NCERT Solutions for Class 8 Maths Chapter 10:
Students can aim for excellent exam scores by using the NCERT Solutions for Class 8 Chapter 10 Exponents and Powers. To comprehend the topics taught in this chapter, consult the NCERT Class 8 Mathematics Exercise Solutions and rehearse the questions and answers based on the CBSE syllabus.
We can refer to an expression as a power if it shows the same factor being multiplied repeatedly. In the case of 42, for instance, the base is the number 4, and the exponent is the number 2. The number of times the base is used as a factor in an expression is represented by an exponent.
NCERT Solutions for Class 8 Maths Chapter 10 Overview
Students can learn the topics more readily and with greater comprehension thanks to these NCERT Solutions. Experts in the field have created this curriculum by the most recent CBSE syllabus that the Board has mandated.
Students will learn how to write huge numbers more conveniently using exponents and powers, express numbers in standard form, cope with negative exponents, numerous rules of exponents, and much more in Class 8 CBSE Mathematical Chapter 10.
NCERT Solutions for Class 8 Maths Chapter 10 PDF
The primary goal of the NCERT Solutions for Class 8 Mathematical Chapter 10 is to assist students in better understanding the fundamental concepts of powers and exponents. In Class 8, one of the most important chapters is Exponents and Powers.
The class 8 answers PDF for exponents and powers is now available for students to download, giving them useful advice on how to ace their examinations. Students will therefore be able to discover how to get ready for their future exams with the aid of our study tools. The NCERT Solution PDFs that are available here include particular methods that students can use to study Chapter 10.
NCERT Solutions for Class 8 Maths Chapter 10 PDF
NCERT Solutions for Class 8 Maths Chapter 10
Here we have provided NCERT Solutions for Class 8 Maths Chapter 10 for the ease of students so that they can prepare better for their upcoming exams -
1. Evaluate:
(i) 3
-2
(ii) (-4)
-2
(iii) (1/2)
-5
Solution:
(i) 3
-2
= (1/3)
2
= 1/9
(ii)
(-4)
-2
= (1/-4)
2
= 1/16
(iii)
(1/2)
-5
= (2/1)
5
= 2
5
= 32
2. Simplify and express the result in power notation with a positive exponent:
(i) (-4)
4
÷(-4)
8
(ii) (1/2
3
)
2
(iii) -(3)
4
×(5/3)
4
(iv) (3
-7
÷3
-10
)×3
-5
(v) 2
-3
×(-7)
-3
Solution:
(i)
= (-4)
5
/(-4)
8
= (-4)
5-8
= 1/(-4)
3
(ii)
(1/2
3
)
2
= 1
2
/(2
3
)
2
= 1/2
3×2
= 1/2
6
(iii)
-(3)
4
×(5/3)
4
= (-1)
4
×3
4
×(5
4
/3
4
)
= 3
(4-4)
×5
4
= 3
0
×5
4
= 5
4
(iv)
= (3
-7
/3
-10
)× 3
-5
= 3
-7 – (-10)
× 3
-5
= 3
(-7+10)
×3
-5
= 3
3
×3
-5
= 3
(3+-5)
= 3
-2
=1/3
2
(v) 2
-3
×(-7)
– 3
= (2×-7)
-3
(Because a
m
×b
m
= (ab)
m
)
= 1/(2×-7)
3
= 1/(-14)
3
3. Find the value of:
(i) (3
0
+4
-1
)×2
2
(ii) (2
-1
×4
-1
)÷2
– 2
(iii) (1/2)
-2
+(1/3)
-2
+(1/4)
-2
(iv) (3
-1
+4
-1
+5
-1
)
0
(v) {(-2/3)
-2
}
2
Solution:
(i)(3
0
+4
– 1
)×2
2
= (1+(1/4))×2
2
= ((4+1)/4 )×2
2
= (5/4)×2
2
= (5/2
2
)×2
2
= 5×2
(2-2)
= 5×2
0
= 5×1 = 5
(ii)(2
-1
×4
-1
)÷2
-2
= [(1/2)×(1/4)] ÷(1/4)
= (1/2×1/2
2
)÷ 1/4
= 1/2
3
÷1/4
= (1/8)×(4)
= 1/2
(iii)
(1/2)
-2
+(1/3)
-2
+(1/4)
-2
= (2
-1
)
-2
+(3
-1
)
-2
+(4
-1
)
-2
= 2
(-1×-2)
+3
(-1×-2)
+4
(-1×-2)
= 2
2
+3
2
+4
2
= 4+9+16
=29
(iv) (3
-1
+4
-1
+5
-1
)
0
= 1
(v)
{(-2/3)
-2
}
2
= (-2/3)
-2×2
= (-2/3)
-4
= (-3/2)
4
= 81/16
4. Evaluate:
(i) (8
-1
×5
3
)/2
-4
(ii) (5
-1
×2
-2
)×6
-1
Solution:
(i) (8
-1
×5
3
)/2
-4
=
= 2×125 = 250
(ii)
(5
-1
×2
-2
)×6
-1
= (1/10)×1/6
= 1/60
5. Find the value of
m
for which 5
m
÷ 5
-3
= 5
5
Solution:
5
m
÷ 5
-3
= 5
5
5
(m-(-3) )
= 5
5
5
m+3
=5
5
Comparing exponents on both sides, we get
m+3 = 5
m = 5-3
m = 2
6. Evaluate:
(i)
(ii)
Solution:
(i)
= 3-4
= -1
(ii)
=
=
=
= 512/125
7. Simplify the following:
(i)
(ii)
Solution:
(i)
=
=
(ii)
=
=
=
=
= 1×1×3125
= 3125
NCERT Solutions for Class 8 Maths Chapter 10 Ex 10.2
1. Express the following numbers in standard form.
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000
Solution:
(i) 0.0000000000085 = 0.0000000000085×(10
12
/10
12
) = 8.5 ×10
-12
(ii) 0.00000000000942 = 0.00000000000942×(10
12
/10
12
) = 9.42×10
-12
(iii) 6020000000000000 = 6020000000000000×(10
15
/10
15
) = 6.02×10
15
(iv) 0.00000000837 = 0.00000000837×(10
9
/10
9
) = 8.37×10
-9
(v) 31860000000 = 31860000000×(10
10
/10
10
) = 3.186×10
10
2. Express the following numbers in the usual form.
(i) 3.02×10
-6
(ii) 4.5×10
4
(iii) 3×10
-8
(iv) 1.0001×10
9
(v) 5.8×10
12
(vi) 3.61492×10
6
Solution:
(i) 3.02
×
10
-6
= 3.02/10
6
= 0 .00000302
(ii) 4.5
×
10
4
= 4.5
×
10000 = 45000
(iii) 3
×
10
-8
= 3/10
8
= 0.00000003
(iv) 1.0001
×
10
9
= 1000100000
(v) 5.8
×
10
12
= 5.8
×
1000000000000 = 5800000000000
(vi) 3.61492
×
10
6
= 3.61492
×
1000000 = 3614920
3. Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to 1/1000000 m.
(ii) The charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.
(iii) Size of bacteria is 0.0000005 m
(iv) The size of a plant cell is 0.00001275 m
(v) The Thickness of a thick paper is 0.07 mm
Solution:
(i) 1 micron = 1/1000000
= 1/10
6
= 1×10
-6
(ii) The charge of an electron is 0.00000000000000000016 coulombs
= 0.00000000000000000016
×
10
19
/10
19
= 1.6
×
10
-19
coulomb
(iii) Size of bacteria = 0.0000005
= 5/10000000 = 5/10
7
= 5×10
-7
m
(iv) The size of a plant cell is 0.00001275 m
= 0.00001275×10
5
/10
5
= 1.275×10
-5
m
(v) Thickness of a thick paper = 0.07 mm
0.07 mm = 7/100 mm = 7/10
2
= 7×10
-2
mm
4. In a stack, there are 5 books, each having a thickness of 20 mm, and 5 paper sheets, each having a thickness of 0.016 mm. What is the total thickness of the stack?
Solution:
The thickness of one book = 20 mm
Thickness of 5 books = 20×5 = 100 mm
The thickness of one paper = 0.016 mm
Thickness of 5 papers = 0.016×5 = 0.08 mm
Total thickness of a stack = 100+0.08 = 100.08 mm
= 100.08×10
2
/10
2
mm
mm
Benefits of NCERT Solutions for Class 8 Maths Chapter 10
NCERT Solutions for Class 8 Maths Chapter 10 on Exponents and Powers offer several benefits that contribute to a student's understanding and learning experience:
Structured Learning
: NCERT solutions provide a structured approach to solving problems, following the curriculum guidelines set by educational boards. This helps students to stay organized and focused on the topics covered in the chapter.
Concept Clarity
: The solutions help clarify concepts related to exponents and powers by providing step-by-step explanations. This is crucial for students to grasp the underlying principles and applications of exponents in mathematical problems.
Variety of Problems
: NCERT solutions typically cover a variety of problems ranging from basic to advanced levels. This variety helps students to practice different types of questions and enhances their problem-solving skills.
Accuracy and Reliability
: NCERT solutions are prepared by experts in the field, ensuring accuracy in the solutions provided. This reliability is important for students and teachers alike, as they can trust the correctness of the information and solutions provided.
Exam Preparation
: By practicing with NCERT solutions, students can familiarize themselves with the type of questions that are likely to appear in exams. This helps them to prepare effectively and perform well in their assessments.
Self-assessment and Improvement
: NCERT solutions enable students to self-assess their understanding of the chapter. They can identify areas where they may need more practice or further clarification, allowing for targeted improvement.
Encourages Independence
: Using NCERT solutions encourages students to think independently and develop their problem-solving skills. They learn how to approach different types of problems and develop confidence in their mathematical abilities.
Accessible Resource
: NCERT solutions are widely available and accessible, either in textbooks or online platforms. This accessibility ensures that students and teachers can easily refer to them whenever needed, both inside and outside the classroom.
