1 Marks Question

Question 511 Mark

$4^0+5^0+6^0=(4+5+6)^0$

Answer

False.
Solution:
$\text { Here, } 4^0+5^0+6^0$
$=1+1+1$
$=3\left[\therefore a^0=1\right]$
$\text { and }(4+5+6)^0$
$=(15)^0$
$=1$
$\text { Hence, } 4^0+5^0+6^0$
$\neq(4+5+6)^0$

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Question 521 Mark

$\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)$

Answer

$\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{32}$ Solution: Here, $\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{16\times2}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}=\text{a}^{\text{mn}}\big]$ $=\Big(\frac{-1}{4}\Big)^{32}$ $\therefore\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{32}$

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Question 531 Mark

$1^0 \times 0^1=1$

Answer

False.
Solution:
$\therefore1^{0}\times0^{1}$
$=1\times0$
$=0\neq1$

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Question 541 Mark

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

Answer

$a^m \times a^n=a^{m+n}$
$x^m \times x^m=x^{m+m}=x^{2 m}$
$\text { Also, } a^k+a^k=2 a^k$
$\text { So, } x^m+x^m=2 x^m$

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Question 551 Mark

$88880000000=$ ______ $\times 10^{10}$

Answer

Given, $88880000000$
For standard form, $88880000000$
$=8888 \times 10^7 \text { Also, } 8888$
$=8.888 \times 10^3$
$\text { So, } 8.888 \times 10^3 \times 10^7$
$=8.888 \times 10^{10}$
$88880000000=8.888 \times 10^{10}$

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Question 561 Mark

Express the following numbers in standard form:
$24$ billion.

Answer

$24 \text { billion }$
$=24,00,00,00,000$
$=24 \times 10^9$
$=2.4 \times 10^1 \times 10^9$
$=2.4 \times 10^{10}$

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Question 571 Mark

$4^9$ is greater than $16^3$

Answer

True.
Solution:
$\therefore 16^3=\left(4^2\right)^3$
${\left[\therefore 16=4 \times 4=4^2\right]}$
$=4^6$
Now, in $4^9$ and $4^6, 4^9>4^6$ as powers $9>6$

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Question 581 Mark

$a. 3^2 .........1^5$

Answer

$\because 3^2=3 \times 3=9$
So, $9<15$
$\therefore 3^2<15$

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Question 591 Mark

Express the following in usual form:
$8.01 \times 10^7$

Answer

Here, $8.01\times10^{7}=\frac{801}{100}\times10000000$ $=80100000$

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Question 601 Mark

Identify the greater number, in the following:
$7.9 \times 104$ or $5.28 \times 105$

Answer

We have, $7.9 \times 10^4$
$=7.9 \times 10000$
$=79000 \text { and } 5.28 \times 10^5$
$=5.28 \times 100000$
$=528000 \text { So, } 5.28 \times 10^5>7.9 \times 10^4$

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Question 611 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

A number in standard form is written as $a \times 10^k$, where $1 ≤ a ≤ 10$ and $k$ is any integer.

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Question 621 Mark

$8 \times 106 + 2 \times 104 + 5 \times 102 + 9 \times 100 = 8020509$

Answer

Take $LHS$
$=8 \times 10^6+2 \times 10^4+5 \times 10^2+9 \times 10^0$
$=8 \times 1000000+2 \times 10000+5 \times 100+9 \times 1$
${\left[\therefore a^0=1\right]}$
$=8000000+2000+500+9$
$=8020509$
$=\text { RHS }$
Hence, $LHS = RHS.$

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Question 631 Mark

$7^4$ _____ $5^4$

Answer

$7^4$ > $5^4$
Solution:
Obviously, $7^4$ > $5^4$
[$\because$ powers are same, so if base is greater then the number is greater]

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Question 641 Mark

Express the following in single exponential form:
$2^3 \times 3^3$

Answer

We have, $2^3 \times 3^3=(2 \times 3)^3$
$=6^3\left[\because a^m b^m=(a \times b)^m\right]$

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Question 651 Mark

$1$ lakh $= 10 $_____.

Answer

Here, $1$ lakh
$=100000=10^5$
$\therefore 1$ lakh
$=10^5$

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Question 661 Mark

$1$ million $= 10$ _____.

Answer

Here, $1$ million
$=1000000=10^6$
$\therefore 1$ million $=10^6$

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Question 671 Mark

Find the value of:
$\left(-3^5\right)$

Answer

$\left(-3^5\right)=\left(-1^5\right) \times 3^5$
$=1 \times 3 \times 3 \times 3 \times 3 \times 3$
$[\because(-1)=-1, \mathrm{n}$ is odd $]$
$=-243$

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Question 681 Mark

$\Big(\frac{11}{15}\Big)^{4}\times$ (_____)$^5$ = $\Big(\frac{11}{15}\Big)^{9}$

Answer

$\Big(\frac{11}{15}\Big)^{4}\times\Big(\frac{11}{15}\Big)^{5}=\Big(\frac{11}{15}\Big)^{9}$ Solution: $\Rightarrow(\text{x})^{5}=\frac{\Big(\frac{11}{15}\Big)^{9}}{\Big(\frac{11}{15}\Big)^{4}}$ $\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{9}\Big(\frac{11}{15}\Big)^{-4}$ $\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{9-4}$ $\big[\because\text{a}^{\text{m}}\times\text{a}^{\text{-n}}=\text{a}^{\text{m+(-n)}}=\text{a}^{\text{m-n}}\big]$ $\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{5}$ Since, the powers are same $\therefore\text{x}=\frac{11}{15}$ Hence, $\Big(\frac{11}{15}\Big)^{4}\times\Big(\frac{11}{15}\Big)^{5}=\Big(\frac{11}{15}\Big)^{9}$

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Question 691 Mark

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

Answer

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 701 Mark

$4^2$ is greater than $2^4$

Answer

False.
Solution:
$4^2=4 \times 4=16\left[\therefore a^m=a \times a \times a \times \ldots \times a(m\right.$ times $\left.)\right]$
and $2^4=2 \times 2 \times 2 \times 2=16$
So, $4^2=2^4$

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Question 711 Mark

$\Bigg[\Big(\frac{7}{11}\Big)^{3}\Bigg]^{4}=\Big(\frac{7}{11}\Big)$

Answer

$\Bigg[\Big(\frac{7}{11}\Big)^{3}\Bigg]^{4}=\Big(\frac{7}{11}\Big)^{3\times4}$
Solution:
Here, $\Bigg[\Big(\frac{7}{11}\Big)^{3}\Bigg]^{4}=\Big(\frac{7}{11}\Big)^{3\times4}=\Big(\frac{7}{11}\Big)^{12}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}=\text{a}^{\text{mn}}\big]$
$\therefore\bigg[\Big(\frac{7}{11}\Big)^{3}\bigg]=\Big(\frac{7}{11}\Big)^{12}$

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Question 721 Mark

$\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)$

Answer

$\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{32}$
Solution:
Here, $\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{16\times2}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}=\text{a}^{\text{mn}}\big]$
$=\Big(\frac{-1}{4}\Big)^{32}$
$\therefore\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{32}$

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Question 731 Mark

$\Big(\frac{16}{13}\Big)^{10}\div\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}=\Big(\frac{6}{13}\Big)$

Answer

$\Big(\frac{16}{13}\Big)^{10}\div\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}=\Big(\frac{6}{13}\Big)^{0}$

Solution:

Here, $\Big(\frac{16}{13}\Big)^{10}\div\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}$

$=\Big(\frac{6}{13}\Big)^{0}+\Big(\frac{6}{13}\Big)^{5\times2}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}+\text{a}^{\text{mn}}$

$\Big(\frac{6}{13}^{10}+\Big(\frac{6}{13}\Big)^{10}=\Big(\frac{6}{13}\Big)^{10-10}$ $\big[\because\text{a}^{\text{m}}+\text{a}^{\text{n}}=\text{a}^{\text{m-n}},(\text{m}>\text{n})\big]$

$=\Big(\frac{6}{13}\Big)^{0}$

$\therefore\Big(\frac{6}{13}\Big)^{10}+\bigg[\Big(\frac{6}{13}\Big)^{5}\bigg]^{2}=\Big(\frac{6}{13}\Big)^{0}$

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Question 741 Mark

$\Big(\frac{-1}{4}\Big)^{3}\times\Big(\frac{-1}{4}\Big)=\Big(\frac{-1}{4}\Big)^{11}$

Answer

$\Big(\frac{-1}{4}\Big)^{3}\times\Big(\frac{-1}{4}\Big)^{\text{x}}=\Big(\frac{-1}{4}\Big)^{11}$
Solution:
Let $\Big(\frac{-1}{4}\Big)^{3}\times\Big(\frac{-1}{4}\Big)=\Big(\frac{-1}{4}\Big)^{11}$
$\Rightarrow\Big(\frac{-1}{4}\Big)^{\text{x}}=\frac{\Big(\frac{-1}{4}\Big)^{11}}{\Big(\frac{-1}{4}\Big)^{3}}$
$\Rightarrow\Big(\frac{-1}{4}\Big)^{\text{x}}=\Big(\frac{-1}{4}\Big)^{\text{11-3}}$ $\big[\because\text{a}^{\text{m}}+\text{a}^{\text{n}}=\text{a}^{\text{m-n}}\text{m}>\text{n}\big]$
$\Rightarrow\Big(\frac{-1}{4}\Big)^{\text{x}}=\Big(\frac{-1}{4}\Big)^{8}$
Since, base are equal do, ny equating the power the pawers, we get
$\text{x}=8$
$\therefore\Big(-\frac{1}{4}\Big)^{3}\times\Big(-\frac{1}{4}\Big)^{8}=\Big(-\frac{1}{4}\Big)^{11}$

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Question 751 Mark

$\Big(\frac{13}{14}\Big)^{5}\div(\ )^{2}=\Big(\frac{13}{14}\Big)^{3}$

Answer

$\Big(\frac{13}{14}\Big)^{5}\div(\text{x})^{2}=\Big(\frac{13}{14}\Big)^{3}$

Solution:

Let $\Big(\frac{13}{14}\Big)^{5}\div(\text{x})^{2}=\Big(\frac{13}{14}\Big)^{3}$

$\therefore(\text{x})^{2}=\frac{\Big(\frac{13}{14}^{5}\Big)}{\Big(\frac{13}{14}\Big)^{3}}$

$\therefore(\text{x})^{2}=\Big(\frac{13}{14}\Big)^{5-3}$ $\big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\big]$

$\therefore(\text{x})^{2}=\Big(\frac{13}{14}\Big)^{2}$

Since, powers are same.

$\therefore\text{x}=\frac{13}{14}$

Hence, $\Big(\frac{13}{14}\Big)^{5}\div(\text{x})^{2}=\Big(\frac{13}{14}\Big)^{3}$

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Question 761 Mark

$\Big(\frac{11}{15}\Big)^{4}\times$ (_____)⁵=$\Big(\frac{11}{15}\Big)^{9}$

Answer

$\Big(\frac{11}{15}\Big)^{4}\times\Big(\frac{11}{15}\Big)^{5}=\Big(\frac{11}{15}\Big)^{9}$
Solution:
$\Rightarrow(\text{x})^{5}=\frac{\Big(\frac{11}{15}\Big)^{9}}{\Big(\frac{11}{15}\Big)^{4}}$
$\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{9}\Big(\frac{11}{15}\Big)^{-4}$
$\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{9-4}$ $\big[\because\text{a}^{\text{m}}\times\text{a}^{\text{-n}}=\text{a}^{\text{m+(-n)}}=\text{a}^{\text{m-n}}\big]$
$\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{5}$
Since, the powers are same
$\therefore\text{x}=\frac{11}{15}$
Hence, $\Big(\frac{11}{15}\Big)^{4}\times\Big(\frac{11}{15}\Big)^{5}=\Big(\frac{11}{15}\Big)^{9}$

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Question 771 Mark

Identify the greater number, in the following:
7.9 × 104 or 5.28 × 105

Answer

We have, 7.9 x 10⁴
= 7.9 × 10000
= 79000 and 5.28 × 10⁵
= 5.28 × 100000
= 528000 So, 5.28 × 10⁵ > 7.9 × 10⁴

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Question 781 Mark

Identify the greater number, in the following:
2⁹ or 9²

Answer

We have, 2⁹
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 512 and 9²
= 9 × 9
= 81
So, 2⁹ > 9²

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Question 791 Mark

Identify the greater number, in the following:
2⁶ or 6²

Answer

We have, 2⁶
= 2 × 2 × 2 × 2 × 2 × 2
= 64 and 6² = 6 × 6
= 36 So, 2⁶ > 6²

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Question 801 Mark

Find the value of:
(8⁰ - 2⁰) × (8⁰ + 2⁰)

Answer

(8⁰ - 2⁰) × (8⁰ + 2⁰)
= (1 - 1) × (1 + 1)
$\Big[\because\text{a}^{0}=1\Big]$
= 0 × 2 =0

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Question 811 Mark

Find the value of:
7⁷ ÷ 7⁷

Answer

$7^{7}\div7^{7}=\frac{7^{7}}{7^{7}}=7^{7-7}$
$\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}}\Big]$
$=7^{0}=1$

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Question 821 Mark

Find the value of:
(-7)^{2 × 7 - 6 - 8}

Answer

(-7)^{2 × 7 - 6 - 8}
= (-7)^14-14 $\Big[\because\text{a}^{0}=1\Big]$
= (-7)⁰
= 1

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Question 831 Mark

Find the value of:
7⁰

Answer

$\Big[\because\text{a}^{0}=1\Big]$
7⁰ = 1

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Question 841 Mark

Find the value of:
-(-4)⁴

Answer

-(-4)⁴
= [(-4) ×(-4) × (-4) × (-4)] $\Big[\because(-1)^{\text{n}} =-1, \text{n is even}\Big]$
= -[(-1)⁴ × (4 × 4 × 4 × 4)]
= (256) = -256

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Question 851 Mark

Find the value of:
(-3⁵)

Answer

(-3⁵) = (-1⁵) × 3⁵
= 1× 3 × 3 × 3 × 3 × 3 $\Big[\because(-1)=-1, \text{n is odd}\Big]$
= -243

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Question 861 Mark

Find the value of:
2⁵

Answer

We knoe that, aⁿ = a × a × a ×...×a (n times)
2⁵ = 2 × 2 × 2 × 2 × 2 = 32

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Question 871 Mark

Find the value of:
(2⁰ + 3⁰ + 4⁰) (4⁰ - 3⁰ - 2⁰)

Answer

(2⁰ + 3⁰ + 4⁰) (4⁰ - 3⁰ - 2⁰)
= (1 + 1 + 1) (1 + 1 + 1)
$\Big[\because\text{a}^{0}=1\Big]$
= (3) (-1) = -3

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Question 881 Mark

Find the value of:
2 × 3 × 4 ÷ 2⁰ × 3⁰× 4⁰

Answer

2 × 3 × 4 ÷ 2⁰ × 3⁰× 4⁰
= 2 × 3 × 4 + 1 × 1 × 1
$\Big[\because\text{a}^{0}=1\Big]$
$=\frac{2\times3\times4}{1\times1\times1}=2\times3\times4=24$

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Question 891 Mark

Express the following numbers in standard form:
8,19,00,000

Answer

8,19,00,000
= 81900000.00
= 819 × 10s
= 8.19 × 10² × 10⁵
= 8.19 × 10⁷

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Question 901 Mark

Express the following numbers in standard form:
5,83,00,00,00,000

Answer

5,83,00,00,00,000
= 583000000000.00
= 583 × 10⁹
= 5.83 × 10² × 10⁹
= 5.83 × 10¹¹

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Question 911 Mark

Express the following numbers in standard form:
24 billion.

Answer

24 billion
= 24,00,00,00,000
= 24 × 10⁹
= 2.4 × 10¹ × 10⁹
= 2.4 × 10¹⁰

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Question 921 Mark

Express the following in usual form:
8.01 × 10⁷

Answer

Here, $8.01\times10^{7}=\frac{801}{100}\times10000000$
$=80100000$

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Question 931 Mark

Express the following in usual form:
1.75 × 10^-3

Answer

Here, $1.75\times10^{-3}=\frac{175}{100}\times\frac{1}{10}^{3}$
$=\frac{175}{100000}=0.00175$ $\Big[\because\text{a}^{\text{-m}}=\frac{1}{\text{a}^{\text{m}}}\Big]$

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Question 941 Mark

Express the following in single exponential form:

(-3)³ × (-10)³

Answer

We have, (-3)³ × (-10)³
= [(-3) × (-10)]³[$\because$ a^m × b^m = (a × b)^m]
[$\because$ (-3) × (-10) = 30]
= (30)³

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Question 951 Mark

Express the following in single exponential form:
(-11)² × (-2)²

Answer

We have, (-11)² × (-2)²
= [(-11) × (-2)]²[$\because$ a^m × b^m = (a × b)^m]
[$\because$ (-11) × (-2) = 22]
= 22²

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