Exponents Test

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Exponents Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

25⁴ ÷5³= 5^---

A
5

B
3

C
1

D
4

Solution

(25)⁴ can be written as = (5 × 5)⁴
= (5²)⁴
(5²)⁴ can be written as = (5)^{2 × 4} … [∵(a^m)ⁿ= a^mn]
= 5⁸
Then,
= 5⁸ ÷ 5³
= 5^{8 – 3} … [∵a^m÷ aⁿ = a^{m – n}]
= 5⁵
Question 2
1 / -0

120719=

A
(1×10⁵)+(0×10³)+(7×10²)+(1×10¹)+(9×10⁰)

B
(1×10⁵)+(2×10⁴)+(7×10²)+(1×10¹)+(9×10⁰)

C
(1×10⁵)+(2×10⁴)+(0×10³)+(7×10²)+(1×10¹)+(9×10⁰)

D
(1×10⁵)+(2×10⁴)+(0×10³)+(7×10²)+(9×10⁰)

Solution

The expanded form of the number 120719 is,
= (1 × 100000) + (2 × 10000) + (0 × 1000) + (7 × 100) + (1 × 10) + (9 ×1)
Now we can express it using powers of 10 in the exponent form,
= (1 × 10⁵) + (2 × 10⁴) + (0 × 10³) + (7 × 10²) + (1 × 10¹) + (9 × 10⁰)
Question 3
1 / -0

2³ __5²

A
<

B
>

C
=

D
can not determine

Solution

Let us consider LHS = 2³
Expansion of 2³ = 2 × 2 × 2
= 8
Now, consider RHS = 5²
Expansion of 5² = 5 × 5
= 25
By comparing LHS and RHS,
LHS < RHS
2³ < 5²
Question 4
1 / -0

((5²)³ ×5⁴)÷ 5⁷= 5^---

A
4

B
3

C
2

D
1

Solution

(5²)³ can be written as = (5)^{2 × 3} … [∵(a^m)ⁿ= a^mn]
= 5⁶
Then,
= (5⁶× 5⁴) ÷ 5⁷
= (5^{6 + 4}) ÷ 5⁷ … [∵a^m× aⁿ = a^{m + n}]
= 5¹⁰ ÷ 5⁷
= 5^{10 – 7} … [∵a^m÷ aⁿ = a^{m – n}]
= 5³
Question 5
1 / -0

In 3006194,the value of 4 =

A
4×10⁰²

B
4×10⁰³

C
4×10⁰¹

D
4×10⁰⁰

Solution

The expanded form of the number 3006194 is,
= (3 × 1000000) + (0 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) +(9 × 10) + 4
Now we can express it using powers of 10 in the exponent form,
= (3 × 10⁶) + (0 × 10⁵) + (0 × 10⁴) + (6 × 10³) + (1 × 10²) + (9 × 10¹) + (4 × 10⁰)
Question 6
1 / -0

108 × 192= 2⁸×3--

A
3

B
4

C
5

D
6

Solution

The factors of 108 = 2 × 2 × 3 × 3 × 3
= 2² × 3³
The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
= 2⁶ × 3
Then,
= (2² × 3³) × (2⁶ × 3)
= 2^{2 + 6} × 3^{3 + 3} … [∵a^m× aⁿ = a^{m + n}]
= 2⁸× 3⁶
Question 7
1 / -0

10 × 10¹¹ =100¹¹

A
True

B
False

C
can not determine

D
Incorrect question

Solution

Let us consider Left Hand Side (LHS) = 10 × 10¹¹
= 10^{1 + 11} … [∵a^m× aⁿ = a^{m + n}]
= 10¹²
Now, consider Right Hand Side (RHS) = 100¹¹
= (10 × 10)¹¹
= (10^{1 + 1})¹¹
= (10²)¹¹
= (10)^{2 × 11} … [∵(a^m)ⁿ= a^mn]
= 10²²
By comparing LHS and RHS,
LHS ≠ RHS
Hence, the given statementis false.
Question 8
1 / -0

279404= (2×10^---)+(7×10⁴)+(9×10³)+(4×10²)+(0×10¹)+(4×10⁰)

A
5

B
6

C
4

D
3

Solution

The expanded form of the number 279404 is,
= (2 × 100000) + (7 × 10000) + (9 × 1000) + (4 × 100) + (0 × 10) + (4 ×1)
Now we can express it using powers of 10 in the exponent form,
= (2 × 10⁵) + (7 × 10⁴) + (9 × 10³) + (4 × 10²) + (0 × 10¹) + (4 × 10⁰)
Question 9
1 / -0

729 × 64= 3⁶ ×2^---

A
7

B
6

C
5

D
4

Solution

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3
= 3⁶
The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2
= 2⁶
Then,
= (3⁶ × 2⁶)
= 3⁶ × 2⁶
Question 10
1 / -0

2806196=

A
(2×10⁶)+(8×10⁵)+(0×10⁴)+(1×10²)+(9×10¹)+(6×10⁰)

B
(2×10⁶)+(8×10⁵)+(0×10⁴)+(6×10³)+(1×10²)+(9×10¹)+(6×10⁰)

C
(2×10⁶)+(8×10⁵)+(0×10⁴)+(6×10³)+(9×10¹)+(6×10⁰)

D
(2×10⁶)+(8×10⁵)+(6×10³)+(1×10²)+(9×10¹)+(6×10⁰)

Solution

The expanded form of the number 2806196 is,
= (2 × 1000000) + (8 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) +(9 × 10) + 6
Now we can express it using powers of 10 in the exponent form,
= (2 × 10⁶) + (8 × 10⁵) + (0 × 10⁴) + (6 × 10³) + (1 × 10²) + (9 × 10¹) + (6 × 10⁰)
Question 11
1 / -0

(8 ×10)⁴ +(6 × 10)³ +(0 × 10)² +(4 × 10)¹ +(5 × 10)⁰=

A
86535

B
86425

C
86045

D
86355

Solution

The expanded form is,
= (8 × 10000) + (6 × 1000) + (0 × 100) + (4 × 10) + (5 × 1)
= 80000 + 6000 + 0 + 40 + 5
= 86045
Question 12
1 / -0

2³ ×3² __6⁵

A
<

B
>

C
=

D
can not determine

Solution

Solution:-
Let us consider LHS = 2³ × 3²
Expansion of 2³ × 3²= 2 × 2 × 2 × 3 × 3
= 72
Now, consider RHS = 6⁵
Expansion of 6⁵ = 6 × 6 × 6 ×6 × 6
= 7776
By comparing LHS and RHS,
LHS < RHS
Question 13
1 / -0

270= 2×3^---×5

A
0

B
1

C
2

D
3

Solution

The factors of 270 = 2 × 3 × 3 × 3 × 5
= 2 × 3³ × 5
Question 14
1 / -0

((2⁵)² ×7³)/(8³ ×7)=__

A
54

B
67

C
86

D
98

Solution

8³ can be written as = (2 × 2 × 2)³
= (2³)³
We have,
= ((2⁵)² × 7³)/ ((2³)³ × 7)
= (2^{5 × 2} × 7³)/ ((2^{3 × 3} × 7) … [∵(a^m)ⁿ= a^mn]
= (2¹⁰× 7³)/ (2⁹ × 7)
= (2^{10 – 9} × 7^{3 – 1}) … [∵a^m÷ aⁿ = a^{m – n}]
= 2 × 7²
= 2 × 7 × 7
= 98
Question 15
1 / -0

3⁰ __(1000)⁰

A
<

B
>

C
=

D
can not determine

Solution

Let us consider LHS = 3⁰
= 1
Now, consider RHS = 1000⁰
= 1
By comparing LHS and RHS,
LHS = RHS
3⁰ = 1000⁰

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Exponents Test - 7

Result

Exponents Test - 7

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Question 1

5

3

1

4

Solution

Question 2

(1×105)+(0×103)+(7×102)+(1×101)+(9×100)

(1×105)+(2×104)+(7×102)+(1×101)+(9×100)

(1×105)+(2×104)+(0×103)+(7×102)+(1×101)+(9×100)

(1×105)+(2×104)+(0×103)+(7×102)+(9×100)

Solution

Question 3

<

>

=

can not determine

Solution

Question 4

4

3

2

1

Solution

Question 5

4×1002

4×1003

4×1001

4×1000

Solution

Question 6

3

4

5

6

Solution

Question 7

True

False

can not determine

Incorrect question

Solution

Question 8

5

6

4

3

Solution

Question 9

7

6

5

4

Solution

Question 10

(2×106)+(8×105)+(0×104)+(1×102)+(9×101)+(6×100)

(2×106)+(8×105)+(0×104)+(6×103)+(1×102)+(9×101)+(6×100)

(2×106)+(8×105)+(0×104)+(6×103)+(9×101)+(6×100)

(2×106)+(8×105)+(6×103)+(1×102)+(9×101)+(6×100)

Solution

Question 11

86535

86425

86045

86355

Solution

Question 12

<

>

=

can not determine

Solution

(1×10⁵)+(0×10³)+(7×10²)+(1×10¹)+(9×10⁰)

(1×10⁵)+(2×10⁴)+(7×10²)+(1×10¹)+(9×10⁰)

(1×10⁵)+(2×10⁴)+(0×10³)+(7×10²)+(1×10¹)+(9×10⁰)

(1×10⁵)+(2×10⁴)+(0×10³)+(7×10²)+(9×10⁰)

4×10⁰²

4×10⁰³

4×10⁰¹

4×10⁰⁰

(2×10⁶)+(8×10⁵)+(0×10⁴)+(1×10²)+(9×10¹)+(6×10⁰)

(2×10⁶)+(8×10⁵)+(0×10⁴)+(6×10³)+(1×10²)+(9×10¹)+(6×10⁰)

(2×10⁶)+(8×10⁵)+(0×10⁴)+(6×10³)+(9×10¹)+(6×10⁰)

(2×10⁶)+(8×10⁵)+(6×10³)+(1×10²)+(9×10¹)+(6×10⁰)