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CBSE 7th Standard Mathematics Exponents and Powers Model Questions Papers Study Material - QB365 Set A

Published on: 31/10/2025

Question Set A

CBSE 7th Standard Mathematics Exponents and Powers Model Questions Papers Study Material - QB365

Download CBSE Class 7th Standard CBSE Mathematics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 7th Standard CBSE Mathematics

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Std: 7th Standard CBSE - Mathematics

chapter: Exponents and Powers - Model Questions Papers

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Questions:

Generic 2

5 x 2 = 10

Using laws of exponents, simplify and write the answer in exponential form: 3² x 3⁴ x 3⁸

Express the following numbers as a power of - 3: -27

Express the following in exponential form: t x t

Find the value of 9³

Express 128 as a power of 2

True or False

5 x 1 = 5

0.0016432001 =1.6432001 x10^-3

(a)

False

(b)

True

2⁶x7⁶=14⁶

(a)

False

(b)

True

\(\left( \frac { 7 }{ 3 } \right) ^{ 2 }\times \left( \frac { 7 }{ 3 } \right) ^{ 5 }=\left( \frac { 7 }{ 3 } \right) ^{ 10 }\).

(a)

True

(b)

False

4⁹ is greater than 16³.

(a)

False

(b)

True

10.

2³ > 5²

(a)

True

(b)

False

Multiple Choice Question

5 x 1 = 5

11.

Standard form corresponding to the number 654300100 is

(a)

6.543001 X10⁷

(b)

6.543001 x10⁸

(c)

6.543 x10⁸

(d)

6.543001 x 10⁹

12.

If \(\frac { a }{ b } =\left( \frac { 625 }{ 81 } \right) \div \left( \frac { { 5 }^{ 4 } }{ { 3 }^{ 4 } } \right) \), then the value of \(\left( \frac { a }{ b } \right) ^{ 5 }\) is

(a)

\(\left( \frac { 5 }{ 3 } \right) ^{ 8 }\)

(b)

\(\left( \frac { 3 }{ 5 } \right) ^{ 8 }\)

(c)

(d)

\(\frac { 3 }{ 5 } \)

13.

If \(\frac { p }{ q } =\left( \frac { 5 }{ 6 } \right) ^{ 2 }\div \left( \frac { 5 }{ 6 } \right) ^{ 0 }\), then the value of \(\left( \frac { p }{ q } \right) ^{ 2 }\) is

(a)

\(\frac { 125 }{ 1290 } \)

(b)

\(\frac { 625 }{ 1296 } \)

(c)

\(\frac { 164 }{ 125 } \)

(d)

\(\frac { 169 }{ 144 } \).

14.

\(\left[ \{ \left( \frac { 2 }{ -9 } \right) ^{ 2 }\} ^{ 0 } \right] ^{ 2 }\) is equal to

(a)

(b)

\(\frac { 4 }{ 81 } \)

(c)

\(\frac { 81 }{ 4 } \)

(d)

15.

The value of \(\frac { { 10 }^{ 22 }+{ 10 }^{ 20 } }{ 10^{ 20 } } \) is

(a)

(b)

10⁴²

(c)

101

(d)

10²²

Fill up / 1 Marks

5 x 1 = 5

16.

7⁴ _____ 5⁴

17.

< ,> or = sign. 10000 ...... 10⁵

18.

< ,> or = sign. 3² ...... 15

19.

53700000 = ........ x 10⁷

20.

(6⁴ ÷ 6³) x (1)⁹² x 2³⁶ ÷ 2³² = _______

Generic 3

5 x 3 = 15

21.

Write 8054000000 in standard form

22.

Simplify: \({ \left( \frac { 3 }{ 4 } \right) }^{ 4 }\div { \left( \frac { 6 }{ 8 } \right) }^{ 2 }\times \left( \frac { 1 }{ 2 } \right) \)

23.

If \(\frac { p }{ q } =\left( \frac { 3 }{ 2 } \right) ^{ 2 }\div \left( \frac { 9 }{ 4 } \right) ^{ 0 }\), then find the value of \(\left( \frac { p }{ q } \right) ^{ 3 }\).

24.

If \(\frac { p }{ q } ={ \left( \frac { -2 }{ 3 } \right) }^{ 9 }\div { \left( \frac { -2 }{ 3 } \right) }^{ 8 }\) then find the value of \({ \left( \frac { p }{ q } \right) }^{ 2 }\)

25.

Express each of the following numbers using exponential notations.
1029

Generic 5

3 x 5 = 15

26.

By what number should (- 4)⁵ be divided so that the quotient may be equal to (-4)³?

27.

Using laws of exponents, solve the following: \(\left[ { \left( \frac { -2 }{ 3 } \right) }^{ 4 }\times \left( \frac { 216 }{ 125 } \right) \right] \div \left[ { \left( \frac { 6 }{ 5 } \right) }^{ 2 }\times \left( \frac { 4 }{ 9 } \right) \right] \)

28.

Simplify \(\frac { { 5 }^{ -2 }\times { 3 }^{ -3 }\times (125)^{ 2/3 } }{ (27)^{ -2/3 }\times (32)^{ -1/5 } } \).

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Answers Key:

Generic 2

5 x 2 = 10

3² x 3⁴ x 3⁸ = (3²⁺⁴)x3⁸ [\(\because\)a^m x aⁿ =a^m+n]
=3⁶ x3⁸ =3⁶⁺⁸ =3¹⁴

(-3)³

t x t = t² [since, t is multiplied 2 times]

9³ = 9 x 9 x 9 = 729
Hence, the value of 9³ is 729.

We have, 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁷

Here, base = 2 and exponent = 7, since 2 repeated 7 times.

True or False

5 x 1 = 5

(a)
False

(a)
False

(b)
False

(a)
False

10.

(b)
False

Multiple Choice Question

5 x 1 = 5

11.

12.

13.

(b)
\(\frac { 625 }{ 1296 } \)

14.

(d)
1

15.

Fill up / 1 Marks

5 x 1 = 5

16.

()
7⁴ > 5⁴

17.

()
10000 < 10⁵

18.

()
3² < 15

19.

()
5.37

20.

()
96

Generic 3

5 x 3 = 15

21.

8.054 x 10⁹

22.

\(\frac { 9 }{ 32 } \)

23.

Given, \(\frac { p }{ q } =\left( \frac { 3 }{ 2 } \right) ^{ 2 }\div \left( \frac { 9 }{ 4 } \right) ^{ 0 }\)
∵ \(\left( \frac { 9 }{ 4 } \right) ^{ 0 }=\{ \left( \frac { 3 }{ 2 } \right) ^{ 2 }\} ^{ 0 }=\left( \frac { 3 }{ 2 } \right) ^{ 2\times 0 }=\left( \frac { 3 }{ 2 } \right) ^{ 0 }\)
Now, \(\left( \frac { 3 }{ 2 } \right) ^{ 3 }\div \left( \frac { 3 }{ 2 } \right) ^{ 0 }\)
Let a= \(\frac { 3 }{ 2 } \)
So, \(\frac { p }{ q } =\left( \frac { 3 }{ 2 } \right) ^{ 2-0 }=\left( \frac { 3 }{ 2 } \right) ^{ 2\times 3 }\) [∵ a^m ÷ aⁿ =a^m-n]
Now, \(\left( \frac { p }{ q } \right) ^{ 3 }=\{ \left( \frac { 3 }{ 2 } \right) ^{ 2 }\} ^{ 3 }=\left( \frac { 3 }{ 2 } \right) ^{ 2x3 }=\left( \frac { 3 }{ 2 } \right) ^{ 6 }\)
= \(\frac { 3\times 3\times 3\times 3\times 3\times 3 }{ 2\times 2\times 2\times 2\times 2\times 2 } =\frac { 729 }{ 64 } \).

24.

Given, \(\frac { p }{ q } ={ \left( \frac { -2 }{ 3 } \right) }^{ 9 }\div { \left( \frac { -2 }{ 3 } \right) }^{ 8 }\)
\(\because\) a^m \(\div\)aⁿ =a^m-n
Let \(a=\left( \frac { -2 }{ 3 } \right) \)
So, \(\frac { p }{ q } ={ \left( \frac { -2 }{ 3 } \right) }^{ 9-8 }={ \left( \frac { -2 }{ 3 } \right) }^{ 1 }=\frac { -2 }{ 3 } \)
\(\therefore\) \({ \left( \frac { p }{ q } \right) }^{ 2 }={ \left( \frac { -2 }{ 3 } \right) }^{ 2 }\)
\(=\frac { -2\times \left( -2 \right) }{ 3\times 3 } =\frac { 4 }{ 9 } \)
Hence, the value of \({ \left( \frac { p }{ q } \right) }^{ 2 }\) is \(\frac { 4 }{ 9 } \)

25.

Given, 1029
∵ 1029 = 3 x 7 x 7 x 7 = 3 x 7³

The exponent form of 1029 is 3 x 7³.

Generic 5

3 x 5 = 15

26.

(-4)² or 16

27.

\(\frac { 8 }{ 3 } \)

28.

\(\frac { { 5 }^{ -2 }\times { 3 }^{ -3 }\times (125)^{ 2/3 } }{ (27)^{ -2/3 }\times (32)^{ -1/5 } } \)
∵ 125 = (5)³ = 5 x 5 x 5
So, (125)^2/3 =(5)^{3 x 2/3} =5² and 27=(3)³
∴ (27)^-2/3 = {(3)³}^-2/3 =\({ (3 })^{ 3\times \frac { -2 }{ 3 } }\) =(3)²
32 = 2 x 2 x 2 x 2 x 2 = (2)⁵
So, (32)^-1/5 = {(2)⁵}^-1/5 = \({ (2) }^{ 5\times \frac { (-1) }{ 5 } }\)=(2)^-1
Now \(\frac { { 5 }^{ -2 }\times { 3 }^{ -3 }\times { 5 }^{ 2 } }{ (3)^{ -2 }\times (2)^{ -1 } } \) \(\left[ \because a^{ -m }=\frac { 1 }{ { a }^{ m } } \right] \)
= 5^-2 x 3^-3 x 3² x 2¹ x 5²
=5^-2+2 x 3^-3+2 x 2¹
a^m x aⁿ =a^m+n
=5⁰ x 3^-1 x 2¹ = 1 x \(\frac { 1 }{ 3 } \times 2=\frac { 2 }{ 3 } \).