NCERT Solutions for Class 7 Maths Chapter 11: Exponents and Powers - Exercise 11.1

Exercise 11.1

Refer to exercise 11.1 in the PDF

1. Find the value of:

i) \[{2^6}\]

Ans: The given expression is : \[{2^6}\]

By expanding the term \[{2^6}\] we get,

${2^6} = 2 \times 2 \times 2 \times 2 \times 2 \times 2$

By multiplying,

${2^6} = 64$

Hence, the required value of \[{2^6}\] is $64$.

ii) ${9^3}$

Ans: The given expression is : ${9^3}$

By expanding the term ${9^3}$ we get,

${9^3} = 9 \times 9 \times 9$

By multiplying,

${9^3} = 729$

Hence, the required value of ${9^3}$ is $729$.

iii) ${11^2}$

Ans: The given expression is : ${11^2}$

By expanding the term ${11^2}$ we get,

${11^2} = 11 \times 11$

By multiplying,

${11^2} = 121$

Hence, the required value of ${11^2}$ is $121$.

iv) ${5^4}$

Ans: The given expression is : ${5^4}$

By expanding the term ${5^4}$ we get,

${5^4} = 5 \times 5 \times 5 \times 5$

By multiplying,

${5^4} = 625$

Hence, the required value of ${5^4}$ is $625$.

2. Express the following in exponential form:

i) $6 \times 6 \times 6 \times 6$

Ans: The given expression is : $6 \times 6 \times 6 \times 6$

The number of $6$’s in the given expression is $4$ .

So, the exponent of $6$ is $4$.

Hence, $6 \times 6 \times 6 \times 6 = {6^4}$.

ii) $t \times t$

Ans: The given expression is : $t \times t$

The number of $t$ ’s in the given expression is $2$.

So, the exponent of $t$ is $2$.

Hence, $t \times t = {t^2}$.

iii) $b \times b \times b \times b$

Ans: The given expression is : $b \times b \times b \times b$

The number of $b$ ’s in the given expression is $4$.

So, the exponent of $t$ is $4$.

Hence, $b \times b \times b \times b = {b^4}$.

iv) $5 \times 5 \times 7 \times 7 \times 7$

Ans: The given expression is : $5 \times 5 \times 7 \times 7 \times 7$

The number of $5$ ’s in the given expression is $2$ .

The number of $7$ ’s in the given expression is $3$.

So, the exponent of $5$ is $2$ and the exponent of $7$ is $3$.

Hence, $5 \times 5 \times 7 \times 7 \times 7 = {5^2} \times {7^3}$.

v) $2 \times 2 \times a \times a$

Ans: The given expression is : $2 \times 2 \times a \times a$

The number of $2$ ’s in the given expression is $2$ .

The number of $a$ ’s in the given expression is $2$.

So, the exponent of $2$ is $2$ and the exponent of $a$ is $2$.

Hence, $2 \times 2 \times a \times a = {2^2} \times {a^2}$.

vi) $a \times a \times a \times c \times c \times c \times c \times d$

Ans: The given expression is : $a \times a \times a \times c \times c \times c \times c \times d$

The number of $a$ ’s in the given expression is $3$.

The number of $c$ ’s in the given expression is $4$.

The number of $d$ ’s in the given expression is $1$.

So, the exponent of $a$ is $3$ , the exponent of $c$ is $4$ and the exponent of $d$ is $1$ .

Hence, $a \times a \times a \times c \times c \times c \times c \times d = {a^3} \times {c^4} \times {d^1}$.

3. Express each of the following numbers using the exponential notation:

i) $512$

Ans: By factorization method, we can find,

In the above expanding form of $512$, numbers of $2$’s as a factor is $9$.

So, we can express $512$ as term of $2$ .

$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

The exponent of $2$ will be $9$ . 

Hence, $512 = {2^9}$.

ii) $343$

Ans: By factorization method we can find,

In the above expanding form of $343$ , numbers of $7$’s as factor is $3$.

So, we can express $343$ as term of $7$ .

$343 = 7 \times 7 \times 7$

The exponent of $7$ will be $3$ .

Hence, $343 = {7^3}$.

iii) $729$

Ans: By factorization method we can find,

In the above expanding form of $729$, numbers of $3$’s as the factor is $6$.

So, we can express $729$ as term of $3$ .

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$

The exponent of $3$ will be $6$ .

Hence, $729 = {3^6}$.

iv) $3125$

Ans: By factorization method, we can find,

In the above expanding form of $3125$, numbers of $5$’s as a factor is $5$.

So, we can express $3125$ as term of $5$ .

$3125 = 5 \times 5 \times 5 \times 5 \times 5$

The exponent of $5$ will be $5$ .

Hence, $3125 = {5^5}$.

4. Identify the greater number, wherever possible, in each of the following:

i) ${4^3}$ and ${3^4}$

Ans: ${4^3}$ can be written as,

${4^3} = 4 \times 4 \times 4$

${4^3} = 64$

${3^4}$can be written as,

${3^4} = 3 \times 3 \times 3 \times 3$

${3^4} = 81$

Since, $81 > 64$

Thus, ${3^4} > {4^3}$.

Hence, ${3^4}$ is greater than ${4^3}$.

ii) ${5^3}$ and ${3^5}$

Ans: ${5^3}$ can be written as,

${5^3} = 5 \times 5 \times 5$

${5^3} = 125$

${3^5}$ can be written as,

${3^5} = 3 \times 3 \times 3 \times 3 \times 3$

${3^5} = 243$

Since, $243 > 125$

Thus, ${3^5} > {5^3}$.

Hence, ${3^5}$ is greater than ${5^3}$.

iii) ${2^8}$ and ${8^2}$

Ans: ${2^8}$ can be written as,

${2^8} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

${2^8} = 256$

${8^2}$ can be written as,

${8^2} = 8 \times 8$

${8^2} = 64$

Since, $256 > 64$

Thus, ${2^8} > {8^2}$.

Hence, ${2^8}$ is greater than ${8^2}$.

iv) ${100^2}$ and ${2^{100}}$

Ans: ${100^2}$ can be written as,

${100^2} = 100 \times 100$

${100^2} = 10000$

${2^{100}}$can be written as,

${2^{100}} = 2 \times 2 \times 2 \times 2 \times 2 \times ....14{\text{ times}} \times .... \times 2$

${2^{100}} = 16384 \times ... \times 2$

Since, $16384 \times ... \times 2 > 10000$

Thus, ${2^{100}} > {100^2}$.

Hence, ${2^{100}}$ is greater than ${100^2}$.

v) ${2^{10}}$ and ${10^2}$

Ans: ${2^{10}}$can be written as,

${2^{10}} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

${2^{10}} = 1024$

${10^2}$ can be written as,

${10^2} = 10 \times 10$

${10^2} = 100$

Since, $1024 > 100$

Thus, ${2^{10}} > {10^2}$.

Hence, ${2^{10}}$ is greater than ${10^2}$.

5. Express each of the following as product of powers of their prime factors:

i) $648$

Ans: The given expression is $648$.

By factorization method we can find the prime factors,

So, $648$ can be written as

$648 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$

It can be express as exponent of $2$ and $3$ .

$648 = {2^3} \times {3^4}$

Hence, the required expression is $648 = {2^3} \times {3^4}$.

ii) $405$

Ans: The given expression is $405$.

By factorization method we can find the prime factors,

So, $405$ can be written as

$405 = 5 \times 3 \times 3 \times 3 \times 3$

Now, it can be express as

$405 = 5 \times {3^4}$

Hence, the required expression is $405 = 5 \times {3^4}$.

iii) $540$

Ans: The given expression is $540$.

By factorization method we can find the prime factors,

So, $540$ can be written as

$540 = 2 \times 2 \times 3 \times 3 \times 3 \times 5$

Now, it can be express as

$540 = {2^2} \times {3^3} \times 5$

Hence, the required expression is $540 = {2^2} \times {3^3} \times 5$.

iv) $3600$

Ans: The given expression is $3600$.

By factorization method we can find the prime factors,

So, $3600$ can be written as

$3600 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5$

Now, it can be express as

$3600 = {2^4} \times {3^2} \times {5^2}$

Hence, the required expression is $3600 = {2^4} \times {3^2} \times {5^2}$.

6. Simplify:

i) $2 \times {10^3}$

Ans: The given expression is: $2 \times {10^3}$

By expanding the given expression,

$2 \times {10^3} = 2 \times 10 \times 10 \times 10$

$2 \times {10^3} = 2000$

Hence, the required solution is $2 \times {10^3} = 2000$.

ii) ${7^2} \times {2^2}$

Ans: The given expression is: ${7^2} \times {2^2}$

By expanding the given expression,

${7^2} \times {2^2} = 7 \times 7 \times 2 \times 2$

${7^2} \times {2^2} = 196$

Hence, the required solution is ${7^2} \times {2^2} = 196$.

iii) ${2^3} \times 5$

Ans: The given expression is: ${2^3} \times 5$

By expanding the given expression,

${2^3} \times 5 = 2 \times 2 \times 2 \times 5$

${2^3} \times 5 = 40$

Hence, the required solution is ${2^3} \times 5 = 40$.

iv) $3 \times {4^4}$

Ans: The given expression is: $3 \times {4^4}$

By expanding the given expression,

$3 \times {4^4} = 3 \times 4 \times 4 \times 4 \times 4$

$3 \times {4^4} = 768$

Hence, the required solution is $3 \times {4^4} = 768$.

v) $0 \times {10^2}$

Ans: The given expression is: $0 \times {10^2}$

By expanding the given expression,

$0 \times {10^2} = 0 \times 10 \times 10$

$0 \times {10^2} = 0$

Hence, the required solution is $0 \times {10^2} = 0$.

vi) ${5^3} \times {3^3}$

Ans: The given expression is: ${5^3} \times {3^3}$

By expanding the given expression,

${5^3} \times {3^3} = 5 \times 5 \times 5 \times 3 \times 3 \times 3$

${5^3} \times {3^3} = 675$

Hence, the required solution is ${5^3} \times {3^3} = 675$.

vii) ${2^4} \times {3^2}$

Ans: The given expression is: ${2^4} \times {3^2}$

By expanding the given expression,

${2^4} \times {3^2} = 2 \times 2 \times 2 \times 2 \times 3 \times 3$

${2^4} \times {3^2} = 144$

Hence, the required solution is ${2^4} \times {3^2} = 144$.

viii) ${3^2} \times {10^4}$

Ans: The given expression is: ${3^2} \times {10^4}$

By expanding the given expression,

${3^2} \times {10^4} = 3 \times 3 \times 10 \times 10 \times 10 \times 10$

${3^2} \times {10^4} = 90,000$

Hence, the required solution is ${3^2} \times {10^4} = 90,000$.

7. Simplify:

i) ${\left( { - 4} \right)^3}$

Ans: The given expression is: ${\left( { - 4} \right)^3}$

By expanding the given expression,

${\left( { - 4} \right)^3} = \left( { - 4} \right) \times \left( { - 4} \right) \times \left( { - 4} \right)$

${\left( { - 4} \right)^3} =  - 64$

Hence, the required solution is${\left( { - 4} \right)^3} =  - 64$.

ii) $\left( { - 3} \right) \times {\left( { - 2} \right)^3}$

Ans: The given expression is: $\left( { - 3} \right) \times {\left( { - 2} \right)^3}$

By expanding the given expression,

$\left( { - 3} \right) \times {\left( { - 2} \right)^3} = \left( { - 3} \right) \times \left( { - 2} \right) \times \left( { - 2} \right) \times \left( { - 2} \right)$

$\left( { - 3} \right) \times {\left( { - 2} \right)^3} = 24$

Hence, the required solution is$\left( { - 3} \right) \times {\left( { - 2} \right)^3} = 24$.

iii) ${\left( { - 3} \right)^2} \times {\left( { - 5} \right)^2}$

Ans: The given expression is: ${\left( { - 3} \right)^2} \times {\left( { - 5} \right)^2}$

By expanding the given expression,

${\left( { - 3} \right)^2} \times {\left( { - 5} \right)^2} = \left( { - 3} \right) \times \left( { - 3} \right) \times \left( { - 5} \right) \times \left( { - 5} \right)$

${\left( { - 3} \right)^2} \times {\left( { - 5} \right)^2} = 225$

Hence, the required solution is${\left( { - 3} \right)^2} \times {\left( { - 5} \right)^2} = 225$.

iv) ${\left( { - 2} \right)^3} \times {\left( { - 10} \right)^3}$

Ans: The given expression is: ${\left( { - 2} \right)^3} \times {\left( { - 10} \right)^3}$

By expanding the given expression,

${\left( { - 2} \right)^3} \times {\left( { - 10} \right)^3} = 2 \times 2 \times 2 \times 10 \times 10 \times 10$

${\left( { - 2} \right)^3} \times {\left( { - 10} \right)^3} = 8000$

Hence, the required solution is${\left( { - 2} \right)^3} \times {\left( { - 10} \right)^3} = 8000$.

8. Compare the following numbers:

i) $2.7 \times {10^{12}};1.5 \times {10^8}$

Ans: The given expressions are :$2.7 \times {10^{12}}$and $1.5 \times {10^8}$.

Since, ${10^{12}} > {10^8}$

On comparing the exponents of base 10,
$2.7 \times {10^{12}} > 1.5 \times {10^8}$

Hence, $2.7 \times {10^{12}}$ is greater than $1.5 \times {10^8}$.

ii) $4 \times {10^{14}};3 \times {10^{17}}$

Ans: The given expressions are :$4 \times {10^{14}}$and $3 \times {10^{17}}$.

Since, ${10^{17}} > {10^{14}}$

On comparing the exponents of base 10,
$3 \times {10^{17}} > 4 \times {10^{14}}$

Hence, $3 \times {10^{17}}$ is greater than $4 \times {10^{14}}$.

NCERT Solutions for Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

Opting for the NCERT solutions for Ex 11.1 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.1 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 7 Maths Chapter 11 Exercise 11.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 7 Maths Chapter 11 Exercise 11.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 7 Maths Chapter 11 Exercise 11.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

Class 7 Maths Chapter 11: Exercises Breakdown

CBSE Class 7 Maths Chapter 11 Other Study Materials

Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Important Related Links for NCERT Class 7 Maths

Access these essential links for NCERT Class 7 Maths, offering comprehensive solutions, study guides, and additional resources to help students master language concepts and excel in their exams.