In this chapter, we provide RS Aggarwal Solutions for Class 7 Chapter 2 Fractions CCE Test Paper for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 7 Chapter 2 Fractions CCE Test Paper Maths pdf, free RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper Maths book pdf download. Now you will get step by step solution to each question.
Textbook | NCERT |
Class | Class 7 |
Subject | Maths |
Chapter | Chapter 2 |
Chapter Name | Fractions |
Exercise | CCE Test Paper |
RS Aggarwal Class 7 Solutions Chapter 2 Fractions CCE Test Paper Download PDF
Question 1.
Solution:
(i) A number of the form ab, where a and b are rational numbers, is called a natural number.
Here, a is the numerator and b is the denominator.
(a) 2/3 is a fraction with 2 as the numerator and 3 as the denominator.
(b) 1/25 is a fraction with 12 as the numerator and 5 as the denominator.
(ii) A fraction whose denominator is a whole number other than 10, 100, 1000, etc., is called a vulgar faction.
Examples : 2/5 and 4/15.
(iii) A fraction whose numerator is greater than or equal to its denominator is called an improper fraction.
Example : 1/13 and 41/35
Question 2.
Solution:
Required number to be added
Hence, the required number is 8/25
Question 3.
Solution:
Question 4.
Solution:
Question 5.
Solution:
Question 6.
Solution:
Question 7.
Solution:
Question 8.
Solution:
Question 9.
Solution:
Mark (√) against the correct answer in each of the following:
Question 10.
Solution:
(d) 58
58 is a vulgar fraction, because its denominator is other than 10,100, 1000, etc.
Question 11.
Solution:
(c) 4663
A fraction ab is said to be irreducible or in its lowest terms if the HCF of a and b is 1
46 = 2 x 23 x 1
63 = 3 x 3 x 21 x 1
Clearly, the HCF of 46 and 63 is 1.
Hence, 4663 is an irreducible fraction.
Question 12.
Solution:
(d) None of these
Reciprocal of 135 = Reciprocal of 85 = 58
Question 13.
Solution:
Question 14.
Solution:
Question 15.
Solution:
Question 16.
Solution:
(b) 33 km
Distance covered by the car on
Question 17.
Solution:
Question 18.
Solution:
(i) False.
By cross multiplication, we have:
9 x 24 = 216 and 13 x 16 = 208
However, 216 > 208
916 > 1324
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