, second part = Rs. (7/9×351) =  Rs. 273


Question 19:

Find the ratio of the price of pencil to that of ball pen, if pencils cost Rs 16 per score and ball pens cost Rs 8.40 per dozen.

Answer 19:

We have
 Cost of 1 score of pencils = Rs. 16
 Since 1 score = 20 items,
 Cost of one pencil = Rs. (16/20) = Rs. 0.8
 Cost of 1 dozen ball pens = Rs. 8.40
 Since 1 dozen =12 items,
 Cost of one ball pen = Rs. (8.40/12) = Rs. 0.7
 So, price of pencil : price of ball pen = 0.8 : 0.7 =0.8/0.7 = 8/7

Price of pencil : price of ball pen = 8 : 7


Question 20:

In a class, one out of every six students fails. If there are 42 students in the class, how many pass?

Answer 20:

We have
One out of every six student fails, which means that 1/6th of the total students fail in the class.
And total number of students in the class = 42.
Therefore, the number of students who fail = (1/6×  42) = 7.
So, the number of students who pass = (Total students −-the number of students who fail) = 42 −- 7 = 35.

The document RD Sharma Solutions (Part - 2) - Ex - 9.1, Ratio And Proportion, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
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FAQs on RD Sharma Solutions (Part - 2) - Ex - 9.1, Ratio And Proportion, Class 7, Math - RD Sharma Solutions for Class 7 Mathematics

1. What are RD Sharma Solutions and how can they help in preparing for the Class 7 Math exam?
Ans. RD Sharma Solutions are a comprehensive set of solutions for the RD Sharma textbook, which is widely used by students for mathematics in India. These solutions provide step-by-step explanations and solutions to all the problems in the textbook, making it easier for students to understand and practice the concepts. By using RD Sharma Solutions, students can strengthen their understanding of the subject, practice different types of problems, and enhance their problem-solving skills, thus helping them prepare effectively for the Class 7 Math exam.
2. What is the importance of the Ratio and Proportion chapter in Class 7 Math?
Ans. The Ratio and Proportion chapter in Class 7 Math is of great importance as it lays the foundation for understanding various mathematical concepts in higher classes. Ratios and proportions are used in various real-life scenarios, such as cooking, mixing ingredients, and scaling of maps or blueprints. Understanding this chapter helps students develop a strong numerical sense and enables them to solve complex problems involving comparisons, scaling, and proportionality. Hence, it is crucial for students to grasp the concepts of ratio and proportion for their overall mathematical development.
3. How can I use RD Sharma Solutions to effectively study the Ratio and Proportion chapter?
Ans. To effectively study the Ratio and Proportion chapter using RD Sharma Solutions, follow these steps: 1. Read the chapter from the textbook to understand the concepts and examples provided. 2. Refer to the corresponding RD Sharma Solution for the chapter, which will provide step-by-step solutions to the exercises and additional practice problems. 3. Start with the solved examples in the RD Sharma Solution to gain clarity on the concepts and problem-solving techniques. 4. Practice solving the exercises on your own, referring to the solution whenever you get stuck or want to verify your answer. 5. If you encounter any difficulty, revisit the corresponding concepts in the textbook or refer to the solution for a better understanding. 6. Regularly practice and solve additional problems from the RD Sharma Solution to strengthen your grasp of the chapter.
4. How can Ratio and Proportion be applied in real-life situations?
Ans. Ratio and Proportion have several real-life applications, including: 1. Cooking: Recipes often require ingredients in specific ratios, and understanding ratios helps in scaling recipes according to the required serving size. 2. Maps and Blueprints: Ratio and Proportion are used to scale down large areas onto a map or blueprint, ensuring accurate representations. 3. Finance: Understanding and comparing ratios helps in financial planning, budgeting, and investment decisions. 4. Sports: Ratios and proportions are used to analyze and compare statistics in sports, such as batting averages, win-loss ratios, and scoring rates. 5. Enlargements and Reductions: In art and design, ratio and proportion are used to scale up or down images or structures while maintaining their proportions.
5. Are RD Sharma Solutions available online, and can they be accessed for free?
Ans. Yes, RD Sharma Solutions for Class 7 Math are available online, and there are various websites and educational platforms that provide these solutions. However, not all sources offer free access to the complete solutions. Some websites may provide limited access or require a subscription or purchase for full access to the solutions. It is recommended to explore different platforms and websites to find the most suitable option for accessing RD Sharma Solutions for free or at an affordable cost.
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