Exponents And Powers(7th Grade > Mathematics ) Questions and answers for exam Preparation

7th Grade > Mathematics

EXPONENTS AND POWERS MCQs

Total Questions : 102 | Page 3 of 11 pages

x = y^{a}, y = z^{b}, z = x^{c}

then find abc. If a = 12, b =

\frac{1}{288}

, then find the value of c. [4 MARKS]

Discuss Question

:
Steps: 1 Marks
Application: 1 Mark
Value of abc: 1 Mark
Value of c: 1 Mark
It is given that

x = y^{a}, y = z^{b}, z = x^{c}

x = y^{a}

= (z^{b})^{a}

(∵ y = z^{b})

= z^{a b}

= (x^{c})^{a b}

(∵ x^{c} = z)

x = x^{a b c}

∵ ((a^{m})^{n} = a^{m n})

\Rightarrow x^{1} = x^{a b c}

As bases are equal, exponents are equal.

\Rightarrow a b c = 1

Now, it is also given that,
a = 12, b =

\frac{1}{288}

\times

\times

c = 1
c =

\frac{1}{a \times b}

On substituting the values we get:
c =

\frac{1}{12 \times \frac{1}{288}}

c =

\frac{288}{12}

c =

24

Hence, the value of c is

24

Question 22. Find the difference between

4^{5}

and

5^{4}

? [2 MARKS]

Discuss Question

:
Steps: 1 Mark
Answer: 1 Mark

4^{5} = (4) \times (4) \times (4) \times (4) \times (4)

\Rightarrow 4^{5} = 1024

5^{4} = (5) \times (5) \times (5) \times (5)

\Rightarrow 5^{4} = 625

The difference between

4^{5}

and

5^{4}

4^{5} - 5^{4} = 1024 - 625

\Rightarrow 4^{5} - 5^{4} = 399

The difference between

4^{5}

and

5^{4}

399

Question 23. Is

3^{0} = (1000)^{0}

true or false? Justify your answer. [2 MARKS]

Discuss Question

:
Concept: 1 Mark
Answer:1 Mark
Consider,

3^{0} = (1000)^{0}

L.H.S. is

3^{0} = 1

[

∵ a^{0} = 1

]
and R.H.S.

(1000)^{0} = 1

[

∵ a^{0} = 1

]

\Rightarrow

L.H.S. = R.H.S.

\Rightarrow 3^{0} = (1000)^{0}

∴

It is true.
We should rememberthat any non-zero number raised to the power 0 is always equal to 1.

Question 24. Which of the numbers 726 and 938, when expressed as factors has greater power of 2?
[2 MARKS]

Discuss Question

:
Steps: 1 Mark
Answer: 1 Mark
The prime factorisation of 726 and 938 are given below.

726 = (2) \times (3) \times (11) \times (11)

\Rightarrow 726 = 2^{1} \times 3^{1} \times 11^{2}

938 = (2) \times (7) \times (67)

\Rightarrow 938 = 2^{1} \times 7^{1} \times 67

Both have the same powers of 2 which is 1.

Question 25. In a race Karan covers 1m in

1^{s t}

second, in the next second he covers twice the distance he covered in

1^{s t}

second, in

3^{r d}

second he covers twice the distance which he covered in

2^{n d}

second and he continues to run like this till six seconds. After that he slows down, as in

7^{t h}

second he covers only half the distance which he covered in

6^{t h}

second and in

8^{t h}

second he covers half the distance which he covered in

7^{t h}

second and he continuous in that manner. If Karan completes the race in 9 seconds, find the length of given track. [4 MARKS]

Discuss Question

:
Application: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
Here are the distances covered by Karan in each second

1^{s t}

second = 1m

2^{n d}

second =

1 \times 2

=2m
Hence, the length of the track is 2 m.

3^{r d}

second =

2 \times 2

= 4m

4^{t h}

second =

4 \times 2

= 8m

5^{t h}

second =

8 \times 2

=16m

6^{t h}

second =

16 \times 2

=32m

7^{t h}

second =

32 \div 2

= 16m

8^{t h}

second =

16 \div 2

= 8m

9^{t h}

second =

8 \div 2

= 4m
Length of the given track = (1+2+4+8+16+32+16+8+4) m

\Rightarrow

Length of the given track = 91 m
Hence, the length of the track is 91 m.

Question 26. Write

(8) \times (8) \times (8) \times (8) \times (8)

in the exponential form, taking 2 as base. [2 MARKS]

Discuss Question

:
Concept: 1 Mark
Answer: 1 Mark
Here weknowthat for any non zero number if the base is same and the number is multiplied,then the exponents are added.

(8) \times (8) \times (8) \times (8) \times (8)

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

= 2^{3 + 3 + 3 + 3 + 3}

∵ a^{b} \times a^{c} = a^{b + c}

= 2^{15}

Question 27. Express each of the following numbers using the exponential notation:
(i) 512
(ii) 343
(iii) 729

Discuss Question

:
Each option: 1 Mark
i)

512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{9}

ii)

343 = 7 \times 7 \times 7 = 7^{3}

iii)

729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^{6}

Question 28. Simplify: [4 MARKS]
i)

{(\frac{5}{8})}^{- 7} \times {(\frac{8}{5})}^{- 5}

ii)

\frac{(2^{5})^{2} \times 7^{3}}{8^{3} \times 7}

Discuss Question

:
Each: [2 Marks]
i)

{(\frac{5}{8})}^{- 7} \times {(\frac{8}{5})}^{- 5}

= \frac{5^{- 7}}{8^{- 7}} \times \frac{8^{- 5}}{5^{- 5}}

= \frac{5^{- 7}}{5^{- 5}} \times \frac{8^{- 5}}{8^{- 7}}

= 5^{(- 7) - (- 5)} \times 8^{(- 5) - (- 7)}

(

∵ \frac{a^{m}}{a^{n}} = a^{m - n}

)

= 5^{- 2} \times 8^{2} = \frac{8^{2}}{5^{2}} = \frac{64}{25}

ii)

\frac{(2^{5})^{2} \times 7^{3}}{8^{3} \times 7}

\frac{(2^{5})^{2} \times 7^{3}}{8^{3} \times 7}

\frac{(2^{5 \times 2}) \times 7^{3}}{(2^{3})^{3} \times 7}

(∵ (a^{m})^{n} = a^{m \times n})

\frac{2^{10} \times 7^{3}}{2^{9} \times 7}

2^{10 - 9} \times 7^{3 - 1} = 2 \times 7^{2}

2 \times 49 = 98

Question 29. A ball is released from a height of

x

meters. Given that it bounces back to half of its original height. Find the product of heights after eight bounces. [4 MARKS]

Discuss Question

:
Formula: 1 Mark
Application: 1 Mark
Answer: 2Marks
Given that
The ball is released from a height of

x m

Height reached after first bounce

= \frac{x}{2}

Height reached after second bounce

= \frac{x}{4} = \frac{x}{2^{2}}

Height reached after third bounce

= \frac{x}{8} = \frac{x}{2^{3}}

Height reached after fourth bounce

= \frac{x}{16} = \frac{x}{2^{4}}

Height reached after fifth bounce

= \frac{x}{32} = \frac{x}{2^{5}}

Height reached after sixth bounce

= \frac{x}{64} = \frac{x}{2^{6}}

Height reached after seventh bounce

= \frac{x}{128} = \frac{x}{2^{7}}

Height reached after eight bounce

= \frac{x}{256} = \frac{x}{2^{8}}

∴

Product of first 9 heights

= x \times \frac{x}{2} \times \frac{x}{2^{2}} \times \frac{x}{2^{3}} \times \frac{x}{2^{4}} \times \frac{x}{2^{5}} \times \frac{x}{2^{6}} \times \frac{x}{2^{7}} \times \frac{x}{2^{8}}

= \frac{x^{(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1)}}{2^{(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8)}}

(∵ a^{b} \times a^{c} = a^{b + c})

= \frac{x^{9}}{2^{36}}

The product of heights after 8 bounces

= \frac{x^{9}}{2^{36}}

Question 30. When a number is multiplied with 25 it gives 78125. Find the number and express it as a power of 5. What is the product if the number is multiplied by