True False[1 Marks ]

Question 11 Mark

x^m ÷ y^m = (x ÷ y)^m, where x and y are non-zero rational numbers and m is a positive integer.

Answer

True.
Solution:
If x and y are rational numbers, then $\frac{\text{x}}{\text{y}^{\text{m}}}=\frac{\text{x}^{\text{m}}}{\text{y}^{\text{m}}}$
or $\text{x}^{\text{m}}\div\text{y}^{\text{m}}=(\text{x}\div\text{y})^{\text{m}}$

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Question 21 Mark

x^m× xⁿ = x^m + n, where x is a non-zero rational number and m,n are positive integers.

Answer

True.
Solution:
If x is a rational number and m and n are positive integers, then
a^m × aⁿ
= a^m+n

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Question 31 Mark

x^m + x^m = x^2m, where x is a non-zero rational number and m is a positive integer

Answer

False.
Solution:
a^m × aⁿ = a^m+n
x^m × x^m = x^m+m = x^2m
Also, a^k + a^k = 2 a^k
So, x^m + x^m= 2 x^m

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Question 41 Mark

x^m × y^m = (x × y)^2m, where x and y are non-zero rational numbers and m is a positive integer.

Answer

False.
Solution:
If a and b are rational numbers, then
x^m× y^m = (ab)^m
x^m × y^m = (xy)^m = (x × y)^m
Hence, $\text{x}^{\text{m}}\times\text{y}^{\text{m}}\neq(\text{x}\times\text{y}^{2\text{m}})$

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Question 51 Mark

x⁰ × x⁰ = x⁰ ÷ x⁰ is true for all non-zero values of x.

Answer

False.
Solution:
A number in standard form is written as a × 10^k, where 1 ≤ a ≤ 10 and k is any integer.

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Question 61 Mark

$\frac{-3^{100}}{5}=\frac{-3^{100}}{-5^{100}}$

Answer

True.
Solution:
$\Big(\frac{-3}{5}\Big)=\Big(\frac{-1\times3}{5}\Big)^{100}$ $\big[\because-3=1\times3\big]$
$\frac{(-1)^{100}\times3^{100}}{5^{100}}$ $\big[\because(\text{a}\times\text{b}^{\text{m}})=\text{a}^{\text{m}}\times\text{b}^{\text{m}}\big]$
$\frac{1\times3^{100}}{5^{100}}$ $\big[\because(-1)^{\text{n}}=1,\text{ if }\text{n}\text{ is even}\big]$
$\frac{3^{100}}{5^{100}}$
Now, taking RHS, we have $\frac{-3^{100}}{-5^{100}}=\frac{3^{10}}{5^{100}}$ [$\because$ if both numerator and denominator have negative sign, then it is cacelled out]
$\therefore$ LHS = RHS
Hence, $\frac{-3^{100}}{-5^{100}}=\frac{3^{10}}{5^{100}}$

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Question 71 Mark

$\Big(\frac{7}{3}\Big)^{2}\times\Big(\frac{7}{3}\Big)^{5}=\Big(\frac{7}{3}\Big)^{10}$

Answer

False.
Solution:
Here,$\Big(\frac{7}{3}\Big)^{2}\times\Big(\frac{7}{3}\Big)^{5}=\Big(\frac{7}{3}\Big)^{2+5}$
$=\Big(\frac{7}{3}\Big)^{7}$ $\big[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}=\text{a}^{\text{m+n}}\big]$
Here, $\Big(\frac{7}{3}\Big)^{2}\times\Big(\frac{7}{3}\Big)^{5}\neq\Big(\frac{7}{3}\Big)^{10}$

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Question 81 Mark

$\Big(\frac{5}{8}\Big)^{9}\div\Big(\frac{5}{8}\Big)^{4}=\Big(\frac{5}{8}\Big)^{4}$

Answer

False.

Solution:

Here,$\Big(\frac{5}{8}\Big)^{9}\div\Big(\frac{5}{8}\Big)^{4}=\Big(\frac{5}{8}\Big)^{9-4}=\Big(\frac{5}{8}\Big)^{5}$

$\big[\because\text{a}^{\text{m}}+\text{a}^{\text{m}}=\text{a}^{\text{m-n}}\big]$

Hence, $\Big(\frac{5}{8}\Big)^{9}\div\Big(\frac{5}{8}\Big)^{4}\neq\Big(\frac{5}{8}\Big)^{4}$

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Question 91 Mark

$\Big(\frac{4}{3}\Big)^{5}\times\Big(\frac{5}{7}\Big)^{5}=\Big(\frac{4}{3}+\frac{5}{7}\Big)^{5}$

Answer

False.
Solution:
Here,$\Big(\frac{4}{3}\Big)^{5}\times\Big(\frac{5}{7}\Big)^{5}=\Big(\frac{4}{3}\times\frac{5}{7}\Big)^{5}$$\big[\because\text{a}^{\text{m}}\times\text{b}^{\text{m}}=(\text{ab})^{\text{m}}\big]$
and $\Bigg[\Big(\frac{4}{3}\Big)^{5}+\Big(\frac{5}{7}\Big)^{5}\Bigg]^{5}$ $\Big[=\Big(\frac{\text{a}}{\text{b}}\Big)+\Big(\frac{\text{c}}{\text{d}}\Big)=\frac{\text{a}}{\text{b}}\times\frac{\text{d}}{\text{c}}\Big]$
$=\Big(\frac{4}{3}\times\frac{5}{7}\Big)^{5}$
Hence, $\Big(\frac{4}{3}\times\frac{5}{7}\Big)^{5}\neq\Big(\frac{4}{3}\times\frac{7}{5}\Big)^{5}$

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Question 101 Mark

$\Big(\frac{2}{5}\Big)^{3}\div\Big(\frac{5}{2}\Big)^{3}=1$

Answer

False.
Solution:
Here, $\Big(\frac{2}{5}\Big)^{3}\div\Big(\frac{5}{2}\Big)^{3}$ $\big[\because=\Big(\frac{\text{a}}{\text{b}}\Big)+\Big(\frac{\text{c}}{\text{d}}\Big)=\frac{\text{a}}{\text{b}}\times\frac{\text{d}}{\text{c}}\big]$
$=\Big(\frac{2}{5}\Big)^{3}\times\Big(\frac{2}{5}\Big)^{3}$ $\big[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}=\text{a}\text{}^{\text{m+n}}\big]$
$\Big(\frac{2}{5}\Big)^{3+3}=\Big(\frac{5}{2}\Big)^{6}$
Here, $\Big(\frac{2}{5}\Big)^{3}\div\Big(\frac{5}{2}\Big)^{3}\neq1$

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Question 111 Mark

One million = 10⁷

Answer

False.
Solution:
One million = 10 lakhs
= 1000000 = 10⁶
Hence, $10^{6}\neq10^{7}$

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Question 121 Mark

One hour = 60² seconds.

Answer

True.
Solution:
1h = 60 min
= 60 × 60s
= 60²s

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Question 131 Mark

In the standard form, a large number can be expressed as a decimal number between 0 and 1, multiplied by a power of 10

Answer

False.
Solution:
A number in standard form is written as a × 10^k, where 1 ≤ a ≤ 10 and k is any integer.

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Question 141 Mark

876543 = 8 × 105 + 7 × 104 + 6 × 103 + 5 × 102 + 4 × 101 + 3 × 10⁰

Answer

True.
Solution:
Take RHS = 8 × 10⁵ + 7 × 10⁴ + 6 × 10³ + 5 × 10² + 4 × 10¹ + 3 × 10⁰
= 8 × 100000 + 7 × 10000 + 6 × 1000 + 5 ×100 + 4 × 10 + 3 × 1
[$\therefore$ a⁰ = 1]
= 800000 + 70000 + 6000 + 500 + 40 + 3
= 876543
= LHS
Hence, RHS = LHS.

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Question 151 Mark

8 × 106 + 2 × 104 + 5 × 102 + 9 × 100 = 8020509

Answer

True.
Solution:
Take LHS
= 8 × 10⁶ + 2 × 10⁴ + 5 × 10² + 9 × 10⁰
= 8 × 1000000 + 2 × 10000 + 5 × 100 + 9 × 1
[$\therefore$ a⁰ = 1]
= 8000000 + 2000 + 500 + 9
= 8020509
= RHS
Hence, LHS = RHS.

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Question 161 Mark

600060 = 6 × 10⁵ + 6 × 10²

Answer

False.
Solution:
Take RHS = 6 × 10⁵ + 6 × 10²
= 6 × 100000 + 6 × 100
$=600000+600=600600\neq\text{LHS}$
Hence, $\text{RHS}\neq\text{LHS}$

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Question 171 Mark

5⁰ × 25⁰ × 125⁰ = (5⁰)⁶

Answer

True.

Solution:

Here, 5⁰ × 25⁰ × 125⁰

= 5⁰ × (5 × 5)⁰ × (5 × 5 × 5)⁰

[$\therefore$ 25 = 5 × 5 and 125 = 5 × 5 × 5]

= 5⁰ × 5⁰ × 5⁰ × 5⁰ × 5⁰× 5⁰ [$\therefore$ a^m × b^m = a^mb^m]

= (5⁰)⁶

Hence, 5⁰× 25⁰ × 125⁰

= (5⁰)⁶

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Question 181 Mark

4⁹ is greater than 16³

Answer

True.
Solution:
$\therefore$ 16³ =(4²)³
[$\therefore$ 16 = 4 × 4 = 4²]
= 4⁶
Now, in 4⁹ and 4⁶, 4⁹ > 4⁶ as powers 9 > 6

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Question 191 Mark

4² is greater than 2⁴

Answer

False.
Solution:
4²= 4 × 4 = 16 [$\therefore$ a^m = a × a × a ×…× a (m times)]
and 2⁴ =2 × 2 × 2 × 2 = 16
So, 4² = 2⁴

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Question 201 Mark

4⁰ + 5⁰ + 6⁰ = (4 + 5 + 6)⁰

Answer

False.

Solution:

Here, 4⁰ + 5⁰ + 6⁰

= 1 + 1 + 1

= 3 [$\therefore$ a⁰ = 1]

and (4+ 5+ 6)⁰

= (15)⁰

= 1

Hence, 4⁰ + 5⁰ + 6⁰

$\neq(4+5+6)^{0}$

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Question 211 Mark

4 × 10⁵ + 3 × 10⁴ + 2 × 10³ + 1 × 10⁰ = 432010

Answer

False.
Solution:
Take LHS
= 4 × 10⁵+ 3 × 10⁴ + 2 × 10³ + 1 × 10⁰
= 4 × 100000 + 3 × 10000 + 2 × 1000 + 1 × 1 [$\therefore$ a⁰ = 1]
= 400000 + 30000 + 2000 + 1
$=432001\neq\text{RHS}$
Hence, $\text{LHS}\neq\text{RHS}$

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Question 221 Mark

(-3)⁴= -12

Answer

False.
Solution:
(-3)⁴ = (-3) × (-3) × (-3) × (-3)
$=81\neq-12$

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Question 231 Mark

34 > 43

Answer

True.
Solution:
$\therefore3^{4}=3\times3\times3\times3=81$
and 4³ = 4 x 4 x 4 = 64
81 > 64
Hence, 3⁴ > 4³

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Question 241 Mark